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Courses/University_of_California_Davis/Chem_110B%3A_Physical_Chemistry_II/Text/18%3A_Partition_Functions_and_Ideal_Gases	Sorry, but all current instructors at UCD use the commercial textbook for this course. In chemistry, we are typically concerned with a collection of molecules. However, if the molecules are reasonably far apart as in the case of a dilute gas, we can approximately treat the system as an ideal gas system and ignore the intermolecular forces. The present chapter deals with systems in which intermolecular interactions are ignored. In ensemble theory, we are concerned with the ensemble probability density, i.e., the fraction of members of the ensemble possessing certain characteristics such as a total energy E, volume V, number of particles N or a given chemical potential μ and so on. The molecular partition function enables us to calculate the probability of finding a collection of molecules with a given energy in a system. The equivalence of the ensemble approach and a molecular approach may be easily realized if we treat part of the molecular system to be in equilibrium with the rest of it and consider the probability distribution of molecules in this subsystem (which is actually quite large compared to systems containing a small number of molecules of the order of tens or hundreds). 18.1: The Translational Partition Function of a Monatomic Ideal Gas 18.2: Most Atoms Are in the Ground Electronic State at Room Temperature 18.3: The Energy of a Diatomic Molecule Can Be Approximated as a Sum of Separate Terms 18.4: Most Molecules Are in the Ground Vibrational State at Room Temperature 18.5: Most Molecules Are in Excited Rotational States at Ordinary Temperatures 18.6: Rotational Partition Functions Contain a Symmetry Number For a homonuclear diatomic molecule, rotating the molecule by 180° brings the molecule into a configuration which is indistinguishable from the original configuration. This leads to an overcounting of the accessible states. To correct for this, we divide the partition function by σ , which is called the symmetry number, which is equal to the distinct number of ways by which a molecule can be brought into identical configurations by rotations. 18.7: The Vibrational Partition Function 18.8: The Rotational Partition Function Depends on the Shape of the Molecule 18.9: Molar Heat Capacities 18.E: Partition Functions and Ideal Gases (Exercises)
Courses/Lumen_Learning/First_Year_Seminar_(Lumen)/07%3A_Module_5-_Academics_and_Advisement/7.01%3A_What_Do_You_Enjoy_Studying	Dr. Patricia Munsch There is a tremendous amount of stress placed on college students regarding their choice of major. Everyday, I meet with students regarding their concern about choosing right major; the path that will lead to a fantastic, high-paying position in a growth industry. There is a hope that one decision, your college major, will have a huge impact on the rest of your life. Students shy away from subject areas they enjoy due to fear that such coursework will not lead to a job. I am disappointed in this approach. As a counselor I always ask—what do you enjoy studying? Based on this answer it is generally easy to choose a major or a family of majors. I recognize the incredible pressure to secure employment after graduation, but forcing yourself to choose a major that you may not have any actual interest in because a book or website mentioned the area of growth may not lead to the happiness you predict. Working in a college setting I have the opportunity to work with students through all walks of life, and I do believe based on my experience, that choosing a major because it is listed as a growth area alone is not a good idea. Use your time in college to explore all areas of interest and utilize your campus resources to help you make connections between your joy in a subject matter and the potential career paths. Realize that for most people, in most careers, the undergraduate major does not lead to a linear career path. As an undergraduate student I majored in Political Science, an area that I had an interest in, but I added minors in Sociology and Women’s Studies as my educational pursuits broadened. Today, as a counselor, I look back on my coursework with happy memories of exploring new ideas, critically analyzing my own assumptions, and developing an appreciation of social and behavioral sciences. So to impart my wisdom in regards to a student’s college major, I will always ask, what do you enjoy studying? Once you have determined what you enjoy studying, the real work begins. Students need to seek out academic advisement. Academic advisement means many different things; it can include course selection, course completion for graduation, mapping coursework to graduation, developing opportunities within your major and mentorship. As a student I utilized a faculty member in my department for semester course selection, and I also went to the department chairperson to organize two different internships to explore different career paths. In addition, I sought mentorship from club advisors as I questioned my career path and future goals. In my mind I had a team of people providing me support and guidance, and as a result I had a great college experience and an easy transition from school to work. I recommend to all students that I meet with to create their own team. As a counselor I can certainly be a part of their team, but I should not be the only resource. Connect with faculty in your department or in your favorite subject. Seek out internships as you think about the transition from college to workplace. Find mentors through faculty, club advisors, or college staff. We all want to see you succeed and are happy to be a part of your journey. As a counselor I am always shocked when students do not understand what courses they need to take, what grade point average they need to maintain, and what requirements they must fulfill in order to reach their goal—graduation! Understand that as a college student it is your responsibility to read your college catalog and meet all of the requirements for graduation from your college. I always suggest that students, starting in their first semester, outline or map out all of the courses they need to take in order to graduate. Of course you may change your mind along the way, but by setting out your plan to graduation you are forcing yourself to learn what is required of you. I do this exercise in my classes and it is by far the most frustrating for students. They want to live in the now and they don’t want to worry about next semester or next year. However, for many students that I see, the consequence of this decision is a second semester senior year filled with courses that the student avoided during all the previous semesters. If you purposefully outline each semester and the coursework for each you can balance your schedule, understand your curriculum, and feel confident that you will reach your goal. CC licensed content, Shared previously Foundations of Academic Success: Words of Wisdom. Authored by : Thomas C. Priester. Located at : http://textbooks.opensuny.org/foundations-of-academic-success/ . Project : Open SUNY Textbooks. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
Courses/Indiana_Tech/EWC%3A_CHEM_1020_-_General_Chemistry_I_(Budhi)/10%3A_Liquids_Solids_and_Intermolecular_Forces/10.5%3A_Vaporization_and_Vapor_Pressure	Learning Objectives To know how and why the vapor pressure of a liquid varies with temperature. To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present. To understand that the relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation. Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid. Evaporation and Condensation Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy ( KE ) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure $\PageIndex{1}$), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ($E_0$) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than $E_0$. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than $E_0$ has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure. To understand the causes of vapor pressure, consider the apparatus shown in Figure $\PageIndex{2}$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $\PageIndex{2}$), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with $KE > E_0$ will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase. As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure $\PageIndex{2}$). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure $\PageIndex{3}$. Equilibrium Vapor Pressure Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid. If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure $\PageIndex{4}$) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile. The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure $\PageIndex{4}$. Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release. Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions. A Video Discussing Vapor Pressure and Boiling Points. Video Source: Vapor Pressure & Boiling Point(opens in new window) [youtu.be] The exponential rise in vapor pressure with increasing temperature in Figure $\PageIndex{4}$ allows us to use natural logarithms to express the nonlinear relationship as a linear one. \[ \boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1} \] where $\ln P$ is the natural logarithm of the vapor pressure, $ΔH_{vap}$ is the enthalpy of vaporization , $R$ is the universal gas constant [8.314 J/(mol•K)], $T$ is the temperature in kelvins, and $C$ is the y-intercept, which is a constant for any given line. Plotting $\ln P$ versus the inverse of the absolute temperature ($1/T$) is a straight line with a slope of −Δ H vap / R . Equation $\ref{Eq1}$, called the Clausius–Clapeyron Equation , can be used to calculate the $ΔH_{vap}$ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine $ΔH_{vap}$ is to measure the vapor pressure of a liquid at two temperatures and insert the values of $P$ and $T$ for these points into Equation $\ref{Eq2}$, which is derived from the Clausius–Clapeyron equation: \[ \ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2} \] Conversely, if we know Δ H vap and the vapor pressure $P_1$ at any temperature $T_1$, we can use Equation $\ref{Eq2}$ to calculate the vapor pressure $P_2$ at any other temperature $T_2$, as shown in Example $\PageIndex{1}$. A Video Discussing the Clausius-Clapeyron Equation. Video Link: The Clausius-Clapeyron Equation(opens in new window) [youtu.be] Example $\PageIndex{1}$: Vapor Pressure of Mercury The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table: T (°C) 80.0 100 120 140 P (torr) 0.0888 0.2729 0.7457 1.845 From these data, calculate the enthalpy of vaporization (Δ H vap ) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.) Given: vapor pressures at four temperatures Asked for: Δ H vap of mercury and vapor pressure at 160°C Strategy: Use Equation $\ref{Eq2}$ to obtain Δ H vap directly from two pairs of values in the table, making sure to convert all values to the appropriate units. Substitute the calculated value of Δ H vap into Equation $\ref{Eq2}$ to obtain the unknown pressure ( P 2 ). Solution: A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation $\ref{Eq1}$ and find the value of Δ H vap from the slope of the line, an alternative approach is to use Equation $\ref{Eq2}$ to obtain Δ H vap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C ( T 1 ) and 120.0°C ( T 2 ) into Equation $\ref{Eq2}$ gives \[\begin{align} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \\[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \\[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \\[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align} \nonumber \] B We can now use this value of Δ H vap to calculate the vapor pressure of the liquid ( P 2 ) at 160.0°C ( T 2 ): \[ \ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber \] Using the relationship $e^{\ln x} = x$, we have \[\begin{align} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \\[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \\[4pt] P_{2} &= 4.21 Torr \end{align} \nonumber \] At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect. Exercise $\PageIndex{1}$: Vapor Pressure of Nickel The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr? Answer 1896°C Boiling Points As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $\PageIndex{4}$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C) Mt. Everest, Nepal/Tibet 29028 240 70.0 Bogota, Colombia 11490 495 88.0 Denver, Colorado 5280 633 95.0 Washington, DC 25 759 100.0 Dead Sea, Israel/Jordan −1312 799 101.4 Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $\PageIndex{1}$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure increases, the boiling point of a liquid increases and vice versa. Example $\PageIndex{2}$: Boiling Mercury Use Figure $\PageIndex{4}$ to estimate the following. the boiling point of water in a pressure cooker operating at 1000 mmHg the pressure required for mercury to boil at 250°C Given: Data in Figure $\PageIndex{4}$, pressure, and boiling point Asked for: corresponding boiling point and pressure Strategy: To estimate the boiling point of water at 1000 mmHg, refer to Figure $\PageIndex{4}$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C. Solution: A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C. Exercise $\PageIndex{2}$: Boiling Ethlyene Glycol Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $\PageIndex{4}$ to estimate the following. the normal boiling point of ethylene glycol the pressure required for diethyl ether to boil at 20°C. Answer a 200°C Answer b 450 mmHg Summary Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation , generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation . Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium . Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation . This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point .
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/08%3A_Quantum_Teleportation/8.63%3A_Quantum_Correlations_Illustrated_with_Photons	A two-stage atomic cascade emits entangled photons (A and B) in opposite directions with the same circular polarization according to observers in their path. The experiment involves the measurement of photon polarization states in the vertical/horizontal measurement basis, and allows for the rotation of the right-hand detector through an angle θ, in order to explore the consequences of quantum mechanical entanglement. PA stands for polarization analyzer and could simply be a calcite crystal. The entangled two-photon polarization state is written in the circular and linear polarization bases, \[ \| \Psi \rangle = \frac{1}{ \sqrt{2}} \left[ \| L \rangle_A \|L \rangle_B + \|R \rangle_A \|R \rangle_B \right] = \frac{1}{ \sqrt{2}} \left[ \|V \rangle_A \|V \rangle_B - \|H \rangle_A \|H \rangle_B \right] \nonumber \] The vertical (eigenvalue +1) and horizontal (eigenvalue -1) polarization states for the photons in the measurement plane are given below. Θ is the angle of the measuring PA. \[ \begin{matrix} V( \theta) = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} & H( \theta) = \begin{pmatrix} - \sin \theta \\ \cos \theta \end{pmatrix} & V(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} & H(0) = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{matrix} \nonumber \] If photon A has vertical polarization photon B also has vertical polarization, the probability that photon B has vertical polarization when measured at an angle θ giving a composite eigenvalue of +1 is, \[ \left( V( \theta)^T V(0) \right)^2 \rightarrow \cos ( \theta)^2 \nonumber \] If photon A has vertical polarization photon B also has vertical polarization, the probability that photon B has horizontal polarization when measured at an angle θ giving a composite eigenvalue of -1 is, \[ \left( H( \theta)^T V(0) \right)^2 \rightarrow \sin ( \theta)^2 \nonumber \] Therefore the overall quantum correlation coefficient or expectation value is: \[ E( \theta) = \left( V( \theta)^T V(0) \right)^2 - \left( H( \theta)^T V(0) \right)^2 \text{ simplify } \rightarrow \cos (2 \theta) \nonumber \] Now it will be shown that a local-realistic, hidden-variable model can be constructed which is in agreement with the quantum calculations for 0, 45 and 90 degrees, but not for 30 and 60 degrees (highlighted). \[ \begin{matrix} \text{E(0 deg) = 1} & \text{E(30 deg) = 0.5} & \text{E(45 deg) = 0} & \text{E(60 deg) = -0.5} & \text{E(90 deg) = -1} \end{matrix} \nonumber \] If objects have well-defined properties independent of measurement, the results for θ = 0 degrees and θ = 90 degrees require that the photons carry the following instruction sets, where the hexagonal vertices refer to θ values of 0, 30, 60, 90, 120, and 150 degrees. There are eight possible instruction sets, six of the type on the left and two of the type on the right. The white circles represent vertical polarization with eigenvalue +1 and the black circles represent horizontal polarization with eigenvalue -1. In any given measurement, according to local realism, both photons (A and B) carry identical instruction sets, in other words the same one of the eight possible sets. The problem is that while these instruction sets are in agreement with the 0 and 90 degree quantum calculations, with expectation values of +1 and -1 respectively, they can't explain the 30 degree predictions of quantum mechanics. The figure on the left shows that the same result should be obtained 4 times with joint eigenvalue +1, and the opposite result twice with joint eigenvalue of -1. For the figure on the right the opposite polarization is always observed giving a joint eigenvalue of -1. Thus, local realism predicts an expectation value of 0 in disagreement with the quantum result of 0.5. \[ \frac{6(1-1+1+1-1+1)+2(-1-1-1-1-1-1)}{8} = 0 \nonumber \] If we look at 60 degrees, E(60 deg) = 0,we reach the same conclusion: local realism disagrees with quantum result of -0.5. \[ \frac{^(-1-1+1-1-1+1)+2(1+1+1+1+1+1)}{8} = 0 \nonumber \] Instruction sets for 45 degrees, E(45 deg) = 0, are shown below. The vertices are 0, 45, 90 and 135 degrees. The instruction sets are constructed so that they satisfy the 0 and 90 degree quantum expectation values of 1 and -1, respectively. It is clear that they also satisfy the 45 degree result yielding an expectation value of zero: (1-1+1-1) = (-1+1-1+1) = (1-1+1-1) = (-1+1-1+1) = 0. This exercise illustrates Bell's theorem: no local hidden-variable theory can reproduce all the predictions of quantum mechanics for entangled composite systems . As the quantum predictions are confirmed experimentally, the local hidden-variable approach to reality must be abandoned. This analysis is based on "Simulating Physics with Computers" by Richard Feynman, published in the I nternational Journal of Theoretical Physics (volume 21, pages 481-485), and Julian Brown's Quest for the Quantum Computer (pages 91-100). Feynman used the experiment outlined above to establish that a local classical computer could not simulate quantum physics. A local classical computer manipulates bits which are in well-defined states, 0s and 1s, shown above graphically in white and black. However, these classical states are incompatible with the quantum mechanical analysis which is consistent with experimental results. This two-photon experiment demonstrates that simulation of quantum physics requires a computer that can manipulate 0s and 1s, superpositions of 0 and 1, and entangled superpositions of 0s and 1s. Simulation of quantum physics requires a quantum computer!
Courses/City_College_of_San_Francisco/Chemistry_101A/Topic_F%3A_Molecular_Structure/10%3A_Orbitals_and_Bonding_Theories/10.4%3A_Molecular_Orbital_Theory	Learning Objectives To use molecular orbital theory to predict bond order To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period. None of the approaches we have described so far can adequately explain why some compounds are colored and others are not, why some substances with unpaired electrons are stable, and why others are effective semiconductors. These approaches also cannot describe the nature of resonance. Such limitations led to the development of a new approach to bonding in which electrons are not viewed as being localized between the nuclei of bonded atoms but are instead delocalized throughout the entire molecule. Just as with the valence bond theory, the approach we are about to discuss is based on a quantum mechanical model. Previously, we described the electrons in isolated atoms as having certain spatial distributions, called orbitals , each with a particular orbital energy . Just as the positions and energies of electrons in atoms can be described in terms of atomic orbitals (AOs), the positions and energies of electrons in molecules can be described in terms of molecular orbitals (MOs) A particular spatial distribution of electrons in a molecule that is associated with a particular orbital energy. —a spatial distribution of electrons in a molecule that is associated with a particular orbital energy. As the name suggests, molecular orbitals are not localized on a single atom but extend over the entire molecule. Consequently, the molecular orbital approach, called molecular orbital theory is a delocalized approach to bonding. Although the molecular orbital theory is computationally demanding, the principles on which it is based are similar to those we used to determine electron configurations for atoms. The key difference is that in molecular orbitals, the electrons are allowed to interact with more than one atomic nucleus at a time. Just as with atomic orbitals, we create an energy-level diagram by listing the molecular orbitals in order of increasing energy. We then fill the orbitals with the required number of valence electrons according to the Pauli principle. This means that each molecular orbital can accommodate a maximum of two electrons with opposite spins. Molecular Orbitals Involving Only ns Atomic Orbitals We begin our discussion of molecular orbitals with the simplest molecule, H 2 , formed from two isolated hydrogen atoms, each with a 1 s 1 electron configuration. As discussed previously , electrons can behave like waves. In the molecular orbital approach, the overlapping atomic orbitals are described by mathematical equations called wave functions . The 1 s atomic orbitals on the two hydrogen atoms interact to form two new molecular orbitals, one produced by taking the sum of the two H 1 s wave functions, and the other produced by taking their difference : \[ \begin{matrix} MO(1)= & AO(atom\; A) & +& AO(atomB) \\ MO(1)= & AO(atom\; A) & -&AO(atomB) \end{matrix} \label{Eq1} \] The molecular orbitals created from Equation $\ref{Eq1}$ are called linear combinations of atomic orbitals (LCAOs) Molecular orbitals created from the sum and the difference of two wave functions (atomic orbitals) . A molecule must have as many molecular orbitals as there are atomic orbitals. Adding two atomic orbitals corresponds to constructive interference between two waves, thus reinforcing their intensity; the internuclear electron probability density is increased . The molecular orbital corresponding to the sum of the two H 1 s orbitals is called a σ 1 s combination (pronounced “sigma one ess”) (part (a) and part (b) in Figure $\PageIndex{1}$). In a sigma (σ) orbital, A bonding molecular orbital in which the electron density along the internuclear axis and between the nuclei has cylindrical symmetry , the electron density along the internuclear axis and between the nuclei has cylindrical symmetry; that is, all cross-sections perpendicular to the internuclear axis are circles. The subscript 1 s denotes the atomic orbitals from which the molecular orbital was derived: The ≈ sign is used rather than an = sign because we are ignoring certain constants that are not important to our argument. \[ \sigma _{1s} \approx 1s\left ( A \right) + 1s\left ( B \right) \label{Eq2} \] Conversely, subtracting one atomic orbital from another corresponds to destructive interference between two waves, which reduces their intensity and causes a decrease in the internuclear electron probability density (part (c) and part (d) in Figure $\PageIndex{1}$). The resulting pattern contains a node where the electron density is zero. The molecular orbital corresponding to the difference is called $ \sigma _{1s}^{} $ (“sigma one ess star”). In a sigma star (σ) orbital An antibonding molecular orbital in which there is a region of zero electron probability (a nodal plane) perpendicular to the internuclear axis. , there is a region of zero electron probability, a nodal plane, perpendicular to the internuclear axis: \[ \sigma _{1s}^{\star } \approx 1s\left ( A \right) - 1s\left ( B \right) \label{Eq3} \] A molecule must have as many molecular orbitals as there are atomic orbitals. The electron density in the σ 1 s molecular orbital is greatest between the two positively charged nuclei, and the resulting electron–nucleus electrostatic attractions reduce repulsions between the nuclei. Thus the σ 1 s orbital represents a bonding molecular orbital. A molecular orbital that forms when atomic orbitals or orbital lobes with the same sign interact to give increased electron probability between the nuclei due to constructive reinforcement of the wave functions . In contrast, electrons in the $ \sigma _{1s}^{\star } $ orbital are generally found in the space outside the internuclear region. Because this allows the positively charged nuclei to repel one another, the $ \sigma _{1s}^{\star } $ orbital is an antibonding molecular orbital (a molecular orbital that forms when atomic orbitals or orbital lobes of opposite sign interact to give decreased electron probability between the nuclei due to destructive reinforcement of the wave functions) . Antibonding orbitals contain a node perpendicular to the internuclear axis; bonding orbitals do not. Energy-Level Diagrams Because electrons in the σ 1 s orbital interact simultaneously with both nuclei, they have a lower energy than electrons that interact with only one nucleus. This means that the σ 1 s molecular orbital has a lower energy than either of the hydrogen 1 s atomic orbitals. Conversely, electrons in the $ \sigma _{1s}^{\star } $ orbital interact with only one hydrogen nucleus at a time. In addition, they are farther away from the nucleus than they were in the parent hydrogen 1 s atomic orbitals. Consequently, the $ \sigma _{1s}^{\star } $ molecular orbital has a higher energy than either of the hydrogen 1 s atomic orbitals. The σ 1 s (bonding) molecular orbital is stabilized relative to the 1 s atomic orbitals, and the $ \sigma _{1s}^{\star } $ (antibonding) molecular orbital is destabilized . The relative energy levels of these orbitals are shown in the energy-level diagram (a schematic drawing that compares the energies of the molecular orbitals (bonding, antibonding, and nonbonding) with the energies of the parent atomic orbitals) in Figure $\PageIndex{2}$ A bonding molecular orbital is always lower in energy (more stable) than the component atomic orbitals, whereas an antibonding molecular orbital is always higher in energy (less stable). To describe the bonding in a homonuclear diatomic molecule (a molecule that consists of two atoms of the same element) such as H 2 , we use molecular orbitals; that is, for a molecule in which two identical atoms interact, we insert the total number of valence electrons into the energy-level diagram (Figure $\PageIndex{2}$). We fill the orbitals according to the Pauli principle and Hund’s rule : each orbital can accommodate a maximum of two electrons with opposite spins, and the orbitals are filled in order of increasing energy. Because each H atom contributes one valence electron, the resulting two electrons are exactly enough to fill the σ 1 s bonding molecular orbital. The two electrons enter an orbital whose energy is lower than that of the parent atomic orbitals, so the H 2 molecule is more stable than the two isolated hydrogen atoms. Thus molecular orbital theory correctly predicts that H 2 is a stable molecule. Because bonds form when electrons are concentrated in the space between nuclei, this approach is also consistent with our earlier discussion of electron-pair bonds. Bond Order in Molecular Orbital Theory In the Lewis electron structures, the number of electron pairs holding two atoms together was called the bond order . In the molecular orbital approach, bond order One-half the net number of bonding electrons in a molecule. is defined as one-half the net number of bonding electrons: \[ bond\; order=\dfrac{number\; of \; bonding\; electrons-number\; of \; antibonding\; electrons}{2} \label{Eq4} \] To calculate the bond order of H 2 , we see from Figure $\PageIndex{2}$ that the σ 1 s (bonding) molecular orbital contains two electrons, while the $ \sigma _{1s}^{\star } $ (antibonding) molecular orbital is empty. The bond order of H 2 is therefore \[ \dfrac{2-0}{2}=1 \label{Eq5} \] This result corresponds to the single covalent bond predicted by Lewis dot symbols. Thus molecular orbital theory and the Lewis electron-pair approach agree that a single bond containing two electrons has a bond order of 1. Double and triple bonds contain four or six electrons, respectively, and correspond to bond orders of 2 and 3. We can use energy-level diagrams such as the one in Figure $\PageIndex{2}$ to describe the bonding in other pairs of atoms and ions where n = 1, such as the H 2 + ion, the He 2 + ion, and the He 2 molecule. Again, we fill the lowest-energy molecular orbitals first while being sure not to violate the Pauli principle or Hund’s rule . Part (a) in Figure $\PageIndex{3}$ shows the energy-level diagram for the H 2 + ion, which contains two protons and only one electron. The single electron occupies the σ 1 s bonding molecular orbital, giving a (σ 1 s ) 1 electron configuration. The number of electrons in an orbital is indicated by a superscript. In this case, the bond order is (1-0)/2=1/2 Because the bond order is greater than zero, the H 2 + ion should be more stable than an isolated H atom and a proton. We can therefore use a molecular orbital energy-level diagram and the calculated bond order to predict the relative stability of species such as H 2 + . With a bond order of only 1/2 the bond in H 2 + should be weaker than in the H 2 molecule, and the H–H bond should be longer. As shown in Table $\PageIndex{1}$, these predictions agree with the experimental data. Part (b) in Figure $\PageIndex{3}$ is the molecular orbital energy-level diagram for He 2 + . This ion has a total of three valence electrons. Because the first two electrons completely fill the σ 1 s molecular orbital, the Pauli principle states that the third electron must be in the $ \sigma _{1s}^{\star} $ antibonding orbital, giving a $ \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{1} $ electron configuration. This electron configuration gives a bond order of (2-1)/2=1/2. As with H 2 + , the He 2 + ion should be stable, but the He–He bond should be weaker and longer than in H 2 . In fact, the He 2 + ion can be prepared, and its properties are consistent with our predictions (Table $\PageIndex{1}$). Molecule or Ion Electron Configuration Bond Order Bond Length (pm) Bond Energy (kJ/mol) H2+ (σ1s)1 1/2 106 269 H2 (σ1s)2 1 74 436 He2+ $ \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{1} $ 1/2 108 251 He2 $ \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{2} $ 0 not observed not observed Finally, we examine the He 2 molecule, formed from two He atoms with 1 s 2 electron configurations. Part (c) in Figure $\PageIndex{3}$ is the molecular orbital energy-level diagram for He 2 . With a total of four valence electrons, both the σ 1 s bonding and $ \sigma _{1s}^{\star } $ antibonding orbitals must contain two electrons. This gives a $ \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{1} $ electron configuration, with a predicted bond order of (2 − 2) ÷ 2 = 0, which indicates that the He 2 molecule has no net bond and is not a stable species. Experiments show that the He 2 molecule is actually less stable than two isolated He atoms due to unfavorable electron–electron and nucleus–nucleus interactions. In molecular orbital theory, electrons in antibonding orbitals effectively cancel the stabilization resulting from electrons in bonding orbitals . Consequently, any system that has equal numbers of bonding and antibonding electrons will have a bond order of 0, and it is predicted to be unstable and therefore not to exist in nature. In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory is able to accommodate systems with an odd number of electrons, such as the H 2 + ion. Molecular Orbital Theory: https://youtu.be/XgtOG0ezw78 In contrast to Lewis electron structures and the valence bond approach, molecular orbital theory can accommodate systems with an odd number of electrons. Example $\PageIndex{1}$ Use a molecular orbital energy-level diagram, such as those in Figure $\PageIndex{3}$, to predict the bond order in the He 2 2 + ion. Is this a stable species? Given: chemical species Asked for: molecular orbital energy-level diagram, bond order, and stability Strategy: Combine the two He valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for the system. Determine the total number of valence electrons in the He 2 2 + ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and predict whether the species is stable. Solution: A Two He 1 s atomic orbitals combine to give two molecular orbitals: a σ 1 s bonding orbital at lower energy than the atomic orbitals and a $ \sigma _{1s}^{\star } $ antibonding orbital at higher energy. The bonding in any diatomic molecule with two He atoms can be described using the following molecular orbital diagram: B The He 2 2 + ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can also view He 2 2+ as being formed from two He + ions, each of which has a single valence electron in the 1 s atomic orbital. We can now fill the molecular orbital diagram: The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ 1 s ) orbital, giving a (σ 1 s ) 2 electron configuration. To avoid violating the Pauli principle, the electron spins must be paired. C So the bond order is $ \dfrac{2-0}{2} =1 $ He 2 2 + is therefore predicted to contain a single He–He bond. Thus it should be a stable species. Exercise $\PageIndex{1}$ Use a molecular orbital energy-level diagram to predict the valence-electron configuration and bond order of the H 2 2− ion. Is this a stable species? Answer H 2 2− has a valence electron configuration of $ \left (\sigma _{1s} \right)^{2}\left (\sigma _{1s}^{\star } \right)^{2} $ with a bond order of 0. It is therefore predicted to be unstable. So far, our discussion of molecular orbitals has been confined to the interaction of valence orbitals, which tend to lie farthest from the nucleus. When two atoms are close enough for their valence orbitals to overlap significantly, the filled inner electron shells are largely unperturbed; hence they do not need to be considered in a molecular orbital scheme. Also, when the inner orbitals are completely filled, they contain exactly enough electrons to completely fill both the bonding and antibonding molecular orbitals that arise from their interaction. Thus the interaction of filled shells always gives a bond order of 0, so filled shells are not a factor when predicting the stability of a species. This means that we can focus our attention on the molecular orbitals derived from valence atomic orbitals. A molecular orbital diagram that can be applied to any homonuclear diatomic molecule with two identical alkali metal atoms (Li 2 and Cs 2 , for example) is shown in part (a) in Figure $\PageIndex{4}$, where M represents the metal atom. Only two energy levels are important for describing the valence electron molecular orbitals of these species: a σ ns bonding molecular orbital and a σ * ns antibonding molecular orbital. Because each alkali metal (M) has an ns 1 valence electron configuration, the M 2 molecule has two valence electrons that fill the σ ns bonding orbital. As a result, a bond order of 1 is predicted for all homonuclear diatomic species formed from the alkali metals (Li 2 , Na 2 , K 2 , Rb 2 , and Cs 2 ). The general features of these M 2 diagrams are identical to the diagram for the H 2 molecule in Figure $\PageIndex{2}$. Experimentally, all are found to be stable in the gas phase, and some are even stable in solution. Similarly, the molecular orbital diagrams for homonuclear diatomic compounds of the alkaline earth metals (such as Be 2 ), in which each metal atom has an ns 2 valence electron configuration, resemble the diagram for the He 2 molecule in part (c) in Figure $\PageIndex{3}$ As shown in part (b) in Figure $\PageIndex{4}$, this is indeed the case. All the homonuclear alkaline earth diatomic molecules have four valence electrons, which fill both the σ ns bonding orbital and the σ ns * antibonding orbital and give a bond order of 0. Thus Be 2 , Mg 2 , Ca 2 , Sr 2 , and Ba 2 are all expected to be unstable, in agreement with experimental data. In the solid state, however, all the alkali metals and the alkaline earth metals exist as extended lattices held together by metallic bonding. At low temperatures, $Be_2$ is st able. Example $\PageIndex{2}$ Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Na 2 − ion. Given: chemical species Asked for: molecular orbital energy-level diagram, valence electron configuration, bond order, and stability Strategy: Combine the two sodium valence atomic orbitals to produce bonding and antibonding molecular orbitals. Draw the molecular orbital energy-level diagram for this system. Determine the total number of valence electrons in the Na 2 − ion. Fill the molecular orbitals in the energy-level diagram beginning with the orbital with the lowest energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and predict whether the species is stable. Solution: A Because sodium has a [Ne]3 s 1 electron configuration, the molecular orbital energy-level diagram is qualitatively identical to the diagram for the interaction of two 1 s atomic orbitals. B The Na 2 − ion has a total of three valence electrons (one from each Na atom and one for the negative charge), resulting in a filled σ 3 s molecular orbital, a half-filled σ 3 s * and a $ \left ( \sigma _{3s} \right)^{2}\left ( \sigma _{3s}^{\star } \right)^{1} $ electron configuration. C The bond order is (2-1) ÷ 2=1/2 With a fractional bond order, we predict that the Na 2 − ion exists but is highly reactive. Exercise $\PageIndex{2}$ Use a qualitative molecular orbital energy-level diagram to predict the valence electron configuration, bond order, and likely existence of the Ca 2 + ion. Answer Ca 2 + has a $ \left ( \sigma _{4s} \right)^{2}\left ( \sigma _{4s}^{\star } \right)^{1} $ electron configurations and a bond order of 1/2 and should exist. Molecular Orbitals Formed from ns and np Atomic Orbitals Atomic orbitals other than ns orbitals can also interact to form molecular orbitals. Because individual p , d , and f orbitals are not spherically symmetrical, however, we need to define a coordinate system so we know which lobes are interacting in three-dimensional space. Recall that for each np subshell, for example, there are np x , np y , and np z orbitals. All have the same energy and are therefore degenerate, but they have different spatial orientations. \[ \sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \label{Eq6}\] Just as with ns orbitals, we can form molecular orbitals from np orbitals by taking their mathematical sum and difference. When two positive lobes with the appropriate spatial orientation overlap, as illustrated for two np z atomic orbitals in part (a) in Figure $\PageIndex{5}$, it is the mathematical difference of their wave functions that results in constructive interference, which in turn increases the electron probability density between the two atoms. The difference therefore corresponds to a molecular orbital called a $ \sigma _{np_{z}} $ bonding molecular orbital because, just as with the σ orbitals discussed previously, it is symmetrical about the internuclear axis (in this case, the z -axis): \[ \sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) \label{Eq6a}\] The other possible combination of the two np z orbitals is the mathematical sum: \[ \sigma _{np_{z}}=np_{z}\left ( A \right)+np_{z}\left ( B \right) \label{Eq7}\] In this combination, shown in part (b) in Figure $\PageIndex{5}$, the positive lobe of one np z atomic orbital overlaps the negative lobe of the other, leading to destructive interference of the two waves and creating a node between the two atoms. Hence this is an antibonding molecular orbital. Because it, too, is symmetrical about the internuclear axis, this molecular orbital is called a $ \sigma _{np_{z}}=np_{z}\left ( A \right)-np_{z}\left ( B \right) $ antibonding molecular orbital . Whenever orbitals combine, the bonding combination is always lower in energy (more stable) than the atomic orbitals from which it was derived, and the antibonding combination is higher in energy (less stable). Overlap of atomic orbital lobes with the same sign produces a bonding molecular orbital, regardless of whether it corresponds to the sum or the difference of the atomic orbitals. The remaining p orbitals on each of the two atoms, np x and np y , do not point directly toward each other. Instead, they are perpendicular to the internuclear axis. If we arbitrarily label the axes as shown in Figure $\PageIndex{6}$, we see that we have two pairs of np orbitals: the two np x orbitals lying in the plane of the page, and two np y orbitals perpendicular to the plane. Although these two pairs are equivalent in energy, the np x orbital on one atom can interact with only the np x orbital on the other, and the np y orbital on one atom can interact with only the np y on the other. These interactions are side-to-side rather than the head-to-head interactions characteristic of σ orbitals. Each pair of overlapping atomic orbitals again forms two molecular orbitals: one corresponds to the arithmetic sum of the two atomic orbitals and one to the difference. The sum of these side-to-side interactions increases the electron probability in the region above and below a line connecting the nuclei, so it is a bonding molecular orbital that is called a pi (π) orbital (a bonding molecular orbital formed from the side-to-side interactions of two or more parallel np atomic orbitals) . The difference results in the overlap of orbital lobes with opposite signs, which produces a nodal plane perpendicular to the internuclear axis; hence it is an antibonding molecular orbital, called a pi star (π) orbital An antibonding molecular orbital formed from the difference of the side-to-side interactions of two or more parallel np atomic orbitals, creating a nodal plane perpendicular to the internuclear axis. . \[ \pi _{np_{x}}=np_{x}\left ( A \right)+np_{x}\left ( B \right) \label{Eq8}\] \[ \pi ^{\star }_{np_{x}}=np_{x}\left ( A \right)-np_{x}\left ( B \right) \label{Eq9}\] The two np y orbitals can also combine using side-to-side interactions to produce a bonding $ \pi _{np_{y}} $ molecular orbital and an antibonding $ \pi _{np_{y}}^{\star } $ molecular orbital. Because the np x and np y atomic orbitals interact in the same way (side-to-side) and have the same energy, the $ \pi _{np_{x}} $ and $ \pi _{np_{y}} $molecular orbitals are a degenerate pair, as are the $ \pi _{np_{x}}^{\star } $ and $ \pi _{np_{y}}^{\star } $ molecular orbitals. Figure $\PageIndex{7}$ is an energy-level diagram that can be applied to two identical interacting atoms that have three np atomic orbitals each. There are six degenerate p atomic orbitals (three from each atom) that combine to form six molecular orbitals, three bonding and three antibonding. The bonding molecular orbitals are lower in energy than the atomic orbitals because of the increased stability associated with the formation of a bond. Conversely, the antibonding molecular orbitals are higher in energy, as shown. The energy difference between the σ and σ molecular orbitals is significantly greater than the difference between the two π and π* sets. The reason for this is that the atomic orbital overlap and thus the strength of the interaction are greater for a σ bond than a π bond, which means that the σ molecular orbital is more stable (lower in energy) than the π molecular orbitals. Although many combinations of atomic orbitals form molecular orbitals, we will discuss only one other interaction: an ns atomic orbital on one atom with an np z atomic orbital on another. As shown in Figure $\PageIndex{8}$, the sum of the two atomic wave functions ( ns + np z ) produces a σ bonding molecular orbital. Their difference ( ns − np z ) produces a σ* antibonding molecular orbital, which has a nodal plane of zero probability density perpendicular to the internuclear axis. Second Row Diatomic Molecules If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N 2 , O 2 , and F 2 . When we draw a molecular orbital diagram for a molecule, there are four key points to remember: The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the law of conservation of orbitals). As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized. The interaction between atomic orbitals is greatest when they have the same energy. The number of molecular orbitals is always equal to the total number of atomic orbitals we started with. We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F 2 . We use the diagram in part (a) in Figure $\PageIndex{9}$; the n = 1 orbitals (σ 1 s and σ 1 s * ) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ 2 s and σ 2 s * molecular orbitals are much lower in energy than the molecular orbitals derived from the 2 p atomic orbitals because of the large difference in energy between the 2 s and 2 p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2 p orbitals on each F is $ \sigma _{2p_{z}} $ and the next most stable are the two degenerate orbitals, $ \pi _{2p_{x}} $ and $ \pi _{2p_{y}} $ . For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $ \sigma ^{\star }_{2p_{z}} $ orbital is higher in energy than either of the degenerate $ \pi _{2p_{x}}^{\star } $ and $ \pi _{2p_{y}}^{\star } $ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy. Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F 2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ 2 s and σ 2 s * orbitals, 2 fill the $ \sigma _{2p_{z}} $ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π * orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F 2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F 2 is predicted to have a stable F–F single bond, in agreement with experimental data. We now turn to a molecular orbital description of the bonding in O 2 . It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O 2 , we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure $\PageIndex{9}$. We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ 2 s and σ 2 s * orbitals, 2 more to fill the $ \sigma _{2p_{z}} $ orbital, and 4 to fill the degenerate $ \pi _{2p_{x}}^{\star } $ and $ \pi _{2p_{y}}^{\star } $ orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π * orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K. None of the other bonding models can predict the presence of two unpaired electrons in O 2 . Chemists had long wondered why, unlike most other substances, liquid O 2 is attracted into a magnetic field. As shown in Figure $\PageIndex{10}$, it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O 2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed. The magnetic properties of O 2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H 2 O, CO 2 , and N 2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H 2 O, CO 2 , and N 2 , have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O 2 with organic compounds to give H 2 O, CO 2 , and N 2 would require that at least one of the electrons on O 2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier . Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke! For period 2 diatomic molecules to the left of N 2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the $ \sigma _{2p_{z}} $ molecular orbital is slightly higher in energy than the degenerate $ \pi ^{\star }_{np_{x}} $ and $ \pi ^{\star }_{np_{y}} $ orbitals. The difference in energy between the 2 s and 2 p atomic orbitals increases from Li 2 to F 2 due to increasing nuclear charge and poor screening of the 2 s electrons by electrons in the 2 p subshell. The bonding interaction between the 2 s orbital on one atom and the 2 pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ 2 s orbital and increases the energy of the $ \sigma _{2p_{z}} $ orbital. Thus for Li 2 , Be 2 , B 2 , C 2 , and N 2 , the $ \sigma _{2p_{z}} $ orbital is higher in energy than the $ \sigma _{3p_{z}} $ orbitals, as shown in Figure $\PageIndex{11}$ Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure $\PageIndex{11}$). Thus for example, the $ \sigma _{2p_{z}} $ molecular orbital is at a lower energy than the $ \pi _{2p_{x,y}} $ pair. Completing the diagram for N 2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N 2 versus 141.2 pm in F 2 ), and the bond energy is much greater for N 2 than for F 2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N 2 bond is much shorter and stronger than the F 2 bond, consistent with what we would expect when comparing a triple bond with a single bond. Example $\PageIndex{3}$ Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S 2 , a bright blue gas at high temperatures. Given: chemical species Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons Strategy: Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S 2 . Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S 2 . Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule. Calculate the bond order and describe the bonding. Solution: A Sulfur has a [Ne]3 s 2 3 p 4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure $\PageIndex{9}$ and Figure $\PageIndex{11}$, we need to know how close in energy the 3 s and 3 p atomic orbitals are because their energy separation will determine whether the $ \pi _{3p_{x,y}} $ or the $ \sigma _{3p_{z}} $ > molecular orbital is higher in energy. Because the ns – np energy gap increases as the nuclear charge increases (Figure $\PageIndex{11}$), the $ \sigma _{3p_{z}} $ molecular orbital will be lower in energy than the $ \pi _{3p_{x,y}} $ pair. B The molecular orbital energy-level diagram is as follows: Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. C Ten valence electrons are used to fill the orbitals through $ \pi _{3p_{x}} $ and $ \pi _{3p_{y}} $, leaving 2 electrons to occupy the degenerate $ \pi ^{\star }_{3p_{x}} $ and $ \pi ^{\star }_{3p_{y}} $ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S 2 is $ \left ( \sigma _{3s} \right)^{2}\left ( \sigma ^{\star }_{3s} \right)^{2}\left ( \sigma _{3p_{z}} \right)^{2}\left ( \pi _{3p_{x,y}} \right)^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right)^{2} $ with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond. Exercise $\PageIndex{3}$ Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O 2 2− ). Answer $ \left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \sigma _{2p_{z}} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right)^{4} $ bond order of 1; no unpaired electrons Molecular Orbitals for Heteronuclear Diatomic Molecules Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χ B > χ A ), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $\PageIndex{12}$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond. A molecular orbital energy-level diagram is always skewed toward the more electronegative atom. An Odd Number of Valence Electrons: NO Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O 2 with N 2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O 2 to produce NO 2 , which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals. Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $\PageIndex{13}$) shows that the general pattern is similar to that for the O 2 molecule (Figure $\PageIndex{11}$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2 p atomic orbitals, the 11th electron must occupy one of the degenerate π * orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N 2 and O 2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot. Note that electronic structure studies show the ground state configuration of $\ce{NO}$ to be $ \left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{2} $ in order of increasing energy. Hence, the $ \pi _{2p_{x,y}}$ orbitals are lower in energy than the $\sigma _{2p_{z}}$ orbital. This is because the $\ce{NO}$ molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond ( Figure $\PageIndex{11}$). Molecular orbital theory can also tell us something about the chemistry of $NO$. As indicated in the energy-level diagram in Figure $\PageIndex{13}$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $NO$ is easily oxidized to the $NO^+$ cation, which is isoelectronic with $N_2$ and has a bond order of 3, corresponding to an N≡O triple bond. Molecular Orbital Bonding for Second Row Elements: https://youtu.be/A_5Xa3sK_YE Nonbonding Molecular Orbitals Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $\PageIndex{14}$ that the 1 s orbital of atomic hydrogen is closest in energy to the 3 p orbitals of chlorine. Consequently, the filled Cl 3 s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1 s and Cl 3 p orbitals. Of the three p orbitals, only one, designated as 3 p z , can interact with the H 1 s orbital. The 3 p x and 3 p y atomic orbitals have no net overlap with the 1 s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3 s , 3 p x , and 3 p y orbitals do not change when HCl forms, they are called nonbonding molecular orbitals . A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3 p z than to the H 1 s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $ H^{\delta +} -- Cl^{\delta -} $. Electrons in nonbonding molecular orbitals have no effect on bond order. Example $\PageIndex{4}$ Use a “skewed” molecular orbital energy-level diagram like the one in Figure $\PageIndex{12}$ to describe the bonding in the cyanide ion (CN − ). What is the bond order? Given: chemical species Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order Strategy: Calculate the total number of valence electrons in CN − . Then place these electrons in a molecular orbital energy-level diagram like Figure $\PageIndex{12}$ in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and describe the bonding in CN − . Solution: A The CN − ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure $\PageIndex{12}$ fills the five lowest-energy orbitals, as shown here: Because χ N > χ C , the atomic orbitals of N (on the right) are lower in energy than those of C. B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN − ion has a triple bond, analogous to that in N 2 . Exercise $\PageIndex{4}$ Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl − ). What is the bond order? Answer: All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1. Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods. Summary Molecular orbital theory, a delocalized approach to bonding, can often explain a compound’s color, why a compound with unpaired electrons is stable, semiconductor behavior, and resonance, none of which can be explained using a localized approach. A molecular orbital (MO) is an allowed spatial distribution of electrons in a molecule that is associated with a particular orbital energy. Unlike an atomic orbital (AO), which is centered on a single atom, a molecular orbital extends over all the atoms in a molecule or ion. Hence the molecular orbital theory of bonding is a delocalized approach. Molecular orbitals are constructed using linear combinations of atomic orbitals (LCAOs) , which are usually the mathematical sums and differences of wave functions that describe overlapping atomic orbitals. Atomic orbitals interact to form three types of molecular orbitals. Orbitals or orbital lobes with the same sign interact to give increased electron probability along the plane of the internuclear axis because of constructive reinforcement of the wave functions. Consequently, electrons in such molecular orbitals help to hold the positively charged nuclei together. Such orbitals are bonding molecular orbitals , and they are always lower in energy than the parent atomic orbitals. Orbitals or orbital lobes with opposite signs interact to give decreased electron probability density between the nuclei because of destructive interference of the wave functions. Consequently, electrons in such molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. These orbitals are called antibonding molecular orbitals , and they are always higher in energy than the parent atomic orbitals. Some atomic orbitals interact only very weakly, and the resulting molecular orbitals give essentially no change in the electron probability density between the nuclei. Hence electrons in such orbitals have no effect on the bonding in a molecule or ion. These orbitals are nonbonding molecular orbitals , and they have approximately the same energy as the parent atomic orbitals. A completely bonding molecular orbital contains no nodes (regions of zero electron probability) perpendicular to the internuclear axis, whereas a completely antibonding molecular orbital contains at least one node perpendicular to the internuclear axis. A sigma (σ) orbital (bonding) or a sigma star (σ) orbital (antibonding) is symmetrical about the internuclear axis. Hence all cross-sections perpendicular to that axis are circular. Both a pi (π) orbital (bonding) and a pi star (π) orbital (antibonding) possess a nodal plane that contains the nuclei, with electron density localized on both sides of the plane. The energies of the molecular orbitals versus those of the parent atomic orbitals can be shown schematically in an energy-level diagram . The electron configuration of a molecule is shown by placing the correct number of electrons in the appropriate energy-level diagram, starting with the lowest-energy orbital and obeying the Pauli principle; that is, placing only two electrons with opposite spin in each orbital. From the completed energy-level diagram, we can calculate the bond order , defined as one-half the net number of bonding electrons. In bond orders, electrons in antibonding molecular orbitals cancel electrons in bonding molecular orbitals, while electrons in nonbonding orbitals have no effect and are not counted. Bond orders of 1, 2, and 3 correspond to single, double, and triple bonds, respectively. Molecules with predicted bond orders of 0 are generally less stable than the isolated atoms and do not normally exist. Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules , molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O 2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules , using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry.
Courses/Los_Medanos_College/Chemistry_6_and_Chemistry_7_Combined_Laboratory_Manual_(Los_Medanos_College)/01%3A_Experiments/1.08%3A_Experiment_608_Six_Unknown_Solutions_1_1_2	0 1 2 3 Student Name NaN Laboratory Date: Date Report Submitted: ___________________________ Student ID NaN Experiment Number and Name Experiment 608: Six Unknown Solutions Experiment 60 8 : S ix Unknown Solutions Section 1: Purpose and Summary Determine if a chemical reaction has taken place when two solutions are combined. Determine the identity of six unknown solutions. In this experiment, students are provided with a set of six solutions: Cu(NO 3 ) 2 , CuSO 4 , BaCl 2 , KIO 3 , Na 2 C 2 O 4 , and NaCl, and the same set of solutions in a set of ‘unidentified’ numbered/lettered bottles. Students will conduct different reactions that involve combination of two known solutions and generate experimental data that will help them identify the set of unknown solutions. Section 2: Safety Precautions and Waste Disposal Safety Precautions: Use of eye protection is recommended for all experimental procedures. Wash your hands after working with metal solutions. Do not eat in the laboratory. Waste Disposal: While you are doing the experiment, collect all waste into a beaker. When you are finished with the experiment, pour the contents of the waste beaker into the inorganic waste container in the fume hood. Section 3: Procedure To determine whether a reaction occurs or not, observe any bubble formation (gas evolution) in the solution, a change in color of the solution, or formation of a precipitate. Record your observations at the time of mixing. If there is a precipitate, write ‘ppt’, and indicate the color. Where there is no evidence of reaction, write ‘NR’. See the suggested data tables after each section below. Part 1: Known solutions 0 Obtain solutions containing the following compounds: Cu(NO3)2, CuSO4, BaCl2, KIO3, Na2C2O4, and NaCl. Obtain a clear plastic or white ceramic well plate and clean thoroughly with tap water. Rinse well with laboratory water. Shake off excess water to dry the plate. Label the top and left side of your well plate with the compounds according to the data table. In each of the wells, combine 2 to 3 drops of each pair of known solutions listed in the data table. Stir mixture with a toothpick or wooden splint. Record your observations. Dispose of reaction mixtures into a designated waste beaker. Clean your well plate thoroughly, rinse with laboratory water, and dry. 0 1 2 3 4 5 NaN NaCl KIO3 Cu(NO3)2 Na2C2O4 BaCl2 CuSO4 NaN NaN NaN NaN NaN BaCl2 NaN NaN NaN NaN X Na2C2O4 NaN NaN NaN X X Cu(NO3)2 NaN NaN X X X KIO3 NaN X X X X Part 2 : Unknown Solutions 0 Obtain a set of unknown solutions. Record the number of the set you used. This is very important! With no identification of which set you used, your work cannot be graded. Record your Unknown Set # tested here: __________ Label the top and left side of your well plate with the compounds according to the data table. In each of the wells, combine 2 to 3 drops of each pair of unknown solutions listed in the data table. Stir mixture with a toothpick or wooden splint. Record your observations. Dispose of reaction mixtures into a designated waste beaker. Clean your well plate thoroughly, rinse with laboratory water, and dry. 0 1 2 3 4 5 NaN F E D C B A NaN NaN NaN NaN NaN B NaN NaN NaN NaN X C NaN NaN NaN X X D NaN NaN X X X E NaN X X X X Section 4 : Identification of Unknown Solutions Based on your data and observations above, determine the identity of each of your unknown solutions. Clearly explain each step of your reasoning. 0 1 2 Solution Identity Experimental basis of your identification A NaN NaN B NaN NaN C NaN NaN D NaN NaN E NaN NaN F NaN NaN Post-lab Questions : Write the net ionic equation for each of the reactions that occurred in this experiment.
Courses/Lumen_Learning/Book%3A_United_States_History%3A_Reconstruction_to_the_Present_(Lumen)/19%3A_Module_17-_The_Conservative_Turn-_America_from_the_1980s_to_the_Present/19.12%3A_Assignment-_American_Conservatism	The 1960s was a liberal decade, the 1970s was a mess, and the 1980s was a conservative decade. In 1964, while helping the Goldwater campaign, a young up-and-coming figure in the Republican Party named Ronald Reagan gave a speech titled “ A Time for Choosing ” (if you don’t want to read it there you can watch it here ). This speech became a foundational address in the modern Republican Party, put Reagan on the fast-track to the White House, and defined American Conservatism from then to the present. After reading (or watching) the speech answer the following in a paragraph: What are the main points of the speech? If it’s a time for choosing, what are the choices? How did this speech set the tone for the current political right? Public domain content Assignment: American Conservatism. Authored by : Chris Thomas. Provided by : Reynolds Community College. License : Public Domain: No Known Copyright
Courses/SUNY_Oneonta/Organic_Chemistry_with_a_Biological_Emphasis_(SUNY_Oneonta)/19%3A_Appendix_I-_Index_of_enzymatic_reactions_by_pathway/19.02%3A_Amino_acid_catabolism	Sections/problems listed with an asterisk () do not discuss the exact reaction indicated, but do discuss a closely related reaction. Transaminase (EC 2.6.1.1, EC 2.6.1.2) Section 14.4E Carbamoyl phosphate synthetase (EC 6.3.4.16) P12. 4 Urea cycle: Ornithine transcarbamylase (EC 2.1.3.3) Section 12.2A,B Argininosuccinate synthetase (EC 6.3.4.5) Section 12.6, Section 14.1D Arginosuccinate lyase (EC 4.3.2.1) Section 14.1D Arginase (EC 3.5.3.1) P12.16 Glycine (glycine cleavage system) Glycine dehydrogenase (decarboxylating) (EC 1.4.4.2) P16.4b Aminomethyltransferase (EC 2.1.2.10) C11.5 Alanine (to pyruvate and glutamate): Alanine transaminase (EC 2.6.1.2) Section 14.4E Serine (to pyruvate): Serine dehydratase (EC 4.3.1.17) Section 13.1B, Section 14.4F (to glycine): Serine hydroxymethyltransferase (EC 2.1.2.1) Section 11.6D, Section 14.4D Cysteine (to pyruvate) Cysteine dioxygenase (EC 1.13.11.20) not covered Aspartate aminotransferase (2.6.1.1) Section 14.4E Threonine pathway 1 (to glycine and acetyl CoA): Threonine dehydrogenase (EC 1.1.1.103) Section 16.4D* Glycine C -acetyltransferase (EC 2.3.1.29) Section 14.4D pathway 2 (to glycine and acetaldehyde): Threonine aldolase (EC 4.1.2.5) Section 14.4D, P14.10a pathway 3 (to succinyl-CoA via propionyl CoA): Threonine dehydratase (EC 4.3.1.19) P14.2 2-oxobutanoate dehydrogenase (EC 1.2.4.4) Section 16.12B Propionyl-CoA carboxylase (EC 6.4.1.3) Section 13.5D, P13.2 Methylmalonyl-CoA mutase (EC 5.4.99.2) not covered Tryptophan (to glutaryl-CoA) Tryptophan 2,3-dioxygenase (EC 1.13.11.11) not covered Arylformamidase (EC 3.5.1.19) Chapter 12* Kynurenine 3-monooxygenase (EC 1.14.13.9) Section 16.10 Kynurenimase (EC 3.7.1.3) P14.10b 3-hydroxyanthranilate 3,4-dioxygenase (EC 1.13.11.6) not covered Aminocarboxymuconate-semialdehyde decarboxylase (EC 4.1.1.45) Section 13.5C, P13.5a 2-aminomuconate semialdehyde dehydrogenase (EC 1.2.1.32) Section 16.4D, P16.2a 2-aminomuconate deaminase (EC 3.5.99.5) Section 11.6, P11.1 2-oxoglutarate dehydrogenase (EC 1.2.4.2) Section 16.12B Asparagine (to aspartate) Asparaginase (EC 3.5.1.1) Chapter 12* Aspartate (to oxaloacetate): Aspartate transaminase (EC 2.6.1.1) Section 14.4E (to fumarate): Aspartate-ammonia lyase (EC 4.3.1.1) P14.3 Glutamine (to glutamate) Glutaminase (EC 3.5.1.2) Chapter 12 Glutamate (to alpha-ketoglutarate): Glutamate dehydrogenase (EC 1.4.1.2) Section 16.4D Arginine (to glutamate): Arginase (EC 3.5.3.1) P12.16b Ornithine transaminase (EC 2.6.1.13) Section 14.4E* Glutamate semialdehyde dehydrogenase (EC 1.2.1.41) Section 16.4D Histidine (to glutamate): Histidine ammonia-lyase (EC 4.3.1.3) C14.5 Urocanate hydratase (EC 4.2.1.49) C16.1 Imidazolonepropionase (EC 3.5.2.7) Chapter 12* Formimidoylglutamase (EC 3.5.3.8) P11.13b Valine, isoleucine, leucine: Branched chain amino acid transaminase (EC 2.6.1.42) Section 14.4E* Branched chain ketoacid dehydrogenase complex (EC 1.2.4.4) Section 16.12B* Acyl CoA dehydrogenase (eg. EC 1.3.99.13) Section 16.5C, Section 17.3C valine (to succinyl CoA): Enoyl-CoA hydratase (EC 4.2.1.17) P16.9 3-hydroxyisobutyryl-CoA hydrolase (EC 3.1.2.4) P16.9 3-hydroxyisobutyrate dehydrogenase (EC 1.1.1.31) P16.9 Methylmalonate-semialdehyde dehydrogenase (1.2.1.27) P16.9 Propionyl-CoA carboxylase (EC 6.4.1.3) Section 13.5D, P13.2 Methylmalonyl-CoA mutase (EC 5.4.99.2) not covered isoleucine (to succinyl CoA and acetyl CoA): Enoyl-CoA hydratase (EC 4.2.1.17) P16.9 Methyl-hydroxybutyryl CoA dehydrogenase (EC 1.1.1.178) Section 16.4* 3-ketoacyl-CoA thiolase (EC 2.3.1.16) Section 13.4B* Propionyl-CoA carboxylase (EC 6.4.1.3) Section 13.5D, P13.2 Methylmalonyl-CoA mutase (EC 5.4.99.2) not covered leucine (to acetyl CoA) Methylcrotonoyl-CoA carboxylase (EC 6.4.1.4) Section 13.5* Methylglutaconyl-CoA hydratase (EC 4.2.1.18) Section 14.1* Hydroxymethylglutaryl-CoA lyase (EC 4.1.3.4) Section 13.3C* Methionine (to cysteine and succinyl-CoA): Methionine adenosyltransferase (EC 2.5.1.6) E9.1 Methyltransferase (eg. 2.1.1.37) Section 9.1A Adenosylhomocysteinase (EC 3.3.1.1) Section 14.1A Cystathionine beta-synthase (EC 4.2.1.22) Section 14.4G* Cystathionine gamma-lyase (EC 4.4.1.1) Section 14.4G 2-oxobutanoate dehydrogenase (EC 1.2.4.4) Section 16.12B* Propionyl-CoA carboxylase (EC 6.4.1.3) Section 13.5D, P13.2 Methylmalonyl-CoA mutase (EC 5.4.99.2) not covered Lysine (to glutaryl-CoA) Saccharopine dehydrogenase (EC 1.5.1.8) P11.6, P16.7a Saccharopine reductase (EC 1.5.1.10) P11.7, P16.7b Aminoadipate-semialdehyde dehydrogenase (EC 1.2.1.31) Section 16.4D* 2-aminoadipate transaminase (EC2.6.1.39) Section 14.4E* 2-oxoglutarate dehydrogenase (EC 1.2.4.2) Section 16.12B* Phenylalanine (to tyrosine): Phenylalanine hydroxylase (EC 1.14.16.1) not covered Tyrosine (to fumarate and acetoacetate): Tyrosine transaminase (EC 2.6.1.5) Section 14.4E* 4-hydroxyphenylpyruvate dioxygenase (EC 1.13.11.27) not covered Homogentisate 1,2-dioxygenase (EC 1.13.11.5) not covered Maleylacetoacetate isomerase (EC 5.2.1.2) Section 14.2A* Fumarylacetoacetase (EC 3.7.1.2) Section 13.4B* Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Courses/CSU_San_Bernardino/CHEM_2100%3A_General_Chemistry_I_(Mink)/05%3A_Thermochemistry/5.07%3A_Summary	Energy is the capacity to supply heat or do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics). The thermal energy of matter is due to the kinetic energies of its constituent atoms or molecules. Temperature is an intensive property of matter reflecting hotness or coldness that increases as the average kinetic energy increases. Heat is the transfer of thermal energy between objects at different temperatures. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule (J). Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes. Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed in the process using known mathematical relations. Calorimeters are designed to minimize energy exchange between their contents and the external environment. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food. If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, q for the change is called the enthalpy change with the symbol Δ H , or for reactions occurring under standard state conditions at 298 K. The value of Δ H for a reaction in one direction is equal in magnitude, but opposite in sign, to Δ H for the reaction in the opposite direction, and Δ H is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar and 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.
Courses/BethuneCookman_University/B-CU%3A_CH-345_Quantitative_Analysis/Book%3A_Analytical_Chemistry_2.1_(Harvey)/06%3A_Equilibrium_Chemistry/6.11%3A_Some_Final_Thoughts_on_Equilibrium_Calculations	In this chapter we developed several tools to evaluate the composition of a system at equilibrium. These tools differ in how precisely they allow us to answer questions involving equilibrium chemistry. They also differ in how easy they are to use. An important part of having several tools to choose from is knowing when to each is most useful. If you need to know whether a reaction if favorable or you need to estimate a solution’s pH, then a ladder diagram usually will meet your needs. On the other hand, if you require a more accurate or more precise estimate of a compound’s solubility, then a rigorous calculation that includes activity coefficients is necessary. A critical part of solving an equilibrium problem is to know what equilibrium reactions to include. The need to include all relevant reactions is obvious, and at first glance this does not appear to be a significant problem—it is, however, a potential source of significant errors. The tables of equilibrium constants in this textbook, although extensive, are a small subset of all known equilibrium constants, which makes it easy to overlook an important equilibrium reaction. Commercial and freeware computational programs with extensive databases are available for equilibrium modeling, two examples of which are Visual Minteq (Windows only) and CurTiPot (for Excel); Visual Minteq can model acid–base, solubility, complexation, and redox equilibria; CurTiPot is limited to acid–base equilibria. Both programs account for the effect of activity. The R package CHNOSZ is used to model aqueous geochemistry systems and the properities of proteins. An integrated set of tools for thermodynamic calculations in aqueous geochemistry and geobiochemistry. Functions are provided for writing balanced reactions to form species from user-selected basis species and for calculating the standard molal properties of species and reactions, including the standard Gibbs energy and equilibrium constant. Calculations of the non-equilibrium chemical affinity and equilibrium chemical activity of species can be portrayed on diagrams as a function of temperature, pressure, or activity of basis species; in two dimensions, this gives a maximum affinity or predominance diagram. The diagrams have formatted chemical formulas and axis labels, and water stability limits can be added to Eh-pH, oxygen fugacity- temperature, and other diagrams with a redox variable. The package has been developed to handle common calculations in aqueous geochemistry, such as solubility due to complexation of metal ions, mineral buffers of redox or pH, and changing the basis species across a diagram ("mosaic diagrams"). CHNOSZ also has unique capabilities for comparing the compositional and thermodynamic properties of different proteins. Finally, a consideration of equilibrium chemistry can only help us decide if a reaction is favorable; however, it does not guarantee that the reaction occurs. How fast a reaction approaches its equilibrium position does not depend on the reaction’s equilibrium constant because the rate of a chemical reaction is a kinetic, not a thermodynamic, phenomenon. We will consider kinetic effects and their application in analytical chemistry in Chapter 13 .
Courses/Tennessee_State_University/CHEM_4210%3A_Inorganic_Chem_II_(Siddiquee)/03%3A_d-Block_Metal_Chemistry-_Coordination_Complexes/3.03%3A_Ligand_Field_Theory	10.3.1: Ligand Field Theory - Molecular Orbitals for an Octahedral Complex Crystal field theory is successful at providing some general insights into the differing energy levels of d orbitals in coordination complexes. That knowledge can help us to understand some of the magnetic properties of these compounds, which are determined by the number of unpaired electrons. It can also help us to understand some of the absorbed wavelengths of light in the UV-visible spectrum, which in some cases depends on the energy difference between different sets of d orbitals. 10.3.2: Orbital Splitting and Electron Spin 10.3.3: Ligand Field Stabilization Energy 10.3.4: Tetrahedral Complexes Tetrahedral geometry is even more common in chemistry than square planar geometry. Assessing the orbital interactions in tetrahedral geometry is somewhat more complicated, however, and it is common to proceed directly to a group theory approach. 10.3.5: Square-Planar Complexes
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/05_Instructors_Manual__Molecular_and_Atomic_Spectroscopy/01_In-class_Questions%3A_General_Background_on_Molecular_Spectroscopy	Electromagnetic Radiation What is the relationship between the energy (E) and frequency ( $\nu$ ) of electromagnetic radiation? Students usually remember the equation that E = h $\nu$ without any prompting on my part and determine that there is a direct proportionality between the two. What is the relationship between the energy and wavelength ( $\lambda$ ) of electromagnetic radiation? Students usually remember that c = $\lambda\nu$ without any prompting on my part and determine that there the energy and wavelength of radiation are inversely proportional. Write the types of radiation observed in the electromagnetic spectrum going from high to low energy. Also include what types of processes occur in atoms or molecules for each type of radiation. The students’ individual ability to identify all the different types of electromagnetic radiation and rank them in energy usually varies widely. Within a group most are able to generate a complete or close to complete list and rank those that they are most familiar with. One of the most perplexing to most students is where to put microwave radiation in the energy ranking. Identifying the types of processes that occur in atoms or molecules for each type of radiation presents more difficulties. What type of process do you already know about in molecules and what radiation produces them? Within groups they can determine that UV/VIS involves transitions of valence electrons and IR corresponds to molecular vibrations. Many are familiar with the idea of a nuclear spin flip from their organic chemistry course, although they may or may not remember that RF radiation is used to excite nuclear spin flips. Some know that it is possible to rotationally excite molecules, although they often do not know that rotational excitation occurs in the microwave region of the spectrum. It is uncommon for them to know what processes occur with $\gamma$ -rays and X-rays. Many are not familiar with the idea of an electron spin flip in paramagnetic substances and that it occurs in the microwave region of the spectrum. At this point, I briefly discuss the difference between absorbance and emission. I also discuss how different spectroscopic methods are of different utility for compound identification and compound quantification. Some techniques (e.g., NMR spectroscopy) are useful for interpretation and identification, whereas others (e.g., IR spectroscopy) are useful for identification but not that amenable to interpretation and instead require use of a computer library to determine the best match. I then go over the basic design of an absorption spectrophotometer and present them with the following series of questions on Beer’s Law. Beer’s Law What factors influence the absorbance that you would measure for a sample? Is each factor directly or inversely proportional to the absorbance? Students quickly realize that the absorbance relates to the concentration and that it is a direct proportion. They often do not think of path length as a variable, I suspect because they are given specific cuvettes to use in any prior measurement they have performed and then don’t think the path length is something they could adjust. What would be the effect of increasing the path length? This is usually sufficient for them to see that there ought to be a direct relationship between path length and absorbance. Some students are familiar with the concept of an extinction coefficient from other courses, but rarely do they have an exact understanding of the meaning of the extinction coefficient. I indicate that molar absorptivity is another term for the extinction coefficient and at this point we can write Beer’s Law on the board. How would you measure a spectrum and draw an example of a UV/VIS absorbance spectrum of a chemical species? Someone in each group usually has enough prior experience and knows that recording a spectrum involves measuring the absorbance as the wavelength is scanned. If so, they can draw a spectrum where the absorbance varies with wavelength so that there are regions of high absorbance and regions of low absorbance. Some may think of an atomic (line) spectrum whereas others think of a molecular absorbance spectrum, and I clarify that they are different but that we can consider the nature of the extinction coefficient using either of them. I also prompt them to consider the concentration and path length when someone records a spectrum, and they realize that both are fixed. Explain why the absorbance is high in some regions and low in others (or that lines in an atomic emission spectrum have different intensities). They can usually rationalize that chemical species have the ability to absorb some wavelengths of light and not others, and when pushed on the differences in intensities of lines in an atomic emission spectrum, that some transitions must have a higher likelihood of occurrence than others. They sometimes wonder whether the difference in intensities reflects differences in the detector response, so it is important to point out that it is a fundamental process taking place in the chemical species. At this point, we can now discuss how the extinction coefficient or molar absorptivity is a measure of the probability that a particular wavelength of light can be absorbed. We discuss the aspect of energy transitions and that different transitions within a chemical species have different probabilities of occurrence. I introduce the idea of selection rules and that it is appropriate to talk about the degree to which a transition is allowed. I also introduce the idea that there are some transitions that are not allowed or forbidden. If you wanted to measure the concentration of a particular species in a sample, describe the procedure you would use to do so. The groups’ first response to this is often rather superficial. They tend to think more of putting the sample into a cuvette, measuring the absorbance and somehow equating that with concentration without explicitly stating that you first need to select a wavelength to use and prepare a standard curve. Referring back to the spectra from the problem above that are still on the board, I point out that the analyst must set a wavelength. Which wavelength would you choose? They usually see right away that $\lambda$ max is the preferable one and that with the highest molar absorptivity would provide the largest response. I also push them to examine how $\lambda$ max would provide the lowest possible detection limits of any of the wavelengths. Can you imagine a situation where you would not use $\lambda$ max for the analysis? Most groups quickly realize that you would need a different wavelength if the sample had another substance in it that absorbed at $\lambda$ max . Having selected the proper wavelength, how would you relate the absorbance of the sample with an unknown concentration to the actual concentration? At this point, they realize the need to examine a sample with a known concentration and some students realize that they will need a standard curve with several concentrations whereas others may think only one known concentration is acceptable. We can then discuss the concept of a blank solution and examine how a standard curve ought to be a linear plot that goes through the origin. We also examine how the slope of the standard curve can be used to determine the molar absorptivity. Suppose a small amount of stray radiation (P S ) always leaked into your instrument and made it to your detector. This stray radiation would add to your measurements of P o and P. Would this cause any deviations to Beer's law? Explain. It is helpful to draw a picture on the board that shows a basic design of the spectrophotometer and indicates P o , P and P s . Consider the situation of a sample with a high concentration and another sample with a low concentration of analyte, and think about the relative magnitudes of the different terms at these different conditions. It can also be useful to indicate on the board the way in which the stray radiation gets incorporated into the expression for the absorbance [A = log(P o + P s )/(P + P s )], and to examine these terms at the extremes of high and low concentrations. At this point, the groups can usually rationalize that the (P + P s ) term will approach P s or a constant as the concentration of analyte is increased. When asked to draw the standard curve that would be observed, they can draw one that shows a negative deviation at higher concentrations. The derivation of Beer's Law assumes that the molecules absorbing radiation don't interact with each other (remember that these molecules are dissolved in a solvent). If the analyte molecules interact with each other, they can alter their ability to absorb the radiation. Where would this assumption break down? Guess what this does to Beer's law? Groups usually realize that the molecules are more likely to interact with each other at high concentration. Beer's law also assumes purely monochromatic radiation. Describe an instrumental set up that would allow you to shine monochromatic radiation on your sample. Is it possible to get purely monochromatic radiation using your set up? Guess what this does to Beer's law. What is meant by “purely monochromatic radiation”? We have not yet discussed the specific details of a monochromator, but based on earlier discussion related to selecting a $\lambda$ max value, they already know that some form of wavelength selection device is necessary. They are familiar with the ability of a prism to disperse radiation. I point out that gratings are more commonly used and that we will discuss gratings in more detail later in the course. With a drawing of a prism on the board, and prompted as to how they would direct only one wavelength on a sample, they realize that it will be necessary to use a slit that blocks out the unwanted wavelengths, but that the radiation passing through the device will never be purely monochromatic. At this point, without explaining it further, I indicate that polychromatic light will lead to negative deviations from Beer’s Law, especially at higher concentrations. Is there a disadvantage to reducing the slit width? What varies as one goes from a wide to a narrow slit width? The groups realize that a wide slit width gives more power (and I point out how we will equate the number of photons with power) and a wider range of wavelengths, whereas a narrow slit width gives fewer photons and a smaller range of wavelengths. Do you want high or low source power? Based on our prior discussion of the effect of stray radiation they usually realize that higher power is desirable. This is also a useful time to further introduce the presence of noise and discuss how the signal-to-noise ratio is an important consideration in spectroscopic measurements. They realize that this argues for the use of wide slits. Is there any situation where you would want to use a small slit width? The groups can usually figure out that the ability to distinguish two nearby peaks is improved with smaller slit widths. This allows us to discuss what is meant by “resolution” in spectroscopic measurements. Finally, we can examine Figure 1.5 to look at the effect of polychromatic radiation on the deviation from Beer’s law. I also draw Figure 1.6 on the board and ask what wavelength they would use for the analysis and to justify why. They readily appreciate that the broader region is better for use because of the prior discussion about deviations to Beer’s Law and because slight changes in the setting of the monochromator will have less significant effects on the measured absorbance. Consider the relative error that would be observed for a sample as a function of the transmittance or absorbance. Is there a preferable region in which to measure the absorbance? What do you think about measuring absorbance values above 1? Examine separately the extent of error that would occur at low and high concentration. Groups can usually determine that the error at the extremes of concentration is more pronounced and there must be some mid-range absorbance measurements where the error is minimized. Determine the percent transmittance that gives an absorbance value of 1 and consider the likelihood that negative deviations to Beer’s law occur in this region? They can figure out that an absorbance of 1 equals only 10% transmittance and realize that negative deviations are likely to occur, enhancing the error of the measurement and reducing the number of significant figures that could be measured. I also discuss with them whether it would ever be acceptable to use a non-linear standard curve. It is ever acceptable to extrapolate a standard curve to higher concentrations? They have enough understanding to determine that the possible onset of negative deviations to Beer’s Law means that one cannot reliably extrapolate to higher concentrations. What are some examples of matrix effects and what undesirable effect could each have that would compromise the absorbance measurement for a sample with an unknown concentration? After describing to the class what is meant by a matrix effect, groups are given a few minutes to discuss this question and I have them report out on what they identified. In the aggregate the class is usually to arrive at important variables such as pH, the possibility that other species might absorb at $\lambda$ max , and the possibility that another species interacting with the analyte could alter the value of $\lambda$ max . Prompting may be required to have them consider scatter from suspended particulate matter and that the solvent may have an effect as well. Are there any special constraints that must be considered in selecting a buffer? They can usually identify that the buffer can’t absorb at $\lambda$ max .
Courses/Lumen_Learning/Book%3A_US_History_II_(Lumen)/30%3A_The_Sixties_Reader/30.07%3A_Primary_Source-_National_Organization_for_Women_Statement_of_Purpose_(1966)	The National Organization for Women was founded in 1966 by prominent American feminists, including Betty Friedan, Shirley Chisolm, and others. The organization’s “statement of purpose” laid out the goals of the organization and the targets of its feminist vision. We, men and women, who hereby constitute ourselves as the National Organization for Women, believe that the time has come for a new movement toward true equality for all women in America, and toward a fully equal partnership of the sexes, as part of the world-wide revolution of human rights now taking place within and beyond our national borders. The purpose of NOW is to take action to bring women into full participation in the mainstream of American society now, exercising all the privileges and responsibilities thereof in truly equal partnership with men. We believe the time has come to move beyond the abstract argument, discussion and symposia over the status and special nature of women which has raged in America in recent years; the time has come to confront, with concrete action, the conditions that now prevent women from enjoying the equality of opportunity and freedom of which is their right, as individual Americans, and as human beings. NOW is dedicated to the proposition that women, first and foremost, are human beings, who like all other people in our society, must have the chance to develop their fullest human potential. We believe that women can achieve such equality only by accepting to the full the challenges and responsibilities they share with all other people in our society, as part of the decision-making mainstream of American political, economic and social life. We organize to initiate or support action, nationally, or in any part of this nation, by individuals or organizations, to break through the silken curtain of prejudice and discrimination against women in government, industry, and professions, the churches, the political parties, the judiciary, the labor unions, in education, science, medicine, law, religion and every other field of importance in American society. Enormous changes taking place in our society make it both possible and urgently necessary to advance the unfinished revolution of women toward true equality now. With a life span lengthened to nearly 75 years it is no longer either necessary or possible for women to devote the greatest part of their lives to child-rearing; yet childbearing and rearing which continues to be a most important part of most women’s lives — still is used to justify barring women from equal professional and economic participation and advance. … Despite all the talk about the status of American women in recent years, the actual position of women in the United States has declined, and is declining, to an alarming degree throughout the 1950’s and ’60s. Although 46.4% of all American women between the ages of 18 and 65 now work outside the home, the overwhelming majority — 75% — are in routine clerical, sales, or factory jobs, or they are household workers, cleaning women, hospital attendants. About two-thirds of Negro women workers are in the lowest paid service occupations. Working women are becoming increasingly — not less — concentrated on the bottom of the job ladder. As a consequence, full-time women workers today earn on the average only 60% of what men earn, and that wage gap has been increasing over the past twenty-five years in every major industry group. In 1964, of all women with a yearly income, 89% earned under $5,000 a year; behalf of all full-time year round women workers earned less than $3,690; only 1.4% of full-time year round women workers had an annual income of $10,000 or more. Further, with higher education increasingly essential in today’s society, too few women are entering and finishing college or going on to graduate or professional school. Today, women earn only one in three of the B.A.’s and M.A’s granted, and one in ten of the Ph.D.’s. In all the professions considered of importance to society, and in the executive ranks of industry and government, women are losing ground. Where they are present it is only a token handful. Women comprise less than 1% of federal judges; less than 4% of all lawyers; 7% of doctors. Yet women represent 51% of the U.S. population. And, increasingly men are replacing women in the top positions in secondary and elementary schools, in social work, and in libraries — once thought to be women’s fields. Official pronouncements of the advance in the status of women hide not only the reality of this dangerous decline, but the fact that nothing is being done to stop it. … Discrimination in employment on the basis of sex is now prohibited by federal law, in Title VII of the Civil Rights Act of 1964. …The Commission has not made clear its intention to enforce the law with the same seriousness on behalf of women as of other victims of discrimination. … Until now, too few women’s organizations and official spokesmen have been willing to speak out against these dangers facing women. Too many women have been restrained by the fear of being called “feminist.” There is no civil rights movement to speak for women, as there has been for Negroes and other victims of discrimination. The National Organization for Women must therefore begin to speak. We believe that the power of American law, and the protection guaranteed by the U.S. Constitution to the civil rights of all individuals, must be effectively applied and enforced to isolate and remove patterns of sex discrimination, to ensure equality of opportunity in employment and education, and equality of civil and political rights and responsibilities on behalf of women, as well as for Negroes and other deprived groups. We realize that women’s problems are linked to many broader questions of social justice; their solution will require concerted action by many groups. Therefore, convinced that human rights for all are indivisible, we expect to give active support to the common cause of equal rights for all those who suffer discrimination and deprivation, and we call upon other organizations committed to such goals to support our efforts toward equality for women. … We believe that this nation has a capacity at least as great as other nations, to innovate new social institutions which will enable women to enjoy true equality of opportunity and responsibility in society, without conflict with their responsibilities as mothers and homemakers…. Above all, we reject the assumption that these problems are the unique responsibility of each individual woman, rather than a basic social dilemma which society must solve. True equality of opportunity and freedom of choice for women requires such practical, and possible innovations as a nationwide network of child-care center which will make in unnecessary for women to retire completely from society until their children are grown, and national programs to provide retraining for women who have chosen the care for their own children full-time. … In the interest of the human dignity of women, we will protest, and endeavor to change, the false image of women now prevalent in the mass media, and in the texts, ceremonies, laws, and practices of our major social institutions. Such images perpetuate contempt for women by society and by women for themselves. We are similarly opposed to all policies and practices — in church, state, college, factory, or office which, in the guise of protectiveness, not only deny opportunities but also foster in women self-denigration, dependence, and evasion of responsibility, undermine their confidence in their own abilities and foster contempt for women. … We believe that women will do most to create a new image of women by acting now, and by speaking out in behalf of their own equality, freedom, and human dignity — not in pleas for special privilege, nor in enmity toward men, who are also victims of the current, half-equality between the sexes — but in an active, self-respecting partnership with men. By so doing, women will develop confidence in their own ability to determine actively, in partnership with men, the conditions of their life, their choices, their future and their society. [Source: National Organization for Women, “Statement of Purpose” (October 29, 1966). Available online via The National Organization for Women ( http://now.org/about/history/statement-of-purpose/) .] CC licensed content, Original Revision and Adaptation. Authored by : Kimlisa Duchicela. License : CC BY: Attribution CC licensed content, Shared previously The American Yawp Reader. Located at : http://www.americanyawp.com/reader.html . License : CC BY-SA: Attribution-ShareAlike
Courses/Lafayette_College/CHEM_212_213%3A_Inorganic_Chemistry_(Nataro)/03%3A_Solid_state/3.21%3A_Lattice_Energies_and_Solubility	Lattice energies can also help predict compound solubilities. Let's consider a Born-Haber cycle for dissolving a salt in water. We can imagine this as the sum of two processes: (1) the vaporization of the salt to produce gaseous ions, characterized by the lattice enthalpy, and (2) the hydration of those ions to produce the solution. The enthalpy change for the overall process is the sum of those two steps. We know that the entropy change for dissolution of a solid is positive, so the solubility depends on the enthalpy change for the overall process. Here we need to consider the trends in both the lattice energy E L and the hydration energy E H . The lattice energy depends on the sum of the anion and cation radii (r + + r - ), whereas the hydration energy has separate anion and cation terms. Generally the solvation of small ions (typically cations) dominates the hydration energy because of the 1/r 2 dependence. \[E_{L} \alpha \frac{1}{r_{+}+ r_{-}}\] \[E_{H} \alpha \frac{1}{r_{+}^{2}} + \frac{1}{r_{-}^{2}}\] For salts that contain large anions, E L doesn't change much as r + changes. That is because the anion dominates the r + + r - term in the denominator of the formula for E L . On the other hand, E H changes substantially with r + , especially for small cations. As a result, sulfate salts of small divalent cations, such as MgSO 4 (epsom salts), are soluble, whereas the lower hydration energy of Ba 2+ in BaSO 4 makes that salt insoluble (K sp = 10 -10 ). 0 Left: EL diagram for sulfate salts. The large SO42- ion is size-mismatched to small cations such as Mg2+, which have large hydration energies, resulting soluble salts. With larger cations such as Ba2+, which have lower EH, the lattice energy exceeds the solvation enthalpy and the salts are insoluble.. Right: In the case of small anions such as F- and OH-, the lattice energy dominates with small cations such as transition metal ions (TMn+), Mg2+, and Li+. Anion-cation size mismatch occurs with larger cations, such as Cs+ and Ba2+, which make soluble fluoride salts. For small anions, E L is more sensitive to r + , whereas E H does not depend on r + as strongly. For fluorides and hydroxides, LiF is slightly soluble whereas CsF is very soluble, and Mg(OH) 2 is insoluble whereas Ba(OH) 2 is very soluble. Putting both trends together, we see that low solubility is most often encountered when the anion and cation match well in their sizes , especially when one or both are multiply charged . 0 Space-filling models showing the van der Waals surfaces of Ba2+ and SO42-. The similarity in size of the two ions contributes to the low solubility of BaSO4 in water. Combining all our conclusions about solubility, we note the following trends: 1) Increasing size mismatch between the anion and cation leads to greater solubility, so CsF and LiI are the most soluble alkali halides. 2) Increasing covalency leads to lower solubility in the salts (due to larger E L . For example, AgF, AgCl, AgBr, and AgI exhibit progressively lower solubility because of increasing covalency. $\ce{AgF > AgCl > AgBr > AgI}$ 3) Increasing the charge on the anion lowers the solubility because the increase in E L is large relative to the increase in E H . 4) Small, polyvalent cations (having large E H ) make soluble salts with large, univalent anions such as I - , NO 3 - , ClO 4 - , PF 6 - , and acetate. Examples: Salts of transition metal and lanthanide ions Ln 3+ : Nitrate salts are soluble, but oxides and hydroxides are insoluble. Fe 3+ : Perchlorate is soluble, but sulfate is insoluble. 5) Multiple charged anions such as O 2- , S 2- , PO 4 3- , and SO 4 2- make insoluble salts with most M 2+ , M 3+ , and M 4+ metals.
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/04%3A_Protein_Structure/4.8%3A_H._Protein_Aggregates_and_Disease/H2._Prions_and_Disease	The normal cellular form of this protein, PrPc is highly conserved in mammals, and is widely expressed in embryogenesis. Techniques exist to delete or make ineffectual genes in mice. When a mice knockout of the PrPc (i.e. the gene for the protein was deleted in all cells) was made, the mice appeared normal. More recent data suggests however, that these mice had altered circadian rhythms and sleep patterns, which suggest a possible link to Fatal Familial Insomnia. The PrPc is a normal membrane protein in neurons. It is anchored to the membrane through a glycosyl-phosphatidyl inositol link, with the protein chain on the outside of the neuronal plasma membrane. The PrPc (without the PI link) is water soluble, protease sensitive, and consists of 42% alpha helix and 3% beta sheet. Jmol : Updated Prion Protein, Mad Cow Disease, and Mutations Jmol14 (Java) \| JSMol (HTML5) The problem in the transmissible spongiform encephalopathies (TSE's) is that amyloid-like protein aggregates form which appear to be neurotoxic. The protein found in the plaques (in cases other that those that are inherited) has the same primary sequence as the PrPc but a different secondary and presumably tertiary structure. The protein found in the plaques, called the PrPsc (the scrapie form of the the normal protein) is insoluble in aqueuous solution, protease resistant, and has a high beta sheet content (43%) and lower alpha helix content (30%) than the normal version of the protein PrPc. Figure: Cartoon Models of PrPc and PrPsc A genetic, inheritable form of disease also exists, in which a mutant form of the PrPc occurs, whose normal structure is destabilized by the mutation. The aggregates caused by the mutant form of the disease are understandable in light of the other diseases which we discussed above. The question is how does the normal PrPc form PrPsc . Evidence shows that if radiolabeled PrPc from scrapie free cells is added to unlabeled PrPsc from scapie infected cells, the PrPc is converted to PrPsc! It appears that the PrPc protein has two forms not that much different in energy, one composed of mostly alpha helix and the other of beta sheet. A dimer of PrP c .PrPsc might form, which destabilizes the PrPc causing a conformational shift to the PrPsc form, which would then aggregate. Exposure to the PrPsc form would then catalyze the conversion of normal PrPc to PrPsc . Hence, it would be transmissible by contact with just the PrPsc form of the protein. Likewise species specificity could be explained if only dimers of PrP c .PrPsc formed from proteins of the same species could occur. The inherited form of the disease would be explained since the mutant form of the normal protein would more easily form the beta structure found in the aggregate. It has recently been found that the very same mutation in PrPc, Asp178Asn can cause two different diseases - CJD and FFI. Which disease you get depends on if you have 1 of two naturally occurring, nonharmful variants at amino acid 129 of the normal PrPc gene. If you have a Met at that position, and acquire the Asp178Asn mutation, you get CJD. If, on the other hand, you have a Val at amino acid 129 and acquire the Asp178Asn mutation, you get FFI. This disease was first observed in 1986 and has been reported in five families in the world. It occurs in the late 50's, equally in men and women. It is characterized by a progressive loss of the ability to sleep and disrupted circadian rhythms. The brain shows neuronal loss. It is known that amino acids 129 and 178 occur at the start of alpha helices, as predicted from propensity calculations. Chronic exposure to micromolar levels of synthetic fragment 106-126 of PrPc kills hippocampal neurons. This peptide also has the greatest tendency to aggregate synthetic PrPc peptides. A series of recent studies have expanded on our knowledge of prion structure. Nelson et al. have obtained the crystal structure of a fibril aggregate made of a short peptide (7 amino acids) from the yeast prion protein Sup35. As presumably occurs in amyloid fibers, these crystals show beta-sheet structures stacked vertically to produce fibril structures. The unit of stacking appears to be pairs of beta-sheets, with the inner side amino acids of one member of the pair interacting with the inner amino acid side chains of the other member of the pair, in a process which excludes water. Similar studies by Ritter et al, using NMR and fluorescence, found pair of beta sheets to be the motif of the fibril. Using fluorescence, they identified two regions, each 15 amino acids, important in collapse to molten-globule like state for nucleation of fibril formation. Kuru killed many members of the Fore tribe in New Guinea until the cannibalistic practice of eating dead relatives was stopped. Analysis of the genes for the prion protein in the Fore tribe and other ethnic groups in the world show two version differing by just one amino acid in all people (remember that a single gene is represented in both maternal and paternal chromosomes. That these two forms exist through the world suggest that they have been selected for by evolution and confer some biological advantage. People who have just one form of the protein are more susceptible to the development of prion diseases. Mead and Collinge have shown that about 75% of older Fore women (who had lived through cannibalistic practices) had two different prion genes, compared to about 15% of women from other ethnic groups. This high percentage suggests that these women were protected from the disease, leading through natural selection to a high percentage of heterozygotes in this defined population. The general presence of two forms of the prion gene (which probably offers protection from prion disease) suggests that cannibalism might have been widespread in our early ancestors. There appears to be one main difference between the formation of amyloid fibers from prion proteins and others such as mutant lysozymes. If you add mutant lysozyme to normal lysozyme, the amyloid fibers contain only the mutant protein. However, if you incubate mutant prion proteins with normal prions, the normal proteins become pathological. QED - Protein Aggregates are not just test tube artifacts, but rather matters of life and death. CDC: Prion Disease Jsmol: Protopedia - Prions
Courses/can/CHEM_231%3A_Organic_Chemistry_I_Textbook/10%3A_Organohalides/10.03%3A_Preparing_Alkyl_Halides_from__Alkanes_-_Radical_Halogenation	As we saw briefly in Section 6.7, simple alkyl halides can sometimes be prepared by the radical reaction of an alkane with Cl 2 or Br 2 in the presence of ultraviolet light. The detailed mechanism is shown in Figure $\PageIndex{1}$ for chlorination. Radical substitution reactions require three kinds of steps: initiation, propagation , and termination . Once an initiation step has started the process by producing radicals, the reaction continues in a self-sustaining cycle. The cycle requires two repeating propagation steps in which a radical, the halogen, and the alkane yield alkyl halide product plus more radical to carry on the chain. The chain is occasionally terminated by the combination of two radicals. Unfortunately, alkane halogenation is a poor synthetic method for preparing alkyl halides because mixtures of products invariably result. For example, chlorination of methane does not stop cleanly at the monochlorinated stage but continues to give a mixture of dichloro, trichloro, and even tetrachloro products. The situation is even worse for chlorination of alkanes that have more than one kind of hydrogen. Chlorination of butane, for instance, gives two monochlorinated products in a 30 : 70 ratio in addition to multiply chlorinated products such as dichlorobutane, trichlorobutane, and so on. As another example, 2-methylpropane yields 2-chloro-2-methylpropane and 1-chloro-2-methylpropane in a 35 : 65 ratio, along with more highly chlorinated products. From these and similar reactions, it’s possible to calculate a reactivity order toward chlorination for different kinds of hydrogen atoms in a molecule. Take the butane chlorination, for instance. Butane has six equivalent primary hydrogens (−CH 3 ) and four equivalent secondary hydrogens (−CH 2 −). The fact that butane yields 30% of 1-chlorobutane product means that each one of the six primary hydrogens is responsible for 30% ÷ 6 = 5% of the product. Similarly, the fact that 70% of 2-chlorobutane is formed means that each of the four secondary hydrogens is responsible for 70% ÷ 4 = 17.5% of the product. Thus, a secondary hydrogen reacts 17.5% ÷ 5% = 3.5 times as often as a primary hydrogen. A similar calculation for the chlorination of 2-methylpropane indicates that each of the nine primary hydrogens accounts for 65% ÷ 9 = 7.2% of the product, while the single tertiary hydrogen (R 3 CH) accounts for 35% of the product. Thus, a tertiary hydrogen is 35% ÷ 7.2% = 5 times as reactive as a primary hydrogen toward chlorination. The observed reactivity order of alkane hydrogens toward radical chlorination can be explained by looking at the bond dissociation energies given previously in Table 6.3. The data show that a tertiary C−H bond (400 kJ/mol; 96 kcal/mol) is weaker than a secondary C−H bond (410 kJ/mol; 98 kcal/mol), which is in turn weaker than a primary C−H bond (421 kJ/mol; 101 kcal/mol). Since less energy is needed to break a tertiary C−H bond than to break a primary or secondary C−H bond, the resultant tertiary radical is more stable than a primary or secondary radical. Exercise $\PageIndex{1}$ Draw and name all monochloro products you would expect to obtain from radical chlorination of 2-methylpentane. Which, if any, are chiral? Answer Chiral: 1-chloro-2-methylpentane, 3-chloro-2-methylpentane, 2-chloro-4-methylpentane Achiral: 2-chloro-2-methylpentane, 1-chloro-4-methylpentane Exercise $\PageIndex{2}$ Taking the relative reactivities of primary, secondary, and tertiary hydrogens atoms into account: what product(S) would you expect to obtain from monochloration of 2-methylbutane ? What would the approximate percentage of each product be? (don't forget to take into account the number of each kind of hydrogen) Answer 1-Chloro-2-methylbutane (29%), 1-chloro-3-methylbutane (14%), 2-chloro-2-methylbutane (24%), 2-chloro-3-methylbutane (33%)
Courses/University_of_Connecticut/Chem_2444%3A_(Second_Semester_Organic_Chemistry)_UConn/01%3A_Structure_Determination_(Mass_Spectrometry_IR_and_UV-Vis_Spectroscopy)/1.02%3A_Interpreting_Mass_Spectra	Objectives After completing this section, you should be able to suggest possible molecular formulas for a compound, given the m/z value for the molecular ion, or a mass spectrum from which this value can be obtained. predict the relative heights of the M+·, (M + 1)+·, etc., peaks in the mass spectrum of a compound, given the natural abundance of the isotopes of carbon and the other elements present in the compound. interpret the fragmentation pattern of the mass spectrum of a relatively simple, known compound (e.g., hexane). use the fragmentation pattern in a given mass spectrum to assist in the identification of a relatively simple, unknown compound (e.g., an unknown alkane). Study Notes When interpreting fragmentation patterns, you may find it helpful to know that, as you might expect, the weakest carbon-carbon bonds are the ones most likely to break. You might wish to refer to the table of bond dissociation energies when attempting problems involving the interpretation of mass spectra. This page looks at how fragmentation patterns are formed when organic molecules are fed into a mass spectrometer, and how you can get information from the mass spectrum. The Origin of Fragmentation Patterns When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion and is often given the symbol M + or . The dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process. The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical. The uncharged free radical will not produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump. The ion, X + , will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this: Note The pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element . With an element, each line represents a different isotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up. In the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion . The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak . This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion. Using Fragmentation Patterns This section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page. Example 12.2.1: Pentane Let's have another look at the mass spectrum for pentane: What causes the line at m/z = 57? How many carbon atoms are there in this ion? There cannot be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C 4 H 9 + then? C 4 H 9 + would be [CH 3 CH 2 CH 2 CH 2 ] + , and this would be produced by the following fragmentation: The methyl radical produced will simply get lost in the machine. The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion: The line at m/z = 29 is typical of an ethyl ion, [CH 3 CH 2 ] + : The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process. Example 12.2.2: Pentan-3-one This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this is not produced by the same ion as the same m/z value peak in pentane. If you remember, the m/z = 57 peak in pentane was produced by [CH 3 CH 2 CH 2 CH 2 ] + . If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it. Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH 3 CH 2 CO] + - which is produced by this fragmentation: You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group. Peak Heights and Stability The more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. We'll look at two common examples of this. Carbocations (carbonium ions) Summarizing the most important conclusion from the page on carbocations: Order of stability of carbocations primary < secondary < tertiary Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms. Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by: The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by: You would get the same ion, of course, if the left-hand CH 3 group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation. Acylium ions, [RCO] + Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one. The base peak, at m/z=57, is due to the [CH 3 CH 2 CO] + ion. We've already discussed the fragmentation that produces this. Note The more stable an ion is, the more likely it is to form. The more of a particular ion that is formed, the higher will be its peak height. Using mass spectra to distinguish between compounds Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra. 0 1 2 pentan-2-one NaN CH3COCH2CH2CH3 pentan-3-one NaN CH3CH2COCH2CH3 Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this: [CH 3 CO] + [COCH 2 CH 2 CH 3 ] + That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind: [CH 3 CH 2 CO] + In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one. The two mass spectra look like this: As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentation patterns that can occur. Provided you have a computer data base of mass spectra, any unknown spectrum can be computer analyzed and simply matched against the data base. Exercises
Bookshelves/General_Chemistry/Chemistry_Labs_(Trufan_and_Bouhoutsos-Brown)/02%3A_Exp_2-_Crystal_Growth_part_1/2.04%3A_Discussion	Discussion Write a minimum one-page (12 font, single spaced) discussion on the experiment conducted this week. Address at least one question in each category as fully as possible integrating the collected data, providing explanations for the observed trends and evaluating whether your original assumptions about the experiment were validated by the results. The assignment will be graded on completeness, clarity of the explanations and the meaningful integration of the collected and calculated data. Correct grammar and appropriate format for the chemical formulae and chemical reactions is expected. You may use the outline included at the end of this document on how to build your essay to address each category. (Existing knowledge, research and views) Describe the contents of a solution and indicate the purpose for each. (Existing knowledge, research and views) Describe at least one methods that you could use to determine if a substance is soluble in water. (Existing knowledge, research, and views) Describe the difference between saturated and unsaturated solutions. (Analysis) Consider the steps involved in preparing the saturated alum solution. Classify the solution as saturated or unsaturated in each step and explain your choice. (Analysis) Provide at least one reason for using hot water to dissolve the alum. (Analysis) Provide at least one supported argument for waiting for the solution to cool to room temperature before filtration. (Lab skill) Describe the filtration process in as much detail as you deem necessary. Provide an explanation for the necessity of each step involved. (Existing knowledge, research and views) Solids can be crystalline or amorphous. Describe at least 3 characteristics of crystalline solids. (Analysis) Describe what happens with the saturated alum solution that causes the appearance and growth of the seed crystals. (Lab skill) Describe the process of monitoring and maintaining the seed crystal. (Analysis) Predict what would happen, if you left all your little crystals in the Petri dish instead of transferring a few into the saturated alum solution, and provide a supported argument for your prediction. (Lab skill) Describe what kind of crystals you would select to start your seed crystal growth and what method you would use to transfer the selected crystal from one dish to another. (Experiment design) Using your experience growing the alum seed crystals, propose a protocol for growing crystals from other compounds, such as copper(II) sulfate, sucrose, sodium chloride, etc. Pick one compound to develop your protocol for. (Existing knowledge, research and views) Growing crystals is visually impressive and useful. Describe at least one application for growing crystals. Recommended outline Solution are composed of ………………and ………. In our experiment the solvent is ………… and the solute is ……………. We know that we have a solution because ……………… The solution is saturated when ………………. and it is unsaturated when ……………….. Solubility depends on temperature. At higher temperature, alum ………………. In the process of making the alum solution we started with hot water because …………………... and we cooled the solution before filtration because ………………………… The steps required for filtration are …………………………… In the first step it is important to ……………… In the second step ……………………. (continue for each step). Crystals form when ………………………. The main characteristics of crystals are ………………………. Seed crystals should be …………………………….. and of an approximate size of ……………… Seed crystals are necessary for ……………….. When monitoring the crystal growth
Courses/San_Francisco_State_University/General_Physical_Chemistry_I_(Gerber)/05%3A_Entropy_and_The_Second_Law_of_Thermodynamics/5.09%3A_The_Statistical_Definition_of_Entropy_is_Analogous_to_the_Thermodynamic_Definition	We learned earlier in 20-5 that entropy, $S$, is related to the number of microstates, $W$, in an ensemble with $A$ systems: \[S_{ensemble}=k_B \ln{W} \label{eq1} \] and \[W=\frac{A!}{\prod_j{a_j}!} \label{eq2} \] Combining Equations \ref{eq1} and \ref{eq2} to get: \[\begin{split} S_{ensemble} &= k_B \ln{\frac{A!}{\prod_j{a_j}!}} \\ &= k_B \ln{A!}-k_B\sum_j{\ln{a_j}!} \end{split} \nonumber \] Using Sterling's approximation: \[\ln{A!} \approx A\ln{A}-A \nonumber \] We obtain: \[S_{ensemble} = k_B A \ln{A}-k_BA-k_B\sum_j{a_j\ln{a_j}}+k_B\sum{a_j} \nonumber \] Since: \[A=\sum{a_j} \nonumber \] The expression simplifies to: \[S_{ensemble} = k_B A \ln{A}-k_B\sum_j{a_j\ln{a_j}} \nonumber \] We can make use of the fact that the number of microstates in state $j$ is equal to the total number of microstates multiplied by the probability of finding the system in state $j$, $p_j$: \[a_j=p_jA \nonumber \] Plugging in, we obtain \[\begin{split}S_{ensemble} &= k_B A \ln{A}-k_B\sum_j{p_jA\ln{p_jA}} \\ &= k_B A \ln{A}-k_B\sum_j{p_jA\ln{p_j}}-k_B\sum_j{p_jA\ln{A}} \end{split} \nonumber \] Since $A$ is a constant and the sum of the probabilities of finding the system in state $j$ is always 1: \[\sum{p_j}=1 \nonumber \] The first and last term cancel out: \[S_{ensemble} = -k_BA\sum_j{p_j\ln{p_j}} \nonumber \] We can use that the entropy of the system is the entropy of the ensemble divided by the number of systems: \[S_{system}=S_{ensemble}/A \nonumber \] Dividing by $A$, we obtain: \[S_{system} = -k_B\sum_j{p_j\ln{p_j}} \nonumber \] We can differentiate this equation and dropping the subscript: \[dS = -k_B\sum_j{\left(dp_j+\ln{p_j}dp_j\right)} \nonumber \] Since $\sum_j{p_j}=1$, the derivative $\sum_j{dp_j}=0$: \[dS = -k_B\sum_j{\ln{p_j}dp_j} \nonumber \] In short: \[\sum_j{\ln{p_j}dp_j}=-\frac{\delta q_{rev}}{k_BT} \nonumber \] Plugging in: \[dS = \frac{\delta q_{rev}}{T} \nonumber \]
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Group_Theory/Woodward-Hoffmann_rules	The Woodward-Hoffman rules (R. B. Woodward was one of the U.S.'s most successful and renowned organic chemists; Roald Hoffmann is a theoretical chemist. Hoffmann shared the 1981 Nobel Prize with Kenichi Fukui) allow us to follow the symmetry of the occupied molecular orbitals along a suggested reaction path connecting reactants to products and to then suggest whether a symmetry-imposed additional energy barrier (i.e., above any thermodynamic energy requirement) would be expected. For example, let us consider the disrotatory closing of 1,3-butadiene to produce cyclobutene. The four $p$ orbitals of butadiene, denoted $\pi_1$ through $\pi_4$ in the figure shown below, evolve into the $\sigma$ and $\sigma^{}$ and $\pi$ and $\pi^{}$ orbitals of cyclobutene as the closure reaction proceeds. Along the disrotatory reaction path, a plane of symmetry is preserved (i.e., remains a valid symmetry operation) so it can be used to label the symmetries of the four reactant and four product orbitals. Labeling them as odd (o) or even (e) under this plane and then connecting the reactant orbitals to the product orbitals according to their symmetries, produces the orbital correlation diagram shown below. Figure: Professor Roald Hoffmann (right) The four $\pi$ electrons of 1,3-butadiene occupy the $\pi_1$ and $\pi_2$ orbitals in the ground electronic state of this molecule. Because these two orbitals do not correlate directly to the two orbitals that are occupied ($\sigma$ and $\pi$) in the ground state of cyclobutene, the Woodward-Hoffmann rules suggest that this reaction, along a disrotatory path, will encounter a symmetry-imposed energy barrier. In contrast, the excited electronic state in which one electron is promoted from $\pi_2$ to $\pi_3$ (as, for example, in a photochemical excitation experiment) does correlate directly to the lowest energy excited state of the cyclobutene product molecule, which has one electron in its p orbital and one in its p * orbital. Therefore, these same rules suggest that photo-exciting butadiene in the particular way in which the excitation promotes one electron from $\pi_2$ to $\pi_3$ will produce a chemical reaction to generate cyclobutene.
Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens	The alpha carbon is the first carbon atom attached to a functional group, such as a carbonyl. T he carbonyl group makes the alpha hydrogen (the hydrogen on the alpha carbon) slightly acidic via a resonance stabilization mechanism. This results in several different reactions of note. Acetoacetic Ester Synthesis The acetoacetic ester synthesis allows for the conversion of ethyl acetoacetate into a methyl ketone with one or two alkyl groups on the alpha carbon. Acidity of Alpha Hydrogens & Keto-enol Tautomerism Aldol Reaction A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Alpha Alkylation The alpha alkylation reaction involves an α hydrogen being replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as other Sn2 reactions. Alpha Halogenation A carbonyl containing compound with α hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. Claisen Reactions Because esters can contain αα hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed; the product is a β-keto ester. Deuterium Exchange Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D2O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D2O is required. The end result is the complete exchange of all α hydrogens with deuterium. Enamine Reactions As previously seen, aldehydes and ketones react with 2 o amines to reversibly form enamines. Example Reversible Enamines act as nucleophiles in a fashion similar to enolates. Because of this enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps: 1) Formation of the enamine, 2) Reaction with an eletrophile to form an iminium salt, 3) Hydrolysis of the iminium salt to reform the aldehyde or ketone. Some of the advantages of using an enamine over and enolate are enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. These reactions are generally known as the Stork enamine reaction after Gilbert Stork of Columbia University who originated the work. Typically we use the following 2 o amines for enamine reactions Malonic Ester Synthesis Michael Additions & Robinson Annulation Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942). Synthesis of Enols and Enolates Template:HideTOC
Courses/City_College_of_San_Francisco/Chemistry_101B/03%3A_Acid-Base_Equilibria/3.2%3A_pH_and_pOH	Learning Objectives Explain the characterization of aqueous solutions as acidic, basic, or neutral Express hydronium and hydroxide ion concentrations on the pH and pOH scales Perform calculations relating pH and pOH As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ( K w ). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions. A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm: \[\mathrm{pX=−\log X} \label{1}\] The pH of a solution is therefore defined as shown here, where [H 3 O + ] is the molar concentration of hydronium ion in the solution: \[\mathrm{pH=-\log[H_3O^+]}\label{$\PageIndex{2}$}\] Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression: \[\mathrm{[H_3O^+]=10^{−pH}}\label{$\PageIndex{3}$}\] Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH: \[\mathrm{pOH=-\log [OH^−]}\label{$\PageIndex{4}$}\] or \[\mathrm{[OH^-]=10^{−pOH}} \label{$\PageIndex{5}$}\] Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_w$ expression: \[K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{$\PageIndex{6}$}\] \[-\log K_\ce{w}=\mathrm{-\log([H_3O^+][OH^−])=-\log[H_3O^+] + -\log[OH^-]}\label{$\PageIndex{7}$}\] \[\mathrm{p\mathit{K}_w=pH + pOH} \label{$\PageIndex{8}$}\] At 25 °C, the value of $K_w$ is $1.0 \times 10^{−14}$, and so: \[\mathrm{14.00=pH + pOH} \label{$\PageIndex{9}$}\] The hydronium ion molarity in pure water (or any neutral solution) is $ 1.0 \times 10^{-7}\; M$ at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: \[\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{10}\] \[\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{11}\] And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $ 1.0 \times 10^{-7}\; M$ and hydroxide ion molarities less than $ 1.0 \times 10^{-7}\; M$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $ 1.0 \times 10^{-7}\; M$ and hydroxide ion molarities greater than $ 1.0 \times 10^{-7}\; M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00). When $pH=7$ Solutions are not Neutral Since the autoionization constant $K_w$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the hydronium molarity of pure water at 80 °C is 4.9 × 10 −7 M , which corresponds to pH and pOH values of: \[\begin{align} pH &=-\log[\ce{H_3O^+}] \\[4pt] &= -\log(4.9\times 10^{−7}) \\[4pt] &=6.31 \label{12} \end{align}\] \[\begin{align} pOH &=-\log[\ce{OH^-}]\\[4pt] & =-\log(4.9\times 10^{−7}) \\[4pt] &=6.31 \label{13}\end{align}\] At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 °C) (Table $\PageIndex{1}$). Classification Relative Ion Concentrations pH at 25 °C acidic [H3O+] > [OH−] pH < 7 neutral [H3O+] = [OH−] pH = 7 basic [H3O+] < [OH−] pH > 7 Figure $\PageIndex{1}$ shows the relationships between [H 3 O + ], [OH − ], pH, and pOH, and gives values for these properties at standard temperatures for some common substances. Example $\PageIndex{1}$: Calculation of pH from $\ce{[H_3O^+]}$ What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of $1.2 \times 10^{−3}\; M$? Solution \[\begin{align} pH &=-\log [H_3O^+] \\[4pt] &= -\log(1.2 \times 10^{−3}) \\[4pt] &=−(−2.92) \\[4pt]&=2.92 \end{align}\] Exercise $\PageIndex{1}$ Water exposed to air contains carbonic acid, H 2 CO 3 , due to the reaction between carbon dioxide and water: \[\ce{CO2(aq) + H2O (l) \rightleftharpoons H2CO3(aq)} \nonumber\] Air-saturated water has a hydronium ion concentration caused by the dissolved $\ce{CO_2}$ of $2.0 \times 10^{−6}\; M$, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C. Answer 5.70 Example $\PageIndex{2}$: Calculation of Hydronium Ion Concentration from pH Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline). Solution \[\mathrm{pH=-\log[H_3O^+]=7.3} \nonumber\] \[\mathrm{\log[H_3O^+]=−7.3} \nonumber\] \[\mathrm{[H_3O^+]=10^{−7.3}} \nonumber\] or \[[\ce{H_3O^+}]=\textrm{antilog of} −7.3 \nonumber\] \[[\ce{H_3O^+}]=5\times 10^{−8}\;M \nonumber\] (On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10 −7.3 .) Exercise $\PageIndex{2}$ Calculate the hydronium ion concentration of a solution with a pH of −1.07. Answer 12 M Environmental Science Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO 2 which forms carbonic acid: \[\ce{H2O (l) + CO2(g) ⟶ H2CO3(aq)} \label{14}\] \[\ce{H2CO3(aq) \rightleftharpoons H^+(aq) + HCO3^- (aq)} \label{15}\] Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO 2 , SO 2 , SO 3 , NO, and NO 2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: \[\ce{H2O (l) + SO3(g) ⟶ H2SO4(aq)} \label{16}\] \[\ce{H2SO4(aq) ⟶ H^+(aq) + HSO4^- (aq)} \label{17}\] Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure $\PageIndex{2}$). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. Example $\PageIndex{3}$: Calculation of pOH What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH? Solution Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH − ] = 0.0125 M: \[\mathrm{pOH=-\log[OH^− ]=-\log 0.0125}\] \[=−(−1.903)=1.903\] The pH can be found from the $\ce{pOH}$: \[\mathrm{pH+pOH=14.00}\] \[\mathrm{pH=14.00−pOH=14.00−1.903=12.10}\] Exercise $\PageIndex{3}$ The hydronium ion concentration of vinegar is approximately $4 \times 10^{−3}\; M$. What are the corresponding values of pOH and pH? Answer pOH = 11.6, pH = 14.00 - pOH = 2.4 The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure $\PageIndex{3}$). The pH of a solution may also be visually estimated using colored indicators (Figure $\PageIndex{3}$). Summary The concentration of hydronium ion in a solution of an acid in water is greater than $ 1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of hydroxide ion in a solution of a base in water is greater than $ 1.0 \times 10^{-7}\; M$ at 25 °C. The concentration of H 3 O + in a solution can be expressed as the pH of the solution; $\ce{pH} = -\log \ce{H3O+}$. The concentration of OH − can be expressed as the pOH of the solution: $\ce{pOH} = -\log[\ce{OH-}]$. In pure water, pH = 7.00 and pOH = 7.00 Key Equations $\ce{pH}=-\log[\ce{H3O+}]$ $\ce{pOH} = -\log[\ce{OH-}]$ [H 3 O + ] = 10 −pH [OH − ] = 10 −pOH pH + pOH = p K w = 14.00 at 25 °C Glossary acidic describes a solution in which [H 3 O + ] > [OH − ] basic describes a solution in which [H 3 O + ] < [OH − ] neutral describes a solution in which [H 3 O + ] = [OH − ] pH logarithmic measure of the concentration of hydronium ions in a solution pOH logarithmic measure of the concentration of hydroxide ions in a solution
Bookshelves/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/05%3A_Coordination_Chemistry_I_-_Structures_and_Isomers/Concept_Review_Questions_Chapter_5	Chapter 5 Section 1 & Section 2 1. Explain how the name “chelate” ligand originates. Section 3 1. What are the main factors determining coordination numbers? 2. Is the coordination number 1 common? Explain. 3. For which electron configuration are the coordination numbers 2 and 3 common? 4. What is the structure of methyl lithium? 5. What electron configuration is common for square planar shapes? 6. What is the most common shape for the coordination number 4? 7. Which are the most common shapes for the coordination number 5? 8. What is the Berry pseudo rotation? Describe its mechanism. 9. What are the most common shapes of coordination compounds with the coordination number 7? What are the point groups associated with these shapes. 10. Explain briefly, why molecular coordination compounds with cubic shapes are not known, while square anti-prisms are relatively common. 11. Explain why coordination compounds with high coordination numbers (7 and larger) are not very common. 12. Name an example of a coordination compound with the coordination number 9. Section 4 1. What is the difference between a constitutional isomer and a stereoisomer. 2. Name four types of constitutional isomerism. 3. What is the definition of hydrate isomerism? 4. Why is hydrate isomerism only possible in solid state? 5. What is the definition of ionization isomerism? 6. What is the definition of coordination isomerism? 7. What is the definition of linkage (ambidentate) isomerism? 8. What is the difference between a diasteromer and an enantiomer? 9. What is the definition of cis-trans isomerism? 10. What is the definition of fac-mer isomerism? 11. What is the definition of a propeller complex? 12. What is the definition of a left-handed/right-handed propeller? Dr. Kai Landskron ( Lehigh University ). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate .
Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/14%3A_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives/14.01%3A_Prelude_to_Organic_Acids_and_Bases_and_Some_of_Their_Derivatives	Organic acids have been known for ages. Prehistoric people likely made acetic acid when their fermentation reactions went awry and produced vinegar instead of wine. The Sumerians (2900–1800 BCE) used vinegar as a condiment, a preservative, an antibiotic, and a detergent. Citric acid was discovered by an Islamic alchemist, Jabir Ibn Hayyan (also known as Geber), in the 8th century, and crystalline citric acid was first isolated from lemon juice in 1784 by the Swedish chemist Carl Wilhelm Scheele. Medieval scholars in Europe were aware that the crisp, tart flavor of citrus fruits is caused by citric acid. Naturalists of the 17th century knew that the sting of a red ant’s bite was due to an organic acid that the ant injected into the wound. The acetic acid of vinegar, the formic acid of red ants, and the citric acid of fruits all belong to the same family of compounds—carboxylic acids. Soaps are salts of long-chain carboxylic acids. Prehistoric people also knew about organic bases—by smell if not by name; amines are the organic bases produced when animal tissue decays.
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/03%3A_Lipid_Structure/3.3%3A_Dynamics_of_Membrane_Lipids/Thermodynamics_Review/C._Change_in_Free_Energy_G	The total Δ G can be expressed as the sum of the two contributions showing the effects of the intrinsic stability (Keq) and concentration: Δ G = Δ G stab + Δ G conc ΔG = Δ G o + RTlnQrx = Δ G o + RTln ( [P][Q])/([A][B] ) for the reaction A + B <=> P + Q, where Δ G oreflects the contribution from the relative intrinsic stability of reactants and products) and RTlnQrx reflects the contribution from the relative concentrations of reactants and products (which has nothing to do with stability). Qrx is the reaction quotient which for the reaction A + B <=> P + Q is given by: Q rx = ([P][Q])/([A][B]) Meaning of ΔG Remember that Δ G is the "driving" force for a reaction, analogous to the difference in potential energy for a ball on a hill. Go back to that analogy. if the ball starts at the top of the hill, does it roll down hill? Of course. It goes from high potential energy to low potential energy. The reaction can be written as: Balltop --> Ballbottom for which the change in potential energy, Δ PE = PEbottom -PEtop< 0. If the ball starts at the bottom, will it go to the top? Obviously not. For that reaction, Ballbottom --> Balltop, Δ PE > 0. If the top of the hill was at the same height at the bottom of the hill (obviously an absurd situation), the ball would not move. It would effectively be at equilibrium, a state of no change. For this reaction, Balltop --> Ballbottom, the ΔPE = 0. As the ball starts rolling down the hill, its potential energy gets closer to the potential it would have at the bottom. Hence the ΔPE changes from negative to more and more positive until it gets to the bottom at which case the ΔPE = 0 and movement ceases. If the ΔPE is not 0, the ball will move until the ΔPE = 0. Likewise, for a chemical reaction that favors products, ΔG < 0. The system is not at equilibrium and the reaction will go in the direction of products. As the reaction proceeds, products build up, and there is less of a driving force for reactants to go to products (LeChatilier's Principle), so the ΔG becomes more a more positive until the ΔG = 0 and the reaction is at equilibrium. A reaction that has a ΔG > 0 is likewise not at equilibrium so it will go in the appropriate direction until equilibrium is reached. Hence for the reaction A + B <==> P + Q, if ΔG < 0, the reaction goes toward products P and Q if ΔG = 0, the reaction is at equilibrium and no further change occurs in the concentration of reactants and products. if ΔG > 0, the reaction goes toward reactants A and B. We can not measure easily the actual free energy G of reactants or products, but we can measure ΔG readily. These points are illustrated in the graph below of ΔG vs time for the hypothetical reaction A + B <==> P + Q. (Also notice the two insert graphs - in blue and red - which show, in analogy to the ball on the hill graphs, the values of ΔG at the two points where the perturbation to the equilibrium were made.) Notice the ΔG is constantly changing until the system reaches equilibrium. Initially the equilibrium is perturbed so that the system is not in equilibrium (shown in blue). The perturbation was such that the products are favored. After equilibrium was reached, the system was perturbed again, this time in a fashion to favor the reverse reaction. Notice in this case the ΔG for the reaction as written: A + B <==> P + Q is positive - i.e. it is not in equilibrium. Therefore the reaction (as written) goes backwards to products. It is important to realize that the reported ΔG is for the reaction as written. Now let's apply ΔG = Δ G o + RTln Q = Δ G o + RTln ( [P][Q])/([A][B]) to two reactions we discussed above: HCl(aq) + H2O(l) <==>H3O+(aq) + Cl-(aq) CH3CO2H(aq) + H2O(l) <==> H3O+(aq) + CH3CO2-(aq) Assume that at time t=0, 0.1 mole of HCl and CH3CO2H were added to two different beakers. At this point the forward reaction are favored, but obviously to different extents. The RTln Q would be identical for both acids, since each reactant is present at 0.1 M, but no products yet exist. However, the Δ G ois negative for HCl and positive for acetic acid since HCl is a strong acid. Hence at t=0, Δ G for the HCl reaction is much more negative than for acetic acid. This is summarized in table below. The direction of the arrow shows if products (-->) or reactants (<---) are favored. The size of the arrow shows very approximately to what extent the ΔG term is favored 0 1 2 3 Reaction at t=0 ΔGo RTln Q ΔG HCl(aq) + H2O(l) ---------------> ---------------> -----------------------------> CH3CO2H(aq) + H2O(l) <------------- ---------------> -> Now when equilibrium is reached, no net change occurs in the concentration of reactants and products, and ΔG = 0. In the case of HCl, there is just an infinitesimal amount of HCl left, and 0.1 M of each product, so concentration favors HCl formation. However, the intrinsic relative stability of reactants and products still favors products. In the case of acetic acid, most of the acetic acid remains (0.099 M) with little product (0.001 M) so concentration favors product. However, the intrinsic relative stability of reactants and products still favors reactants. This is summarized in table below. 0 1 2 3 Reaction at equlib. ΔGostab RTln Q ΔG HCl(aq) + H2O(l) ---------------> <--------------- favors neither, = 0 CH3CO2H(aq) + H2O(l) <------------- --------------> favors neither, = 0 Compare the two tables above (one at time t= 0 and the other at equilibrium). Notice: Δ G onever changes, since it has nothing to do with concentration. Only RTln Q changes during the course of a reaction, until equilibrium is achieved. Meaning of Δ G o To get a better meaning of the significance of Δ G o, let's consider the following equations under two different conditions: ΔG = Δ G o + RTln Q = Δ G o + RTln ( [P][Q])/([A][B]) Condition I: Reaction at equilibrium, DG = 0 The equation reduces to: Δ G o = - RTln ( [P]eq[Q]eq)/([A]eq[B]eq) or Δ G o = - RTln Keq = - 2.303RTlog Keq This supports our idea that D G o is independent of concentration since Keq should also be independent of concentration. Condition II: Concentration of all reactants and products is 1 M (standard state, assuming solution reaction) The equation reduces to: ΔG = Δ G o + RTln ( [1][1])/([1][1]) = Δ G o + 2.303RTlog 1 = Δ G o The implies that when all reactants are at this concentration, defined as the standard state (1 M for solutes), the Δ G at that particular moment just happens to be the D G o for the reaction. If one of the reactant or products is H3O+, it would make little biological sense to calculate D G o for the reaction using the standard state of [H3O+] = 1 M, or a pH of -1. Instead, it is assumed the pH = 7, [H3O+] = 10-7 M. A new symbol is used for D G o under these condition, Δ G o ' . H + H --> H2 Does this reaction occur spontaneously? It does. You should remember that individual H atoms are unstable, since they don't have an completed outer shell of electrons - in this case a duet. As they approach, they can interact to form a covalent bond and in the process release energy. The bonded state is a lower energy state than two separated H atoms. This should be clear since energy has to be added to a molecule of H2 to break the bond. ` 2C8H18(l) + 25O2(g) --> 16CO2(g) + 18H2O(g). To carry out this reaction, every C-C, C-H and O-O bond in the reactants must be broken (which requires an input of energy) but a lots of energy is released on formation of C-O and H-O covalent bonds in the products. Is more energy needed to break the bonds in the reactants or is more energy released on formation of bonds in the product? The answer should be clear. The products must be at a lower energy than the reactants since huge amounts of heat and light energy are released on combustion of gasoline and other hydrocarbons. These reactions suggest that energy must be released for a reaction to proceed to any extent in a given direction. Now consider, however, the following reaction: Ba(OH)2.8H2O(s) + 2NH4SCN(s) --> 10H2O( l ) + 2NH3( g ) + Ba(SCN)2( aq+s) When these two solids are mixed, and stirred, a reaction clearly takes places, as evidenced by the formation of a liquid (water) and the smell of ammonia. What is surprising is that heat is not released into the surroundings in this reaction. Rather heat was absorbed from the surroundings turning the beaker so cold that it freezes to a piece of wood (with a layer of water added to the wood) on which it was placed. This reaction seems to violate our idea that a reaction proceeds in a direction in which heat is liberated. Reactions which liberate heat and raise the temperature of the surroundings are called exothermic reactions. Reactions which absorb heat from the surroundings and hence lower the temperature of the surroundings are endothermic reactions.
Courses/American_River_College/Chemistry_305_(S21_Zarzana)/Map%3A_Introductory_Chemistry_(Tro)/17%3A_Radioactivity_and_Nuclear_Chemistry/17.08%3A_Nuclear_Power-_Using_Fission_to_Generate_Electricity	Fission Reactors Fission reactions can be used in the production of electricity if we control the rate at which the fission occurs. The great majority of all electrical generating systems (whether coal burning power plants, hydroelectric plants, or nuclear power plants) is that they follow a reasonably simple design. The electricity is produced by spinning a coil of wire inside a magnetic field. When a fluid (air, steam, water) is forced through the pipe, it spins the fan blades, which in turn spin the axle. To generate electricity, the axle of a turbine is attached to the loop of wire in a generator. When a fluid is forced through the turbine, the fan blades turn, the turbine axle turns, and the loop of wire inside the generator turns—thus generating electricity. The essential difference in various kinds of electrical generating systems is the method used to spin the turbine. For a wind generator, the turbine is a windmill. In a geothermal generator, steam from a geyser is forced through the turbine. In hydroelectric generating plants, water falling over a dam passes through the turbine and spins it. In fossil fuel (coal, oil, natural gas) generating plants, the fossil fuel is burned and the heat is used to boil water into steam, and then the steam passes through the turbine to make it spin. In a fission reactor generating plant, a fission reaction is used to boil the water into steam, and the steam passes through the turbine to make it spin. Once the steam is generated by the fission reaction, a nuclear power plant is essentially the same as a fossil fuel plant. Naturally occurring uranium is composed almost totally of two uranium isotopes. It contains more than $99\%$ uranium-238 and less than $1\%$ uranium-235. It is the uranium-235, however, that is fissionable (will undergo fission). In order for uranium to be used as fuel in a fission reactor, the percentage of uranium-235 must be increased, usually to about $3\%$. (Uranium in which the $\ce{U}$-235 content is more than $1\%$ is called enriched uranium .) Once the supply of $\ce{U}$-235 is acquired, it is placed in a series of long cylindrical tubes called fuel rods. These fuel cylinders are bundled together with control rods made of neutron-absorbing material. The amount of $\ce{U}$-235 in all the fuel rods taken together is adequate to carry on a chain reaction, but is less than the critical mass. (In the United States, all public nuclear power plants contain less than a critical mass of $\ce{U}$-235 and therefore, could never produce a nuclear explosion.) The amount of heat generated by the chain reaction is controlled by the rate at which the nuclear reaction occurs. The rate of the nuclear reaction is dependent on how many neutrons are emitted by one $\ce{U}$-235 nuclear disintegration and strike a new $\ce{U}$-235 nucleus to cause another disintegration. The purpose of the control rods is to absorb some of the neutrons and thus stop them from causing further disintegration. The control rods can be raised or lowered into the fuel rod bundle. When the control rods are lowered all the way into the fuel rod bundle, they absorb so many neutrons that the chain reaction essentially stops. When more heat is desired, the control rods are raised so that they catch fewer neutrons, the chain reaction speeds up, and more heat is generated. The control rods are operated in a fail-safe system, so that power is necessary to hold them up; during a power failure, gravity will pull the control rods down into the shut off position. $\ce{U}$-235 nuclei can capture neutrons and disintegrate more efficiently if the neutrons are moving slower than the speed at which they are released. Fission reactors use a moderator surrounding the fuel rods to slow down the neutrons. Water is not only a good coolant, but also a good moderator. A common type of fission reactor has the fuel core submerged in a huge pool of water. You can follow the operation of an electricity-generating fission reactor in the figure below. The reactor core is submerged in a pool of water. The heat from the fission reaction heats the water and the water is pumped into a heat exchanger container where the heated water boils the water in the heat exchanger. The steam from there is forced through a turbine which spins a generator and produces electricity. After the water passes through the turbine, it is condensed back to liquid water and pumped back to the heat exchanger. In the United States, heavy opposition to the use of nuclear energy was mounted in the late 1960's and early 1970's. Every environmentalist organization in the US opposed the use of nuclear energy; the constant pressure from environmentalist groups caused an increase of public fear and, therefore, opposition to nuclear energy. This is not true today; at least one environmental leader has published a paper in favor of nuclear-powered electricity generation. In 1979, a reactor core meltdown at Pennsylvania's Three Mile Island nuclear power plant reminded the entire country of the dangers of nuclear radiation. The concrete containment structure (six feet thick walls of reinforced concrete), however, did what it was designed to do—prevent radiation from escaping into the environment. Although the reactor was shut down for years, there were no injuries or deaths among nuclear workers or nearby residents. Three Mile Island was the only serious accident in the entire history of 103 civilian power plants operating for 40 years in the United States. There has never been a single injury or death due to radiation in any public nuclear power plant in the U.S. The accident at Three Mile Island did, however, frighten the public so that there has not been a nuclear power plant built in the U.S. since the accident. The 103 nuclear power plants operating in the U.S. deliver approximately $19.4\%$ of American electricity with zero greenhouse gas emission. There are 600 coal-burning electric plants in the US delivering $48.5\%$ of American electricity and producing 2 billion tons of $\ce{CO_2}$ annually, accounting for $40\%$ of U.S. $\ce{CO_2}$ emissions and $10\%$ of global emissions. These coal burning plants also produce $64\%$ of the sulfur dioxide emissions, $26\%$ of the nitrous oxide emissions, and $33\%$ of mercury emissions. Fusion Nuclear reactions, in which two or more lighter-mass nuclei join together to form a single nucleus, are called fusion reactions or nuclear fusions. Of particular interest are fusion reactions in which hydrogen nuclei combine to form helium. Hydrogen nuclei are positively charged and repel each other. The closer the particles come, the greater the force of repulsion. In order for fusion reactions to occur, the hydrogen nuclei must have extremely high kinetic energies, so that the velocities can overcome the forces of repulsion. These kinetic energies only occur at extreme temperatures such as those that occur in the cores of the sun and other stars. Nuclear fusion is the power source for the stars, where the necessary temperature to ignite the fusion reaction is provided by massive gravitational pressure. In stars more massive than our sun, fusion reactions involving carbon and nitrogen are possible. These reactions produce more energy than hydrogen fusion reactions. Intensive research is now being conducted to develop fusion reactors for electricity generation. The two major problems slowing the development are: finding a practical means for generating the intense temperature needed, and developing a container that will not melt under the conditions of a fusion reaction. Electricity-producing fusion reactors are still a distant dream. Summary Nuclear fission refers to the splitting of atomic nuclei. Nuclear fusion refers to the joining together of two or more smaller nuclei to form a single nucleus. The fission of $\ce{U}$-235 or $\ce{Pu}$-239 is used in nuclear reactors. 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Courses/College_of_the_Canyons/Chem_201%3A_General_Chemistry_I_OER/08%3A_Thermochemistry/8.09%3A_End_of_Chapter_Problems	A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not? Answer The temperature of 1 gram of burning wood is approximately the same for both a match and a bonfire. This is an intensive property and depends on the material (wood). However, the overall amount of produced heat depends on the amount of material; this is an extensive property. The amount of wood in a bonfire is much greater than that in a match; the total amount of produced heat is also much greater, which is why we can sit around a bonfire to stay warm, but a match would not provide enough heat to keep us from getting cold. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile. Explain the difference between heat capacity and specific heat of a substance. Answer Heat capacity refers to the heat required to raise the temperature of the mass of the substance 1 degree; specific heat refers to the heat required to raise the temperature of 1 gram of the substance 1 degree. Thus, heat capacity is an extensive property, and specific heat is an intensive one. Calculate the heat capacity, in joules and in calories per degree, of the following: 28.4 g of water 1.00 oz of lead Calculate the heat capacity, in joules and in calories per degree, of the following: 45.8 g of nitrogen gas 1.00 pound of aluminum metal Answer a. 47.6 J/°C; 11.38 cal °C −1 ; b. 407 J/°C; 97.3 cal °C −1 How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? Answer $q=mCΔ°T$ $q=(75.0g)\times(\dfrac{0.449\:J}{g\:°C})\times(1,510°K) = 50,800J$ 50,800J ; 12,200cal How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C? Answer 1310 J; 313 cal How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added? Answer ΔT ° = 31.7° C If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase? Answer 7.15 °C A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance? If it is one of the substances found in Table , what is its likely identity? Answer a.) Solve for the specific heat $C$ and compare the values with the chart $q=mCΔ°T$ $2110J=(44.7\:g)(C)(66.4°C)$ $C=\dfrac{2110\:J}{2970\:g\:°C}$ $C=\dfrac{0.711\:J}{g\:°C}$ b.) Silicon A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. What is the specific heat of the substance? If it is one of the substances found in Table , what is its likely identity? Answer a. 0.390 J/g °C; b. Copper is a likely candidate. An aluminum kettle weighs 1.05 kg. What is the heat capacity of the kettle? How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C? How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)? Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 × 10 6 J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)? Answer We assume that the density of water is 1.0 g/cm 3 (1 g/mL) and that it takes as much energy to keep the water at 85 °F as to heat it from 72 °F to 85 °F. We also assume that only the water is going to be heated. Energy required = 7.47 kWh A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers. Would the amount of heat measured for the reaction in Example be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. Answer lesser; more heat would be lost to the coffee cup and the environment and so Δ T for the water would be lesser and the calculated q would be lesser Would the amount of heat absorbed by the dissolution in Example appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. Would the amount of heat absorbed by the dissolution in Example appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer. Answer greater, since taking the calorimeter’s heat capacity into account will compensate for the thermal energy transferred to the solution from the calorimeter; this approach includes the calorimeter itself, along with the solution, as “surroundings”: q rxn = −( q solution + q calorimeter ); since both q solution and q calorimeter are negative, including the latter term ( q rxn ) will yield a greater value for the heat of the dissolution How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat. How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water. Answer The temperature of the coffee will drop 1 degree. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal. What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer. The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C. Answer $5.7 \times 10^2\; kJ$ A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure . The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal? If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure , what is the resulting temperature of the water? Answer 38.5 °C A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter ( Figure ). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? Dissolving 3.0 g of CaCl 2 ( s ) in 150.0 g of water in a calorimeter ( Figure ) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? Answer 2.2 kJ; The heat produced shows that the reaction is exothermic. When 50.0 g of 0.200 M NaCl( aq ) at 24.1 °C is added to 100.0 g of 0.100 M AgNO 3 ( aq ) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl( s ) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced. The addition of 3.15 g of Ba(OH) 2 •8H 2 O to a solution of 1.52 g of NH 4 SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: \[Ba(OH)_2 \cdot 8H_2O_{(s)} + 2NH_4SCN_{(aq)} \rightarrow Ba(SCN)_{2(aq)} + 2NH_{3(aq)} + 10H_2O_{(l)}\] Answer 1.4 kJ The reaction of 50 mL of acid and 50 mL of base described in Example increased the temperature of the solution by 6.9 degrees. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer. If the 3.21 g of NH 4 NO 3 in Example were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer. Answer 22.6. Since the mass and the heat capacity of the solution is approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease of the temperature change. When 1.0 g of fructose, C 6 H 12 O 6 ( s ), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion? When a 0.740-g sample of trinitrotoluene (TNT), C 7 H 5 N 2 O 6 , is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample? Answer 11.7 kJ One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 × 10 3 pounds). The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g? Answer 30% A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g? What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g? Answer 0.24 g A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 × 10 3 Calories if the average number of Calories for fat is 9.1 Calories/g? A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g? Answer 1.4 × 10 2 Calories Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories. Explain how the heat measured in [link] differs from the enthalpy change for the exothermic reaction described by the following equation: \[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\] The enthalpy change of the indicated reaction is for exactly 1 mol HCL and 1 mol NaOH; the heat in the example is produced by 0.0500 mol HCl and 0.0500 mol NaOH. Using the data in the check your learning section of [link] , calculate Δ H in kJ/mol of AgNO 3 ( aq ) for the reaction: \[\ce{NaCl}(aq)+\ce{AgNO3}(aq)⟶\ce{AgCl}(s)+\ce{NaNO3}(aq)\] Calculate the enthalpy of solution (Δ H for the dissolution) per mole of NH 4 NO 3 under the conditions described in [link] . Answer 25 kJ mol −1 Calculate Δ H for the reaction described by the equation. $\ce{Ba(OH)2⋅8H2O}(s)+\ce{2NH4SCN}(aq)⟶\ce{Ba(SCN)2}(aq)+\ce{2NH3}(aq)+\ce{10H2O}(l)$ Calculate the enthalpy of solution (Δ H for the dissolution) per mole of CaCl 2 . Answer 81 kJ mol −1 Although the gas used in an oxyacetylene torch is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table . Considering the conditions for which the tabulated data are reported, suggest an explanation. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions? Answer 5204.4 kJ How much heat is produced by combustion of 125 g of methanol under standard state conditions? How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions? Answer 1.83 × 10 −2 mol What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions? When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions? Answer 802 kJ mol −1 How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed? \[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\hspace{20px}ΔH^\circ_{298}=\mathrm{−58\:kJ}\] If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation? A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents? Answer 15.5 kJ/ºC Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO 2 must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl 2 F 2 (enthalpy of vaporization is 17.4 kJ/mol)? The vaporization reactions for SO 2 and CCl 2 F 2 are $\ce{SO2}(l)⟶\ce{SO2}(g)$ and $\ce{CCl2F}(l)⟶\ce{CCl2F2}(g)$, respectively. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C. Answer 7.43 g Which of the enthalpies of combustion in Table the table are also standard enthalpies of formation? Does the standard enthalpy of formation of H 2 O( g ) differ from Δ H ° for the reaction $\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g)$? Answer No. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO( s ) to Hg( l ) and O 2 ( g ) under standard conditions? How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn 3 O 4 ( s ) at standard state conditions? Answer 459.6 kJ How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe 2 O 3 ( s ) at standard state conditions? The following sequence of reactions occurs in the commercial production of aqueous nitric acid: $\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−907\:kJ}$ $\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\hspace{20px}ΔH=\mathrm{−113\:kJ}$ $\ce{3NO2}+\ce{H2O}(l)⟶\ce{2HNO2}(aq)+\ce{NO}(g)\hspace{20px}ΔH=\mathrm{−139\:kJ}$ Determine the total energy change for the production of one mole of aqueous nitric acid by this process. Answer 495 kJ/mol Both graphite and diamond burn. $\ce{C}(s,\:\ce{diamond})+\ce{O2}(g)⟶\ce{CO2}(g)$ For the conversion of graphite to diamond: $\ce{C}(s,\:\ce{graphite})⟶\ce{C}(s,\:\ce{diamond})\hspace{20px}ΔH^\circ_{298}=\mathrm{1.90\:kJ}$ Which produces more heat, the combustion of graphite or the combustion of diamond? From the molar heats of formation in Appendix G , determine how much heat is required to evaporate one mole of water: $\ce{H2O}(l)⟶\ce{H2O}(g)$ Answer 44.01 kJ/mol Which produces more heat? $\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(s)$ or $\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(g)$ for the phase change $\ce{OsO4}(s)⟶\ce{OsO4}(g)\hspace{20px}ΔH=\mathrm{56.4\:kJ}$ Calculate $ΔH^\circ_{298}$ for the process $\ce{Sb}(s)+\dfrac{5}{2}\ce{Cl2}(g)⟶\ce{SbCl5}(g)$ from the following information: $\ce{Sb}(s)+\dfrac{3}{2}\ce{Cl2}(g)⟶\ce{SbCl3}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−314\:kJ}$ $\ce{SbCl3}(s)+\ce{Cl2}(g)⟶\ce{SbCl5}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−80\:kJ}$ Answer 394 kJ Calculate $ΔH^\circ_{298}$ for the process $\ce{Zn}(s)+\ce{S}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)$ from the following information: $\ce{Zn}(s)+\ce{S}(s)⟶\ce{ZnS}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−206.0\:kJ}$ $\ce{ZnS}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−776.8\:kJ}$ Calculate Δ H for the process $\ce{Hg2Cl2}(s)⟶\ce{2Hg}(l)+\ce{Cl2}(g)$ from the following information: $\ce{Hg}(l)+\ce{Cl2}(g)⟶\ce{HgCl2}(s)\hspace{20px}ΔH=\mathrm{−224\:kJ}$ $\ce{Hg}(l)+\ce{HgCl2}(s)⟶\ce{Hg2Cl2}(s)\hspace{20px}ΔH=\mathrm{−41.2\:kJ}$ Answer 265 kJ Calculate $ΔH^\circ_{298}$ for the process $\ce{Co3O4}(s)⟶\ce{3Co}(s)+\ce{2O2}(g)$ from the following information: $\ce{Co}(s)+\dfrac{1}{2}\ce{O2}(g)⟶\ce{CoO}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−237.9\:kJ}$ $\ce{3Co}(s)+\ce{O2}(g)⟶\ce{Co3O4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−177.5\:kJ}$ Calculate the standard molar enthalpy of formation of NO( g ) from the following data: $\ce{N2}(g)+\ce{2O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{66.4\:kJ}$ $\ce{2NO}(g)+\ce{O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−114.1\:kJ}$ Answer 90.3 mol −1 of NO Using the data in Appendix G , calculate the standard enthalpy change for each of the following reactions: $\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g)$ $\ce{Si}(s)+\ce{2Cl2}(g)⟶\ce{SiCl4}(g)$ $\ce{Fe2O3}(s)+\ce{3H2}(g)⟶\ce{2Fe}(s)+\ce{3H2O}(l)$ $\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g)$ Using the data in Appendix G , calculate the standard enthalpy change for each of the following reactions: $\ce{Si}(s)+\ce{2F2}(g)⟶\ce{SiF4}(g)$ $\ce{2C}(s)+\ce{2H2}(g)+\ce{O2}(g)⟶\ce{CH3CO2H}(l)$ $\ce{CH4}(g)+\ce{N2}(g)⟶\ce{HCN}(g)+\ce{NH3}(g)$; $\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g)$ Answer a. −1615.0 kJ mol −1 ; b. −484.3 kJ mol −1 ; c. 164.2 kJ; d. −232.1 kJ The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each. $\ce{2Ag2O}(s)⟶\ce{4Ag}(s)+\ce{O2}(g)$ $\ce{SnO}(s)+\ce{CO}(g)⟶\ce{Sn}(s)+\ce{CO2}(g)$ $\ce{Cr2O3}(s)+\ce{3H2}(g)⟶\ce{2Cr}(s)+\ce{3H2O}(l)$ $\ce{2Al}(s)+\ce{Fe2O3}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)$ The decomposition of hydrogen peroxide, H 2 O 2 , has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G , determine how much heat is produced by the decomposition of exactly 1 mole of H 2 O 2 under standard conditions. $\ce{2H2O2}(l)⟶\ce{2H2O}(g)+\ce{O2}(g)$ Answer −54.04 kJ mol −1 Calculate the enthalpy of combustion of propane, C 3 H 8 ( g ), for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of propane is −104 kJ/mol. Q5.3.34 Calculate the enthalpy of combustion of butane, C 4 H 10 ( g ) for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of butane is −126 kJ/mol. Answer 2660 kJ mol −1 Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned? The white pigment TiO 2 is prepared by the reaction of titanium tetrachloride, TiCl 4 , with water vapor in the gas phase: $\ce{TiCl4}(g)+\ce{2H2O}(g)⟶\ce{TiO2}(s)+\ce{4HCl}(g)$. How much heat is evolved in the production of exactly 1 mole of TiO 2 ( s ) under standard state conditions? Answer 67.1 kJ Water gas, a mixture of H 2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: $\ce{C}(s)+\ce{H2O}(g)⟶\ce{CO}(g)+\ce{H2}(g)$. Assuming that coke has the same enthalpy of formation as graphite, calculate $ΔH^\circ_{298}$ for this reaction. Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst: \[\ce{2H2}(g)+\ce{CO}(g)⟶\ce{CH3OH}(g).\] Under the conditions of the reaction, methanol forms as a gas. Calculate $ΔH^\circ_{298}$ for this reaction and for the condensation of gaseous methanol to liquid methanol. Calculate the heat of combustion of 1 mole of liquid methanol to H 2 O( g ) and CO 2 ( g ). In the early days of automobiles, illumination at night was provided by burning acetylene, C 2 H 2 . Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC 2 : $\ce{CaC2}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)+\ce{C2H2}(g)$. Calculate the standard enthalpy of the reaction. The $ΔH^\circ_\ce{f}$ of CaC 2 is −15.14 kcal/mol. Answer 122.8 kJ From the data in Table , determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO( g ), CH 4 ( g ), or C 2 H 2 ( g ). The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 × 10 5 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane). Answer 3.7 kg Ethanol, C 2 H 5 OH, is used as a fuel for motor vehicles, particularly in Brazil. Write the balanced equation for the combustion of ethanol to CO 2 ( g ) and H 2 O( g ), and, using the data in Appendix G , calculate the enthalpy of combustion of 1 mole of ethanol. The density of ethanol is 0.7893 g/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol. Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C 8 H 18 ($ΔH^\circ_\ce{f}=\mathrm{−208.4\:kJ/mol}$; density = 0.7025 g/mL). Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B 2 H 6 , produces B 2 O 3 ( s ) and H 2 O( g )], methane [CH 4 , produces CO 2 ( g ) and H 2 O( g )], and hydrazine [N 2 H 4 , produces N 2 ( g ) and H 2 O( g )]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The $ΔH^\circ_\ce{f}$ of B 2 H 6 ( g ), CH 4 ( g ), and N 2 H 4 ( l ) may be found in Appendix G . Answer On the assumption that the best rocket fuel is the one that gives off the most heat, B 2 H 6 is the prime candidate. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions? Ethylene, C 2 H 2 , a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. $\ce{C2H4}(g)+\ce{H2O}(g)⟶\ce{C2H5OH}(l)$ Using the data in the table in Appendix G , calculate Δ H ° for the reaction. Answer 88.2 kJ The oxidation of the sugar glucose, C 6 H 12 O 6 , is described by the following equation: $\ce{C6H12O6}(s)+\ce{6O2}(g)⟶\ce{6CO2}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−2816\:kJ}$ The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body. How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose? How many Calories can be produced by the metabolism of 1.0 g of glucose? Propane, C 3 H 8 , is a hydrocarbon that is commonly used as a fuel. Write a balanced equation for the complete combustion of propane gas. Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O 2 by volume. (Hint: we will see how to do this calculation in a later chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O 2 per liter.) The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, $ΔH^\circ_\ce{f}$ of propane given that $ΔH^\circ_\ce{f}$ of H 2 O( l ) = −285.8 kJ/mol and $ΔH^\circ_\ce{f}$ of CO 2 ( g ) = −393.5 kJ/mol. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water. Answer a. $\ce{C3H8}(g)+\ce{5O2}(g)⟶\ce{3CO2}(g)+\ce{4H2O}(l)$; b. 330 L; c. −104.5 kJ mol −1 ; d. 75.4 °C During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house). Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was 56 °F; at this temperature and a pressure of 1 atm, natural gas has a density of 0.681 g/L. How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [C 3 H 8 : density, 0.5318 g/mL; enthalpy of combustion, 2219 kJ/mol for the formation of CO 2 ( g ) and H 2 O( l )] and the furnace used to burn the LPG has the same efficiency as the gas furnace. What mass of carbon dioxide is produced by combustion of the methane used to heat the house? What mass of water is produced by combustion of the methane used to heat the house? What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains 23% oxygen by mass. The average density of air during the month was 1.22 g/L. How many kilowatt–hours (1 kWh = 3.6 × 10 6 J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house. Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%?
Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/08%3A_Reading%3A_Vocabulary/08.3%3A_Using_Context_Clues	This video offers definition of what context clues are, when it comes to deepening your vocabulary, and examples of how to put them into practice. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ec1/?p=88 Given this understanding of what context clues are, the next video below specifies 4 types of context clues. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ec1/?p=88 CC licensed content, Original Video: Using Context Clues. Provided by : Lumen Learning. License : CC BY: Attribution All rights reserved content Using Context Clues to improve vocabulary. Authored by : LoveYourPencil. Located at : https://youtu.be/85QcLiXBm6A . License : All Rights Reserved . License Terms : Standard YouTube License Four Types of Context Clues. Authored by : XansWorld. Located at : https://youtu.be/hi-hg9Igo7k . License : All Rights Reserved . License Terms : Standard YouTube License
Courses/Prince_Georges_Community_College/CHEM_2000%3A_General_Chemistry_for_Engineers_-_F21/12%3A_Solutions	Template:HideTOC 12.1: Thirsty Solutions- Why You Should Not Drink Seawater 12.2: Types of Solutions and Solubility Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. 12.3: Energetics of Solution Formation 12.4: Solution Equilibrium and Factors Affecting Solubility The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. 12.5: Expressing Solution Concentration Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways. 12.6: Colligative Properties- Freezing Point Depression, Boiling Point Elevation, and Osmosis Colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. 12.7: The Colligative Properties of Strong Electrolyte Solutions Ion-pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH all affect solubility. There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes 12.8: Colloids A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. 12.E: Solutions (Exercises)
Courses/San_Diego_Miramar_College/Chemistry_201%3A_General_Chemistry_II_(Garces)/04%3A_Chemical_Equilibrium/4.04%3A_Expressing_the_Equilibrium_Constant_in_Terms_of_Pressure	Learning Objectives To understand how different phases affect equilibria. When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid. As noted in the previous section, the equilibrium constant expression is actually a ratio of activities. To simplify the calculations in general chemistry courses, the activity of each substance in the reaction is often approximated using a ratio of the molarity of a substance compared to the standard state of that substance. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, the ratio of the molarity to the standard state for substances that are liquids or solids always has a value of 1. For example, for a compound such as CaF 2 (s), the term going into the equilibrium expression is [CaF 2 ]/[CaF 2 ] which cancels to unity. Thus, when the activities of the solids and liquids (including solvents) are incorporated into the equilibrium expression, they do not change the value. Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes: \[\ce{CO2(g) + C(s) \rightleftharpoons 2CO(g)} \label{Eq14.4.1} \] The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows: \[K=\dfrac{a_{\ce{CO}}^2}{a_{\ce{CO2}}a_{C}}=\dfrac{[\ce{CO}]^2}{[\ce{CO2}][1]}=\dfrac{[\ce{CO}]^2}{[\ce{CO_2}]}\label{Eq14.4.2} \] The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases: \[K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.3} \] Incorporating all the constant values into $K′$ or $K_p$ allows us to focus on the substances whose concentrations change during the reaction. Although the activities of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of $\ce{CO}$ and $\ce{CO_2}$, the system described in Equation $\ref{Eq14.4.1}$ will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure $\PageIndex{1}$, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium. Example $\PageIndex{1}$ Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions. $\ce{PCl3(l) + Cl2(g) <=> PCl5(s)}$ $\ce{Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)}$ Given : balanced equilibrium equations. Asked for : expressions for $K$ and $K_p$. Strategy : Find $K$ by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express $K_p$ as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation. Solution This reaction contains a pure solid ($PCl_5$) and a pure liquid ($PCl_3$). Their activities are equal to 1, so when incorporated into the equilibrium constant expression, they do not change the value. So \[K=\dfrac{1}{(1)[Cl_2]} \nonumber \] and \[K_p=\dfrac{1}{(1)P_{Cl_2}} \nonumber \] This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which are each assigned a value of 1 in the equilibrium constant expressions: \[K=\dfrac{(1)[H_2O]^4}{(1)[H_2]^4} \nonumber \] and \[K_p=\dfrac{(1)(P_{H_2O})^4}{(1)(P_{H_2})^4} \nonumber \] Exercise $\PageIndex{1}$ Write the expressions for $K$ and $K_p$ for the following reactions. $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ $ \underset{glucose}{\ce{C6H12O6(s)}} + \ce{6O2(g) <=> 6CO2(g) + 6H2O(g)}$ Answer a $K = [\ce{CO_2}]$ and $K_p = P_{\ce{CO_2}}$ Answer b $K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$ For reactions carried out in solution, the solvent is assumed to be pure, and therefore is assigned an activity equal to 1 in the equilibrium constant expression. The activities of the solutes are approximated by their molarities. The result is that the equilibrium constant expressions appear to only depend upon the concentrations of the solutes. The activities of pure solids, pure liquids, and solvents are defined as having a value of '1'. Often, it is said that these activities are "left out" of equilibrium constant expressions. This is an unfortunate use of words. The activities are not "left out" of equilibrium constant expressions. Rather, because they have a value of '1', they do not change the value of the equilibrium constant when they are multiplied together with the other terms. The activities of the solutes are approximated by their molarities. Summary An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/24%3A_Indistinguishable_Molecules_-_Statistical_Thermodynamics_of_Ideal_Gases/24.09%3A_The_Rotational_Partition_Function_of_A_Diatomic_Ideal_Gas	For a diatomic molecule that is free to rotate in three dimensions, we can distinguish two rotational motions; however, their wave equations are intertwined, and the quantum mechanical result is that there is one set of degenerate rotational energy levels. The energy levels are \[{\epsilon }_{r,J}=\frac{J\left(J+1\right)h^2}{8{\pi }^2I} \nonumber \] with degeneracies $g_J=2J+1$, where $J=0,\ 1,\ 2,\ 3,\dots$. (Recall that $I$ is the moment of inertia , defined as $I=\sum{m_ir^2_i}$, where $r_i$ is the distance of the $i^{th}$ nucleus from the molecule’s center of mass. For a diatomic molecule, $XY$, whose internuclear distance is $r_{XY}$, the values of $r_X$ and $r_Y$ must satisfy the conditions $r_X+r_Y=r_{XY}$ and $m_Xr_X=m_Yr_Y$. From these relationships, it follows that the moment of inertia is $I=\mu r^2_{XY}$, where $\mu$ is the reduced mass.) For heteronuclear diatomic molecules, the rotational partition function is \[z_r=\sum^{\infty }_{J=0}{\left(2J+1\right)}{\mathrm{exp} \left[\frac{J\left(J+1\right)h^2}{8{\pi }^2IkT}\right]\ } \nonumber \] For homonuclear diatomic molecules, there is a complication. This complication occurs in the quantum mechanical description of the rotation of any molecule for which there is more than one indistinguishable orientation in space. When we specify the locations of the atoms in a homonuclear diatomic molecule, like $H_2$, we must specify the coordinates of each atom. If we rotate this molecule by ${360}^{\mathrm{o}}$ in a plane, the molecule and the coordinates are unaffected. If we rotate it by only ${180}^{\mathrm{o}}$ in a plane, the coordinates of the nuclei change, but the rotated molecule is indistinguishable from the original molecule. Our mathematical model distinguishes the ${180}^{\mathrm{o}}$-rotated molecule from the original, unrotated molecule, but nature does not. This means that there are twice as many energy levels in the mathematical model as actually occur in nature. The rotational partition function for a homonuclear diatomic molecule is exactly one-half of the rotational partition function for an “otherwise identical” heteronuclear diatomic molecule. To cope with this complication in general, it proves to be useful to define a quantity that we call the symmetry number for any molecule. The symmetry number is usually given the symbol $\sigma$; it is just the number of ways that the molecule can be rotated into indistinguishable orientations. For a homonuclear diatomic molecule, $\sigma =2$; for a heteronuclear diatomic molecule, $\sigma =1$. Making use of the symmetry number, the rotational partition function for any diatomic molecule becomes \[z_r=\left(\frac{1}{\sigma }\right)\sum^{\infty }_{J=0}{\left(2J+1\right)}{\mathrm{exp} \left[\frac{J\left(J+1\right)h^2}{8{\pi }^2IkT}\right]\ } \label{exact} \] For most molecules at ordinary temperatures, the lowest rotational energy level is much less than $kT$, and this infinite sum can be approximated to good accuracy as the corresponding integral. That is \[z_r \approx \left(\frac{1}{\sigma }\right)\int^{\infty }_{J=0}{\left(2J+1\right){\mathrm{exp} \left[\frac{J\left(J+1\right)h^2}{8{\pi }^2IkT}\right]\ }}dJ \nonumber \] Initial impressions notwithstanding, this integral is easily evaluated. The substitutions $a={h^2}/{8{\pi }^2IkT}$ and $u=J\left(J+1\right)$ yield \[ \begin{align} z_r & \approx \left(\frac{1}{\sigma }\right)\int^{\infty }_{u=0} \mathrm{exp} \left(-au\right) du \\[4pt] & \approx \left(\frac{1}{\sigma }\right)\left(\frac{1}{a}\right)=\frac{8{\pi }^2IkT}{\sigma h^2} \label{approx}\end{align} \] To see that this is a good approximation for most molecules at ordinary temperatures, we calculate the successive terms in the partition function of the hydrogen molecule at $25\ \mathrm{C}$. The results are shown in Table 1. We choose hydrogen because the energy difference between successive rotational energy levels becomes greater the smaller the values of $I$ and $T$. Since hydrogen has the smallest angular momentum of any molecule, the integral approximation will be less accurate for hydrogen than for any other molecule at the same temperature. For hydrogen, summing the first seven terms in the exact calculation (Equation \ref{exact}) gives $z_{\mathrm{rotation}}=1.87989$, whereas the approximate calculation (Equation \ref{approx}) gives $1.70284$. This difference corresponds to a difference of $245\ \mathrm{J}$ in the rotational contribution to the standard Gibbs free energy of molecular hydrogen. J $=\frac{\left(2J+1 \right)}{ \sigma} exp ^{Z_J} \left( - \frac{J \left( J+1 \right) h^2}{8 \pi^2 IkT} \right)$ $\approx \sum^{Z_r} Z_J$ 0 0.50000 0.50000 1 0.83378 1.33378 2 0.42935 1.76313 3 0.10323 1.86637 4 0.01267 1.87904 5 0.00082 1.87986 6 0.00003 1.87989
Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/14%3A_Radioactivity_and_Nuclear_Chemistry/14.4%3A_Applications_of_Nuclear_Chemistry/14.4.01%3A_Detecting_Radioactivity	Learning Objectives Understand how the Geiger counter can be used to quantify the rate of ionization radiation. When alpha, beta or gamma particles collide with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization . Alpha, beta and gamma radiation are broadly referred to as ionizing radiation . A Geiger counter (or Geiger-Müller counter) takes advantage of this in order to detect these particles. In a Geiger tube, the electron produced by ionization of a captive gas travels to the anode and the change in voltage is detected by the attached circuitry. Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter. A simple schematic of a Geiger counter is shown in Figure $\PageIndex{1}$. Although scientists were not aware at the time of the Geiger counter's invention, all of us are subjected to a certain amount of radiation every day. This radiation is called background radiation and comes from a variety of natural and artificial radiation sources. Approximately 82% of background radiation comes from natural sources. These natural sources include: Sources in the earth—including naturally occurring radioactive elements—which are incorporated in building materials, and also in the human body. Sources from space in the form of cosmic rays. Sources in the atmosphere, such as radioactive radon gas released from the earth; and radioactive atoms like carbon-14, produced in the atmosphere by bombardment from high-energy cosmic rays.
Courses/Modesto_Junior_College/Chemistry_143_-_Bunag/Chemistry_143_-_Introductory_Chemistry_(Bunag)/19%3A_Nuclear_Chemistry/19.12%3A_PET_Scans	Does your heart beat faster when you are scared? Do you have a tender feeling in your heart for that "special person"? Sorry to disappoint you, but that is actually simply a response generated by the brain. Science tells us that the seat of emotion is in the brain. Using PET scans and other techniques, scientists look at specific areas of the brain that process and store information dealing with strong emotions. They haven't localized the love site, but they're working on it. PET Scans One of the more interesting and useful approaches to radioisotope use in medicine is positron emission topography (PET) , often referred to as a PET scan. This technique is especially useful in studying the processes in the brain. Many compounds do not enter the brain because of what is called the " blood-brain barrier ", a filtering system to block material from being transported into brain tissue. This mechanism serves to protect the brain from a wide variety of harmful substances. In order to get a good picture of what is happening in the brain, radiolabels are attached to different compounds that will enter the brain. Since the brain uses about $25\%$ of the glucose found in the body, this molecule is often labeled with a positron emitter such as $\ce{F}$-18 (half-life of 109.8 minutes) to study brain function in general. Other labels are attached to specific compounds that will localize in certain areas of the brain to look at specific structures. The PET scanner detects gamma emissions from the collision of a positron with an electron (see figure below). As the positron is released from the nucleus of the atom, it will collide with an electron. This meeting of matter (electron) with antimatter (positron) results in annihilation of both particles, and the release of two gamma emissions that are $180^\text{o}$ apart from one another. The apparatus detects these gamma rays and stores the data in a computer. From this information, a detailed picture of the brain can be developed. One useful application of PET scanning is in the diagnosis of Alzheimer's disease. This debilitating memory loss condition primarily occurs in elderly individuals. A protein known as beta-amyloid gradually forms deposits in the brain called plaque. Memory loss and impaired movement are the result of the plaque growth. The compound known as Pittsburgh compound B is often used to identify areas of plaque in the brain. The radiolabel is $\ce{C}$-11 (half-life of 20.38 minutes). The label attaches to plaque and can be observed using PET scans. The computer translates the amount of isotope into a color scale, with red indicating a high level of radioactivity and yellow indicating somewhat less activity. We can see from the scans that the cognitively healthy individual shows the presence of very little plaque in the brain (see figure above). The individual with Alzheimer's demonstrates high concentrations of beta-amyloid in numerous areas of the brain. Other studies have been done looking at brain function in drug addicts. One of the theories about drug addiction involves the amount of dopamine action in the brain (a chemical that is a part of the system to transport nerve impulses). Studies of dopamine action have been helpful in understanding addictive processes. The figure above shows the binding of chemicals that attach to dopamine receptors. The non-addicted individuals have large numbers of receptors for dopamine. The addicted persons show less binding to these receptors, indicating that fewer receptors are present. Since dopamine is somehow linked with the sense of pleasure, these data may help to bring a better understanding to the biochemical processes in drug addiction. Summary Positron Emission Topography (PET) scans make use of radioisotopes in medicine. As an example, PET scans can be used to detect Alzheimer's Disease and to help scientists learn about addiction.
Courses/Sacramento_City_College/SCC%3A_CHEM_330_-_Adventures_in_Chemistry_(Alviar-Agnew)/03%3A_Atomic_Structure/3.06%3A_Electron_Arrangement-_The_Bohr_Model_(Orbits)	Learning Objectives Know the properties of different types of electromagnetic radiation. Define an energy level in terms of the Bohr model. Discuss how the Bohr model can be used to explain atomic spectra. Describe the arrangement of electrons using the shell model. Electromagnetic waves have an extremely wide range of wavelengths, frequencies, and energies. The highest energy form of electromagnetic waves are gamma (γ) rays and the lowest energy form are radio waves. The figure below shows the electromagnetic spectrum , which is all forms of electromagnetic radiation. On the far left of Figure $\PageIndex{1}$ are the highest energy electromagnetic waves. These are called gamma rays and can be quite dangerous, in large numbers, to living systems. The next lower energy form of electromagnetic waves are called x-rays . Most of you are familiar with the penetration abilities of these waves. They can also be dangerous to living systems. Humans are advised to limit as much as possible the number of medical x-rays they have per year. Next lower, in energy, are ultraviolet rays . These rays are part of sunlight and the upper end of the ultraviolet range can cause sunburn and perhaps skin cancer. The tiny section next in the spectrum is the visible range of light … this section has been greatly expanded in the bottom half of the figure so it can be discussed in more detail. The visible range of electromagnetic radiation are the frequencies to which the human eye responds. Lower in the spectrum are infrared rays and radio waves. The light energies that are in the visible range are electromagnetic waves that cause the human eye to respond when those frequencies enter the eye. The eye sends a signal to the brain and the individual "sees" various colors. The highest energy waves in the visible region cause the brain to see violet and as the energy decreases, the colors change to blue, green, yellow, orange, and red. When the energy of the wave is above or below the visible range, the eye does not respond to them. When the eye receives several different frequencies at the same time, the colors are blended by the brain. If all frequencies of light strike the eye together, the brain sees white and if there are no visible frequencies striking the eye, the brain sees black. The objects that you see around you are light absorbers - that is, the chemicals on the surface of the object will absorb certain frequencies and not others. Your eyes detect the frequencies that strike your eye. Therefore, if your friend is wearing a red shirt, it means the dye in that shirt absorbs every frequency except red and the red frequencies are reflected. If your only light source was one exact frequency of blue light and you shined it on a shirt that was red in sunlight, the shirt would appear black because no light would be reflected. The light from fluorescent types of lights do not contain all the frequencies of sunlight and so clothes inside a store may appear to be a slightly different color than when you get them home. Continuous and Line Spectra Electric light bulbs contain a very thin wire in them that emits light when heated. The wire is called a filament. The particular wire used in light bulbs is made of tungsten. A wire made of any metal would emit light under these circumstances but tungsten was chosen because the light it emits contains virtually every frequency and therefore, the light emitted by tungsten appears white. A wire made of some other element would emit light of some color that was not convenient for our uses. Every element emits light when energized by heating or passing electric current through it. Elements in solid form begin to glow when they are heated sufficiently and elements in gaseous form emit light when electricity passes through them. This is the source of light emitted by neon signs and is also the source of light in a fire. Each Element Has a Unique Spectrum The light frequencies emitted by atoms are mixed together by our eyes so that we see a blended color. Several physicists, including Angstrom in 1868 and Balmer in 1875, passed the light from energized atoms through glass prisms in such a way that the light was spread out so they could see the individual frequencies that made up the light. The emission spectrum (or atomic spectrum ) of a chemical element is the unique pattern of light obtained when the element is subjected to heat or electricity. When hydrogen gas is placed into a tube and electric current passed through it, the color of emitted light is pink. But when the color is spread out, we see that the hydrogen spectrum is composed of four individual frequencies. The pink color of the tube is the result of our eyes blending the four colors. Every atom has its own characteristic spectrum; no two atomic spectra are alike. The image below shows the emission spectrum of iron. Because each element has a unique emission spectrum, elements can be defined using them. You may have heard or read about scientists discussing what elements are present in the sun or some more distant star, and after hearing that, wondered how scientists could know what elements were present in a place no one has ever been. Scientists determine what elements are present in distant stars by analyzing the light that comes from stars and finding the atomic spectrum of elements in that light. If the exact four lines that compose hydrogen's atomic spectrum are present in the light emitted from the star, that element contains hydrogen. Bohr's Explanation of Line Spectra In 1913, the Danish physicist Niels Bohr proposed a model of the electron cloud of an atom in which electrons orbit the nucleus and were able to produce atomic spectra. Understanding Bohr's model requires some knowledge of electromagnetic radiation (or light). Bohr's key idea in his model of the atom is that electrons occupy definite orbitals that require the electron to have a specific amount of energy. In order for an electron to be in the electron cloud of an atom, it must be in one of the allowable orbitals and it must have the precise energy required for that orbit. Orbits closer to the nucleus would require smaller amounts of energy for an electron and orbits farther from the nucleus would require the electrons to have a greater amount of energy. The possible orbits are known as energy levels (n) . One of the weaknesses of Bohr's model was that he could not offer a reason why only certain energy levels or orbits were allowed. Figure $\PageIndex{5}$: Niels Bohr with Albert Einstein at Paul Ehrenfest's home in Leiden (December 1925). Bohr hypothesized that the only way electrons could gain or lose energy would be to move from one energy level to another, thus gaining or losing precise amounts of energy. The energy levels are quantized , meaning that only specific amounts are possible. It would be like a ladder that had rungs only at certain heights. The only way you can be on that ladder is to be on one of the rungs and the only way you could move up or down would be to move to one of the other rungs. Suppose we had such a ladder with 10 rungs. Other rules for the ladder are that only one person can be on a rung in normal state and the ladder occupants must be on the lowest rung available. If the ladder had five people on it, they would be on the lowest five rungs. In this situation, no person could move down because all the lower rungs are full. Bohr worked out rules for the maximum number of electrons that could be in each energy level in his model and required that an atom in its normal state (ground state) had all electrons in the lowest energy levels available. Under these circumstances, no electron could lose energy because no electron could move down to a lower energy level. In this way, Bohr's model explained why electrons circling the nucleus did not emit energy and spiral into the nucleus. Figure $\PageIndex{6}$ The energy levels (n= 1,2,3...) of the electrons can be viewed as rungs on a ladder. The evidence used to support Bohr's model came from the atomic spectra. He suggested that an atomic spectrum is made by the electrons in an atom moving energy levels. Ground States and Excited States The electrons typically have the lowest energy possible, called the ground state . If the electrons are given energy (through heat, electricity, light, etc.) the electrons in an atom could absorb energy by jumping to a higher energy level, or excited state . The electrons then give off the energy in the form of a piece of light, called a photon , they had absorbed to fall back to a lower energy level. The energy emitted by electrons dropping back to lower energy levels would always be precise amounts of energy because the differences in energy levels were precise. This explains why you see specific lines of light when looking at an atomic spectrum - each line of light matches a specific "step down" that an electron can take in that atom. This also explains why each element produces a different atomic spectrum. Because each element has different acceptable energy levels for their electrons, the possible steps each element's electrons can take differ from all other elements. Based on the wavelengths of the spectral lines, Bohr was able to calculate the energies that the hydrogen electron would have in each of its allowed energy levels. He then mathematically showed which energy level transitions correspond to the spectral lines in the atomic emission spectrum (see below). He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level $\left( n=2 \right)$. This is called the Balmer series ( Figure $\PageIndex{8}$ ). Transitions ending in the ground state $\left( n=1 \right)$ are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. The transitions called the Paschen series and the Brackett series both result in spectral lines in the infrared region because the energies are too small. Bohr's model was a tremendous success in explaining the spectrum of the hydrogen atom. Unfortunately, when the mathematics of the model was applied to atoms with more than one electron, it was not able to correctly predict the frequencies of the spectral lines. While Bohr's model represented a great advancement in the atomic model and the concept of electron transitions between energy levels is valid, improvements were needed in order to fully understand all atoms and their chemical behavior. Different metal electrons emit different wavelengths of light to return to their respective ground states, so the flame colors are varied. These flames can be used to produce atomic emmision spectra of the elements combusted. Using known values of emission spectra, one can perform a flame test on un unknown substance, gather an emmision spectrum from it, and determine which elements are in the unknown substance. For example, in the case of copper ion, there are multiple different "paths" that the excited electrons can follow to emit photon of certain discrete energy. This produces multiple spectra lines because each discrete energy level difference will yield a specific wavelength of light, which determines the color. Building Atoms: Main Shells An electron shell is the outside part of an atom around the atomic nucleus. It is a group of atomic orbitals with the same value of the principal quantum number $n$. Electron shells have one or more electron subshells, or sublevels. The name for electron shells comes from the Bohr model, in which groups of electrons were believed to go around the nucleus at certain distances, so that their orbits formed "shells". An electron shell may be thought of as an orbit followed by electrons around an atom nucleus. Because each shell can contain only a fixed number of electrons, each shell is associated with a particular range of electron energy, and thus each shell must fill completely before electrons can be added to an outer shell. The electrons in the outermost shell determine the chemical properties of the atom (see Valence shell). For an explanation of why electrons exist in these shells see electron configuration. Figure $\PageIndex{10}$ A shell diagram of lithium (left) and Sodium (right) The electron shells are labeled K, L, M, N, O, P, and Q; or 1, 2, 3, 4, 5, 6, and 7; going from innermost shell outwards. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells. This makes them more important in determining how the atom reacts chemically and behaves as a conductor, because the pull of the atom's nucleus upon them is weaker and more easily broken. In this way, a given element's reactivity is highly dependent upon its electronic configuration. Drawing Shell Models Video $\PageIndex{1}$ How to draw the shell model for sulfur. Note: The number of electrons that can occupy each energy level are 2 (first level), 8 (2nd level), 18 (3rd level), and 32 (4th level) based on the formula: # of electrons = 2(n) 2 , wherein n = principle energy level. Table $\PageIndex{1}$ shows the number of electrons that fill each shell for neutral atoms of several elements. As mentioned earlier, the innermost shell (corresponding to lowest energy) is filled first and only a fixed number of electrons is allowed in each shell. The only electron in hydrogen (Z=1) goes to the first shell. In lithium atom (Z=3), the two electrons fill the first shell, and the third electron goes to the second shell. An argon atom (Z=18) has 18 electrons. The 10 electrons fill the first and second shells, and the remaining 8 electrons go to the third shell. The electron configuration for elements pass argon are covered in more detail in section 3.7. Element Symbol Atomic Number* (Z) First Shell n=1 (2 electrons allowed) Second Shell n=2 (8 electrons allowed) Third Shell n=3 (18 electrons allowed) H 1 1 NaN NaN He 2 2 NaN NaN C 6 2 4.0 NaN N 7 2 5.0 NaN Na 11 2 8.0 1.0 Mg 12 2 8.0 2.0 Cl 17 2 8.0 7.0 Ar 18 2 8.0 8.0 Note:* In a neutral atom he number of protons is equal to the number of electrons. Summary Electromagnetic radiation has a wide spectrum, including gamma rays, X-rays, UV rays, visible light, IR radiation, microwaves, and radio waves. The different colors of light differ in their frequencies (or wavelengths). Bohr's model suggests each atom has a set of unchangeable energy levels and electrons in the electron cloud of that atom must be in one of those energy levels. Bohr's model suggests that the atomic spectra of atoms is produced by electrons gaining energy from some source, jumping up to a higher energy level, then immediately dropping back to a lower energy level and emitting the energy different between the two energy levels. The existence of the atomic spectra is support for Bohr's model of the atom. Bohr's model was only successful in calculating energy levels for the hydrogen atom. The shell model is a good representation of electron arrangement only for elements 1-18.
Courses/University_of_San_Diego/Fall_2024_Chem_220_Analytical_Chemistry_David_De_Haan/03%3A_Basic_Analytical_Tools/3.03%3A_Preparing_Solutions	Preparing a solution of known concentration is perhaps the most common activity in any analytical lab. The method for measuring out the solute and the solvent depend on the desired concentration and how exact the solution’s concentration needs to be known. Pipets and volumetric flasks are used when we need to know a solution’s exact concentration; graduated cylinders, beakers, and/or reagent bottles suffice when a concentrations need only be approximate. Two methods for preparing solutions are described in this section. Preparing Stock Solutions A stock solution is prepared by weighing out an appropriate portion of a pure solid or by measuring out an appropriate volume of a pure liquid, placing it in a suitable flask, and diluting to a known volume. Exactly how one measure’s the reagent depends on the desired concentration unit. For example, to prepare a solution with a known molarity you weigh out an appropriate mass of the reagent, dissolve it in a portion of solvent, and bring it to the desired volume. To prepare a solution where the solute’s concentration is a volume percent, you measure out an appropriate volume of solute and add sufficient solvent to obtain the desired total volume. Describe how to prepare the following three solutions: (a) 500 mL of approximately 0.20 M NaOH using solid NaOH; (b) 1 L of 150.0 ppm Cu 2 + using Cu metal; and (c) 2 L of 4% v/v acetic acid using concentrated glacial acetic acid (99.8% w/w acetic acid). Solution (a) Because the desired concentration is known to two significant figures, we do not need to measure precisely the mass of NaOH or the volume of solution. The desired mass of NaOH is \[\frac {0.20 \text{ mol NaOH}} {\text{L}} \times \frac {40.0 \text{ g NaOH}} {\text{mol NaOH}} \times 0.50 \text{ L} = 4.0 \text{ g NaOH} \nonumber\] To prepare the solution, place 4.0 grams of NaOH, weighed to the nearest tenth of a gram, in a bottle or beaker and add approximately 500 mL of water. (b) Since the desired concentration of Cu 2 + is given to four significant figures, we must measure precisely the mass of Cu metal and the final solution volume. The desired mass of Cu metal is \[\frac {150.0 \text{ mg Cu}} {\text{L}} \times 1.000 \text{ M } \times \frac {1 \text{ g}} {1000 \text{ mg}} = 0.1500 \text{ g Cu} \nonumber\] To prepare the solution, measure out exactly 0.1500 g of Cu into a small beaker and dissolve it using a small portion of concentrated HNO 3 . To ensure a complete transfer of Cu 2 + from the beaker to the volumetric flask—what we call a quantitative transfer —rinse the beaker several times with small portions of water, adding each rinse to the volumetric flask. Finally, add additional water to the volumetric flask’s calibration mark. (c) The concentration of this solution is only approximate so it is not necessary to measure exactly the volumes, nor is it necessary to account for the fact that glacial acetic acid is slightly less than 100% w/w acetic acid (it is approximately 99.8% w/w). The necessary volume of glacial acetic acid is \[\frac {4 \text{ mL } \ce{CH3COOH}} {100 \text{ mL}} \times 2000 \text{ mL} = 80 \text{ mL } \ce{CH3COOH} \nonumber\] To prepare the solution, use a graduated cylinder to transfer 80 mL of glacial acetic acid to a container that holds approximately 2 L and add sufficient water to bring the solution to the desired volume. Provide instructions for preparing 500 mL of 0.1250 M KBrO 3 . Answer Preparing 500 mL of 0.1250 M KBrO 3 requires \[0.5000 \text{ L} \times \frac {0.1250 \text{ mol } \ce{KBrO3}} {\text{L}} \times \frac {167.00 \text{ g } \ce{KBrO3}} {\text{mol } \ce{KBrO3}} = 10.44 \text{ g } \ce{KBrO3} \nonumber\] Because the concentration has four significant figures, we must prepare the solution using volumetric glassware. Place a 10.44 g sample of KBrO 3 in a 500-mL volumetric flask and fill part way with water. Swirl to dissolve the KBrO 3 and then dilute with water to the flask’s calibration mark. Preparing Solutions by Dilution Solutions are often prepared by diluting a more concentrated stock solution. A known volume of the stock solution is transferred to a new container and brought to a new volume. Since the total amount of solute is the same before and after dilution, we know that \[C_o \times V_o = C_d \times V_d \label{2.1}\] where $C_o$ is the stock solution’s concentration, $V_o$ is the volume of stock solution being diluted, $C_d$ is the dilute solution’s concentration, and $V_d$ is the volume of the dilute solution. Again, the type of glassware used to measure $V_o$ and $V_d$ depends on how precisely we need to know the solution’s concentration. Note that Equation \ref{2.1} applies only to those concentration units that are expressed in terms of the solution’s volume, including molarity, formality, normality, volume percent, and weight-to-volume percent. It also applies to weight percent, parts per million, and parts per billion if the solution’s density is 1.00 g/mL. We cannot use Equation \ref{2.1} if we express concentration in terms of molality as this is based on the mass of solvent, not the volume of solution. See Rodríquez-López, M.; Carrasquillo, A. J. Chem. Educ. 2005 , 82 , 1327-1328 for further discussion. A laboratory procedure calls for 250 mL of an approximately 0.10 M solution of NH 3 . Describe how you would prepare this solution using a stock solution of concentrated NH 3 (14.8 M). Solution Substituting known values into Equation \ref{2.1} \[14.8 \text{ M} \times V_o = 0.10 \text{ M} \times 250 \text{ mL} \nonumber\] and solving for $V_o$ gives 1.7 mL. Since we are making a solution that is approximately 0.10 M NH 3 , we can use a graduated cylinder to measure the 1.7 mL of concentrated NH 3 , transfer the NH 3 to a beaker, and add sufficient water to give a total volume of approximately 250 mL. Although usually we express molarity as mol/L, we can express the volumes in mL if we do so both for both $V_o$ and $V_d$. To prepare a standard solution of Zn 2 + you dissolve a 1.004 g sample of Zn wire in a minimal amount of HCl and dilute to volume in a 500-mL volumetric flask. If you dilute 2.000 mL of this stock solution to 250.0 mL, what is the concentration of Zn 2 + , in μg/mL, in your standard solution? Answer The first solution is a stock solution, which we then dilute to prepare the standard solution. The concentration of Zn 2 + in the stock solution is \[\frac {1.004 \text{ g } \ce{Zn^{2+}}} {500.0 \text{ mL}} \times \frac {10^6 \: \mu \text{g}} {\text{g}} = 2008 \: \mu \text{g } \ce{Zn^{2+}} \text{/mL} \nonumber\] To find the concentration of the standard solution we use Equation \ref{2.1} \[\frac {2008 \: \mu \text{g } \ce{Zn^{2+}}} {\text{mL}} \times 2.000 \text{ mL} = C_d \times 250.0 \text{ mL} \nonumber\] where C d is the standard solution’s concentration. Solving gives a concentration of 16.06 μg Zn 2 + /mL. As shown in the following example, we can use Equation \ref{2.1} to calculate a solution’s original concentration using its known concentration after dilution. A sample of an ore was analyzed for Cu 2 + as follows. A 1.25 gram sample of the ore was dissolved in acid and diluted to volume in a 250-mL volumetric flask. A 20 mL portion of the resulting solution was transferred by pipet to a 50-mL volumetric flask and diluted to volume. An analysis of this solution gives the concentration of Cu 2 + as 4.62 μg/mL. What is the weight percent of Cu in the original ore? Solution Substituting known values (with significant figures appropriate for pipets and volumetric flasks) into Equation \ref{2.1} \[(C_{\ce{Cu}})_o \times 20.00 \text{ mL} = 4.62 \: \mu \text{g/mL } \ce{Cu^{2+}} \times 50.00 \text{ mL} \nonumber\] and solving for $(C_{\ce{Cu}})_o $ gives the original concentration as 11.55 μg/mL Cu 2 + . To calculate the grams of Cu 2 + we multiply this concentration by the total volume \[\frac {11.55 \mu \text{g } \ce{Cu^{2+}}} {\text{mL}} \times 250.0 \text{ mL} \times \frac {1 \text{ g}} {10^6 \: \mu \text{g}} = 2.888 \times 10^{-3} \text{ g } \ce{Cu^{2+}} \nonumber\] The weight percent Cu is \[\frac {2.888 \times 10^{-3} \text{ g } \ce{Cu^{2+}}} {1.25 \text{ g sample}} \times 100 = 0.231 \text{% w/w } \ce{Cu^{2+}} \nonumber\]
Courses/Siena_Heights_University/Organic_Chemistry_I_(Siena_Heights_University)/1%3A_Chapter_1_Structure_Determines_Properties/1.05%3A_Formal_Charges	Objectives After completing this section, you should be able to calculate the formal charge of an atom in an organic molecule or ion. identify and recognize the bonding patterns for atoms of carbon, hydrogen, oxygen, nitrogen and the halogens that have a formal charge of zero. Key Terms Make certain that you can define, and use in context, the key term below. valence electrons bonding and non-bonding electrons formal charge carbocations Study Notes It is more important that students learn to easily identify atoms that have formal charges of zero, than it is to actually calculate the formal charge of every atom in an organic compound. Students will benefit by memorizing the "normal" number of bonds and non-bonding electrons around atoms whose formal charge is equal to zero. A formal charge compares the number of electrons around a "neutral atom" (an atom not in a molecule) versus the number of electrons around an atom in a molecule. Formal charge is assigned to an atom in a molecule by assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules: Non-bonding electrons are assigned to the atom on which they are located. Bonding electrons are divided equally between the two bonded atoms, so one electron from each bond goes to each atom. The formal charge of each atom in a molecule can be calculated using the following equation: Formal Charge = (# of valence electrons in free atom) - (# of lone-pair electrons) - (1/2 # of bond pair electrons) Eqn. 2.3.1 To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH 3 ) whose Lewis structure is as follows: A neutral nitrogen atom has five valence electrons (it is in group 15). From the Lewis structure, the nitrogen atom in ammonia has one lone pair and three bonds with hydrogen atoms. Substituting into Equation 2.3.1, we obtain Formal Charge of N = (5 valence e-) - (2 lone pair e-) - (1/2 x 6 bond pair e-) = 0 A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule has no non-bonding electrons and one bond. Using Equation 2.3.1 to calculate the formal charge on hydrogen, we obtain Formal Charge of H = (1 valence e-) - (0 lone pair e-) - (1/2 x 2 bond pair e-) = 0 The sum of the formal charges of each atom must be equal to the overall charge of the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall neutral charge of the NH 3 molecule. Typically, the structure with the most formal charges of zero on atoms is the more stable Lewis structure. In cases where there MUST be positive or negative formal charges on various atoms, the most stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges for polyatomic ions. Example 2.3.1 Calculate the formal charges on each atom in the NH 4 + ion. Given: chemical species Asked for: formal charges Strategy: Identify the number of valence electrons in each atom in the NH 4 + ion. Use the Lewis electron structure of NH 4 + to identify the number of bonding and non-bonding electrons associated with each atom and then use Equation 2.3.1 to calculate the formal charge on each atom. Solution: The Lewis electron structure for the NH 4 + ion is as follows: The nitrogen atom in ammonium has zero non-bonding electrons and 4 bonds. Using Equation 2.3.1, the formal charge on the nitrogen atom is therefore Formal Charge of N = (5 valence e-) - (0 lone pair e-) - (1/2 x 8 bond pair e-) = +1 f o r m a l c h a r g e ( N ) = 5 − ( 0 + 8 2 ) = 0 Each hydrogen atom in has one bond and zero non-bonding electrons. The formal charge on each hydrogen atom is therefore Formal Charge of H = (1 valence e-) - (0 lone pair e-) - (1/2 x 2 bond pair e-) = 0 f o r m a l c h a r g e ( H ) = 1 − ( 0 + 2 2 ) = 0 The formal charges on the atoms in the NH 4 + ion are thus Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1, which is the same a s the overall charge of the ammonium polyatomic ion. Exercise $\PageIndex{1}$ Write the formal charges on all atoms in BH 4 − . Answer . Common bonding patterns in organic structures The calculation method reviewed above for determining formal charges on atoms is an essential starting point for a novice organic chemist, and works well when dealing with small structures. But this method becomes unreasonably time-consuming when dealing with larger structures. It would be exceptionally tedious to determine the formal charges on each atom in 2'-deoxycytidine (one of the four nucleoside building blocks that make up DNA) using equation 2.3.1. And yet, organic chemists, and especially organic chemists dealing with biological molecules, are expected to draw the structure of large molecules such as this on a regular basis. Clearly, you need to develop the ability to quickly and efficiently draw large structures and determine formal charges. Fortunately, this only requires some practice with recognizing common bonding patterns. Carbon Carbon, the most important element for organic chemists. In the structures of methane, methanol, ethane, ethene, and ethyne, there are four bonds to the carbon atom. And each carbon atom has a formal charge of zero. In other words, carbon is tetravalent , meaning that it commonly forms four bonds. Carbon is tetravalent in most organic molecules, but there are exceptions. Later in this chapter and throughout this book are examples of organic ions called ‘carbocations’ and carbanions’, in which a carbon atom has a positive or negative formal charge, respectively. Carbocations occur when a carbon has only three bonds and no lone pairs of electrons. Carbocations have only 6 valence electrons and a formal charge of +1. Carbanions occur when the carbon atom has three bonds plus one lone pair of electrons. Carbanions have 8 valence electrons and a formal charge of -1. Two other possibilities are carbpon radicals and carbenes, both of which have a formal charge of zero. A carbon radical has three bonds and a single, unpaired electron. Carbon radicals have 7 valence electrons and a formal charge of zero. Carbenes are a highly reactive species, in which a carbon atom has two bonds and one lone pair of electrons, giving it a formal charge of zero. You may encounter carbenes in more advanced chemistry courses, but they will not be discussed any further in this book. You should certainly use the methods you have learned to check that these formal charges are correct for the examples given above. More importantly, you will need, before you progress much further in your study of organic chemistry, to simply recognize these patterns (and the patterns described below for other atoms) and be able to identify carbons that bear positive and negative formal charges by a quick inspection. Hydrogen The common bonding pattern for hydrogen is easy: hydrogen atoms in organic molecules typically have only one bond, no unpaired electrons and a formal charge of zero. The exceptions to this rule are the proton, H + , the hydride ion, H - , and the hydrogen radical, H . . The proton is a hydrogen with no bonds and no lone pairs and a formal charge of +1. The hydride ion is a is a hydrogen with no bonds, a pair of electrons, and a formal charge of -1. The hydrogen radical is a hydrogen atom with no bonds, a single unpaired electron and a formal charge of 0. Because this book concentrates on organic chemistry as applied to living things, however, we will not be seeing ‘naked’ protons and hydrides as such, because they are too reactive to be present in that form in aqueous solution. Nonetheless, the idea of a proton will be very important when we discuss acid-base chemistry, and the idea of a hydride ion will become very important much later in the book when we discuss organic oxidation and reduction reactions. As a rule, though, all hydrogen atoms in organic molecules have one bond, and no formal charge. Oxygen The common arrangement of oxygen that has a formal charge of zero is when the oxygen atom has 2 bonds and 2 lone pairs. Other arrangements are oxygen with 1 bond and 3 lone pairs, that has a -1 formal charge, and oxygen with 3 bonds and 1 lone pair that has a formal charge of +1. All three patterns of oxygen fulfill the octet rule. If it has two bonds and two lone pairs, as in water, it will have a formal charge of zero. If it has one bond and three lone pairs, as in hydroxide ion, it will have a formal charge of-1. If it has three bonds and one lone pair, as in hydronium ion, it will have a formal charge of +1. When we get to our discussion of free radical chemistry in chapter 17, we will see other possibilities, such as where an oxygen atom has one bond, one lone pair, and one unpaired (free radical) electron, giving it a formal charge of zero. For now, however, concentrate on the three main non-radical examples, as these will account for virtually everything we see until chapter 17. Nitrogen Nitrogen has two major bonding patterns, both of which fulfill the octet rule: If a nitrogen has three bonds and a lone pair, it has a formal charge of zero. If it has four bonds (and no lone pair), it has a formal charge of +1. In a fairly uncommon bonding pattern, negatively charged nitrogen has two bonds and two lone pairs. Phosphorus and Sulfur Two third row elements are commonly found in biological organic molecules: phosphorus and sulfur. Although both of these elements have other bonding patterns that are relevant in laboratory chemistry, in a biological context sulfur almost always follows the same bonding/formal charge pattern as oxygen, while phosphorus is present in the form of phosphate ion (PO 4 3- ), where it has five bonds (almost always to oxygen), no lone pairs, and a formal charge of zero. Remember that elements in the third row of the periodic table have d orbitals in their valence shell as well as s and p orbitals, and thus are not bound by the octet rule. Halogens The halogens (fluorine, chlorine, bromine, and iodine) are very important in laboratory and medicinal organic chemistry, but less common in naturally occurring organic molecules. Halogens in organic compounds usually are seen with one bond, three lone pairs, and a formal charge of zero. Sometimes, especially in the case of bromine, we will encounter reactive species in which the halogen has two bonds (usually in a three-membered ring), two lone pairs, and a formal charge of +1. These rules, if learned and internalized so that you don’t even need to think about them, will allow you to draw large organic structures, complete with formal charges, quite quickly. Once you have gotten the hang of drawing Lewis structures, it is not always necessary to draw lone pairs on heteroatoms, as you can assume that the proper number of electrons are present around each atom to match the indicated formal charge (or lack thereof). Occasionally, though, lone pairs are drawn if doing so helps to make an explanation more clear. Using Formal Charges to Distinguish between Lewis Structures As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO 2 . Both structures conform to the rules for Lewis electron structures. CO 2 1. C is less electronegative than O, so it is the central atom. 2. C has 4 valence electrons and each O has 6 valence electrons, for a total of 16 valence electrons. 3. Placing one electron pair between the C and each O gives O–C–O, with 12 electrons left over. 4. Dividing the remaining electrons between the O atoms gives three lone pairs on each atom: This structure has an octet of electrons around each O atom but only 4 electrons around the C atom. 5. No electrons are left for the central atom. 6. To give the carbon atom an octet of electrons, we can convert two of the lone pairs on the oxygen atoms to bonding electron pairs. There are, however, two ways to do this. We can either take one electron pair from each oxygen to form a symmetrical structure or take both electron pairs from a single oxygen atom to give an asymmetrical structure: Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO 2 are as follows: Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do not represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound. Note The Lewis structure with the set of formal charges closest to zero is usually the most stable . Example 2.3.2 The thiocyanate ion (SCN − ), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons. Given: chemical species Asked for: Lewis electron structures, formal charges, and preferred arrangement Strategy: A Use the step-by-step procedure to write two plausible Lewis electron structures for SCN − . B Calculate the formal charge on each atom using Equation 2.3.1. C Predict which structure is preferred based on the formal charge on each atom and its electronegativity relative to the other atoms present. Solution: A Possible Lewis structures for the SCN − ion are as follows: B We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore 6 - (6 + 2/2) = -1. In (b), the sulfur atom has a formal charge of 0. In (c), the sulfur atom has a formal charge of +1. Continuing with the nitrogen, we observe that in (a) the nitrogen atom shares three bonding pairs and has one lone pair and has a total of 5 valence electrons. The formal charge on the nitrogen atom is therefore 5 - (2 + 6/2) = 0. In (b), the nitrogen atom has a formal charge of -1. In (c), the nitrogen atom has a formal charge of -2. C Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2. Exercise $\PageIndex{2}$ Salts containing the fulminate ion (CNO − ) are used in explosive detonators. Draw three Lewis electron structures for CNO − and use formal charges to predict which is more stable. (Note: N is the central atom.) Answer The second structure is predicted to be the most stable. Exercises Draw the Lewis structure of each of the molecules listed below. CH 3 + , NH 2 − , CH 3 − , NH 4 + , BF 4 − In each case, use the method of calculating formal charge described to satisfy yourself that the structures you have drawn do in fact carry the charges shown. Answer: Contributors Dr. Dietmar Kennepohl FCIC (Professor of Chemistry, Athabasca University ) Prof. Steven Farmer ( Sonoma State University ) William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/17%3A_Aromatic_Compounds/17.01%3A_Introduction-_The_Discovery_of_Benzene	The 100 Year Mystery of Benzene It took humans over 100 years to determine and confirm the structure of benzene. Why did it take so long? Why was there such a curiosity? The 1:1 ratio of carbon to hydrogen in the empirical formula and low chemical reactivity of benzene were a paradox to chemists in the early 1800's. In 1825, Michael Faraday isolated an oily residue of gas lamps. Faraday called this liquid "bicarburet of hydrogen" and measured the boiling point to be 80°C. Additionally, Faraday determined the empirical formula to be CH. About nine years later, Eilhard Mitscherlich synthesized the same compound from benzoic acid and lime (CaO). During the mid to late 1800's, several possible structures (shown below) were proposed for benzene. It was not until the 1930's that Kekule's structure was confirmed by X-ray and electron diffraction. During the end of Kekule's career he revealed that the structure came to him in a vision after enjoying a glass or two of wine by the fire in his favorite chair. His inspiration for the structure of benzene was derived from an ouroboros in the flames.
Courses/Indiana_Tech/Chemistry_2300_(Budhi)/13%3A_Carbonyl_Compounds_III-_Reactions_at_the_-_Carbon/13.12%3A_The_Malonic_Ester_Synthesis-_A_Way_to_Synthesize_a_Carboxylic_Acid	Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2. reaction 1: reaction 2: reaction 3: reaction 4: A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester. However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions. Malonic ester synthesis provides a more convenient alternative to convert 3 to 4 . Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6 . R 3 , R 4 = identical or different alkyl groups eg: reaction 1: reaction 2: reaction 1 (repeat): reaction 2 (repeat): reaction 3: reaction 4: Malonic Ester Synthesis Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester. Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides. After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step. Mechanism 1) Saponification 2) Decarboxylation 3) Tautomerization All of the steps together form the Malonic ester synthesis. \[RX \rightarrow RCH_2CO_2H\] Example Gamini Gunawardena from the OChemPal site ( Utah Valley University )
Courses/American_River_College/CHEM_309%3A_Applied_Chemistry_for_the_Health_Sciences/12%3A_Gases/12.06%3A_The_Combined_Gas_Law	Learning Objectives Understand mathematical relationship between pressure, volume, and temperature give the amount of gas is constant. Combined gas law The laws relating to pressure $P$, volume $V$, and temperature $T$ for a constant amount of a gas are the following: If amount of gas is constant: \[\dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}}\nonumber\] This is a combination of Boyle's law, Charles law and the Gay-Lussac's law. 1. If T 1 = T 2 : then $P_{1} V_{1}=P_{2} V_{2}$, this is the Boyle's law. 2. If P 1 = P 2 : then $\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}$, this is the Charles's law. 3. If V 1 = V 2 : then $\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}$, this is the Gay Lussac's law. The combined gas law allows calculating the effect of varying two parameters on the third. Example $\PageIndex{1}$ A weather balloon contains $212 \mathrm{~L}$ of helium at $25.0^{\circ} \mathrm{C}$ and $750. \mathrm{~mm} \mathrm{Hg}$. What is the volume of the balloon when it ascends to an altitude where the temperature is $-40.0{ }^{\circ} \mathrm{C}$ and $540. \mathrm{~mm} \mathrm{Hg}$, assuming the quantity of gas remains the same? Solution Given and desired parameters (temperatures must be converted to Kelvin scale): \[\begin{array}{lll} \mathrm{P}_{1}=750. \mathrm{~mm} \mathrm{Hg}, & \mathrm{V}_{1}=212 \mathrm{~L}, & \mathrm{~T}_{1}=25.0^{\circ} \mathrm{C}+273.15=298.2 \mathrm{~K} \\ \mathrm{P}_{2}=540. \mathrm{~mm} \mathrm{Hg}, & \mathrm{V}_{2}=? & \mathrm{~T}_{2}=-40.0^{\circ} \mathrm{C}+273.15=233.2 \mathrm{~K} \end{array}\nonumber\] Formula: \[\dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}}, \nonumber\] rearrange the formula to isolate the desired parameter: \[V_{2}=\dfrac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}. \nonumber\] Calculations: \[V_{2}=\dfrac{750. \cancel{\mathrm{~mm} \mathrm{Hg}} \times 212 \mathrm{~L} \times 233.2 \cancel{\mathrm{~K}}}{298.2 \cancel{\mathrm{~K}} \times 540. \cancel{\mathrm{~mm} \mathrm{Hg}}}=230. \mathrm{~L}. \nonumber\] Summary The combined gas law is a combination of Boyle's, Charles', and the Gay-Lussac's law.
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_II_(Wade)/19%3A_Ketones_and_Aldehydes/19.05%3A_Nucleophilic_Addition_Reactions_of_Ketones_and_Aldehydes	Carbonyls are Electrophiles Before we consider in detail the reactivity of aldehydes and ketones, we need to look back and remind ourselves of what the bonding picture looks like in a carbonyl. Carbonyl carbons are sp 2 hybridized, with the three sp 2 orbitals forming soverlaps with orbitals on the oxygen and on the two carbon or hydrogen atoms. These three bonds adopt trigonal planar geometry. The remaining unhybridized 2p orbital on the central carbonyl carbon is perpendicular to this plane, and forms a ‘side-by-side’ pbond with a 2p orbital on the oxygen. The carbon-oxygen double bond is polar: oxygen is more electronegative than carbon, so electron density is higher on the oxygen side of the bond and lower on the carbon side. Recall that bond polarity can be depicted with a dipole arrow, or by showing the oxygen as holding a partial negative charge and the carbonyl carbon a partial positive charge. A third way to illustrate the carbon-oxygen dipole is to consider the two main resonance contributors of a carbonyl group: the major form, which is what you typically see drawn in Lewis structures, and a minor but very important contributor in which both electrons in the pbond are localized on the oxygen, giving it a full negative charge. The latter depiction shows the carbon with an empty 2p orbital and a full positive charge. The result of carbonyl bond polarization, however it is depicted, is straightforward to predict. The carbon, because it is electron-poor, is an electrophile: it is a great target for attack by an electron-rich nucleophilic group. Because the oxygen end of the carbonyl double bond bears a partial negative charge, anything that can help to stabilize this charge by accepting some of the electron density will increase the bond’s polarity and make the carbon more electrophilic. Very often a general acid group serves this purpose, donating a proton to the carbonyl oxygen. The same effect can also be achieved if a Lewis acid, such as a magnesium ion, is located near the carbonyl oxygen. Nucelophilic Addition to a Carbonyl When a nucleophile reacts with the carbonyl carbon of an aldehyde or ketone, there is no leaving group – the incoming nucleophile simply ‘pushes’ the electrons in the pi bond up to the oxygen. After the carbonyl has reacted with the nucleophile, the negatively charged oxygen has the capacity to act as a nucleophile. The nucleophile can be charged or neutral. However, most commonly the oxygen acts as a base, abstracting a proton from a nearby acid group in the solvent or enzyme active site. The nucelophiiles studied in this chapter are water (H 2 O), cyanide (CN - ), Grignard reagent (RMgX), amines (and ammonia), hydrazine (N 2 H 4 ), alcohols (ROH), and phosphorus ylides (R 3 P=CRH). The generic mechanism for a strong nucleophile generally occurs under basic conditions as shown below. A closer look at the tetrahedral intermediate shows us that if the carbonyl reforms, then the original aldehyde or ketone is reformed. It is possible for the nucleophile to repeatedly add and leave the carbonyl group. Protonation of the tetrahedral intermediate to form the nucleophilic additon product is favored by the low activation energy of proton transfer reactions. For reaction with weak nucleophiles generally occurs under acidic conditions to increase the electrophilicity of the carbonyl group as resonance form below illustrates. While the net result of the reaction is similar, the mechanism is slightly different due to the order of the proton transfer reactions. Relative Reactivity of Carbonyl Compounds to Nucleophilic Addition In general aldehydes are more reactive than ketones because of the lack of stabilizing alkly groups. The primary carbocation formed in the in the polarizing resonance structure of an aldehyde (discussed above) is less stable and therefore more reactive than the secondary carbocation formed by a ketone. Exercise 4. Compare the mechanisms of an S N 2 reaction between 2-bromobutane and cyanide tetrahedral complex formation between 2-butanone and cyanide. Answer 4.
Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/01%3A_Atomic_Structure/1.05%3A_Unit_1_Practice_Problems	Unit 1A Exercise 1 What is true about the Rutherford atom model? It explains why atoms do not send out electromagnetic radiation permanently It explains why alpha particles get scattered by atoms It explains the atomic spectrum of the hydrogen atom It explains the wave-particle dualism Answer b) It explains why alpha particles get scattered by atoms Exercise 2 The Bohr radius (the radius of the electron orbit for the H atom in its ground state) is 5.29 x 10 -11 m. Calculate the radius of an electron in the third shell of the H atom according to the Bohr atom model. Answer r 3 = 3 2 x 5.29 x 10 -11 m = 47.61 x 10 -11 m. Exercise 3 The electron in an H atom undergoes an electronic transition from the 3 rd to the 2 nd shell. What frequency does the light that is emitted have? The energy of the electron in the first shell is -2.18 x 10 -18 J. Answer Energy of electron in the 3 rd shell: E 3 = -2.18 x 10 -18 J / 3 2 = -0.24 x 10 -18 J Energy of electron in the 2 nd shell: E 2 = -2.18 x 10 -18 J / 2 2 = -0.545 x 10 -18 J Energy difference between the two electrons: E 3 -E 2 = 0.305 x 10 -18 J Frequency of emitted light: n=(E 3 -E 2 )/h = 0.305 x 10 -18 J/ 6.63 x 10 -34 Js = 4.60 x 10 14 s - Exercise 4 What is the mass of a photon with a wavelength of 400 nm that travels though space? Answer λ = h/mc --> m = h/λc = 6.63 x 10 - 34 Js / (400 x 10 -9 m x 3.00 x 10 8 m/s) = 5.525 x 10 -36 kg. Exercise 5 Two objects are moving at the same speed. Which (if any) of the following statements are true? The DeBroglie wavelength of the heavier object is longer than that of the lighter one. If one object has twice as much mass as the other, its wavelength is one-half of the other Doubling the speed of one of the objects will have the same effect on its wavelength as doubling its mass. Answer a) If one object has twice as much mass as the other, its wavelength is one-half of the other b) Doubling the speed of one of the objects will have the same effect on its wavelength as doubling its mass Exercise 6 The power of a red laser with a wavelength of 630 nm) is 1.00 Watt (1.00 Js). How many photons per second does the laser emit? Answer Unit 1B Exercise 7 Which of the following waves would you consider to be standing matter wave? The vibration of a drum. Sound traveling through open air. A tsunami. None of the above. Answer a) The vibration of a drum. Exercise 8 What are orbitals (more than one answer can be correct)? Wave functions that describe the electrons as three-dimensional standing matter waves in an atom. Spaces inside an atom in which the electron travels as a classical particle. Solutions of the Schrödinger equation for the hydrogen atom. Answer a) Wave functions that describe the electrons as three-dimensional standing matter waves in an atom. c) Solutions of the Schrödinger equation for the hydrogen atom. Exercise 9 Which quantum numbers l are allowed when the quantum number n is 4? Answer l can be 3,2,1,0 Exercise 10 Which quantum numbers m are allowed when the quantum number l is 3? Answer -3,-2,-1,0,+1,+2,+3 Exercise 11 What is true about the following wave function? \[\Psi=\frac{1}{\sqrt{2 \pi}} \quad \frac{\sqrt{6}}{2} \cos \theta \quad \frac{4}{81 \sqrt{6} a_0 3 / 2}\left[6-\frac{r}{a_0}\right] \frac{r}{a_0} e^{-r / 3 a_0} \nonumber\] Its does not have angular nodes Its does not have spherical nodes Its amplitude is 0 at the nucleus The wave function represents an s orbital Answer c) Its amplitude is 0 at the nucleus Exercise 12 The angular part of the wavefunction of an orbital has the following form: \[\Theta \Phi(x, y, z)=\frac{1}{4} \sqrt{\frac{15}{\pi}} \frac{\left(x^2-y^2\right)}{r^2} \nonumber\] Which planes are the planar nodes in this orbital? Answer The wavefunction is 0 for x=y and x=-y. So it is two planes that bisect the x and the y axis. Unit 1C Exercise 13 Order the following orbitals with respect to their penetration abilities: 4s, 4p, 4d, 4f Answer 4s > 4p > 4d > 4f Exercise 14 What is the effective nuclear charge on an electron in an He + ion? Answer Exercise 15 Calculate the orbital energy of a 3p electron in a sulfur atom using the Slater rules. Answer (1s 2 ) (2s 2 2p 6 ) (3s 2 3p 4 ) Exercise 16 Calculate the orbital energy of a 3d electron in a palladium atom using the Slater rules. Answer (1s 2 ) (2s 2 2p 6 ) (3s 2 3p 6 )(3d 10 )(4s 2 4p 6 )(4d 8 )(5s 2 ) Exercise 17 Calculate how much higher the first ionization energy of an oxygen atom is compared to a fluorine atom. Use the Slater rules to answer the question. Answer O: (1s 2 ) (2s 2 2p 4 ) Dr. Kai Landskron ( Lehigh University ). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate .
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/Contextual_Modules/Optical_Properties_of_Gold_Nanoparticles/03_Laboratory_Experiment/1._Safety_and_Waste_Disposal	Wear gloves and eye protection. Sodium citrate may cause irritation to skin, eyes, and respiratory tract. Tetrachloroauric acid causes eye and skin burns. Harmful if swallowed or inhaled. All reactions should be conducted under a hood. All leftover chemicals must be collected in the waste bottle in the waste hood. All glassware for the AuNP synthesis must be free of contaminants and should be washed in aqua regia prior to use and rinsed thoroughly with nanopure water. Aqua regia, once prepared, can be used for approximately 4 weeks. If nanopure water is not available, commercially purchased distilled water can be substituted. When weighing gold chloride and sodium citrate, it is common practice to use a folded weighing paper and a spatula as the folded weighing paper gives better control for transferring small quantities. Students should receive instruction in proper use of pipets. Chemical waste should be properly disposed in heavy metal hazardous waste containers.
Courses/Woodland_Community_College/WCC%3A_Chem_2A_-_Introductory_Chemistry_I/08%3A_Gases_Liquids_and_Solids/8.01%3A_States_of_Matter_and_Their_Changes	Learning Objectives Review the states of matter and their properties Describe how change in temperature will affect the state of matter. Previously, you were introduced to the three states, also called phases, of matter; solid, liquid, and gas. A phase is a certain form of matter that includes a specific set of physical properties. That is, the atoms, the molecules, or the ions that make up the phase do so in a consistent manner throughout the phase. Science recognizes three stable phases: the solid phase , in which individual particles can be thought of as in contact and held in place; the liquid phase , in which individual particles are in contact but moving with respect to each other; and the gas phase , in which individual particles are separated from each other by relatively large distances (see Figure $\PageIndex{1}$). The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the attractive forces between molecules, called intermolecular forces . The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces draw the particles together. A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces. We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO 2 , as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure $\PageIndex{2}$. Energy Changes That Accompany Phase Changes Phase changes are always accompanied by a change in the enthalpy, $\Delta H$, of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; the $\Delta H$ is positive (endothermic). Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; the $\Delta H$ is negative (exothermic). The energy change associated with each common phase change is shown in Figure $\PageIndex{2}$. $\Delta H$ is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state. Previously, we defined the enthalpy changes associated with various chemical and physical processes. The molar enthalpy of fusion ($ΔH_{fus}$), is the energy required to convert a solid to a liquid, a process known as fusion (or melting). As noted above, the process of melting requires energy and therefore, the ($ΔH_{fus}$) is positive. The reverse process of freezing would release energy making the ($ΔH_{fus}$) negative. The molar enthalpy of vaporization ($ΔH_{vap}$), is the energy required to convert a liquid to a gas, known as vaporization. Melting points, enthalpies of fusion, boiling points, and enthalpies of vaporization for selected compounds are listed in Table $\PageIndex{1}$. Substance Melting Point (°C) ΔHfus (kJ/mol) Boiling Point (°C) ΔHvap (kJ/mol) N2 −210.0 0.71 −195.8 5.6 HCl −114.2 2.00 −85.1 16.2 Br2 −7.2 10.60 58.8 30.0 CCl4 −22.6 2.56 76.8 29.8 CH3CH2OH (ethanol) −114.1 4.93 78.3 38.6 CH3(CH2)4CH3 (n-hexane) −95.4 13.10 68.7 28.9 H2O 0 6.01 100 40.7 Na 97.8 2.60 883 97.4 NaF 996 33.40 1704 176.1 The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid).
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.13%3A_Summary_of_Reactions	No stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines. Preparation of alkynes Dehydrohalogenation of vicinal dihalides (Section 9.2) Alkylation of acetylide anions (Section 9.8) Reactions of alkynes Addition of HCl and HBr (Section 9.3) Addition of Cl 2 and Br 2 (Section 9.3) Hydration (Section 9.4) (1) Mercuric sulfate catalyzed (2) Hydroboration–oxidation Reduction (Section 9.5) (1) Catalytic hydrogenation (2) Lithium in liquid ammonia Conversion into acetylide anions (Section 9.7)
Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/28%3A_Grammar%3A_Sentence_Structure/27.5%3A_Apostrophes_and_Quotation_Marks	Apostrophes Possession With possessives, the apostrophe is used in combination with an s to represent that a word literally or conceptually possesses what follows it. Singular words whether or not they end in s , are made possessive by adding an apostrophe + s . For plural words, we typically indicate possession simply by adding the apostrophe without an additional s . However, a plural that does not end in an s (e.g., bacteria ), we would add an apostrophe + s . a student’s paper one hour’s passing Illinois’s law interviewees’ answers her professors’ office (an office shared by two of her professors; if it were just one professor we would write her professor’s office ) Note: Practices vary from style to style, so be sure to check the rules in your course’s discipline for this. Contractions A contraction is a shortened phrase. He will becomes he’ll , are not becomes aren’t , would have becomes would’ve , and it is becomes it’s . In all of these cases, the apostrophe stands in for the missing letters. You may find yourself being steered away from using contractions in your papers. While you should write to your teacher’s preference, keep in mind that leaving out contractions can often make your words sound over formal and stilted. (And you shouldn’t eliminate contractions in your papers just to up your word count!) Your versus You’re Your versus You’re Your v. you’re Its v. it’s Their v. they’re All three of these pairs are the same kind of pair: a possessive pronoun and a contracted version of a pronoun + to be ( you’re = you are ; it’s = it is ; they’re = they are ). These are easy to mix up (especially its / it’s ) because—as we’ve learned—an apostrophe + s indicates possession. The best way to use these correctly is to remember that possessive pronouns never have an apostrophe: if there’s an apostrophe with a pronoun, it’s a contraction, not a possessive. Acronyms and Numbers In technical writing, acronyms and numbers are frequently pluralized with the addition of an apostrophe s , but this is falling out of favor, and there is typically no need to put an apostrophe in front of the s . Therefore, SSTs (sea surface temperatures) is more acceptable than SST’s when your intention is simply to pluralize. Ideally, use the apostrophe before the s with an acronym or a number only to show possession (i.e., “an 1860’s law”; “DEP’s testing”) or when confusion would otherwise result (“mind your p ’s and q ’s”). When talking about a specific decade the 1920s should be shortened to the ’20s . Notice that the apostrophe curls away from the numbers, indicating that the missing characters originally appeared prior to the apostrophe. Practice Read the following passage. Identify any errors with apostrophes. Type the corrected words in the text frame below: Thanks to NASAs’ team of sniffers, led by George Aldrich, astronauts can breathe a little bit easier. Aldrich is the “chief sniffer” at the White Sands Test Facility in New Mexico. His’s job is to smell items before they can be flown in the space shuttle. Aldrich explained that smells change in space and that once astronauts are up there, their stuck with whatever smells are onboard with them. In space, astronauts aren’t able to open the window for extra ventilation. He also said that its important not to introduce substances that will change the delicate balance of the climate of the International Space Station and the space shuttle. [practice-area rows=”4″][/practice-area] [reveal-answer q=”61337″]Show Answer[/reveal-answer] [hidden-answer a=”61337″]Here is the passage with the errors in bold: Thanks to NASAs’ team of sniffers, led by George Aldrich, astronauts can breathe a little bit easier. Aldrich is the “chief sniffer” at the White Sands Test Facility in New Mexico. His’s job is to smell items before they can be flown in the space shuttle. Aldrich explained that smells change in space and that once astronauts are up there, their stuck with whatever smells are onboard with them. In space, astronauts aren’t able to open the window for extra ventilation. He also said that its important not to introduce substances that will change the delicate balance of the climate of the International Space Station and the space shuttle. NASAs’ should be NASA’s . His’s doesn’t need the apostrophe + s . In fact, possessive pronouns don’t require apostrophes at all. His’s should be His . Their is a possessive pronoun; the correct word is they’re , which is a contraction of the words they are . Its is a possessive pronoun; the correct word is it’s , which is a contraction of the words it is . The contraction aren’t is used correctly in the passage. [/hidden-answer] Quotation Marks There are three typical ways quotation marks are used. The first is pretty self-explanatory: you use quotation marks when you’re making a direct quote. He said “I’ll never forget you.” It was the best moment of my life. Yogi Berra famously said, “A nickel ain’t worth a dime anymore.” If you’re just writing an approximation of something a person said, you would not use quotation marks: She told me about Pizza the three-toed sloth yesterday. He said that he would be late today. The second is when you’re calling attention to a word. For example: I can never say “Worcestershire” correctly. How do you spell “definitely”? Note: It is this course’s preference to use italics in these instances: I can never say Worcestershire correctly. How do you spell definitely ? However, using quotes is also an accepted practice. The last use is scare quotes. This is the most misused type of quotation marks. People often think that quotation marks mean emphasis. Buy some “fresh” chicken today! We’ll give it our “best” effort. Employees “must” wash their hands before returning to work. However, when used this way, the quotation marks insert a silent “so-called” into the sentence, which is often the opposite of the intended meaning. Where do Quotation Marks Go? Despite what you may see practiced, the fact is that the period and comma always go inside the quotation marks. (The rules in British English are different, which may be where some of the confusion arises.) Correct: The people of the pine barrens are often called “pineys.” Incorrect: The people of the pine barrens are often called “pineys”. The semicolon, colon, dash, question mark, and exclamation point can fall insider outside of the quotation marks, depending on whether the punctuation is a part of the original quote: This measurement is commonly known as “dip angle”; dip angle is the angle formed between a normal plane and a vertical. Built only 50 years ago, Shakhtinsk—“minetown”—is already seedy. When she was asked the question “Are rainbows possible in winter?” she answered by examining whether raindrops freeze at temperatures below 0 °C. (Quoted material has its own punctuation.) Did he really say “Dogs are the devil’s henchmen”? (The quote is a statement, but the full sentence is a question.) Practice Has the following passage been punctuated correctly? Type any corrections in the text frame below: Gabrielly and Marcelo both knew a lot of “fun facts” that they liked to share with each other. Yesterday Gabrielly said to Marcelo, “Did you know that wild turkeys can run up to twenty-five miles per hour?” “Well, an emu can run twice that speed,” Marcelo responded. “Did you know that there’s a dinosaur-themed park in Poland called JuraPark Bałtów”? Gabrielly asked. Marcelo then told her about “Rusik, the first Russian police sniffer cat, who helped search for illegal cargoes of fish and caviar”. [practice-area rows=”4″][/practice-area] [reveal-answer q=”443726″]Show Answer[/reveal-answer] [hidden-answer a=”443726″]There are five sets of quotation marks in this passage. Let’s look at each set. The first set, around fun facts , may or may not be appropriate. If the intent is to emphasize the facts, then the quotes are incorrect. However, if you want to indicate that the facts aren’t actually fun (and possibly annoying), the quotes are appropriate. The second and third sets are used correctly, and their surrounding punctuation is also correct. Remember, commas always go inside quotation marks. The fourth set starts correctly; however, the question mark at the end should be inside the quotation marks, since the quote is a question. “Did you know that there’s a dinosaur-themed park in Poland called JuraPark Bałtów?” Gabrielly asked. The fifth set surrounds an approximation of what Marcelo said. This means no quotation marks are needed. However, even if the quotes were needed, the sentence would still be incorrect: periods always go inside quotation marks. Marcelo then told her about Rusik, the first Russian police sniffer cat, who helped search for illegal cargoes of fish and caviar. Marcelo then said, “Rusik, the first Russian police sniffer cat, helped search for illegal cargoes of fish and caviar.” [/hidden-answer] CC licensed content, Original Revision and Adaptation. Provided by : Lumen Learning. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Original Icons. Provided by : Lumen Learning. License : CC BY: Attribution Quotation Practice Activity. Provided by : Lumen Learning. License : CC BY: Attribution CC licensed content, Shared previously Style For Students Online. Authored by : Joe Schall. Provided by : The Pennsylvania State University. Located at : https://www.e-education.psu.edu/styleforstudents/ . Project : Penn State's College of Earth and Mineral Sciences' OER Initiative. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Public domain content George Aldrich (errors added). Provided by : NASA. Located at : http://spaceflight.nasa.gov/shuttle/support/people/galdrich.html . Project : Behind the Scenes: Meet the People. License : Public Domain: No Known Copyright
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/The_Live_Textbook_of_Physical_Chemistry_(Peverati)/26%3A_Introduction_to_Molecules/26.03%3A_Solving_the_Electronic_Eigenvalue_Problem	Once we have invoked the Born-Oppenheimer approximation, we can attempt to solve the electronic TISEq in Equation 27.2.7 . However, for molecules with more than one electron, we need to—once again—keep in mind the antisymmetry of the wave function. This obviously means that we need to write the electronic wave function as a Slater determinant (i.e., all molecules but $\mathrm{H}_2^+$ and a few related highly exotic ions). Once this is done, we can work on approximating the Hamiltonian, a task that is necessary because the presence of the electron-electron repulsion term forbids its analytic treatment. Similarly to the many-electron atom case, the simplest approximation to solve the molecular electronic TISEq is to use the variational method and to neglect the electron-electron repulsion. As we noticed in the previous chapter, this approximation is called the Hartree-Fock method. The Hartree-Fock Method The main difference when we apply the variational principle to a molecular Slater determinant is that we need to build orbitals (one-electron wave functions) that encompass the entire molecule. This can be done by assuming that the atomic contributions to the molecular orbitals will closely resemble the orbitals that we obtained for the hydrogen atom. The total molecular orbital can then be built by linearly combine these atomic contributions. This method is called linear combination of atomic orbitals (LCAO) . A consequence of the LCAO method is that the atomic orbitals on two different atomic centers are not necessarily orthogonal, and Equation 26.2.4 cannot be simplified easily. If we replace each atomic orbital $\psi(\mathbf{r})$ with a linear combination of suitable basis functions $f_i(\mathbf{r})$: \[ \psi(\mathbf{r}) = \sum_i^m c_{i} f_i(\mathbf{r}), \label{27.3.1} \] we can then use the following notation: \[ \displaystyle H_{ij} = \int \phi_i^* {\hat H} \phi_j d\mathbf{\tau}\;, \qquad \displaystyle S_{ij} = \int \phi_i^* \phi_jd\mathbf{\tau}, \label{27.3.2} \] to simplify Equation 26.2.4 to: \[ E[\Phi] = \dfrac{\sum_{ij} c_i^* c_j H_{ij}}{\sum_{ij} c_i^* c_j S_{ij}}. \label{27.3.3} \] Differentiating this energy with respect to the expansion coefficients $c_i$ yields a non-trivial solution only if the following “secular determinant” equals zero: \[ \begin{vmatrix} H_{11}-ES_{11} & H_{12}-ES_{12} & \cdots & H_{1m}-ES_{1m}\\\ H_{21}-ES_{21} & H_{22}-ES_{22} & \cdots & H_{2m}-ES_{2m}\\\ \vdots & \vdots & \ddots & \vdots\\\ H_{m1}-ES_{m1} & H_{m2}-ES_{m2} & \cdots & H_{mm}-ES_{mm} \end{vmatrix}=0 \label{27.3.4} \] where $m$ is the number of basis functions used to expand the atomic orbitals. Solving this set of equations with a Hamiltonian where the electron-electron correlation is neglected results is non-trivial, but possible. The reason for the complications comes from the fact that even if we are neglecting the direct interaction between electrons, each of them interact with the nuclei through an interaction that is screened by the average field of all other electrons, similarly to what we saw for the helium atom. This means that the Hamiltonian itself and the value of the coefficients $c_i$ in the wave function mutually depend on each other. A solution to this problem can be achieved numerically using specialized computer programs that use a cycle called the self-consistent-field (SCF) procedure . Starting from an initial guess of the coefficients, an approximated Hamiltonian operator is built from them and used to solve Equation \ref{27.3.4}. This solution gives updated values of the coefficients, which can then be used to create an improved version of the approximated Hamiltonian. This procedure is repeated until both the coefficients and the operator do not change anymore. From this final solution, the energy of the molecule is then calculated.
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_332_--_Organic_Chemistry_II_(Lund)/12%3A_Reactions_at_the_-Carbon_Part_I	12.0: Prelude to Reactions at the α-carbon, part I There is a connection between the killer platypus in Australia and the 'hunting magic' in the Amazon, and it has to do with the structure and reactivity of what organic chemists refer to as the α -carbon: the carbon atom positioned adjacent to a carbonyl or imine group in an organic molecule. 12.1: Review of Acidity at the α-Carbon Let's review what we learned in section 7.6 about the acidity of a proton on an a-carbon and the structure of the relevant conjugate base, the enolate ion. Remember that this acidity can be explained by the fact that the negative charge on the enolate conjugate base is delocalized by resonance to both the α -carbon and the carbonyl oxygen. 12.2: Isomerization at the α-Carbon Enolate ions are the key reactive intermediates in many biochemical isomerization reactions. Isomerizations can involve either the interconversion of constitutional isomers, in which bond connectivity is altered, or of stereoisomers, where the stereochemical configuration is changed. Enzymes that interconvert constitutional isomers are usually called isomerases, while those that interconvert the configuration of a chiral carbon are usually referred to as racemases or epimerases. 12.3: Aldol Addition Along with Claisen condensation reactions, which we will study in the next chapter, aldol additions are responsible for most of the carbon-carbon bond forming events that occur in a living cell. Because biomolecules are built upon a framework of carbon-carbon bonds, it is difficult to overstate the importance of aldol addition and Claisen condensation reactions in the chemistry of living things! 12.4: α-Carbon Reactions in the Synthesis Lab - Kinetic vs. Thermodynamic Alkylation Products While aldol addition reactions are widespread in biochemical pathways as a way of forming carbon-carbon bonds, synthetic organic chemists working the lab also make use of aldol-like reactions for the same purpose. 12.5: Nitrosation In organic chemistry, nitrosamines are organic compounds with a nitroso group (-N=O) bonded to an amine. Most nitrosamines are carcinogenic. There are significant levels of nitrosamines in many foodstuffs, especially meat and cheese products preserved with nitrite compounds. 12.E: Reactions at the α-Carbon, Part I (Exercises) 12.S: Reactions at the α-Carbon, Part I (Summary)
Courses/CSU_Chico/CSU_Chico%3A_CHEM_451_-_Biochemistry_I/CHEM_451_Test/04%3A_Protein_Structure/4.1%3A_Amino_Acids/A8._Reactions_of_Histidine	Histidine is one of the strongest bases at physiological pH's. The nitrogen atom in a secondary amine might be expected to be a stronger nucleophile than a primary amine through electron release to that N in a secondary amine. Opposing this effect is the steric hindrance by the two attached Cs of the N on attach on an electrophile . However, in His, this steric effect is minimized since the 2Cs are restrained by the ring. With a pKa of about 6.5, this amino acid is one of the strongest available bases at physiological pH (7.0). Hence, it can often cross-react with many of the reagents used to modify Lys side chains. His reacts with reasonably high selectivity with diethyl pyrocarbonate. Figure: REACTIONS OF HISTIDINE Figure: Where is the H on His? Where is the Charge?
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Introduction_to_Lasers/03_Applications/01_Common_Devices	Barcode Readers, CD Players and Laser Pointers Take advantage of the brightness and highly directional properties of laser radiation and often employ diode lasers. Supermarket barcode readers a - As the light beam is scanned across the barcode, the white and black regions produce a modulation in the reflected light intensity. Lasers enable supermarket scanners to read barcodes rapidly and when presented over a range of angles. CD (and DVD) players b , c - Lasers are used to both encode and read information on CDs and DVDs. A smaller beam focus enables a greater density of information to be stored. Blu-ray technology uses a shorter wavelength than earlier technologies and hence provides greater information storage. http://www.fotosearch.com/photos-images/laser.html Applications References http://www.explainthatstuff.com/barcodescanners.html www.explainthatstuff.com/cdplayers.html http://micro.magnet.fsu.edu/electrom.../cd/index.html
Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/01%3A_Chemistry	1.1: Science and Technology - The Roots of Knowledge 1.2: Science- Reproducible, Testable, Tentative, Predictive, and Explanatory 1.3: Solving Society’s Problems - Scientific Research How did chemistry develop? What is happening in the field of chemistry today? What can I do with a chemistry degree? All of these are good questions and they should be asked by students interested in chemistry. Research in chemistry (or any other field, for that matter) is interesting and challenging. But there are different directions a person can take as they explore research opportunities. 1.4: Chemistry - A Study of Matter and Its Changes 1.5: Classification of Matter 1.6: The Measurement of Matter 1.7: Density 1.8: Energy - Heat and Temperature
Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/Organic_Laboratory_Experiments/03%3A_Inorganic_Laboratory_Experiments/3.04%3A_Preparation_of_some_Cr(III)_and_Cu(II)_Oxalate_Complexes	Theory Oxalic acid, (1,2-ethanedioic acid), in the form of the dianion, functions as a bidentate ligand with many transition metal ions. For example, the Fe(III) complex is the most widely used compound in chemical actinometry (see the CHEM2111 manual). The Calcium complex is the deposit that is responsible for cataracts and kidney stones. Chromium is an abundant element in the earth's crust. The metal is used for plating and in chrome steels. The chromates [chromium(VI)] have many industrial uses as pigments, catalysts and fungicides. Chromium(III) is a common and stable oxidation state that displays significant kinetic inertness. The most common geometry is octahedral with other shapes being quite rare. Copper is not as abundant (68ppm compared to 122ppm for $\ce{Cr}$). It has been known since prehistoric times and is used extensively in wiring. The usual stereochemistry is either 4 coordinate or distorted 6 coordinate with 4 short bonds and 2 long bonds in accordance with the Jahn-Teller theorem. The solid state structure of the complex with 2 oxalates depends on the cation used: for Na+ and K+ both are 6 coordinate with links from the carbonyl O to a neighbouring copper. Preparation of $\ce{K3[Cr(C2O4)3].3H2O}$ Note The discovery of this compound is credited to Dr Wilton George Turner who in the London winter of 1830-1831 isolated it by accident. Turner was born in Clarendon, Jamaica in 1810 and died in the Turks Islands in 1855. He obtained a PhD from the University of Geissen in 1838 (with Justus von Liebig). It was described thus: This beautiful salt crystallizes in thin elongated prisms, which appear black by reflection, blue by transmitted light, and green when reduced to powder. Its solution is green and red at the same time, except by candlelight, when it is of a pure red. Suspend 5.5 g of oxalic acid dihydrate in 10 cm 3 of cold water in a 600 cm 3 beaker. Add, in small portions with stirring, 1.8 g of potassium dichromate. CARE ! $\ce{Cr(VI)}$ is a suspected carcinogen The orange mixture soon warms up spontaneously, almost to boiling point, and a vigorous evolution of gas occurs. When the reaction has ceased, decant one half of the hot green-black liquid into a 100 cm 3 beaker. (Save the other half for the next preparation.) . Stir in 1.1 g of potassium oxalate monohydrate and when all the solid has dissolved add 2 cm 3 of ethanol. Heat the reaction mixture on a water bath until solid begins to deposit from solution. Then add, with stirring, 5 cm 3 of ethanol and cool in an ice bath. When cold, filter off the green-black solid and wash with three 5 cm 3 portions of cold 2:1 ethanol/water and then with 10 cm 3 ethanol ( No flames ). The product is dried in air and the yield calculated as a percentage based on chromium. The overall reaction is: \[\ce{K2Cr2O7 + 7H2C2O4 + 2K2C2O4 → 2K3[Cr(C2O4)3].3H2O + 6CO2 + H2O}\nonumber \] Preparation of trans- $\ce{K[Cr(C2O4)2 (H2O)2].3H2O}$ The solution saved from the first part of the experiment contains an equilibrium mixture of the cis- and trans- isomers of the diaquabis(oxalato)chromate(III) ion. The potassium salt of the trans- isomer is purple and the cis- isomer is blue-gray. The lower solubility of the trans- isomer results in its preferential crystallization. Slowly evaporate the solution saved, to about one half its bulk ( do not boil ) and allow further evaporation to occur spontaneously at room temperature. By this stage, crystals should have been deposited, which are filtered off and washed with ethanol. The product is air-dried and the yield calculated as a percentage based on chromium. The yield is small but it is more important to isolate the pure trans- isomer than to maximize the yield. Do not evaporate the solution to dryness. Show your product to a demonstrator. If at the end of the lab day no product has formed, cover container and store carefully in your cupboard for the next lab session. Preparation of $\ce{K2[Cu(C2O4)2].2H2O}$ Prepare two solutions: Dissolve 3.1 g of potassium oxalate monohydrate in 15 cm 3 of water Dissolve 2.0 g of $\ce{CuSO4.5H2O}$ in 10 cm 3 of water. Heat both solutions to about 60 °C and then slowly add the copper(II) solution to the oxalate solution with stirring. (If you reverse the order of mixing the crystals formed are usually found to be smaller). Cool the mixture in ice water and filter off the blue crystals via a weighed No.3 sintered glass crucible. Wash with 2 x 5 cm 3 portions of ice-cold water and 10 cm 3 of ethanol (NO FLAMES !!). Air dry the product and determine the yield. Show the product to your demonstrator! (KEEP THE $\ce{K2[Cu(C2O4)2].2H2O}$ FOR NEXT WEEK'S LAB) Write an equation for the preparation and calculate the percentage yield based on the quantity of copper sulfate crystals used. Questions Draw the structures of the cis- and trans- diaquabisoxalatochromate(III) ions. The trioxalatochromate ion is also a mixture of isomers. Explain and draw the isomers. Draw in perspective, the structure of the copper complex. Dissolve a small quantity of $\ce{K2Cr2O7}$ in dilute $\ce{NaOH}$. Note and explain your observations.
Courses/University_of_South_Carolina__Upstate/CHEM_U109%3A_Chemistry_of_Living_Things_-_Mueller/07%3A_Energy_and_Chemical_Processes	Energy is a vital component of the world around us. Nearly every physical and chemical process, including all the chemical reactions discussed in previous chapters, occurs with a simultaneous energy change. In this chapter, we will explore the nature of energy and how energy and chemistry are related. 7.1: Energy and Its Units Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules. 7.2: Heat Heat transfer is related to temperature change. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat. 7.3: Phase Changes There is an energy change associated with any phase change. There is an energy change associated with any phase change. 7.4: Bond Energies and Chemical Reactions Atoms are held together by a certain amount of energy called bond energy. Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively. 7.5: The Energy of Biochemical Reactions Energy to power the human body comes from chemical reactions. 7.6: Energetics and Kinetics If we want to know if a reaction is likely proceed and how far it is likely to go, we can look at its energetics (or thermodynamics). If the products end at lower energy than the reactants, they are more stable and thus likely to form. It is also important to look at the speed of a reaction, or its kinetics. Here we will again look at reaction energy diagrams; this time to summarize both the energetics and the kinetics of a reaction --- how far and how fast it will go. 7.7: Equilibrium Many reactions are reversible, or can go both backward and forward. This process is sometimes referred to as equilibrium. How far the reaction goes forward or backward in the equilibrium process is related to the energy of the reaction. 7.8: Chapter Summary To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms in the following summary and ask yourself how they relate to the topics in the chapter. 7.E: Exercises Problems and Solutions to accompany the chapter.
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/22%3A_Arenes_Electrophilic_Aromatic_Substitution/22.12%3A_Some_Conjugated_Cyclic_Polyenes	Azulene There are several compounds that possess some measure of aromatic character typical of benzene, but do not possess a benzenoid ring. Appropriately, they have $\left( 4n + 2 \right)$ $\pi$ electrons and are classified as nonbenzenoid aromatic compounds ( Section 21-9 ). An example is azulene, which is isomeric with naphthalene and has a five- and a seven-membered ring fused through adjacent carbons: As the name implies, it is deep blue. It is less stable than naphthalene, to which it isomerizes quantitatively on heating above $350^\text{o}$ in the absence of air: Azulene has a significant polarity, with the five-membered ring negative and the seven-membered ring positive. The structure can be represented as a hybrid of neutral and ionic structures: The polarization that has the five-membered ring negative and the seven-membered ring positive corresponds to ionic structures that have six (i.e., $4n + 2$) electrons in both the five- and seven-membered rings ( Section 21-9B ). In keeping with its aromatic character and unsymmetrical charge distribution, azulene undergoes certain typical electrophilic substitution reactions at the 1 and 3 positions. Thus Friedel-Crafts acylation leads to a mixture of 1-ethanoylazulene and 1,3-diethanoylazulene: Furthermore, in the presence of strong acids the 1 position is protonated to give a derivative of the relatively stable cycloheptatrienyl (tropylium) ion ( Section 21-9B ): Cyclooctatetraene 1,3,5,7-Cyclooctatetraene (or simply cyclooctatetraene) is a bright-yellow, nonplanar, nonaromatic compound ( Section 21-9A ). Apparently the resonance energy of a planar structure is insufficient to overcome the unfavorable angle strain of a planar structure would have with its $\ce{C-C-C}$ bond angles of $135^\text{o}$. Cyclooctatetraene normally assumes a "tub" structure with alternating single and double bonds: There is, however, nmr evidence that indicates that the tub form is in rapid equilibrium with a very small amount of the planar form at room temperature. There is about a $15$-$\text{kcal mol}^{-1}$ energy difference between the two forms. The dication, $\ce{C_8H_8^{2+}}$, and the dianion of cyclooctatetraene, $\ce{C_8H_8^{2-}}$, which have $\left( 4n + 2 \right)$ $\pi$ electrons, appear to exist in planar conformation. Cyclooctatetraene can be prepared readily by polymerization of ethyne in the presence of nickel cyanide: It could be manufactured on a large scale, but no large-scale commercial uses of the substance have yet been developed. The chemistry of cyclooctatetraene is interesting and unusual. Particularly noteworthy is the way in which it undergoes addition reactions to form products that appear to be derived from the bicyclic isomer, bicyclo[4.2.0]2,4,7-octatriene, $8$. In fact, there is an electrocyclic equilibrium between cyclooctatetraene and $8$ ( Section 21-10D ) and, although the position of equilibrium lies far on the side of cyclooctatetraene, $8$ is more reactive and leads to the observed addition products: Treatment of the bridged dichloride with strong bases causes elimination of hydrogen chloride and formation of chlorocyclooctatetraene: The diverse ways in which cyclooctatetraene can react with a give reagent under different conditions is well illustrated by the variety of products obtained with mercuric ethanoate in ethanoic acid, methanol, and water: Efforts to prepare "pentalene", a bridged analog of cyclooctatetraene, have not been very successful so far. A substance that appears to be a methylpentalene has been characterized at $-180^\text{o}$ by its spectral properties. On warming to $-105^\text{o}$ it forms a dimer. Annulenes Cyclooctatetraene is nonplanar. One reason is that the angle strain is severe in the planar form. Is it possible that larger-ring conjugated polyalkenes may have strainless planar structures? Models show that a strainless structure can be achieved with two or more of the double bonds only in trans configurations, and then only with a large enough ring that the "inside" hydrogens do not interfere with one another. In discussing compounds of this type, it will be convenient to use the name [n]annulene to designate the simple conjugated cyclic polyalkenes, with n referring to the number of carbons in the ring - benzene being [6]annulene. The simplest conjugated cyclic polyolefin that could have a strainless planar ring containing trans double bonds, except for interferences between the inside hydrogens, is [10]annulene: Inside-hydrogen interferences are likely to be of some importancee in all annulenes up to [30]annulene. Many annulenes have been synthesized by F. Sondheimer. We have mentioned already ( Section 22-3C ) the large differences in nmr chemical shifts between the inside and outside hydrogens of [18]annulene - a substance which with 18 $\pi$ electrions should be aromatic by the $4n + 2$ rule. These differences are observed only at low temperatures. The proton nmr spectrum of [18]annulene at room temperature is a single resonance, which indicates that the inside $\left( \ce{H}_a \right)$ and outside $\left( \ce{H}_b \right)$ hydrogens are equilibrating rapidly. This can take place only if cis-trans interconversion occurs about the double bonds (marked $c$ and $t$): At low temperatures, this equilibrium is slow enough that separate groups of resonances for the inside and outside hydrogens can be discerned in an nmr experiment (see Sections 9-10C and 27-2 ). A theoretical prediction that has been borne out by experiment is that an annulene with $4n$ $\pi$ electrons should have a paramagnetic circulation of electrons - that is, opposite in direction to that shown in Figure 22-4 for benzene. For example, [16]annulene, which has $4n$ electrons, is not very stable and exists as a very rapidly interconverting mixture of two configurational isomers: At very low temperatures $\left( -155^\text{o} \right)$, the proton nmr spectrum shows the inner hydrogens at $\delta 12.9$-$10.5$ and the outer hydrogens at $\delta 5.7$-$6.4$, which is in exactly the opposite order to the shifts with [18]annulene and the other known $\left[ 4n + 2 \right]$ $\pi$-electron annulenes. The annulenes generally are not stable compounds, but the $\left[ 4n + 2 \right]$annulenes clearly show typical aromatic reactions. For instance [18]annulene has been converted to the nitro, ethanoyl, bromo, and carbaldehyde derivatives by electrophilic substitution reactions.
Courses/Pasadena_City_College/CHEM_001A%3A_General_Chemistry_and_Chemical_Analysis/26%3A_Transition_Metals_and_Coordination_Compounds/26.06%3A_Applications_of_Coordination_Compounds	Chelating Agents for Poising (EDTA) Biomolecules (Porphorins and Hemoglogin) Drugs and Health (cis-Platin) Learning Objectives To become familiar with some of the roles of transition-metal complexes in biological systems. In this section, we describe several systems that illustrate the roles transition metals play in biological systems. Our goal is for you to understand why the chemical properties of these elements make them essential for life. We begin with a discussion of the strategies organisms use to extract transition metals from their environment. The section continues with a brief discussion of the use of transition metals in reactions that involve the transfer of electrons, reactions of small molecules such as O 2 , Lewis-acid catalysis, and the generation of reactive organic radicals. Uptake and Storage of Transition Metals There are three possible dietary levels for any essential element: deficient, optimal, and toxic, in order of increasing concentration in the diet. If the concentration of an essential element in the diet is too low, an organism must be able to extract the element from the environment and concentrate it. If the concentration of an essential element in the diet is too high, an organism must be able to limit its intake to avoid toxic effects. Moreover, organisms must be able to switch off the uptake process rapidly if dietary levels rise suddenly, and they must be able to store essential elements for future use. Three distinct steps are involved in transition metal uptake. First, the metal must be “mobilized” from the environment and brought into contact with a cell in a form that can be absorbed. Second, the metal must be transported across the cell membrane into the cell. Third, the element must be transported to its point of utilization within a cell or to other cells within the organism. In our discussion, we focus on the uptake, transport, and storage of iron, which illustrates the most important points. Because iron deficiency (anemia) is the most widespread nutritional deficiency known in humans, the uptake of iron is especially well understood. Iron complexes in biological systems. Iron(III) forms very stable octahedral complexes with hydroxamate and catecholate ligands. The solubility of metal ions such as Fe 3 + , which form highly insoluble hydroxides, depends on the pH and the presence of complexing agents. In an oxygen-containing atmosphere, iron exists as Fe(III) because of the positive reduction potential of Fe 3 + (Fe 3 + + e − → Fe 2 + ; E° = +0.77 V). Because ferric hydroxide [Fe(OH) 3 ] is highly insoluble (K sp ≈ 1 × 10 −39 ), the equilibrium concentration of Fe 3 + (aq) at pH 7.0 is very low, about 10 −18 M. You would have to drink 2 × 10 13 L of iron-saturated water per day (roughly 5 mi 3 ) to consume the recommended daily intake of Fe for humans, which is about 1 mg/day. Animals such as humans can overcome this problem by consuming concentrated sources of iron, such as red meat, but microorganisms cannot. Consequently, most microorganisms synthesize and secrete organic molecules called siderophores to increase the total concentration of available iron in the surrounding medium. Siderophores are generally cyclic compounds that use bidentate ligands, such as the hydroxamate and catecholate groups shown here, to bind Fe 3 + in an octahedral arrangement. Typical siderophores are ferrichrome, a cyclic peptide produced by fungi, and enterobactin, a cyclic ester produced by bacteria (Figure $\PageIndex{1}$). Attaching the three iron ligands to a cyclic framework greatly increases the stability of the resulting Fe 3 + complex due to the chelate effect described in Section 23.4. The formation constants for the Fe 3 + complexes of ferrichrome and enterobactin are about 10 32 and 10 40 , respectively, which are high enough to allow them to dissolve almost any Fe(III) compound. Siderophores increase the [Fe 3 + ] in solution, providing the bacterium that synthesized them (as well as any competitors) with a supply of iron. In addition, siderophores neutralize the positive charge on the metal ion and provide a hydrophobic “wrapping” that enables the Fe 3 + –siderophore complex to be recognized by a specific protein that transports it into the interior of a cell. Once it is inside a cell, the iron is reduced to Fe 2 + , which has a much lower affinity for the siderophore and spontaneously dissociates. In contrast, multicellular organisms can increase the concentration of iron in their diet by lowering the pH in the gastrointestinal tract. At pH 1.0 (the approximate pH of the stomach), most Fe(III) salts dissolve to form Fe 3 + (aq), which is absorbed by specific proteins in the intestinal wall. A protein called transferrin forms a complex with iron(III), allowing it to be transported to other cells. Proteins that bind tightly to Fe(III) can also be used as antibacterial agents because iron is absolutely essential for bacterial growth. For example, milk, tears, and egg white all contain proteins similar to transferrin, and their high affinity for Fe 3 + allows them to sequester iron, thereby preventing bacteria from growing in these nutrient-rich media. Iron is released from transferrin by reduction to Fe 2 + , and then it is either used immediately (e.g., for the synthesis of hemoglobin) or stored in a very large protein called ferritin for future use (Figure $\PageIndex{2}$). Ferritin uses oxygen to oxidize Fe 2 + to Fe 3 + , which at neutral pH precipitates in the central cavity of the protein as a polymeric mixture of Fe(OH) 3 and FePO 4 . Because a fully loaded ferritin molecule can contain as many as 4500 Fe atoms, which corresponds to about 25% Fe by mass, ferritin is an effective way to store iron in a highly concentrated form. When iron is needed by a cell, the Fe 3 + is reduced to the much more soluble Fe 2 + by a reductant such as ascorbic acid (vitamin C). The structure of ferritin contains channels at the junctions of the subunits, which provide pathways for iron to enter and leave the interior of a molecule. Metalloproteins and Metalloenzymes A protein that contains one or more metal ions tightly bound to amino acid side chains is called a metalloprotein; some of the most common ligands provided by amino acids are shown here. A metalloprotein that catalyzes a chemical reaction is a metalloenzyme. Thus all metalloenzymes are metalloproteins, but the converse is not true. Recent estimates suggest that more than 40% of all known enzymes require at least one metal ion for activity, including almost all the enzymes responsible for the synthesis, duplication, and repair of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). Electron-Transfer Proteins Proteins whose function is to transfer electrons from one place to another are called electron-transfer proteins. Because they do not catalyze a chemical reaction, electron-transfer proteins are not enzymes; they are biochemical reductants or oxidants consumed in an enzymatic reaction. The general reaction for an electron-transfer protein is as follows: \[M^{n+} + e^- \rightleftharpoons M^{(n−1)+} \label{23.14} \] Because many transition metals can exist in more than one oxidation state, electron-transfer proteins usually contain one or more metal ions that can undergo a redox reaction. Incorporating a metal ion into a protein has three important biological consequences: The protein environment can adjust the redox potential (E 0 ′), of the metal ion over a rather large potential range, whereas the redox potential of the simple hydrated metal ion [M n + (aq)], is essentially fixed. The protein can adjust the structure of the metal complex to ensure that electron transfer is rapid. The protein environment provides specificity, ensuring that the electron is transferred to only the desired site. Three important classes of metalloproteins transfer electrons: blue copper proteins, cytochromes, and iron–sulfur proteins, which generally transfer electrons at high (> 0.20 V), intermediate (±0 V), and low (−0.20 to −0.50 V) potentials, respectively (Table 23.12). Although these electron-transfer proteins contain different metals with different structures, they are all designed to ensure rapid electron transfer to and from the metal. Thus when the protein collides with its physiological oxidant or reductant, electron transfer can occur before the two proteins diffuse apart. For electron transfer to be rapid, the metal sites in the oxidized and reduced forms of the protein must have similar structures. Protein Metal Center M/e− Transferred Reduction Potential (V) iron–sulfur proteins* [Fe(SR)4]2− 1 Fe −0.1 to +0.1 iron–sulfur proteins* [(RS)2FeS2Fe(SR)2]2− 2 Fe −0.2 to −0.4 iron–sulfur proteins* [Fe3S4(SR)3]3− 3 Fe −0.1 to −0.2 iron–sulfur proteins* [Fe4S4(SR)4]2− 4 Fe −0.3 to −0.5 cytochromes Fe-heme (low spin) 1 Fe ~0 blue copper proteins† [Cu(Im)2(SR)(SR2)]− 1 Cu ≥ +0.20 * A sulfur bound to an organic group is represented as SR. * A sulfur bound to an organic group is represented as SR. * A sulfur bound to an organic group is represented as SR. * A sulfur bound to an organic group is represented as SR. † See Figure $\PageIndex{2b}$: for the structure of imidazole (Im). † See Figure $\PageIndex{2b}$: for the structure of imidazole (Im). † See Figure $\PageIndex{2b}$: for the structure of imidazole (Im). † See Figure $\PageIndex{2b}$: for the structure of imidazole (Im). Blue Copper Proteins Blue copper proteins were first isolated from bacteria in the 1950s and from plant tissues in the early 1960s. The intense blue color of these proteins is due to a strong absorption band at a wavelength of about 600 nm. Although simple Cu 2 + complexes, such as [Cu(H 2 O) 6 ] 2+ and [Cu(NH 3 ) 4 ] 2+ , are also blue due to an absorption band at 600 nm, the intensity of the absorption band is about 100 times less than that of a blue copper protein. Moreover, the reduction potential for the Cu 2 + /Cu + couple in a blue copper protein is usually +0.3 to +0.5 V, considerably more positive than that of the aqueous Cu 2 + /Cu + couple (+0.15 V). The copper center in blue copper proteins has a distorted tetrahedral structure, in which the copper is bound to four amino acid side chains (Figure $\PageIndex{3}$). Although the most common structures for four-coordinate Cu 2 + and Cu + complexes are square planar and tetrahedral, respectively, the structures of the oxidized (Cu 2 + ) and reduced (Cu + ) forms of the protein are essentially identical. Thus the protein forces the Cu 2 + ion to adopt a higher-energy structure that is more suitable for Cu + , which makes the Cu 2 + form easier to reduce and raises its reduction potential. Moreover, by forcing the oxidized and reduced forms of the metal complex to have essentially the same structure, the protein ensures that electron transfer to and from the copper site is rapid because only minimal structural reorganization of the metal center is required. Kinetics studies on simple metal complexes have shown that electron-transfer reactions tend to be slow when the structures of the oxidized and reduced forms of a metal complex are very different, and fast when they are similar. You will see that other metal centers used for biological electron-transfer reactions are also set up for minimal structural reorganization after electron transfer, which ensures the rapid transfer of electrons. Cytochromes The cytochromes (from the Greek cytos, meaning “cell”, and chroma, meaning “color”) were first identified in the 1920s by spectroscopic studies of cell extracts. Based on the wavelength of the maximum absorption in the visible spectrum, they were classified as cytochromes a (with the longest wavelength), cytochromes b (intermediate wavelength), and cytochromes c (shortest wavelength). It quickly became apparent that there was a correlation between their spectroscopic properties and other physical properties. For examples, cytochromes c are generally small, soluble proteins with a reduction potential of about +0.25 V, whereas cytochromes b are larger, less-soluble proteins with reduction potentials of about 0 V. All cytochromes contain iron, and the iron atom in all cytochromes is coordinated by a planar array of four nitrogen atoms provided by a cyclic tetradentate ligand called a porphyrin. The iron–porphyrin unit is called a heme group. The structures of a typical porphyrin (protoporphyrin IX) and its iron complex (protoheme) are shown here. In addition to the four nitrogen atoms of the porphyrin, the iron in a cytochrome is usually bonded to two additional ligands provided by the protein, as shown in Figure $\PageIndex{4}$. A cytochrome. Shown here is protoporphyrin IX and its iron complex, protoheme. In contrast to the blue copper proteins, two electron configurations are possible for both the oxidized and reduced forms of a cytochrome, and this has significant structural consequences. Thus Fe 2 + is d 6 and can be either high spin (with four unpaired electrons) or low spin (with no unpaired electrons). Similarly, Fe 3 + is d 5 and can also be high spin (with five unpaired electrons) or low spin (with one unpaired electron). In low-spin heme complexes, both the Fe 2 + and the Fe 3 + ions are small enough to fit into the “hole” in the center of the porphyrin; hence the iron atom lies almost exactly in the plane of the four porphyrin nitrogen atoms in both cases. Because cytochromes b and c are low spin in both their oxidized and reduced forms, the structures of the oxidized and reduced cytochromes are essentially identical. Hence minimal structural changes occur after oxidation or reduction, which makes electron transfer to or from the heme very rapid. Electron transfer reactions occur most rapidly when minimal structural changes occur during oxidation or reduction. Iron–Sulfur Proteins Although all known bacteria, plants, and animals use iron–sulfur proteins to transfer electrons, the existence of these proteins was not recognized until the late 1950s. Iron–sulfur proteins transfer electrons over a wide range of reduction potentials, and their iron content can range from 1 to more than 12 Fe atoms per protein molecule. In addition, most iron–sulfur proteins contain stoichiometric amounts of sulfide (S 2− ). These properties are due to the presence of four different kinds of iron–sulfur units, which contain one, two, three, or four iron atoms per Fe–S complex (Figure $\PageIndex{5}$). In all cases, the Fe 2 + and Fe 3 + ions are coordinated to four sulfur ligands in a tetrahedral environment. Due to tetrahedral coordination by weak-field sulfur ligands, the iron is high spin in both the Fe 3 + and Fe 2 + oxidation states, which results in similar structures for the oxidized and reduced forms of the Fe–S complexes. Consequently, only small structural changes occur after oxidation or reduction of the Fe–S center, which results in rapid electron transfer. Reactions of Small Molecules Although small molecules, such as O 2 , N 2 , and H 2 , do not react with organic compounds under ambient conditions, they do react with many transition-metal complexes. Consequently, virtually all organisms use metalloproteins to bind, transport, and catalyze the reactions of these molecules. Probably the best-known example is hemoglobin, which is used to transport O 2 in many multicellular organisms. Under ambient conditions, small molecules, such as O 2 , N 2 , and H 2 , react with transition-metal complexes but not with organic compounds. Oxygen Transport Many microorganisms and most animals obtain energy by respiration, the oxidation of organic or inorganic molecules by O 2 . At 25°C, however, the concentration of dissolved oxygen in water in contact with air is only about 0.25 mM. Because of their high surface area-to-volume ratio, aerobic microorganisms can obtain enough oxygen for respiration by passive diffusion of O 2 through the cell membrane. As the size of an organism increases, however, its volume increases much more rapidly than its surface area, and the need for oxygen depends on its volume. Consequently, as a multicellular organism grows larger, its need for O 2 rapidly outstrips the supply available through diffusion. Unless a transport system is available to provide an adequate supply of oxygen for the interior cells, organisms that contain more than a few cells cannot exist. In addition, O 2 is such a powerful oxidant that the oxidation reactions used to obtain metabolic energy must be carefully controlled to avoid releasing so much heat that the water in the cell boils. Consequently, in higher-level organisms, the respiratory apparatus is located in internal compartments called mitochondria, which are the power plants of a cell. Oxygen must therefore be transported not only to a cell but also to the proper compartment within a cell. Three different chemical solutions to the problem of oxygen transport have developed independently in the course of evolution, as indicated in Table $\PageIndex{2}$. Mammals, birds, reptiles, fish, and some insects use a heme protein called hemoglobin to transport oxygen from the lungs to the cells, and they use a related protein called myoglobin to temporarily store oxygen in the tissues. Several classes of invertebrates, including marine worms, use an iron-containing protein called hemerythrin to transport oxygen, whereas other classes of invertebrates (arthropods and mollusks) use a copper-containing protein called hemocyanin. Despite the presence of the hem- prefix, hemerythrin and hemocyanin do not contain a metal–porphyrin complex. Protein Source M per Subunit M per O2 Bound Color (deoxy form) Color (oxy form) hemoglobin mammals, birds, fish, reptiles, some insects 1 Fe 1 Fe red-purple red hemerythrin marine worms 2 Fe 2 Fe colorless red hemocyanin mollusks, crustaceans, spiders 2 Cu 2 Cu colorless blue Myoglobin and Hemoglobin Myoglobin is a relatively small protein that contains 150 amino acids. The functional unit of myoglobin is an iron–porphyrin complex that is embedded in the protein (Figure 26.8.1). In myoglobin, the heme iron is five-coordinate, with only a single histidine imidazole ligand from the protein (called the proximal histidine because it is near the iron) in addition to the four nitrogen atoms of the porphyrin. A second histidine imidazole (the distal histidine because it is more distant from the iron) is located on the other side of the heme group, too far from the iron to be bonded to it. Consequently, the iron atom has a vacant coordination site, which is where O 2 binds. In the ferrous form (deoxymyoglobin), the iron is five-coordinate and high spin. Because high-spin Fe 2 + is too large to fit into the “hole” in the center of the porphyrin, it is about 60 pm above the plane of the porphyrin. When O 2 binds to deoxymyoglobin to form oxymyoglobin, the iron is converted from five-coordinate (high spin) to six-coordinate (low spin; Figure 26.8.2). Because low-spin Fe 2 + and Fe 3 + are smaller than high-spin Fe 2 + , the iron atom moves into the plane of the porphyrin ring to form an octahedral complex. The O 2 pressure at which half of the molecules in a solution of myoglobin are bound to O 2 (P 1 /2 ) is about 1 mm Hg (1.3 × 10 −3 atm). A vacant coordination site at a metal center in a protein usually indicates that a small molecule will bind to the metal ion, whereas a coordinatively saturated metal center is usually involved in electron transfer. Hemoglobin consists of two subunits of 141 amino acids and two subunits of 146 amino acids, both similar to myoglobin; it is called a tetramer because of its four subunits. Because hemoglobin has very different O 2 -binding properties, however, it is not simply a “super myoglobin” that can carry four O 2 molecules simultaneously (one per heme group). The shape of the O 2 -binding curve of myoglobin (Mb; Figure $\PageIndex{7}$) can be described mathematically by the following equilibrium: \[MbO_2 \rightleftharpoons Mb + O_ 2 \label{26.8.1a} \] \[K_{diss}=\dfrac{[Mb][O_2]}{[MbO_2]} \label{26.8.1b} \] In contrast, the O 2 -binding curve of hemoglobin is S shaped (Figure $\PageIndex{8}$). As shown in the curves, at low oxygen pressures, the affinity of deoxyhemoglobin for O 2 is substantially lower than that of myoglobin, whereas at high O 2 pressures the two proteins have comparable O 2 affinities. The physiological consequences of the unusual S-shaped O 2 -binding curve of hemoglobin are enormous. In the lungs, where O 2 pressure is highest, the high oxygen affinity of deoxyhemoglobin allows it to be completely loaded with O 2 , giving four O 2 molecules per hemoglobin. In the tissues, however, where the oxygen pressure is much lower, the decreased oxygen affinity of hemoglobin allows it to release O 2 , resulting in a net transfer of oxygen to myoglobin. The S-shaped O 2 -binding curve of hemoglobin is due to a phenomenon called cooperativity, in which the affinity of one heme for O 2 depends on whether the other hemes are already bound to O 2 . Cooperativity in hemoglobin requires an interaction between the four heme groups in the hemoglobin tetramer, even though they are more than 3000 pm apart, and depends on the change in structure of the heme group that occurs with oxygen binding. The structures of deoxyhemoglobin and oxyhemoglobin are slightly different, and as a result, deoxyhemoglobin has a much lower O 2 affinity than myoglobin, whereas the O 2 affinity of oxyhemoglobin is essentially identical to that of oxymyoglobin. Binding of the first two O 2 molecules to deoxyhemoglobin causes the overall structure of the protein to change to that of oxyhemoglobin; consequently, the last two heme groups have a much higher affinity for O 2 than the first two. Oxygen is not unique in its ability to bind to a ferrous heme complex; small molecules such as CO and NO bind to deoxymyoglobin even more tightly than does O 2 . The interaction of the heme iron with oxygen and other diatomic molecules involves the transfer of electron density from the filled t 2g orbitals of the low-spin d 6 Fe 2 + ion to the empty π* orbitals of the ligand. In the case of the Fe 2 + –O 2 interaction, the transfer of electron density is so great that the Fe–O 2 unit can be described as containing low-spin Fe 3 + (d 5 ) and O 2 − . We can therefore represent the binding of O 2 to deoxyhemoglobin and its release as a reversible redox reaction: \[Fe^{2+} + O_2 \rightleftharpoons \ce{Fe^{3+}–O_2^−} \label{26.8.2} \] As shown in Figure $\PageIndex{9}$, the Fe–O 2 unit is bent, with an Fe–O–O angle of about 130°. Because the π* orbitals in CO are empty and those in NO are singly occupied, these ligands interact more strongly with Fe 2 + than does O 2 , in which the π* orbitals of the neutral ligand are doubly occupied. Although CO has a much greater affinity for a ferrous heme than does O 2 (by a factor of about 25,000), the affinity of CO for deoxyhemoglobin is only about 200 times greater than that of O 2 , which suggests that something in the protein is decreasing its affinity for CO by a factor of about 100. Both CO and NO bind to ferrous hemes in a linear fashion, with an Fe–C(N)–O angle of about 180°, and the difference in the preferred geometry of O 2 and CO provides a plausible explanation for the difference in affinities. As shown in Figure $\PageIndex{9}$, the imidazole group of the distal histidine is located precisely where the oxygen atom of bound CO would be if the Fe–C–O unit were linear. Consequently, CO cannot bind to the heme in a linear fashion; instead, it is forced to bind in a bent mode that is similar to the preferred structure for the Fe–O 2 unit. This results in a significant decrease in the affinity of the heme for CO, while leaving the O 2 affinity unchanged, which is important because carbon monoxide is produced continuously in the body by degradation of the porphyrin ligand (even in nonsmokers). Under normal conditions, CO occupies approximately 1% of the heme sites in hemoglobin and myoglobin. If the affinity of hemoglobin and myoglobin for CO were 100 times greater (due to the absence of the distal histidine), essentially 100% of the heme sites would be occupied by CO, and no oxygen could be transported to the tissues. Severe carbon-monoxide poisoning, which is frequently fatal, has exactly the same effect. Thus the primary function of the distal histidine appears to be to decrease the CO affinity of hemoglobin and myoglobin to avoid self-poisoning by CO. Hemerythrin Hemerythrin is used to transport O 2 in a variety of marine invertebrates. It is an octamer (eight subunits), with each subunit containing two iron atoms and binding one molecule of O 2 . Deoxyhemerythrin contains two Fe 2 + ions per subunit and is colorless, whereas oxyhemerythrin contains two Fe 3 + ions and is bright reddish violet. These invertebrates also contain a monomeric form of hemerythrin that is located in the tissues, analogous to myoglobin. The binding of oxygen to hemerythrin and its release can be described by the following reaction, where the HO 2 − ligand is the hydroperoxide anion derived by the deprotonation of hydrogen peroxide (H 2 O 2 ): \[\ce{2Fe^{2+} + O2 + H^{+} <=> 2Fe^{3+}–O2H} \label{23.17} \] Thus O 2 binding is accompanied by the transfer of two electrons (one from each Fe 2 + ) and a proton to O 2 . Hemocyanin Hemocyanin is used for oxygen transport in many arthropods (spiders, crabs, lobsters, and centipedes) and in mollusks (shellfish, octopi, and squid); it is responsible for the bluish-green color of their blood. The protein is a polymer of subunits that each contain two copper atoms (rather than iron), with an aggregate molecular mass of greater than 1,000,000 amu. Deoxyhemocyanin contains two Cu + ions per subunit and is colorless, whereas oxyhemocyanin contains two Cu 2 + ions and is bright blue. As with hemerythrin, the binding and release of O 2 correspond to a two-electron reaction: \[\ce{2Cu^{+} + O2 <=> Cu^{2+}–O2^{2−}–Cu^{2+}} \label{23.18} \] Although hemocyanin and hemerythrin perform the same basic function as hemoglobin, these proteins are not interchangeable. In fact, hemocyanin is so foreign to humans that it is one of the major factors responsible for the common allergies to shellfish. Myoglobin, hemoglobin, hemerythrin, and hemocyanin all use a transition-metal complex to transport oxygen. Enzymes Involved in Oxygen Activation Many of the enzymes involved in the biological reactions of oxygen contain metal centers with structures that are similar to those used for O 2 transport. Many of these enzymes also contain metal centers that are used for electron transfer, which have structures similar to those of the electron-transfer proteins discussed previously. In this section, we briefly describe two of the most important examples: dioxygenases and methane monooxygenase. Dioxygenases are enzymes that insert both atoms of O 2 into an organic molecule. In humans, dioxygenases are responsible for cross-linking collagen in connective tissue and for synthesizing complex organic molecules called prostaglandins, which trigger inflammation and immune reactions. Iron is by far the most common metal in dioxygenases; and the target of the most commonly used drug in the world, aspirin, is an iron enzyme that synthesizes a specific prostaglandin. Aspirin inhibits this enzyme by binding to the iron atom at the active site, which prevents oxygen from binding. Methane monooxygenase catalyzes the conversion of methane to methanol. The enzyme is a monooxygenase because only one atom of O 2 is inserted into an organic molecule, while the other is reduced to water: \[\ce{CH_4 + O_2 + 2e^- + 2H^+ \rightarrow CH_3OH + H_2O} \label{23.19} \] Because methane is the major component of natural gas, there is enormous interest in using this reaction to convert methane to a liquid fuel (methanol) that is much more convenient to ship and store. Because the C–H bond in methane is one of the strongest C–H bonds known, however, an extraordinarily powerful oxidant is needed for this reaction. The active site of methane monooxygenase contains two Fe atoms that bind O 2 , but the details of how the bound O 2 is converted to such a potent oxidant remain unclear. Metal Ions as Lewis Acids Reactions catalyzed by metal ions that do not change their oxidation states during the reaction are usually group transfer reactions, in which a group such as the phosphoryl group (−PO 3 2− ) is transferred. These enzymes usually use metal ions such as Zn 2 + , Mg 2 + , and Mn 2 + , and they range from true metalloenzymes, in which the metal ion is tightly bound, to metal-activated enzymes, which require the addition of metal ions for activity. Because tight binding is usually the result of specific metal–ligand interactions, metalloenzymes tend to be rather specific for a particular metal ion. In contrast, the binding of metal ions to metal-activated enzymes is largely electrostatic in nature; consequently, several different metal ions with similar charges and sizes can often be used to give an active enzyme. Metalloenzymes generally contain a specific metal ion, whereas metal-activated enzymes can use any of several metal ions of similar size and charge. A metal ion that acts as a Lewis acid can catalyze a group transfer reaction in many different ways, but we will focus on only one of these, using a zinc enzyme as an example. Carbonic anhydrase is found in red blood cells and catalyzes the reaction of CO 2 with water to give carbonic acid. \[\ce{CO_2(g) + H_2O(l) \rightleftharpoons H^{+}(aq) + HCO^{-}3(aq)} \label{23.20} \] Although this reaction occurs spontaneously in the absence of a catalyst, it is too slow to absorb all the CO 2 generated during respiration. Without a catalyst, tissues would explode due to the buildup of excess CO 2 pressure. Carbonic anhydrase contains a single Zn 2 + ion per molecule, which is coordinated by three histidine imidazole ligands and a molecule of water. Because Zn 2 + is a Lewis acid, the pK a of the Zn 2 + –OH 2 unit is about 8 versus 14 for pure water. Thus at pH 7–8, a significant fraction of the enzyme molecules contain the Zn 2 + –OH − group, which is much more reactive than bulk water. When carbon dioxide binds in a nonpolar site next to the Zn 2 + –OH − unit, it reacts rapidly to give a coordinated bicarbonate ion that dissociates from the enzyme: \[\ce{Zn^{2+}–OH^{-} + CO_2 \rightleftharpoons Zn^{2+}–OCO_2H^- \rightleftharpoons Zn^{2+} + HCO_3^{-}} \label{23.21} \] The active site of carbonic anhydrase. Thus the function of zinc in carbonic anhydrase is to generate the hydroxide ion at pH 7.0, far less than the pH required in the absence of the metal ion. Enzymes That Use Metals to Generate Organic Radicals An organic radical is an organic species that contains one or more unpaired electrons. Chemists often consider organic radicals to be highly reactive species that produce undesirable reactions. For example, they have been implicated in some of the irreversible chemical changes that accompany aging. It is surprising, however, that organic radicals are also essential components of many important enzymes, almost all of which use a metal ion to generate the organic radical within the enzyme. These enzymes are involved in the synthesis of hemoglobin and DNA, among other important biological molecules, and they are the targets of pharmaceuticals for the treatment of diseases such as anemia, sickle-cell anemia, and cancer. In this section, we discuss one class of radical enzymes that use vitamin B 12 . Vitamin B 12 was discovered in the 1940s as the active agent in the cure of pernicious anemia, which does not respond to increased iron in the diet. Humans need only tiny amounts of vitamin B 12 , and the average blood concentration in a healthy adult is only about 3.5 × 10 −8 M. The structure of vitamin B 12 , shown in Figure $\PageIndex{10}$, is similar to that of a heme, but it contains cobalt instead of iron, and its structure is much more complex. In fact, vitamin B 12 has been called the most complex nonpolymeric biological molecule known and was the first naturally occurring organometallic compound to be isolated. When vitamin B 12 (the form present in vitamin tablets) is ingested, the axial cyanide ligand is replaced by a complex organic group. The cobalt–carbon bond in the enzyme-bound form of vitamin B 12 and related compounds is unusually weak, and it is particularly susceptible to homolytic cleavage: \[\ce{CoCH_2R <=> Co^{2+⋅} + ⋅CH_2R} \label{23.22} \] Homolytic cleavage of the Co 3 + –CH 2 R bond produces two species, each of which has an unpaired electron: a d 7 Co 2 + derivative and an organic radical, ·CH 2 R, which is used by vitamin B 12 -dependent enzymes to catalyze a wide variety of reactions. Virtually all vitamin B 12 -catalyzed reactions are rearrangements in which an H atom and an adjacent substituent exchange positions: In the conversion of ethylene glycol to acetaldehyde, the initial product is the hydrated form of acetaldehyde, which rapidly loses water: The enzyme uses the ·CH 2 R radical to temporarily remove a hydrogen atom from the organic substrate, which then rearranges to give a new radical. Transferring the hydrogen atom back to the rearranged radical gives the product and regenerates the ·CH 2 R radical. The metal is not involved in the actual catalytic reaction; it provides the enzyme with a convenient mechanism for generating an organic radical, which does the actual work. Many examples of similar reactions are now known that use metals other than cobalt to generate an enzyme-bound organic radical. Nearly all vitamin B 12 -catalyzed reactions are rearrangements that occur via a radical reaction. Summary Three separate steps are required for organisms to obtain essential transition metals from their environment: mobilization of the metal, transport of the metal into the cell, and transfer of the metal to where it is needed within a cell or an organism. The process of iron uptake is best understood. To overcome the insolubility of Fe(OH) 3 , many bacteria use organic ligands called siderophores, which have high affinity for Fe(III) and are secreted into the surrounding medium to increase the total concentration of dissolved iron. The iron–siderophore complex is absorbed by a cell, and the iron is released by reduction to Fe(II). Mammals use the low pH of the stomach to increase the concentration of dissolved iron. Iron is absorbed in the intestine, where it forms an Fe(III) complex with a protein called transferrin that is transferred to other cells for immediate use or storage in the form of ferritin. Proteins that contain one or more tightly bound metal ions are called metalloproteins, and metalloproteins that catalyze biochemical reactions are called metalloenzymes. Proteins that transfer electrons from one place to another are called electron-transfer proteins. Most electron-transfer proteins are metalloproteins, such as iron–sulfur proteins, cytochromes, and blue copper proteins that accept and donate electrons. The oxidized and reduced centers in all electron-transfer proteins have similar structures to ensure that electron transfer to and from the metal occurs rapidly. Metalloproteins also use the ability of transition metals to bind small molecules, such as O 2 , N 2 , and H 2 , to transport or catalyze the reactions of these small molecules. For example, hemoglobin, hemerythrin, and hemocyanin, which contain heme iron, nonheme iron, and copper, respectively, are used by different kinds of organisms to bind and transfer O 2 . Other metalloenzymes use transition-metal ions as Lewis acids to catalyze group transfer reactions. Finally, some metalloenzymes use homolytic cleavage of the cobalt–carbon bond in derivatives of vitamin B 12 to generate an organic radical that can abstract a hydrogen atom and thus cause molecular rearrangements to occur. Key Takeaway Organisms have developed strategies to extract transition metals from the environment and use the metals in electron-transfer reactions, reactions of small molecules, Lewis-acid catalysis, and the generation of reactive organic radicals. Conceptual Problems What are the advantages of having a metal ion at the active site of an enzyme? Why does the structure of the metal center in a metalloprotein that transfers electrons show so little change after oxidation or reduction? Structure and Reactivity In enzymes, explain how metal ions are particularly suitable for generating organic radicals. A common method for treating carbon-monoxide poisoning is to have the patient inhale pure oxygen. Explain why this treatment is effective.
Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.06%3A_Mechanisms_of_Phase_Two	Phase one of glycolysis is setting the stage. An initial investment of ATP is needed to get the system ready to produce more ATP; it's like a farmer who sows sunflower seeds in the spring to get a bigger crop of sunflower seeds in the fall. Two molecules of glyceraldehyde-3-phosphate are produced in phase one of glycolysis. Both of those molecules enter into phase two. The first step in this phase is an oxidation reaction. In organic chemistry, the term oxidation suggests that a carbon atom is getting fewer bonds to hydrogen, or more bonds to oxygen. (The complementary term, reduction, suggests new carbon-hydrogen bonds are forming, or carbon-oxygen bonds are disappearing.) Notice that the carbonyl carbon, C=O, is becoming a carboxylic carbon, O-C=O. That isn't all that's going on. There is also a phosphorylation step here, but this time ATP is not required. The source of the phosphate group is a simple phosphate ion. Furthermore, a molecule of NADH is produced. You might remember that NAD + can pick up a hydride ion (that's right, H - instead of H + ) to become NADH. In this case, the hydride ion is coming from the aldehyde that is converted into a carboxyloid, the phosphoric anhydride group in BPG. This first step has consequences for energy-packaging pathways further downstream. NADH is the starting material for oxidative phosphorylation, an elegant process in which electrons are passed from one metal ion to another within membrane-bound proteins; as the electrons move across the membrane, they draw oppositely-charged protons along with them. A proton gradient builds up, with protons on one side of the membrane outnumbering those on the other; this osmotic pressure is relieved when the protons find a channel to pour back through the membrane, but as they do so they turn a molecular millwheel that drives the production of more ATP. Remember, oxidative phosphorylation, along with glycolysis and the citric acid cycle, is one of the three pathways that together make up the process of respiration. There is also a more immediate energy-packaging result. That first step produces 1,3-bisphosphoglycerate, which is primed to deliver a phosphate to a molecule of ADP. In addition to ATP, a molecule of 3-phosphoglycerate is left behind. A slight modification of the 3-phosphoglycerate ensues. In this step, the phosphate group migrates from the 3-position to the 2-position, resulting in 2-phosphoglycerate. Subsequently, the 2-phosphoglycerate undergoes a dehydration, the loss of water. You might recall that dehydrations sometimes occur after aldol reactions: the O=C-CH-C-OH loses a proton at the alpha position and a hydroxide at the beta position to give the enone group, O=C-C=C, and water, HOH. That step is often driven by the conjugated system that results. In this case, the conjugated product is phosphoenolpyruvate. The final step in glycolysis is the loss of the phosphate group from phosphoenolpyruvate. This phosphate group is transferred to another molecule of ADP, forming ATP. The ATP can then be used to power processes elsewhere in the cell. Note that this is the second molecule of ATP produced during phase two. Since each molecule of glucose produces two three-carbon sugars that enter into phase two, a total of four molecules of ATP are produced per glucose. Remember, phase one required the consumption of two molecules of glucose, so this yield represents a doubling of the initial investment of ATP. It's like putting two dollars in the bank and getting four dollars back out. That's the end of glycolysis, but that's not the end of the story. So far, glucose has only been broken down to a three-carbon sugar. When we think of respiration, we think of glucose breaking down all the way to carbon dioxide. That part continues in the tricarboxylic acid cycle, or TCA cycle.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Mechanics__in_Chemistry_(Simons_and_Nichols)/10%3A_Angular_Momentum_and_Group_Symmetries_of_Electronic_Wavefunctions	Electronic wavefunctions must also possess proper symmetry. These include angular momentum and point group symmetries 10.1: Angular Momentum Symmetry and Strategies for Angular Momentum Coupling Any acceptable model or trial wavefunction for a multi-electron systme should be constrained to also be an eigenfunction of the symmetry operators that commute with H. If the atom or molecule has additional symmetries (e.g., full rotation symmetry for atoms, axial rotation symmetry for linear molecules and point group symmetry for nonlinear polyatomics), the trial wavefunctions should also conform to these spatial symmetries. 10.2: Electron Spin Angular Momentum Individual electrons possess intrinsic spin characterized by angular momentum quantum numbers $S$ and $m_s$. The proper spin eigenfunctions must be constructed from antisymmetric (i.e., determinental) wavefunctions. 10.3: Coupling of Angular Momenta The simple "vector coupling" method applies to any angular momenta. If the angular momenta are "equivalent" in the sense that they involve indistinguishable particles that occupy the same orbital shell, the Pauli principle eliminates some of the expected term symbols. For linear molecules, the orbital angular momenta of electrons are not vector coupled, but the electrons' spin angular momenta are vector coupled. For non-linear polyatomic molecules, on spin angular momenta is vector coupled. 10.4: Atomic Term Symbols and Wavefunctions When coupling non-equivalent angular momenta (e.g., a spin and an orbital angular momenta or two orbital angular momenta of non-equivalent electrons), one vector couples using the fact that the coupled angular momenta range from the sum of the two individual angular momenta to the absolute value of their difference. 10.5: Atomic Configuration Wavefunctions To express, in terms of Slater determinants, the wavefunctions corresponding to each of the states in each of the levels, one proceeds as follows in this section. 10.6: Inversion Symmetry One more quantum number, that relating to the inversion (i) symmetry operator can be used in atomic cases because the total potential energy V is unchanged when all of the electrons have their position vectors subjected to inversion (i.e., ir=-r ). This quantum number is straightforward to determine. 10.7: Review of Atomic Cases
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/14%3A_Spectroscopy/14.01%3A_Vocabulary	Electromagnetic radiation —light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another are explained best by describing light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20 th century. 1 Nevertheless, the dual models of wave and particle behavior provide a useful description for electromagnetic radiation. Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. In a vacuum electromagnetic radiation travels at the speed of light, c , which is 2.997 × 10 8 m/s. When electromagnetic radiation moves through a medium other than a vacuum its velocity, v , is less than the speed of light in a vacuum. The difference between v and c is sufficiently small (<0.1%) that the speed of light to three significant figures, 3.00 × 10 8 m/s, is accurate enough for most purposes. The oscillations in the electric and magnetic fields are perpendicular to each other, and to the direction of the wave’s propagation. Figure $\PageIndex{1}$ shows an example of plane-polarized electromagnetic radiation, consisting of a single oscillating electric field and a single oscillating magnetic field. An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. 2 For example, the amplitude of the oscillating electric field at any point along the propagating wave is \[A_\ce{t} = A_\ce{e}\sin(2πνt + \phi)\] where A t is the magnitude of the electric field at time t , A e is the electric field’s maximum amplitude , ν is the wave’s frequency —the number of oscillations in the electric field per unit time—and $\phi$ is a phase angle , which accounts for the fact that A t need not have a value of zero at t = 0. The identical equation for the magnetic field is \[A_\ce{t} =A_\ce{m}\sin(2πνt + \phi)\] where A m is the magnetic field’s maximum amplitude. Units Other properties also are useful for characterizing the wave behavior of electromagnetic radiation. The wavelength , λ , is defined as the distance between successive maxima (Figure $\PageIndex{1}$). For ultraviolet and visible electromagnetic radiation the wavelength is usually expressed in nanometers (1 nm = 10 –9 m), and for infrared radiation it is given in microns (1 μm = 10 –6 m). The relationship between wavelength and frequency is \[λ = \dfrac{c}{ν}\] Another unit useful unit is the wavenumber , $\tilde{ν}$, which is the reciprocal of wavelength \[\tilde{ν} = \dfrac{1}{λ}\] Wavenumbers are frequently used to characterize infrared radiation, with the units given in cm –1 . Example $\PageIndex{1}$ In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line? Solution The frequency and wavenumber of the sodium D line are \[ν = \dfrac{c}{λ} = \mathrm{\dfrac{3.00×10^8\: m/s}{589×10^{−9}\: m} = 5.09×10^{14}\: s^{−1}}\] \[\tilde{ν} = \dfrac{1}{λ} = \mathrm{\dfrac{1}{589×10^{−9}\: m} × \dfrac{1\: m}{100\: cm} = 1.70×10^4\: cm^{−1}}\] Exercise $\PageIndex{1}$ Another historically important series of spectral lines is the Balmer series of emission lines form hydrogen. One of the lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line? Click here to review your answer to this exercise. Regions of the Spectrum The frequency and wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the electromagnetic spectrum —based on the type of atomic or molecular transition that gives rise to the absorption or emission of photons (Figure $\PageIndex{2}$). The boundaries between the regions of the electromagnetic spectrum are not rigid, and overlap between spectral regions is possible. Above, we defined several characteristic properties of electromagnetic radiation, including its energy, velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. A spectroscopic measurement is possible only if the photon’s interaction with the sample leads to a change in one or more of these characteristic properties. We can divide spectroscopy into two broad classes of techniques. In one class of techniques there is a transfer of energy between the photon and the sample. Table $\PageIndex{1}$ provides a list of several representative examples. Type of Energy Transfer Region of Electromagnetic Spectrum Spectroscopic Techniquea absorption γ-ray Mossbauer spectroscopy NaN X-ray X-ray absorption spectroscopy NaN UV/Vis UV/Vis spectroscopy atomic absorption spectroscopy NaN IR infrared spectroscopy raman spectroscopy NaN Microwave microwave spectroscopy NaN Radio wave electron spin resonance spectroscopy nuclear magnetic resonance spectroscopy emission (thermal excitation) UV/Vis atomic emission spectroscopy photoluminescence X-ray X-ray fluorescence NaN UV/Vis fluorescence spectroscopy phosphorescence spectroscopy atomic fluorescence spectroscopy chemiluminescence UV/Vis chemiluminescence spectroscopy Line Widths A spectral line extends over a range of frequencies, not a single frequency (i.e., it has a nonzero linewidth). There are multiple reasons for this broadening and shifts and only two are discussed below. Lifetime Broadening. This mechanism ordinates directly from the Heisenberg Principle , which the lifetime of an excited state (due to the spontaneous radiative decay) with the uncertainty of its energy $\Delta t$. Since the system is changing in time, it is impossible to estimate the energies of wavefunctions exactly ($\Delta E$. \[ \Delta E \Delta t \ge \dfrac{\hbar}{2}\] Hence, systems with excited-states that have short lifetimes will have a large energy uncertainty and a broad emission. This broadening effect results in an broadened profile. As expected, the lifetime broadening can be experimentally if the decay rates can be artificially suppressed or enhanced. Doppler Broadening . This mechanism is intrinsic to atoms in a gas which are emitting radiation will have a distribution of velocities. Each photon emitted will be "red"- or "blue"-shifted by the Doppler effect depending on the velocity of the atom relative to the observer. The Doppler effect is the change in frequency of a wave (or other periodic event) for an observer moving relative to its source. It is commonly heard when a vehicle sounding a siren or horn approaches, passes, and recedes from an observer. Compared to the emitted frequency, the received frequency is higher during the approach, identical at the instant of passing by, and lower during the recession. Doppler Broadening example with sound, but applies to light too. (left): Stationary sound source produces sound waves at a constant frequency $ u$, and the wave-fronts propagate symmetrically away from the source at a constant speed c (assuming speed of sound, c = 330 m/s), which is the speed of sound in the medium. The distance between wave-fronts is the wavelength. All observers will hear the same frequency, which will be equal to the actual frequency of the source where $ u= u-o$ (right): The same sound source is radiating sound waves at a constant frequency in the same medium. However, now the sound source is moving with a speed so the center of each new wavefront is now slightly displaced to the right. As a result, the wave-fronts begin to bunch up on the right side (in front of) and spread further apart on the left side (behind) of the source. An observer in front of the source will hear a higher frequency. Images used with permission from Wikipedia (credit Lookang ) . If the speeds are small compared to the speed of the wave, the relationship between observed frequency $ u_o$ and emitted frequency $ u$ is approximately \[ u =\left(1+\dfrac{\Delta v}{c}\right)v_0 \] The Doppler effect also applies for spectroscopy. For example, the higher the temperature of the gas, the wider the distribution of velocities in the gas (via the Maxwell-Bolztmann Distribution). Since the spectral line is a combination of all of the emitted radiation, the higher the temperature of the gas, the broader the spectral line emitted from that gas. If the average velocity of a gas is non-zero, then a Doppler shift will be observed that is correlated with the amplitude of this average velocity. Redshift of spectral lines in the optical spectrum of a supercluster of distant galaxies (right), as compared to that of the Sun (left) since the sun is not moving with respect to Earth (or weakly) and the galaxy is moving away. from Wikipedia (Credit Georg Wiora). Doppler broadening is one of the explanations for the broadening of spectral lines, and as such gives an indication for the temperature of observed material. Other causes of velocity distributions may exist, though, for example due to turbulent motion. Doppler broadening can also be used to determine the velocity distribution of a gas given its absorption spectrum. In particular, this has been used to determine the velocity distribution of interstellar gas clouds. Example $\PageIndex{2}$ Find the uncertainty of simultaneously measuring the frequency and wavelength of an emission from an excited molecule, if the wavelength is 430 nm and the excited state lifetime is 0.50 nanoseconds. Soluton 1. Use Heisenberg's uncertainty principle and the relationship between energy and frequency to find the uncertainty of frequency. \[ \Delta E \Delta t \geq \dfrac{h}{4 \pi} \nonumber \] \[ \Delta E = h \Delta u \nonumber\] \[ h \Delta u \Delta t \geq \dfrac{h}{4 \pi} \nonumber\] \[ \Delta u \geq \dfrac{1}{4 \pi \Delta t} \nonumber\] The maximum value for Δt is the lifetime of the excited state. \[ \Delta u \geq \dfrac{1}{4 \pi 0.50 ns \times \dfrac{s}{10^9\ ns}} \nonumber\] \[ \Delta u \geq 1.6 \times 10^8\ s^{-1} \nonumber\] 2. Use the uncertainty of frequency and the relationship between frequency and wavelength to find the uncertainty of the wavelength. \[ \lambda = \dfrac{c}{ u} \nonumber\] \[ \| \Delta \lambda \| = \dfrac{c}{ u^2} \| \Delta u \| \nonumber\] \[ \| \Delta \lambda \| = \dfrac{\lambda ^2 \| \Delta u \| }{c} \nonumber\] \[ \| \Delta \lambda \| = \dfrac{(430\ nm)^2 \times 1.6 \times 10^8\ s^{-1}}{2.998 \times 10^8\ m/s} \times \dfrac{m}{10^9\ nm} \nonumber\] \[ \| \Delta \lambda \| = 2.3 \times 10^{-7}\ nm \nonumber\] Absorption and Emission In absorption spectroscopy a photon is absorbed by an atom or molecule, which undergoes a transition from a lower-energy state to a higher-energy, or excited state (Figure $\PageIndex{3}$). The type of transition depends on the photon’s energy. The electromagnetic spectrum in Figure $\PageIndex{2}$, for example, shows that absorbing a photon of visible light promotes one of the atom’s or molecule’s valence electrons to a higher-energy level. When an molecule absorbs infrared radiation, on the other hand, one of its chemical bonds experiences a change in vibrational energy. When it absorbs electromagnetic radiation the number of photons passing through a sample decreases. The measurement of this decrease in photons, which we call absorbance , is a useful analytical signal. Note that the each of the energy levels in Figure $\PageIndex{3}$ has a well-defined value because they are quantized. Absorption occurs only when the photon’s energy, h ν , matches the difference in energy, ∆ E , between two energy levels. A plot of absorbance as a function of the photon’s energy is called an absorbance spectrum . Figure $\PageIndex{4}$, for example, shows the absorbance spectrum of cranberry juice. When an atom or molecule in an excited state returns to a lower energy state, the excess energy often is released as a photon, a process we call emission (Figure $\PageIndex{4}$). There are several ways in which an atom or molecule may end up in an excited state, including thermal energy, absorption of a photon, or by a chemical reaction. Emission following the absorption of a photon is also called photoluminescence , and that following a chemical reaction is called chemiluminescence . A typical emission spectrum is shown in Figure $\PageIndex{6}$. Selection Rules The transition probability is defined as the probability of particular spectroscopic transition to take place. When an atom or molecule absorbs a photon, the probability of an atom or molecule to transit from one energy level to another depends on two things: the nature of initial and final state wavefunctions and how strongly photons interact with an wavefunction. Transition strengths are used to describe transition probability. Selection rules are utilized to determine whether a transition is allowed or not. Electronic dipole transitions are by far the most important for the topics covered in this module. In an atom or molecule, an electromagnetic wave (for example, visible light) can induce an oscillating electric or magnetic moment. If the frequency of the induced electric or magnetic moment is the same as the energy difference between one wavefunction $\psi_1$ and another wavefunction $\psi_2$, the interaction between an atom or molecule and the electromagnetic field is resonant (which means these two have the same frequency). Typically, the amplitude of this (electric or magnetic) moment is called the transition moment. In quantum mechanics, the transition probability of one molecule from one wavefunction $\psi_1$ to another wavefunction $\psi_2$ is given by $\|\vec{M}_{21}\|^2$, and $\vec{M}_{21}$ is called the transition dipole moment , or transition moment, from $\psi_1$ to $\psi_2$. $\vec{M}_{21}$ can be written as \[\vec{M}_{21}=\int \psi_2\vec{\mu}\psi_1d\tau \label{Select}\] where $\psi_1$ and $\psi_2$ are two different wavefunctions in one molecule, and $\vec{M}_{21}$ is the electric dipole moment operator . If we have a system with n atoms and each has charge $q_n$, and the dipole moment operator is can be written as \[\displaystyle \vec{\mu}=\sum_{n}q_n\vec{r}_n\] the $\vec{r}_{n}$ is the position vector operator for the ith charge. The nature of $\psi_1$ and $\psi_1$ change (e..g, the quantum numbers associated with each wavefunction) $\vec{M}_{21}$. Large values of $\vec{M}_{21}$ signify transitions with strong probabilities and small $\vec{M}_{21}$ values represent weak probabilities. A zero probability for a transition is a forbidden transition. For electronic wavefucntion (either atoms or molecules), the two primary selection rules governing transitions between electronic energy wavefunctions are: $ΔS = 0$ (The Spin Rule) $Δl = \pm 1$ (The Orbital Rule (or Laporte rule)) The spin multiplicity can be calculated from the quantum number $S$ of the total spin or from the number of unpaired electrons (like when determining paramagnetic properties of molecules). The spin-multiplicity is $(2S+1)$, where \[S = \underset{\text{spin quantum #}}{\sum s}\] The Spin Rule says that allowed transitions must involve the promotion of electrons without a change in their spin. The Orbital Rule says that transitions within a given set of p or d orbitals (i.e. those which only involve a redistribution of electrons within a given subshell) are forbidden. The orbital rule can be used to construct Grotrian diagrams, which show the allowed electronic transitions between the energy levels of atoms (Figure $\PageIndex{6}$) by taking into account the specific selection rules related to the system (e..g, the Orbital or Spin Rules). Transitions not permitted by selection rules are said forbidden, which means that theoretically they must not occur. However, in practice they may occur, but very low probabilities (see $\PageIndex{3}$ below). The Beers-Lambert Law The Beer-Lambert law relates the attenuation of light to the properties of the material through which the light is traveling. This page takes a brief look at the Beer-Lambert Law and explains the use of the terms absorbance and molar absorptivity relating to UV-visible absorption spectrometry. For each wavelength of light passing through the spectrometer, the intensity of the light passing through the reference cell is measured. This is usually referred to as $I_o$ - that's $I$ for Intensity. The intensity of the light passing through the sample cell is also measured for that wavelength - given the symbol, $I$. If $I$ is less than $I_o$, then the sample has absorbed some of the light (neglecting reflection of light off the cuvetter surface). A simple bit of math is then done in the computer to convert this into something called the absorbance of the sample - given the symbol, $A$. Note The absorbance of a transition depends on two external assumptions. The absorbance is directly proportional to the concentration ($c$) of the solution of the the sample used in the experiment. The absorbance is directly proportional to the length of the light path ($l$), which is equal to the width of the cuvette. Assumption one relates the absorbance to concentration and can be expressed as \[A \propto c \label{1}\] The absorbance ($A$) is defined via the incident intensity $I_o$ and transmitted intensity $I$ by \[ A=\log_{10} \left( \dfrac{I_o}{I} \right) \label{2}\] Assumption two can be expressed as \[A \propto l \label{3}\] Combining Equations $\ref{1}$ and $\ref{3}$: \[A \propto cl \label{4}\] This proportionality can be converted into an equality by including a proportionality constant. \[A = \epsilon c l \label{5}\] This formula is the common form of the Beer-Lambert Law , although it can be also written in terms of intensities: \[ A=\log_{10} \left( \dfrac{I_o}{I} \right) = \epsilon l c \label{6} \] The constant $\epsilon$ is called molar absorptivity or molar extinction coefficient and is a measure of the probability of the electronic transition. On most of the diagrams you will come across, the absorbance ranges from 0 to 1, but it can go higher than that. An absorbance of 0 at some wavelength means that no light of that particular wavelength has been absorbed. The intensities of the sample and reference beam are both the same, so the ratio I o /I is 1. Log 10 of 1 is zero. Example $\PageIndex{3}$ In a sample with an absorbance of 1 at a specific wavelength, what is the relative amount of light that was absorbed by the sample? Solution This question does not need Beer-Lambert Law (Equation $\ref{5}$) to solve, but only the definition of absorbance (Equation $\ref{2}$) \[ A=\log_{10} \left( \dfrac{I_o}{I} \right) \nonumber \] The relative loss of intensity is \[\dfrac{I-I_o}{I_o} = 1- \dfrac{I}{I_o} \nonumber\] Equation $\ref{2}$ can be rearranged using the properties of logarithms to solved for the relative loss of intensity: \[ 10^A= \dfrac{I_o}{I} \nonumber \] \[ 10^{-A}= \dfrac{I}{I_o} \nonumber\] \[ 1-10^{-A}= 1- \dfrac{I}{I_o} \nonumber\] Substituting in $A=1$ \[ 1- \dfrac{I}{I_o}= 1-10^{-1} = 1- \dfrac{1}{10} = 0.9 \nonumber\] Hence 90% of the light at that wavelength has been absorbed and that the transmitted intensity is 10% of the incident intensity To confirm, substituting these values into Equation $\ref{2}$ to get the absorbance back: \[\dfrac{I_o}{I} = \dfrac{100}{10} =10 \label{7a} \nonumber\] and \[\log_{10} 10 = 1 \label{7b} \nonumber\] You will find that various different symbols are given for some of the terms in the equation - particularly for the concentration and the solution length. The Greek letter epsilon in these equations is called the molar absorptivity - or sometimes the molar absorption coefficient. The larger the molar absorptivity, the more probable the electronic transition. In uv spectroscopy, the concentration of the sample solution is measured in molL -1 and the length of the light path in cm. Thus, given that absorbance is unitless, the units of molar absorptivity are L mol -1 cm -1 . However, since the units of molar absorptivity is always the above, it is customarily reported without units. Example $\PageIndex{4}$ Guanosine has a maximum absorbance of 275 nm. $\epsilon_{275} = 8400 M^{-1} cm^{-1}$ and the path length is 1 cm. Using a spectrophotometer, you find the that $A_{275}= 0.70$. What is the concentration of guanosine? Solution To solve this problem, you must use Beer's Law. \[A = \epsilon lc \nonumber\] 0.70 = (8400 M -1 cm -1 )(1 cm)($c$) Next, divide both side by [(8400 M -1 cm -1 )(1 cm)] $c$ = 8.33x10 -5 mol/L Example $\PageIndex{5}$ There is a substance in a solution (4 g/liter). The length of cuvette is 2 cm and only 50% of the certain light beam is transmitted. What is the extinction coefficient? Solution Using Beer-Lambert Law, we can compute the absorption coefficient. Thus, \[- \log \left(\dfrac{I_t}{I_o} \right) = - \log(\dfrac{0.5}{1.0}) = A = {8} \epsilon \nonumber\] Then we obtain that $\epsilon$ = 0.0376 Example $\PageIndex{6}$ In Example 3 above, what is the molar absorption coefficient if the molecular weight is 100? Solution It can simply obtained by multiplying the absorption coefficient by the molecular weight. Thus, $\epsilon$ = 0.0376 x 100 = 3.76 L·mol - 1 ·cm - 1 The proportion of the light absorbed will depend on how many molecules it interacts with. Suppose you have got a strongly colored organic dye. If it is in a reasonably concentrated solution, it will have a very high absorbance because there are lots of molecules to interact with the light. However, in an incredibly dilute solution, it may be very difficult to see that it is colored at all. The absorbance is going to be very low. Suppose then that you wanted to compare this dye with a different compound. Unless you took care to make allowance for the concentration, you couldn't make any sensible comparisons about which one absorbed the most light. Example $\PageIndex{7}$ In Example $\PageIndex{3}$ above, how much is the beam of light is transmitted when 8 g/liter ? Solution Since we know $\epsilon$, we can calculate the transmission using Beer-Lambert Law. Thus, $log(1) - log(I_t) = 0 - log(I_t)$ = 0.0376 x 8 x 2 = 0.6016 $log(I_t)$ = -0.6016 Therefore, $I_t$ = 0.2503 = 25% Example $\PageIndex{7}$ The absorption coefficient of a glycogen-iodine complex is 0.20 at light of 450 nm. What is the concentration when the transmission is 40 % in a cuvette of 2 cm? Solution It can also be solved using Beer-Lambert Law. Therefore, \[- \log(I_t) = - \log_{10}(0.4) = 0.20 \times c \times 2 \nonumber\] Then $c$ = 0.9948 The Beer-Lambert law Equation $\ref{5}$ can be rearranged to obtain an expression for $\epsilon$ (the molar absorptivity): \[ \epsilon = \dfrac{A}{lc} \label{8}\] Remember that the absorbance of a solution will vary as the concentration or the size of the container varies. Molar absorptivity compensates for this by dividing by both the concentration and the length of the solution that the light passes through. Essentially, it works out a value for what the absorbance would be under a standard set of conditions - the light traveling 1 cm through a solution of 1 mol dm-3. That means that you can then make comparisons between one compound and another without having to worry about the concentration or solution length. Values for molar absorptivity can vary hugely. For example, ethanal has two absorption peaks in its UV-visible spectrum - both in the ultra-violet. One of these corresponds to an electron being promoted from a lone pair on the oxygen into a pi anti-bonding orbital; the other from a $\pi$ bonding orbital into a $\pi$ anti-bonding orbital. Table $\PageIndex{2}$ gives values for the molar absorptivity of a solution of ethanal in hexane. Notice that there are no units given for absorptivity. That's quite common since it assumes the length is in cm and the concentration is mol dm -3 , the units are mol -1 dm 3 cm -1 . electron jump wavelength of maximum absorption (nm) molar absorptivity lone pair to $\pi$ anti-bonding orbital 290 15 $\pi$ bonding to $\pi$ anti-bonding orbital 180 10000 The ethanal obviously absorbs much more strongly at 180 nm than it does at 290 nm. (Although, in fact, the 180 nm absorption peak is outside the range of most spectrometers.) You may come across diagrams of absorption spectra plotting absorptivity on the vertical axis rather than absorbance. However, if you look at the figures above and the scales that are going to be involved, you are not really going to be able to spot the absorption at 290 nm. It will be a tiny little peak compared to the one at 180 nm. To get around this, you may also come across diagrams in which the vertical axis is plotted as log 10 (molar absorptivity). If you take the logs of the two numbers in the table, 15 becomes 1.18, while 10,000 becomes 4. That makes it possible to plot both values easily, but produces strangely squashed-looking spectra! Transition type Typical values of ε /m2mol-1 Spin forbidden and Laporte forbidden 0.1 Spin allowed and Laporte forbidden 1 - 10 Spin allowed and Laporte allowed e.g. charge transfer bands 1,000 - 106
Courses/University_of_Illinois_Springfield/UIS%3A_CHE_269_(Morsch_and_Andrews)/Chapters/Chapter_21%3A_Nucleophilic_Addition/21.17_An_Introduction_to_Carbohydrates	Carbohydrates are the most abundant class of organic compounds found in living organisms. They originate as products of photosynthesis , an endothermic reductive condensation of carbon dioxide requiring light energy and the pigment chlorophyll. \[ nCO_2 + n H_2O + \text{Energy} \rightarrow C_nH_{2n}O_n + nO_2\] As noted here, the formulas of many carbohydrates can be written as carbon hydrates, $C_n(H_2O)_n$, hence their name. The carbohydrates are a major source of metabolic energy, both for plants and for animals that depend on plants for food. Aside from the sugars and starches that meet this vital nutritional role, carbohydrates also serve as a structural material (cellulose), a component of the energy transport compound ATP/ADP , recognition sites on cell surfaces, and one of three essential components of DNA and RNA . The most useful carbohydrate classification scheme divides the carbohydrates into groups according to the number of individual simple sugar units. Monosaccharides contain a single unit; disaccharides contain two sugar units; and polysaccharides contain many sugar units as in polymers - most contain glucose as the monosaccharide unit. Some sugars can undergo a intermolecular cyclization to form a hemi-acetal. The hemiacetal carbon atom (C-1) becomes a new stereogenic center, commonly referred to as the anomeric carbon, and the α and β-isomers are called anomers. Disaccharides made up of other sugars are known, but glucose is often one of the components. Two important examples of such mixed disaccharides are displayed above. Lactose, also known as milk sugar, is a galactose-glucose compound joined as a beta-glycoside. It is a reducing sugar because of the hemiacetal function remaining in the glucose moiety. Many adults, particularly those from regions where milk is not a dietary staple, have a metabolic intolerance for lactose. Infants have a digestive enzyme which cleaves the beta-glycoside bond in lactose, but production of this enzyme stops with weaning. Sucrose, or cane sugar, is our most commonly used sweetening agent. It is a non-reducing disaccharide composed of glucose and fructose joined at the anomeric carbon of each by glycoside bonds (one alpha and one beta). In the formula shown here the fructose ring has been rotated 180º from its conventional perspective. Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry
Courses/can/CHEM_210_General_Chemistry_I_(Puenzo)/10%3A_Chemical_Bonding_I-_Lewis_Structures/10.01%3A_Models_and_Funtionality_of_proteins	X-ray Crystallography is a scientific method used to determine the arrangement of atoms of a crystalline solid in three-dimensional space. This technique takes advantage of the interatomic spacing of most crystalline solids by employing them as a diffraction gradient for x-ray light, which has wavelengths on the order of 1 angstrom (10 -8 cm). Introduction In 1895, Wilhelm Rontgen discovered x- rays. The nature of x- rays, whether they were particles or electromagnetic radiation, was a topic of debate until 1912. If the wave idea was correct, researchers knew that the wavelength of this light would need to be on the order of 1 Angstrom (A) (10 -8 cm). Diffraction and measurement of such small wavelengths would require a gradient with spacing on the same order of magnitude as the light. In 1912, Max von Laue, at the University of Munich in Germany, postulated that atoms in a crystal lattice had a regular, periodic structure with interatomic distances on the order of 1 A. Without having any evidence to support his claim on the periodic arrangements of atoms in a lattice, he further postulated that the crystalline structure can be used to diffract x-rays, much like a gradient in an infrared spectrometer can diffract infrared light. His postulate was based on the following assumptions: the atomic lattice of a crystal is periodic, x- rays are electromagnetic radiation, and the interatomic distance of a crystal are on the same order of magnitude as x- ray light. Laue's predictions were confirmed when two researchers: Friedrich and Knipping, successfully photographed the diffraction pattern associated with the x-ray radiation of crystalline CuSO4⋅5H2OCuSO4⋅5H2O. The science of x-ray crystallography was born. The arrangement of the atoms needs to be in an ordered, periodic structure in order for them to diffract the x-ray beams. A series of mathematical calculations is then used to produce a diffraction pattern that is characteristic to the particular arrangement of atoms in that crystal. X-ray crystallography remains to this day the primary tool used by researchers in characterizing the structure and bonding of organometallic compounds. Proteins are among the many biological molecules that are used for x-ray Crystallography studies. They are involved in many pathways in biology, often catalyzing reactions by increasing the reaction rate. Most scientists use x-ray Crystallography to solve the structures of proteins and to determine the functions of residues, interactions with substrates, and interactions with other proteins or nucleic acids. Proteins can be co-crystallized with these substrates, or they may be soaked into the crystal after crystallization. 3D Structures The tridimensional structure of proteins is fundamental for biological activity. That means that it is important for scientists to learn this structure and the shape of the molecules. Lewis Structures A Lewis Structure is a representation of molecules where all the valence electrons are shown distributed about the bonded atoms as either shared electron pairs (bond pairs) or unshared electron pairs (lone pairs). A shared pair of electrons is represented as a short line (a single bond). Sometimes atoms can share two pairs of electrons, represented by two short lines (a double bond). Atoms can even share three pairs of electrons, represented by three short lines (a triple bond). Pairs of dots are used to represent lone pair electrons.
Bookshelves/Introductory_Chemistry/Fundamentals_of_General_Organic_and_Biological_Chemistry_(LibreTexts)/13%3A_Alkenes_Alkynes_and_Aromatic_Compounds/13.04%3A_Properties_of_Alkenes_and_Alkynes	Learning Objectives To identify the physical properties of alkenes and describe trends in these properties. The physical properties of alkenes are similar to those of the alkanes. The table at the start of the chapter shows that the boiling points of straight-chain alkenes increase with increasing molar mass, just as with alkanes. For molecules with the same number of carbon atoms and the same general shape, the boiling points usually differ only slightly, just as we would expect for substances whose molar mass differs by only 2 u (equivalent to two hydrogen atoms). Like other hydrocarbons, the alkenes are insoluble in water but soluble in organic solvents. Looking Closer: Environmental Note Alkenes occur widely in nature. Ripening fruits and vegetables give off ethylene, which triggers further ripening. Fruit processors artificially introduce ethylene to hasten the ripening process; exposure to as little as 0.1 mg of ethylene for 24 h can ripen 1 kg of tomatoes. Unfortunately, this process does not exactly duplicate the ripening process, and tomatoes picked green and treated this way don’t taste much like vine-ripened tomatoes fresh from the garden. The bright red color of tomatoes is due to lycopene—a polyene. Other alkenes that occur in nature include 1-octene, a constituent of lemon oil, and octadecene (C 18 H 36 ) found in fish liver. Dienes (two double bonds) and polyenes (three or more double bonds) are also common. Butadiene (CH 2 =CHCH=CH 2 ) is found in coffee. Lycopene and the carotenes are isomeric polyenes (C 40 H 56 ) that give the attractive red, orange, and yellow colors to watermelons, tomatoes, carrots, and other fruits and vegetables. Vitamin A, essential to good vision, is derived from a carotene. The world would be a much less colorful place without alkenes.
Courses/Solano_Community_College/Introductory_Chemistry_at_Solano_College_-_Updated_2023_04_03/11%3A_States_of_Matter/11.11%3A_Vapor_Pressure	What causes this toy to move? The drinking duck is a toy that many kids (and adults) enjoy playing with. You can see the drinking duck in action in the video below: The motion of the duck illustrates a physical principle called vapor pressure. As the vapor pressure changes, the liquid in the duck moves up and down, causing the duck to move. Vapor Pressure When a partially filled container of liquid is sealed with a stopper, some liquid molecules at the surface evaporate into the vapor phase . However, the vapor molecules cannot escape from the container. So, after a certain amount of time, the space above the liquid reaches a point where it cannot hold any more vapor molecules. Now, some of the vapor molecules condense back into a liquid. The system reaches the point where the rate of evaporation is equal to the rate of condensation (see figure below). This is considered a dynamic equilibrium between the liquid and vapor phase. A dynamic equilibrium can be illustrated by an equation with a double arrow, meaning that the reaction is occurring in both directions and at the same rate. \[\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_2O} \left( g \right)\nonumber \] The forward direction represents the evaporation process, while the reverse direction represents the condensation process. Because they cannot escape the container, the vapor molecules above the surface of the liquid exert a pressure on the walls of the container. The vapor pressure is a measure of the pressure (force per unit area) exerted by a gas above a liquid in a sealed container. Vapor pressure is a property of a liquid based on the strength of its intermolecular forces. A liquid with weak intermolecular forces evaporates more easily and has a higher vapor pressure. A liquid with stronger intermolecular forces does not evaporate easily, and thus has a lower vapor pressure. For example, diethyl ether is a nonpolar liquid with weak dispersion forces. Its vapor pressure at $20^\text{o} \text{C}$ is $58.96 \: \text{kPa}$. Water is a polar liquid whose molecules are attracted to one another by relatively strong hydrogen bonding. The vapor pressure of water at $20^\text{o} \text{C}$ is only $2.33 \: \text{kPa}$, far less than that of diethyl ether. Vapor Pressure and Temperature Vapor pressure is dependent upon temperature. When the liquid in a closed container is heated, more molecules escape the liquid phase and evaporate. The greater number of vapor molecules strike the container walls more frequently, resulting in an increase in pressure. The table below shows the temperature dependence of the vapor pressure of three liquids. Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures.1 Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures.2 Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures.3 Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures.4 Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures.5 Table $\PageIndex{1}$: Vapor Pressure (in $\text{kPa}$ of Three Liquids at Different Temperatures.6 NaN $0^\text{o} \text{C}$ $20^\text{o} \text{C}$ $40^\text{o} \text{C}$ $60^\text{o} \text{C}$ $80^\text{o} \text{C}$ $100^\text{o} \text{C}$ Water 0.61 2.33 7.37 19.92 47.34 101.33 Ethanol 1.63 5.85 18.04 47.02 108.34 225.75 Diethyl Ether 24.70 58.96 122.80 230.65 399.11 647.87 Notice that the temperature dependence of the vapor pressure is not linear. From $0^\text{o} \text{C}$ to $80^\text{o} \text{C}$, the vapor pressure of water increases by $46.73 \: \text{kPa}$, while it increases by $53.99 \: \text{kPa}$ in only a span of twenty degrees from $80^\text{o} \text{C}$ to $100^\text{o} \text{C}$. Simulation Ever try to boil water on the top of a mountain? Explore the relationship between altitude, vapor pressure and boiling point in this simulation: Does water boil at 100 ° C everywhere? Summary Vapor pressure is a measure of the pressure exerted by a gas above a liquid in a sealed container. Strong intermolecular forces produce a lower rate of evaporation and a lower vapor pressure. Weak intermolecular forces produce a higher rate of evaporation and a higher vapor pressure. As the temperature increases, the vapor pressure increases. Review Define vapor pressure. How do intermolecular forces affect vapor pressure? How does temperature affect vapor pressure?
Courses/Los_Medanos_College/Chemistry_6_and_Chemistry_7_Combined_Laboratory_Manual_(Los_Medanos_College)/01%3A_Experiments/1.14%3A_Experiment_614_Synthesis_of_Aspirin_1_1_2	0 1 2 3 Student Name NaN Laboratory Date: Date Report Submitted: ___________________________ Student ID NaN Experiment Number and Name Experiment 614: Synthesis of Aspirin Experiment 6 1 4 : Synthesis of Aspirin Section 1: Purpose and Summary Conduct a chemical reaction to produce aspirin. Separate the aspirin from the reaction by-products using vacuum filtration. Analyze the aspirin and estimate its purity. Acetylsalicylic acid, commonly known as aspirin, is the most widely used drug in the world today. Its analgesic, antipyretic, and anti-inflammatory properties make it a powerful and effective drug to relive symptoms of pain, fever, and inflammation. Salicylic acid, whose name comes from Salix, the willow family of plants, was derived from willow bark extracts. Hippocrates the ancient Greek physician, as well as Native Americans before Columbus’ time, prepared willow bark teas as headache remedies and other tonics. In modern times, salicylic acid is administered in the form of aspirin which is less irritating to the stomach than salicylic acid. To prepare aspirin, salicylic acid is reacted with an excess of acetic anhydride. A small amount of a strong acid is used as a catalyst which speeds up the reaction. In this experiment, sulfuric acid will be used as the catalyst. The excess acetic anhydride will be quenched (reacted) with the addition of water. Overall, the reaction takes place between a carboxylic acid and an acid anhydride to form an ester. Aspirin is not very soluble in water so the aspirin product will precipitate when water is added. Some of the other compounds, acetic anhydride and acetic acid, dissolve in water, but salicylic acid is only slightly soluble in cold water. Vacuum filtration will separate the crystalline aspirin away from everything else in the reaction mixture except for any salicylic acid that did not react. The aspirin should be analyzed for the presence of any contaminating salicylic acid. In the final part of today’s lab, you will take a small amount of your aspirin and test it with iron(III) chloride (FeCl 3 ). FeCl 3 reacts with phenols (alcohol groups attached to aromatic rings) to produce colored complexes. Notice that salicylic acid contains the phenol functional group but aspirin does not. Therefore, the more salicylic acid that contaminates your aspirin, the darker the color will be with FeCl 3 . Section 2: Safety Precautions and Waste Disposal Safety Precautions: Use of eye protection is recommended for all experimental procedures. Sulfuric acid (H 2 SO 4 ) is highly corrosive. Avoid contact with your eyes, skin, and clothing. In case of contact, rinse with plenty of water. Ask instructor to assist with the cleaning of any spills. Acetic anhydride is a lachrymator (its vapor irritates the eyes causing tears to flow). Keep it in the fume hood. Waste Disposal: The reaction mixtures used in this experiment may be safely disposed of in the sink, followed by copious amount of running water. Solids, aspirin and salicylic acid, should be disposed of in the regular trash container. Section 3: Procedure Part 1: Synthesis of Aspirin 0 1 Prepare a boiling-water bath by filling a 600-mL beaker with about 400 mL of tap water. Put the beaker on the hot plate. Weigh out about 2.1 grams of salicylic acid on a piece of weighing paper. To do this, first tare a piece of weighing paper (This means to put a piece of paper on the balance and press the zero or tare button. Fig.1) Add some salicylic acid a little at a time onto the weighing paper until you have about 2.1 gram of it on the paper. It is OK to weigh a little extra mass (do not return excess salicylic acid to its container as this might contaminate the entire amount if your spatula is not perfectly clean). Record the mass of the solid. Place this solid into a 125-mL Erlenmeyer flask (other sizes may be acceptable). In the fume hood, measure out 4 mL of acetic anhydride in a small graduated cylinder and add it to the flask. From this point on, keep your flask under the hood, because it now contains acetic anhydride (the vapors of acetic anhydride are very irritating). Add about 5 drops of concentrated sulfuric acid. This will be the catalyst for the reaction. Place the flask in the boiling water bath and clamp it in place. (Fig.2) Heat the reaction for at least 15 minutes. Put at least 60 mL of laboratory water into a 150mL beaker (or similar size). Then put this beaker in an ice-water bath. Use this cold water in steps 9, 12, 13, and 14 below. After the reaction has heated for at least 15 minutes, remove it from the boiling water bath. In 1 or 2 mL portions, add about 10mL cold water to the reaction, swirl the reaction between each 1-2mL portions. This water will react and destroy any remaining acetic anhydride. (Fig3.) Put the flask with the reaction into an ice water bath (Fig.4). Crystals of aspirin should form. Chill for at least 10 minutes in order to obtain the maximum amount of crystalline product. Collect the aspirin crystals by vacuum filtration. To prepare a vacuum filtration set up: a) You will need a Buchner funnel, a clean 250-mL vacuum filter flask, a filter adaptor (often a one-hole stopper), a pre-cut filter paper and Beaker #2. WEIGH the dry filter paper. b) Place the filter paper on the Buchner funnel. The filter paper should fit snugly and cover all the small holes in the funnel. c) Using the filter adaptor, place the Buchner funnel on top of a vacuum filter flask. You might need to use a clamp and a ring-stand to keep the setup intact and upright. d) Connect a vacuum tubing hose to the vacuum filter flask and to the vacuum supply. e) Wet the filter paper with a little laboratory water. Turn the vacuum on and confirm that there are no leaks.(Fig.5) Add about 25 mL cold water to the reaction and mix well. Filter the crystals. Use a few milliliters of cold water to transfer as much solid as possible. Disconnect the vacuum temporarily and add 15 mL of cold water to the crystals in order to rinse them. After the crystals have soaked in the cold water for a few seconds, reconnect the vacuum to filter once more. Leave the vacuum on for several minutes to air dry the crystals. With the vacuum on, stir the crystals carefully to help dry them. Do this for about 15 minutes. Turn off the vacuum and weigh just the filter paper with the crystals. If you use a watch glass to carry the filter paper and crystals to the balance, be sure to tare the weight of the watch glass (in other words, don’t include the weight of the watch glass with your filter paper and crystals.) Clean up: Be sure to keep the aspirin for Parts 2 & 3. Throw away the filter paper in the trash. Dispose of the filtrate (the aqueous solution in the vacuum filtration flask) down the drain with plenty of water. In the fume hood, rinse with water all pipets and glassware in contact with acetic anhydride and pour the rinse water down the drain in the fume hood. Mass of Salicylic Acid: ______________ grams Volume of Acetic Anhydride: ______________ mL Mass of filter paper: ______________ grams Mass of crystals and filter paper: ______________ grams Part 2 : Analysis of Aspirin (How pure is it?) 0 1 Obtain 3 test tubes and about 1 mL of ethanol (CH3CH2OH) to all 3 test tubes. Also add 1-2 drops of 1% FeCl3 solution to each test tube. Be sure all 3 test tubes have the same amounts of reagents. NaN Label the first test tube “Aspirin” and add to this test tube a few of your aspirin crystals that you made today. Stir well. Describe the Color: Label the second test tube “Salicylic Acid” and add a few crystals of Salicylic Acid to this second test tube. Stir well. This is your Positive Control. The positive control shows you what to look for when a test is positive: in today’s analysis, the positive control shows you a dark color because it contains unreacted salicylic acid. Describe the Color: The third test tube is your Negative Control. Do not add any crystals to this test tube! The negative control shows you what to look for when a test is negative: in today’s analysis, the negative control shows you a color when no contaminating salicylic acid is present. Describe the Color: Compare the colors among the 3 test tubes and describe their similarities and differences. Describe the similarities and differences in color: (You can use the empty space to the left also). Part 3 : Analysis of Aspirin (How stable is it?) 0 1 Obtain a dry test tube and add about a pea-sized amount of your aspirin crystals (borrow some relatively pure crystals from a classmate if your aspirin was highly contaminated with salicylic acid). Tap the crystals to the bottom of the test tube. Use a test tube clamp to hold the test tube. Ignite a Bunsen burner. Pointing the open end of the test tube away from you, GENTLY heat the bottom of the test tube until the crystals melt. DON”T burn the aspirin.(Fig.5) Remove the test tube from the flame and carefully WAFT any odors from the open end of the test tube towards you. Do you smell anything? (To waft the odors means to “fan” the odors towards you, mixing in plenty of air to dilute the fumes.) If you don’t smell anything, try “pouring” the vapors out of the test tube by holding the test tube horizontally. Any odor detected may indicate that aspirin has decomposed with excessive heat. Aspirin may also decompose slowly upon long-term storage in a humid environment. After the test tube cools, wash the test tube in the sink with soap and water. Describe the Odor: Section 4: Calculations 0 1 Mass of salicylic acid (g) NaN Calculate the number of moles of salicylic acid (mol) (Molar mass = 138 g/mol) NaN Volume of acetic anhydride (mL) NaN Calculate the number of grams of acetic anhydride (g) (Density = 1.08 g/mL) NaN Calculate the number of moles of acetic anhydride (mol) (Molar mass = 102 g/mol) NaN Identify the Limiting Reagent (Name of compound) NaN Theoretical Yield of Aspirin (g) (Molar mass = 180 g/mol) NaN Mass of filter paper and aspirin (g) NaN Mass of filter paper (g) NaN Calculate the mass of aspirin (g) NaN Percent Yield of Aspirin (%) NaN Post Lab Questions: Aspirin is slightly soluble in water: the solubility of aspirin in water is 0.33 grams per 100 mL water at room temperature. In today’s experiment, you rinsed your aspirin with at least 50 mL of water. If the water was at room temperature, how many grams of your aspirin would have dissolved by rinsing it today? Suggest a reason why you were instructed to rinse the aspirin even though it is known that this will cause some of the aspirin to dissolve and be lost. Suppose that another student performed today’s experiment, but they forgot to air dry the aspirin during vacuum filtration. If they weigh the aspirin when it is still damp, how will this affect the percent yield? Suggest a scientific reason why we did not dry the aspirin in an oven today. A desiccant, or drying agent, is often added to a container with medications to prolong the shelf life. Cotton is sometimes used as a desiccant which is often found inside a package of aspirin. Moisture in the air (humidity) can cause aspirin to slowly decompose through a reaction called hydrolysis. The cotton absorbs moisture and delays hydrolysis. Suggest an easy and quick way to determine if your aspirin at home has begun to hydrolyze. Suppose two groups of students performed today’s experiment and obtained different results. One group obtained aspirin in 88% yield which turned dark in the FeCl 3 test. The other group obtained aspirin in 65% yield which produced no color change in the FeCl 3 test. Explain which group was more successful in lab.
Courses/Roosevelt_University/General_Organic_and_Biochemistry_with_Problems_Case_Studies_and_Activities/06%3A_Moles	6.1: The Mole The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be $6.022 \times 10^{23}$, a quantity called Avogadro’s number. 6.2: Formula Mass The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be $6.022 \times 10^{23}$, a quantity called Avogadro’s number. 6.3: Converting between Grams and Moles The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be $6.022 \times 10^{23}$, a quantity called Avogadro’s number. 6.4: Case Study- The Mole Concept in Clinical Laboratory Analysis 6.E: Exercises
Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/10%3A_Acids_and_Bases/10.E%3A_Acids_and_Bases_(Exercises)	10.1: Arrhenius Definition of Acids and Bases Concept Review Exercises Give the Arrhenius definitions of an acid and a base. What is neutralization? Answers Arrhenius acid: a compound that increases the concentration of hydrogen ion (H + ) in aqueous solution; Arrhenius base: a compound that increases the concentration of hydroxide ion (OH − ) in aqueous solution. the reaction of an acid and a base Exercises Give two examples of Arrhenius acids. Give two examples of Arrhenius bases. List the general properties of acids. List the general properties of bases. Name each compound. (For acids, look up the name in Table 10.1.1. For bases, use the rules for naming ionic compounds from Chapter 3.) a. HBr(aq) b. Ca(OH) 2 (aq) c. HNO 3 (aq) d. Fe(OH) 3 (aq) 6. Name each compound. a. HI(aq) b. Cu(OH) 2 (aq) c. H 3 PO 4 (aq) d. CsOH(aq) 7. Write a balanced chemical equation for the neutralization of Ba(OH) 2 (aq) with HNO 3 (aq). 8. Write a balanced chemical equation for the neutralization of H 2 SO 4 (aq) with Cr(OH) 3 (aq). 9. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl . Milk of Magnesia, a suspension of solid Mg(OH) 2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction. 10. Identify the salt produced in each acid-base reaction below. Then, balance the equation. a. 2HCl + Sr(OH) 2 → 2H 2 O + ?? b. KNO 3 ; HNO 3 + KOH → ?? + H 2 O c. HF + Ca(OH) 2 ---> ?? + H 2 O 11. How many moles of sodium hydroxide (NaOH) are needed to neutralize 0.844 mol of acetic acid (HC 2 H 3 O 2 )? (Hint: begin by writing a balanced chemical equation for the process.) 12. How many moles of perchloric acid (HClO 4 ) are needed to neutralize 0.052 mol of calcium hydroxide [Ca(OH) 2 ]? (Hint: begin by writing a balanced chemical equation for the process 13. Hydrazoic acid (HN 3 ) can be neutralized by a base. a. Write the balanced chemical equation for the reaction between hydrazoic acid and calcium hydroxide. b. How many milliliters of 0.0245 M Ca(OH) 2 are needed to neutralize 0.564 g of HN 3 ? 14. Citric acid (H 3 C 6 H 5 O 7 ) has three hydrogen atoms that can form hydrogen ions in solution. a. Write the balanced chemical equation for the reaction between citric acid and sodium hydroxide. b. If an orange contains 0.0675 g of H 3 C 6 H 5 O 7 , how many milliliters of 0.00332 M NaOH solution are needed to neutralize the acid? 15. Magnesium hydroxide [Mg(OH) 2 ] is an ingredient in some antacids. How many grams of Mg(OH) 2 are needed to neutralize the acid in 158 mL of 0.106 M HCl(aq)? It might help to write the balanced chemical equation first. 16. Aluminum hydroxide [Al(OH) 3 ] is an ingredient in some antacids. How many grams of Al(OH) 3 are needed to neutralize the acid in 96.5 mL of 0.556 M H 2 SO 4 (aq)? It might help to write the balanced chemical equation first. 17. Write the balanced chemical equation for the reaction between HBr and Ca(OH) 2 . What volume of 0.010 M HBr solution is be required to neutralize 25 mL of a 0.0100 M Ca(OH) 2 solution? 18. Write the balanced chemical equation for the reaction between HNO 3 and KOH . What volume of 0.5M HNO 3 is required to neutralize 60 mL of 0.4M KOH solution? Answers HCl and HNO 3 (answers will vary) NaOH and Ca(OH) 2 (answers will vary) sour taste, react with metals, react with bases, and turn litmus red bitter taste, feels slippery, react with acids and turn litmus blue a. hydrobromic acid b. calcium hydroxide c. nitric acid d. iron(III) hydroxide 6. a. hydroiodic acid b. cupric hydroxide c. phosphoric acid d. cesium hydroxide 7. 2HNO 3 (aq) + Ba(OH) 2 (aq) → Ba(NO 3 ) 2 (aq) + 2H 2 O 8. 3H 2 SO 4 (aq) + 2Cr(OH) 3 (aq) → Cr 2 (SO 4 ) 3 (aq) + 6H 2 O 9. Mg(OH) 2 + 2HCl --> MgCl 2 + 2H 2 O 10. a. SrCl 2 ; 2HCl + Sr(OH) 2 → 2H 2 O + SrCl 2 b. KNO 3 ; HNO 3 + KOH → KNO 3 + H 2 O c. CaF 2 ; 2HF + Ca(OH) 2 → CaF 2 + 2H 2 O 11. 0.844 mol 12. 0.104 mol 13. Part 1: 2HN 3 (aq) + Ca(OH) 2 → Ca(N 3 ) 2 + 2H 2 O Part 2: 268 mL 14. Part 1: H 3 C 6 H 5 O 7 (aq) + 3NaOH(aq) → Na 3 C 6 H 5 O 7 (aq) + 3H 2 O Part 2: 317.5 mL 15. 0.488 g 16. 2.79 g 17. 2HBr + Ca(OH) 2 → CaBr 2 + 2H 2 O ; 50 mL HBr 18. HNO 3 + KOH → KNO 3 + H 2 O ; 48 mL HNO 3 10.2: Brønsted-Lowry Definition of Acids and Bases Concept Review Exercise Give the definitions of a Brønsted-Lowry acid and a Brønsted-Lowry base. Answer A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. Exercises Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base. HCl(aq) + NH 3 (aq) → NH 4 + (aq) + Cl − (aq) Label each reactant as a Brønsted-Lowry acid or a Brønsted-Lowry base. H 2 O(ℓ) + N 2 H 4 (aq) → N 2 H 5 + (aq) + OH − (aq) Explain why a Brønsted-Lowry acid can be called a proton donor. Explain why a Brønsted-Lowry base can be called a proton acceptor. Write the chemical equation of the reaction of ammonia in water and label the Brønsted-Lowry acid and base. Write the chemical equation of the reaction of methylamine (CH 3 NH 2 ) in water and label the Brønsted-Lowry acid and base. Demonstrate that the dissolution of HNO 3 in water is actually a Brønsted-Lowry acid-base reaction by describing it with a chemical equation and labeling the Brønsted-Lowry acid and base. Identify the Brønsted-Lowry acid and base in the following chemical equation: C 3 H 7 NH 2 (aq) + H 3 O + (aq) → C 3 H 7 NH 3 + (aq) + H 2 O(ℓ) Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in each of the following equations 1. $\ce{NO2- + H2O ⟶ HNO2 + OH-}$ 2. $\ce{HBr + H2O ⟶ H3O+ + Br-}$ 3. $\ce{HS- + H2O ⟶ H2S + OH-}$ 4. $\ce{H2PO4- + OH- ⟶HPO4^2- + H2O}$ 5. $\ce{H2PO4- + HCl ⟶ H3PO4 + Cl-}$ 10. Write the chemical equation for the reaction that occurs when cocaine hydrochloride (C 17 H 22 ClNO 4 ) dissolves in water and donates a proton to a water molecule. (When hydrochlorides dissolve in water, they separate into chloride ions and the appropriate cation.) 11. If codeine hydrobromide has the formula C 18 H 22 BrNO 3 , what is the formula of the parent compound codeine? Answers HCl: Brønsted-Lowry acid; NH 3 : Brønsted-Lowry base H 2 O: Brønsted-Lowry acid; N 2 H 4 : Brønsted-Lowry base A Brønsted-Lowry acid gives away an H + ion—nominally, a proton—in an acid-base reaction. A Brønsted-Lowry base accepts an H + ion (a proton) in an acid-base reaction. NH 3 + H 2 O → NH 4 + + OH − (here NH 3 = Brønsted-Lowry base; H 2 O = Brønsted-Lowry acid) CH 3 NH 2 + H 2 O → CH 3 NH 3 + + OH − (here CH 3 NH 2 = Brønsted-Lowry base; H 2 O = Brønsted-Lowry acid) HNO 3 + H 2 O → H 3 O + + NO 3 − (here HNO 3 = Brønsted-Lowry acid; H 2 O = Brønsted-Lowry base) C 3 H 7 NH 2 (aq) + H 3 O + (aq) → C 3 H 7 NH 3 + (aq) + H 2 O(ℓ) (here H 3 O + = Brønsted-Lowry acid; C 3 H 7 NH 2 = Brønsted-Lowry base) 10.3: Water - Both an Acid and a Base Concept Review Exercises Explain how water can act as an acid. Explain how water can act as a base. Answers Under the right conditions, H 2 O can donate a proton, making it a Brønsted-Lowry acid. Under the right conditions, H 2 O can accept a proton, making it a Brønsted-Lowry base. Exercises Is H 2 O(ℓ) acting as an acid or a base? H 2 O(ℓ) + NH 4 + (aq) → H 3 O + (aq) + NH 3 (aq) Is H 2 O(ℓ) acting as an acid or a base? CH 3 − (aq) + H 2 O(ℓ) → CH 4 (aq) + OH − (aq) In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some C 2 H 3 O 2 − solutions, the following reaction can occur: C 2 H 3 O 2 − (aq) + H 2 O(ℓ) → HC 2 H 3 O 2 (aq) + OH − (aq) Is H 2 O acting as an acid or a base in this reaction? In the aqueous solutions of some salts, one of the ions from the salt can react with water molecules. In some NH 4 + solutions, the following reaction can occur: NH 4 + (aq) + H 2 O → NH 3 (aq) + H 3 O + (aq) Is H 2 O acting as an acid or a base in this reaction? Why is pure water considered neutral? Answers base acid acid base 5. When water ionizes, equal amounts of H + (acid) and OH − (base) are formed, so the solution is neither acidic nor basic: H 2 O(ℓ) → H + (aq) + OH − (aq) 10.4: The Strengths of Acids and Bases Concept Review Exercises Explain the difference between a strong acid or base and a weak acid or base. Explain what is occurring when a chemical reaction reaches equilibrium. Define pH . AnswerS A strong acid or base is 100% ionized in aqueous solution; a weak acid or base is less than 100% ionized. The overall reaction progress stops because the reverse process balances out the forward process. pH is a measure of the hydrogen ion concentration. Exercises Name a strong acid and a weak acid. (Hint: use Table 10.4.1.) Name a strong base and a weak base. (Hint: use Table 10.4.1.) Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: use Table 10.4.1.) HF HC 2 H 3 O 2 HCl HClO 4 Is each compound a strong acid or a weak acid? Assume all are in aqueous solution. (Hint: use Table 10.4.1.) H 2 SO 4 HSO 4 − HPO 4 2 − HNO 3 Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: use Table 10.4.1.) NH 3 NaOH Mg(OH) 2 Cu(OH) 2 Is each compound a strong base or a weak base? Assume all are in aqueous solution. (Hint: use Table 10.4.1.) KOH H 2 O Fe(OH) 2 Fe(OH) 3 Write the chemical equation for the equilibrium process for each weak acid in Exercise 3. Write the chemical equation for the equilibrium process for each weak acid in Exercise 4. Write the chemical equation for the equilibrium process for each weak base in Exercise 5. Write the chemical equation for the equilibrium process for each weak base in Exercise 6. Which is the stronger acid—HCl(aq) or HF(aq)? Which is the stronger base—KOH(aq) or Ni(OH) 2 (aq)? Consider the two acids in Exercise 11. For solutions that have the same concentration, which one would you expect to have a lower pH? Consider the two bases in Exercise 12. For solutions that have the same concentration, which one would you expect to have a higher pH? Consider the list of substances in Table \PageIndex3\PageIndex3.2"The pH Values of Some Common Solutions". What is the most acidic substance on the list that you have encountered recently? Consider the list of substances in Table \PageIndex3\PageIndex3.2"The pH Values of Some Common Solutions". What is the most basic substance on the list that you have encountered recently? Indicate whether solutions with the following pH values are acidic, basic, or neutral: pH = 9.4 pH = 7.0 pH = 1.2 pH = 6.5 Answers strong acid: HCl; weak acid: HC 2 H 3 O 2 (answers will vary) strong base: NaOH; weak base: NH 3 (answers will vary) weak weak strong strong strong weak weak strong weak strong strong weak strong weak weak weak 7. 3a: HF(aq) ⇆ H + (aq) + F − (aq); 3b: HC 2 H 3 O 2 (aq) ⇆ H + (aq) + C 2 H 3 O 2 − (aq) 8. 4b: HSO 4 − (aq) ⇆ H + (aq) + SO 4 2 − (aq); 4c: HPO 4 2 − (aq) ⇆ H + (aq) + PO 4 3 − (aq) 9. 5a: NH 3 (aq) + H 2 O ⇆ NH 4 + (aq) + OH − (aq); 5d: Cu(OH) 2 (aq) ⇆ Cu 2 + (aq) + 2OH − (aq) 10. 6b: H 2 O + H 2 O ⇆ H 3 O + (aq) + OH − (aq); 6c: Fe(OH) 2 (aq) ⇆ Fe 2 + (aq) + 2OH − (aq); 6d: Fe(OH)3(aq) ⇆ Fe 3 + (aq) + 3OH − (aq) 11. HCl(aq) 12. KOH(aq) 13. HCl(aq) 14. KOH(aq) 15. (answers will vary) 16. (answers will vary) 17. 1. basic 2. neutral 3. acidic (strongly) 4. acidic (mildly) 10.5: Buffers Concept Review Exercise Explain how a buffer prevents large changes in pH. Answer A buffer has components that react with both strong acids and strong bases to resist sudden changes in pH. Exercises Describe a buffer. What two related chemical components are required to make a buffer? Can a buffer be made by combining a strong acid with a strong base? Why or why not? Which solute combinations can make a buffer? Assume all are aqueous solutions. HCl and NaCl HNO 2 and NaNO 2 NH 4 NO 3 and HNO 3 NH 4 NO 3 and NH 3 Which solute combinations can make a buffer? Assume all are aqueous solutions. H 3 PO 4 and Na 3 PO 4 NaHCO 3 and Na 2 CO 3 NaNO 3 and Ca(NO 3 ) 2 HN 3 and NH 3 For each combination in Exercise 3 that is a buffer, write the chemical equations for the reactions of the buffer components when a strong acid and a strong base is added. For each combination in Exercise 4 that is a buffer, write the chemical equations for the reaction of the buffer components when a strong acid and a strong base is added. The complete phosphate buffer system is based on four substances: H 3 PO 4 , H 2 PO 4 − , HPO 4 2 − , and PO 4 3 − . What different buffer solutions can be made from these substances? Explain why NaBr cannot be a component in either an acidic or a basic buffer. Explain why Mg(NO 3 ) 2 cannot be a component in either an acidic or a basic buffer. Answers A buffer resists sudden changes in pH. It has a weak acid or base and a salt of that weak acid or base. No. Combining a strong acid and a strong base will produce salt and water. Excess strong acid or strong base will not act as a buffer. not a buffer buffer not a buffer buffer 4. 1. not a buffer 2. buffer 3. not a buffer 4. not buffer 3b: strong acid: H + + NO 2 − → HNO 2 ; strong base: OH − + HNO 2 → H 2 O + NO 2 − ; 3d: strong acid: H + + NH 3 → NH 4 + ; strong base: OH − + NH 4 + → H 2 O + NH 3 4b: strong acid: H + + CO 3 2 − → HCO3 − ; strong base: OH − + HCO 3 − → H 2 O + CO 3 2 − ; Buffers can be made by combining H 3 PO 4 and H 2 PO 4 − , H 2 PO 4 − and HPO 4 2 − , and HPO 4 2 − and PO 4 3 − . NaBr splits up into two ions in solution, Na + and Br − . Na + will not react with any added base knowing that NaOH is a strong base. Br- will not react with any added acid knowing that HBr is a strong acid. Because NaBr will not react with any added base or acid, it does not resist change in pH and is not a buffer. Mg(NO 3 ) 2 includes two types of ions, Mg 2 + and NO 3 − . Mg(OH) 2 is strong base and completely dissociates (100% falls apart), so Mg 2 + will not react with any added base (0% combines with OH − ). HNO 3 is strong acid and completely dissociates (100% falls apart), so NO 3 − will not react with any added acid (0% combines with H + ). Because Mg(NO 3 ) 2 will not react with any added base or acid, it does not resist change in pH and is not a buffer. Additional Exercises The properties of a 1.0 M HCl solution and a 1.0 M HC 2 H 3 O 2 solution are compared. Measurements show that the hydrochloric acid solution has a higher osmotic pressure than the acetic acid solution. Explain why. Of a 0.50 M HNO 3 solution and a 0.50 M HC 2 H 3 O 2 solution, which should have the higher boiling point? Explain why. The reaction of sulfuric acid [H 2 SO 4 (aq)] with sodium hydroxide [NaOH(aq)] can be represented by two separate steps, with only one hydrogen ion reacting in each step. Write the chemical equation for each step. The reaction of aluminum hydroxide [Al(OH) 3 (aq)] with hydrochloric acid [HCl(aq)] can be represented by three separate steps, with only one hydroxide ion reacting in each step. Write the chemical equation for each step. A friend brings you a small sample of an unknown chemical. Assuming that the chemical is soluble in water, how would you determine if the chemical is an acid or a base? A neutral solution has a hydrogen ion concentration of about 1 × 10 −7 M. What is the concentration of the hydroxide ion in a neutral solution? The Lewis definitions of an acid and a base are based on electron pairs, not protons. A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. Use Lewis diagrams to show that H + (aq) + OH − (aq) → H 2 O(ℓ) is an acid-base reaction in the Lewis sense as well as in the Arrhenius and Brønsted-Lowry senses. Given the chemical reaction NH 3 (g) + BF 3 (g) → NH 3 —BF 3 (s) show that the reaction illustrated by this equation is an acid-base reaction if we use the Lewis definitions of an acid and a base (see Exercise 7). The product contains a bond between the N and B atoms. Answers HCl is a strong acid and yields more ions in solution. HC 2 H 3 O 2 is a weak acid and undergoes partial ionization in solution. HNO 3 is a strong acid while HC 2 H 3 O 2 is a weak acid. HNO 3 dissociates 100% and its solution contains more ions. The more ions the solution contains the lower is its vapor pressure; the higher temperature is required for it to boil. H 2 SO 4 + NaOH → NaHSO 4 + H 2 O; NaHSO 4 + NaOH → Na 2 SO 4 + H 2 O Al(OH) 3 + HCl → Al(OH) 2 Cl + H 2 O; Al(OH) 2 Cl + HCl → Al(OH)Cl 2 + H 2 O; Al(OH)Cl 2 + HCl → AlCl3 + H 2 O One way is to add it to NaHCO 3 ; if it bubbles, it is an acid. Alternatively, add the sample to litmus and look for a characteristic color change (red for acid, blue for base). In a neutral solution, [OH - ] = [H + ] = 1.0 x 10 -7 M The O atom is donating an electron pair to the H + ion, making the base an electron pair donor and the acid an electron pair acceptor. The N atom is donating a lone pair to B in BF 3 , Hence NH 3 is the Lewis base and BF 3 is the Lewis acid.
Courses/University_of_Illinois_Springfield/Introduction_to_Organic_Spectroscopy/5%3A_Proton_Nuclear_Magnetic_Resonance_Spectroscopy_(NMR)/5.13%3A_Integration_of_Proton_Spectra	Objectives After completing this section, you should be able to explain what information can be obtained from an integrated 1 H NMR spectrum, and use this information in the interpretation of such a spectrum. use an integrated 1 H NMR spectrum to determine the ratio of the different types of protons present in an organic compound. Study Notes The concept of peak integration is that the area of a given peak in a 1H NMR spectrum is proportional to the number of (equivalent) protons giving rise to the peak. Thus, a peak which is caused by a single, unique proton has an area which measures one third of the area of a peak resulting from a methyl (CH 3 ) group in the same spectrum. In practice, we do not have to measure these areas ourselves: it is all done electronically by the spectrometer, and an integration curve is superimposed on the rest of the spectrum. The integration curve appears as a series of steps, with the height of each step being proportional to the area of the corresponding absorption peak, and consequently, to the number of protons responsible for the absorption. As it can be difficult to decide precisely where to start and stop when measuring integrations, you should not expect your ratios to be exact whole numbers. Signal integration The computer in an NMR instrument can be instructed to automatically integrate the area under a signal or group of signals. This is very useful, because in 1 H-NMR spectroscopy the area under a signal is proportional to the number of hydrogens to which the peak corresponds . The two signals in the methyl acetate spectrum, for example, integrate to approximately the same area, because they both correspond to a set of three equivalent protons. Take a look next at the spectrum of para -xylene (IUPAC name 1,4-dimethylbenzene): This molecule has two sets of protons: the six methyl (H a ) protons and the four aromatic (H b ) protons. When we instruct the instrument to integrate the areas under the two signals, we find that the area under the peak at 2.6 ppm is 1.5 times greater than the area under the peak at 7.4 ppm. This (along with the actual chemical shift values, which we'll discuss soon) tells us which set of protons corresponds to which NMR signal. The integration function can also be used to determine the relative amounts of two or more compounds in a mixed sample. If we have a sample that is a 50:50 (mole/mole) mixture of benzene and acetone, for example, the acetone signal should integrate to the same value as the benzene sample, because both signals represent six equivalent protons. If we have a 50:50 mixture of acetone and cyclopentane, on the other hand, the ratio of the acetone peak area to the cylopentane peak area will be 3:5 (or 6:10), because the cyclopentane signal represents ten protons. Example $\PageIndex{1}$ You take a 1 H-NMR spectrum of a mixed sample of acetone (CH 3 (CO)CH 3 ) and dichloromethane (CH 2 Cl 2 ). The integral ratio of the two signals (acetone : dichloromethane) is 2.3 to 1. What is the molar ratio of the two compounds in the sample?. Solution The acetone:dichloromethane molar ratio is 0.76:1. If the molar ratio was a 1:1 ratio, the signal integration ratio would be 3:1 because acetone has 6 protons while dichloromethane has 2. So, you need to divide the 2.3 integral value by a factor of 3 to get 0.76. Example $\PageIndex{2}$ You take the 1 H-NMR spectrum of a mixed sample of 36% para -xylene and 64% acetone in CDCl 3 solvent. How many peaks do you expect to see? What is the expected ratio of integration values for these peaks? (set the acetone peak integration equal to 1.0) Solution There are three peaks: two from para -xylene and one from acetone. The acetone peak and the para -xylene methyl peak both represent six protons, so the ratio of their integration values is simply 64 to 36 or 1 to 0.56. The ratio of the para -xylene methyl peak to the para -xylene aromatic peak is 6 to 4, or 0.56 to 0.37. So the final integral ratio of acetone:methyl:aromatic signals should be 1 to 0.56 to 0.37. Exercise Exercise $\PageIndex{1}$ Predict how many signals the following molecule would have? Sketch the spectra and estimate the integration of the peaks. Answer There will be two peaks in a ratio (integration) of 2 to 3 of benzene to methyl protons.
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Circular_Dichroism	Circular Dichroism, an absorption spectroscopy, uses circularly polarized light to investigate structural aspects of optically active chiral media. It is mostly used to study biological molecules, their structure, and interactions with metals and other molecules. Introduction Circular Dichroism (CD) is an absorption spectroscopy method based on the differential absorption of left and right circularly polarized light. Optically active chiral molecules will preferentially absorb one direction of the circularly polarized light. The difference in absorption of the left and right circularly polarized light can be measured and quantified. UV CD is used to determine aspects of protein secondary structure. Vibrational CD, IR CD, is used to study the structure of small organic molecules, proteins and DNA. UV/Vis CD investigates charge transfer transitions in metal-protein complexes. Circular Polarization of Light Electromagnetic radiation consists of oscillating electric and magnetic fields perpendicular to each other and the direction of propagation. Most light sources emit waves where these fields oscillate in all directions perpendicular to the propagation vector. Linear polarized light occurs when the electric field vector oscillates in only one plane. In circularly polarized light, the electric field vector rotates around the propagation axis maintaining a constant magnitude. When looked at down the axis of propagation the vector appears to trace a circle over the period of one wave frequency (one full rotation occurs in the distance equal to the wavelength). In linear polarized light the direction of the vector stays constant and the magnitude oscillates. In circularly polarized light the magnitude stays constant while the direction oscillates. As the radiation propagates the electric field vector traces out a helix. The magnetic field vector is out of phase with the electric field vector by a quarter turn. When traced together the vectors form a double helix. Light can be circularly polarized in two directions: left and right. If the vector rotates counterclockwise when the observer looks down the axis of propagation, the light is left circularly polarized (LCP). If it rotates clockwise, it is right circularly polarized (RCP). If LCP and RCP of the same amplitude, they are superimposed on one another and the resulting wave will be linearly polarized. Interaction with Matter As with linear polarized light, circularly polarized light can be absorbed by a medium. An optically active chiral compound will absorb the two directions of circularly polarized light by different amounts \[ \Delta A = A_l - A_r \] This can be extended to the Beer-Lambert Law. The molar absorpitivty of a medium will be different for LCP and RCP. The Beer-Lambert Law can be rewritten as \[A = (\varepsilon_l -\varepsilon_r)cl\] The difference in molar absorptivity is also known as the molar circular dichroism \[\Delta \varepsilon = \varepsilon_l -\varepsilon_r \] The molar circular dichroism is not only wavelength dependent but also depends on the absorbing molecules conformation, which can make it a function of concentration, temperature, and chemical environment. Any absorption of light results in a change in amplitude of the incident wave; absorption changes the intensity of the light and intensity of the square of the amplitude. In a chiral medium the molar absorptivities of LCP and RCP light are different so they will be absorbed by the medium in different amounts. This differential absorption results in the LCP and RCP having different amplitudes which means the superimposed light is no longer linearly polarized. The resulting wave is elliptically polarized. Molar Ellipticity The CD spectrum is often reported in degrees of ellipticity, $\theta$, which is a measure of the ellipticity of the polarization given by: \[tan \theta = \frac{E_l-E_r}{E_l+E_r}\] where E is the magnitude of the electric field vector. The change in polarization is usually small and the signal is often measured in radians where $\theta = \frac{2.303}{4}(A_l - A_r)$ and is a function of wavelength. $\theta$ can be converted to degrees by multiplying by $\frac{180}{\pi}$ which gives $\theta = 32.98 \Delta A$ The historical reported unit of CD experiments is molar ellipticity, $[\theta]$, which removes the dependence on concentration and path length \[[\theta] = 3298 \Delta \varepsilon\] where the 3298 converts from the units of molar absorptivity to the historical units of degrees$\cdot$ cm 2 $\cdot$dmol -1 . Applications Instrumentation Most commercial CD instruments are based on the modulation techniques introduced by Grosjean and Legrand. Light is linearly polarized and passed through a monochromator. The single wavelength light is then passed through a modulating device, usually a photoelastic modulator (PEM), which transforms the linear light to circular polarized light. The incident light on the sample switches between LCP and RCP light. As the incident light swtches direction of polarization the absorption changes and the differention molar absorptivity can be calculated. Biological molecules The most widely used application of CD spectroscopy is identifying structural aspects of proteins and DNA. The peptide bonds in proteins are optically active and the ellipticity they exhibit changes based on the local conformation of the molecule. Secondary structures of proteins can be analyzed using the far-UV (190-250 nm) region of light. The ordered $\alpha$-helices, $\beta$-sheets, $\beta$-turn, and random coil conformations all have characteristic spectra. These unique spectra form the basis for protein secondary structure analysis. It should be noted that in CD only the relative fractions of residues in each conformation can be determined but not specifically where each structural feature lies in the molecule. In reporting CD data for large biomolecules it is necessary to convert the data into a normalized value that is independent of molecular length. To do this the molar ellipticity is divided by the number of residues or monomer units in the molecule. The real value in CD comes from the ability to show conformational changes in molecules. It can be used to determine how similar a wild type protein is to mutant or show the extent of denaturation with a change in temperature or chemical environment. It can also provide information about structural changes upon ligand binding. In order to interpret any of this information the spectrum of the native conformation must be determined. Some information about the tertiary structure of proteins can be determined using near-UV spectroscopy. Absorptions between 250-300 nm are due to the dipole orientation and surrounding environment of the aromatic amino acids, phenylalanine, tyrosine, and tryptophan, and cysteine residues which can form disulfide bonds. Near-UV techniques can also be used to provide structural information about the binding of prosthetic groups in proteins. Metal containing proteins can be studied by visible CD spectroscopy. Visible CD light excites the d-d transitions of metals in chiral environments. Free ions in solution will not absorb CD light so the pH dependence of the metal binding and the stoichiometry can be determined. Vibrational CD (VCD) spectroscopy uses IR light to determine 3D structures of short peptides, nucleic acids, and carbohydrates. VCD has been used to show the shape and number of helices in A-, B-, and Z-DNA. VCD is still a relatively new technique and has the potential to be a very powerful tool. Resolving the spectra requires extensive ab initio calculations, as well as, high concentrations and must be performed in water, which may force the molecule into a nonnative conformation. References Woody, R. W. Circular-Dichroim. Methods in Enzymology 246, 34-71 (1995). Johnson, W. C. Protein secondary structure and circular dichroism: A practical guide. Proteins: Structure, Function, and Genetics 7, 205–214 (1990). Drake, A. F. Polarisation modulation-the measurement of linear and circular dichroism. Journal of Physics E: Scientific Instruments 19, 170–181 (1986). Neidig, M. L., Wecksler, A. T., Schenk, G., Holman, T. R. & Solomon, E. I. Kinetic and spectroscopic studies of N694C lipoxygenase: a probe of the substrate activation mechanism of a nonheme ferric enzyme. Journal of the American Chemical Society 129, 7531–7537 (2007). Polavarapu, P. L. & Zhao, C. X. Vibrational circular dichroism: a new spectroscopic tool for biomolecular structural determination. Fresenius Journal Anaytical Chemistry 366, 727–734 (2000).
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/21%3A_Carboxylic_Acid_Derivatives-_Nucleophilic_Acyl_Substitution_Reactions/21.06%3A_Chemistry_of_Esters	Esters are among the most widespread of all naturally occurring compounds. Many simple esters are pleasant-smelling liquids that are responsible for the fragrant odors of fruits and flowers. For example, methyl butanoate is found in pineapple oil, and isopentyl acetate is a constituent of banana oil. The ester linkage is also present in animal fats and in many biologically important molecules. The chemical industry uses esters for a variety of purposes. Ethyl acetate, for instance, is a commonly used solvent, and dialkyl phthalates are used as plasticizers to keep polymers from becoming brittle. You may be aware that there is current concern about the possible toxicity of phthalates at high concentrations, although a recent assessment by the U.S. Food and Drug Administration found the risk to be minimal for most people, with the possible exception of male infants. Preparation of Esters Esters are usually prepared from carboxylic acids by the methods already discussed. Thus, carboxylic acids are converted directly into esters by S N 2 reaction of a carboxylate ion with a primary alkyl halide or by Fischer esterification of a carboxylic acid with an alcohol in the presence of a mineral acid catalyst. In addition, acid chlorides are converted into esters by treatment with an alcohol in the presence of base (Section 21.4). Reactions of Esters Esters undergo the same kinds of reactions that we’ve seen for other carboxylic acid derivatives, but they are less reactive toward nucleophiles than either acid chlorides or anhydrides. All their reactions are applicable to both acyclic and cyclic esters, called lactones . Conversion of Esters into Carboxylic Acids: Hydrolysis An ester is hydrolyzed, either by aqueous base or aqueous acid, to yield a carboxylic acid plus an alcohol. Ester hydrolysis in basic solution is called saponification , after the Latin word sapo, meaning “soap.” We’ll see in Section 27.2 that soap is in fact made by boiling animal fat with aqueous base to hydrolyze the ester linkages. As shown in Figure $\PageIndex{1}$, ester hydrolysis occurs through a typical nucleophilic acyl substitution pathway in which hydroxide ion is the nucleophile that adds to the ester carbonyl group to give a tetrahedral intermediate. Loss of alkoxide ion then gives a carboxylic acid, which is deprotonated to give the carboxylate ion. Addition of aqueous HCl, in a separate step after the saponification is complete, protonates the carboxylate ion and gives the carboxylic acid. The mechanism shown in Figure $\PageIndex{1}$ is supported by isotope-labeling studies. When ethyl propanoate labeled with 18 O in the ether-like oxygen is hydrolyzed in aqueous NaOH, the 18 O label shows up exclusively in the ethanol product. None of the label remains with the propanoic acid, indicating that saponification occurs by cleavage of the C–OR′ bond rather than the CO–R′ bond. Acid-catalyzed ester hydrolysis can occur by more than one mechanism, depending on the structure of the ester. The usual pathway, however, is just the reverse of a Fischer esterification reaction (Section 21.3). As shown in Figure $\PageIndex{2}$, the ester is first activated toward nucleophilic attack by protonation of the carboxyl oxygen atom, and nucleophilic addition of water then occurs. Transfer of a proton and elimination of alcohol yields the carboxylic acid. Because this hydrolysis reaction is the reverse of a Fischer esterification reaction, Figure $\PageIndex{2}$ is the reverse of Figure $\PageIndex{1}$. Ester hydrolysis is common in biological chemistry, particularly in the digestion of dietary fats and oils. We’ll save a complete discussion of the mechanistic details of fat hydrolysis until Section 29.2 but will note for now that the reaction is catalyzed by various lipase enzymes and involves two sequential nucleophilic acyl substitution reactions. The first is a transesterification reaction in which an alcohol group on the lipase adds to an ester linkage in the fat molecule to give a tetrahedral intermediate that expels alcohol and forms an acyl enzyme intermediate. The second is an addition of water to the acyl enzyme, followed by expulsion of the enzyme to give a hydrolyzed acid and a regenerated enzyme. Exercise $\PageIndex{1}$ Why is the saponification of an ester irreversible? In other words, why doesn’t treatment of a carboxylic acid with an alkoxide ion yield an ester? Answer Reaction of a carboxylic acid with an alkoxide ion gives the carboxylate ion. Conversion of Esters into Amides: Aminolysis Esters react with ammonia and amines to yield amides. The reaction is not often used, however, because it’s usually easier to prepare an amide by starting with an acid chloride ( Section 21.4 ). Conversion of Esters into Alcohols: Reduction Esters are easily reduced by treatment with LiAlH 4 to yield primary alcohols ( Section 17.4 ). The mechanism of ester reduction is similar to that of acid chloride reduction in that a hydride ion first adds to the carbonyl group, followed by elimination of alkoxide ion to yield an aldehyde. Further reduction of the aldehyde gives the primary alcohol. The aldehyde intermediate can be isolated if 1 equivalent of diisobutylaluminum hydride (DIBAH, or DIBAL-H) is used as the reducing agent instead of LiAlH 4 . The reaction has to be carried out at –78 °C to avoid further reduction to the alcohol. Such partial reductions of carboxylic acid derivatives to aldehydes also occur in numerous biological pathways, although the substrate is either a thioester or acyl phosphate rather than an ester. Exercise $\PageIndex{1}$ What product would you expect from the reaction of butyrolactone with LiAlH 4 ? What product would you expect from the reaction of butyrolactone with DIBAH? Answer LiAlH4 gives $\ce{HOCH2CH2CH2CH2OH}$ DIBAH gives $\ce{HOCH2CH2CH2CHO}$. Exercise $\PageIndex{1}$ Show the products you would obtain by reduction of the following esters with LiAlH 4 : Answer $\ce{CH3CH2CH2CH(CH3)CH2OH + CH3OH}$ $\ce{PhOH + PhCH2OH}$ Conversion of Esters into Alcohols: Grignard Reaction Esters react with 2 equivalents of a Grignard reagent to yield a tertiary alcohol in which two of the substituents are identical ( Section 17.5 ). The reaction occurs by the usual nucleophilic substitution mechanism to give an intermediate ketone, which reacts further with the Grignard reagent to yield a tertiary alcohol. Exercise $\PageIndex{1}$ What ester and what Grignard reagent might you start with to prepare the following alcohols? Answer $\ce{Ethyl benzoate + 2 CH3MgBr}$ $\ce{Ethyl acetate + 2 PhMgBr}$ $\ce{Ethyl pentanoate + 2 CH3CH2MgBr}$
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.07%3A_Occurrence_Preparation_and_Properties_of_Nitrogen	Learning Objectives By the end of this section, you will be able to: Describe the properties, preparation, and uses of nitrogen Most pure nitrogen comes from the fractional distillation of liquid air. The atmosphere consists of 78% nitrogen by volume. This means there are more than 20 million tons of nitrogen over every square mile of the earth’s surface. Nitrogen is a component of proteins and of the genetic material (DNA/RNA) of all plants and animals. Under ordinary conditions, nitrogen is a colorless, odorless, and tasteless gas. It boils at 77 K and freezes at 63 K. Liquid nitrogen is a useful coolant because it is inexpensive and has a low boiling point. Nitrogen is very unreactive because of the very strong triple bond between the nitrogen atoms. The only common reactions at room temperature occur with lithium to form Li 3 N, with certain transition metal complexes, and with hydrogen or oxygen in nitrogen-fixing bacteria. The general lack of reactivity of nitrogen makes the remarkable ability of some bacteria to synthesize nitrogen compounds using atmospheric nitrogen gas as the source one of the most exciting chemical events on our planet. This process is one type of nitrogen fixation . In this case, nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. Nitrogen fixation also occurs when lightning passes through air, causing molecular nitrogen to react with oxygen to form nitrogen oxides, which are then carried down to the soil. Chemistry in Everyday Life: Nitrogen Fixation All living organisms require nitrogen compounds for survival. Unfortunately, most of these organisms cannot absorb nitrogen from its most abundant source—the atmosphere. Atmospheric nitrogen consists of N 2 molecules, which are very unreactive due to the strong nitrogen-nitrogen triple bond. However, a few organisms can overcome this problem through a process known as nitrogen fixation, illustrated in Figure $\PageIndex{1}$. Nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. To date, the only known kind of biological organisms capable of nitrogen fixation are microorganisms. These organisms employ enzymes called nitrogenases, which contain iron and molybdenum. Many of these microorganisms live in a symbiotic relationship with plants, with the best-known example being the presence of rhizobia in the root nodules of legumes. Large volumes of atmospheric nitrogen are necessary for making ammonia—the principal starting material used for preparation of large quantities of other nitrogen-containing compounds. Most other uses for elemental nitrogen depend on its inactivity. It is helpful when a chemical process requires an inert atmosphere. Canned foods and luncheon meats cannot oxidize in a pure nitrogen atmosphere, so they retain a better flavor and color, and spoil less rapidly, when sealed in nitrogen instead of air. This technology allows fresh produce to be available year-round, regardless of growing season. There are compounds with nitrogen in all of its oxidation states from 3− to 5+. Much of the chemistry of nitrogen involves oxidation-reduction reactions. Some active metals (such as alkali metals and alkaline earth metals) can reduce nitrogen to form metal nitrides. In the remainder of this section, we will examine nitrogen-oxygen chemistry. There are well-characterized nitrogen oxides in which nitrogen exhibits each of its positive oxidation numbers from 1+ to 5+. When ammonium nitrate is carefully heated, nitrous oxide (dinitrogen oxide) and water vapor form. Stronger heating generates nitrogen gas, oxygen gas, and water vapor. No one should ever attempt this reaction—it can be very explosive. In 1947, there was a major ammonium nitrate explosion in Texas City, Texas, and, in 2013, there was another major explosion in West, Texas. In the last 100 years, there were nearly 30 similar disasters worldwide, resulting in the loss of numerous lives. In this oxidation-reduction reaction, the nitrogen in the nitrate ion oxidizes the nitrogen in the ammonium ion. Nitrous oxide, shown in Figure $\PageIndex{2}$: , is a colorless gas possessing a mild, pleasing odor and a sweet taste. It finds application as an anesthetic for minor operations, especially in dentistry, under the name “laughing gas.” Low yields of nitric oxide, NO, form when heating nitrogen and oxygen together. NO also forms when lightning passes through air during thunderstorms. Burning ammonia is the commercial method of preparing nitric oxide. In the laboratory, the reduction of nitric acid is the best method for preparing nitric oxide. When copper reacts with dilute nitric acid, nitric oxide is the principal reduction product: \[\ce{3 Cu(s) + 8 HNO_3(aq) -> 2 NO(g) + 3 Cu(NO_3)_2(aq) + 4 H2O(l)} \nonumber \] Gaseous nitric oxide is the most thermally stable of the nitrogen oxides and is the simplest known thermally stable molecule with an unpaired electron. It is one of the air pollutants generated by internal combustion engines, resulting from the reaction of atmospheric nitrogen and oxygen during the combustion process. At room temperature, nitric oxide is a colorless gas consisting of diatomic molecules. As is often the case with molecules that contain an unpaired electron, two molecules combine to form a dimer by pairing their unpaired electrons to form a bond. Liquid and solid NO both contain N 2 O 2 dimers, like that shown in Figure $\PageIndex{3}$. Most substances with unpaired electrons exhibit color by absorbing visible light; however, NO is colorless because the absorption of light is not in the visible region of the spectrum. Cooling a mixture of equal parts nitric oxide and nitrogen dioxide to −21 °C produces dinitrogen trioxide, a blue liquid consisting of N 2 O 3 molecules (shown in Figure $\PageIndex{4}$). Dinitrogen trioxide exists only in the liquid and solid states. When heated, it reverts to a mixture of NO and NO 2 . It is possible to prepare nitrogen dioxide in the laboratory by heating the nitrate of a heavy metal, or by the reduction of concentrated nitric acid with copper metal, as shown in Figure $\PageIndex{5}$. Commercially, it is possible to prepare nitrogen dioxide by oxidizing nitric oxide with air. The nitrogen dioxide molecule (illustrated in Figure $\PageIndex{6}$) contains an unpaired electron, which is responsible for its color and paramagnetism. It is also responsible for the dimerization of NO 2 . At low pressures or at high temperatures, nitrogen dioxide has a deep brown color that is due to the presence of the NO 2 molecule. At low temperatures, the color almost entirely disappears as dinitrogen tetraoxide, N 2 O 4 , forms. At room temperature, an equilibrium exists: \[\ce{2 NO_2(g) <=> N2O4(g)} \quad \quad \quad \quad K_P=6.86 \nonumber \] Dinitrogen pentaoxide, N 2 O 5 (illustrated in Figure $\PageIndex{7}$), is a white solid that is formed by the dehydration of nitric acid by phosphorus(V) oxide (tetraphosphorus decoxide): \[\ce{P4O10(s) + 4 HNO3(l) \longrightarrow 4 HPO3(s) + 2 N2O5(s)} \nonumber \] It is unstable above room temperature, decomposing to N 2 O 4 and O 2 . The oxides of nitrogen(III), nitrogen(IV), and nitrogen(V) react with water and form nitrogen-containing oxyacids. Nitrogen(III) oxide, N 2 O 3 , is the anhydride of nitrous acid; HNO 2 forms when N 2 O 3 reacts with water. There are no stable oxyacids containing nitrogen with an oxidation state of 4+; therefore, nitrogen(IV) oxide, NO 2 , disproportionates in one of two ways when it reacts with water. In cold water, a mixture of HNO 2 and HNO 3 forms. At higher temperatures, HNO 3 and NO will form. Nitrogen(V) oxide, N 2 O 5 , is the anhydride of nitric acid; HNO 3 is produced when N 2 O 5 reacts with water: \[\ce{N2O5(s) + H2O(l) \longrightarrow 2 HNO3(aq)} \nonumber \] The nitrogen oxides exhibit extensive oxidation-reduction behavior. Nitrous oxide resembles oxygen in its behavior when heated with combustible substances. N 2 O is a strong oxidizing agent that decomposes when heated to form nitrogen and oxygen. Because one-third of the gas liberated is oxygen, nitrous oxide supports combustion better than air (one-fifth oxygen). A glowing splinter bursts into flame when thrust into a bottle of this gas. Nitric oxide acts both as an oxidizing agent and as a reducing agent. For example: \[\begin{align} &\text{oxidizing agent:} \quad \quad \ce{P_4(s) + 6 NO(g) -> P4O6(s) + 3 N2(g)} \\[4pt][4pt] &\text{reducing agent:} \quad \quad \ce{Cl_2(g) + 2 NO(g) -> 2 ClNO(g)} \end{align} \nonumber \] Nitrogen dioxide (or dinitrogen tetraoxide) is a good oxidizing agent. For example: \[\begin{align} \ce{NO_2(g) + CO(g) &\longrightarrow NO(g) + CO_2(g)} \\[4pt][4pt] \ce{NO_2(g) + 2 HCl(aq) &\longrightarrow NO(g) + Cl_2(g) + H2O(l)} \end{align} \nonumber \]
Courses/Nassau_Community_College/General_Chemistry_II	This text is designed for CHE152 - General Chemistry II. For many students, this course provides the foundation to a career in chemistry, while for others, this may be their only college-level science course. As such, this textbook provides an important opportunity for students to learn the core concepts of chemistry and understand how those concepts apply to their lives and the world around them. The text has been developed to meet the scope and sequence of most general chemistry courses. Front Matter 1: Kinetics 2: Fundamental Equilibrium Concepts 3: Acid-Base Equilibria 4: Equilibria of Other Reaction Classes 6: Thermodynamics 5: Electrochemistry 7: Appendices Back Matter
Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/02%3A_Nomenclature_and_physical_properties_of_organic_compounds/2.05%3A_Functional_groups_containing_mix_of_sp3-_and_sp2-_or_sp-hybridized_heteroatom	Learning Objectives Identify, assign IUPAC name, and draw structure from the IUPAC name of carboxylic acids and their derivatives, including acid halides, acid anhydrides, esters, amides, and nitriles. Predict the changes in the polarity and its effect on the reactivity of carboxylic acids and their derivatives, including acid halides, acid anhydrides, esters, amides, and nitriles. Identify phosphoric acid, anhydrides of phosphoric acids, phosphate anions, and esters of phosphoric acids. What are carboxylic acids and carboxylic acid derivatives? Carboxylic acids have a carbonyl group ($\ce{C=O}$) and a hydroxyl group ($\ce{-OH}$) on the same carbon, i.e., $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-OH}$ group. The carboxylic acid group is represented as $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-OH}$, or as $\ce{-COOH}$. Carboxyl acids have some characteristics of the $\ce{C=O}$ group, some characteristics of the $\ce{-OH}$ group, and some additional characteristics due to the interaction of the two groups. In carboxylic acid derivates , the $\ce{-OH}$ group is replaced with another group, that includes acid halides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-X}$), acid anhydrides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}-R'}}$), easters ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-OR'}$), and amides ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-NH2}$). Nitrile group that has the carbonyl $\ce{O}$ replaced with a $\ce{N}$ and the $\ce{-OH}$ group also replaced with the same $\ce{N}$, i.e., $\ce{R-C≡N}$ group is also classified as a carboxylic acid derivative. The nomenclature and physical characteristics of carboxylic acids and their derivatives are described in the following sections. Carboxylic acids ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-OH}$) Nomenclature of carboxylic acids The IUPAC nomenclature of the carboxylic acids follows the following rules. The longest hydrocarbon chain containing the carboxylic acid group is chosen as the parent name, with the last letter 'e' of its suffix replaced with -oic acid. For example, $\ce{HCOOH}$ is methanoic acid, $\ce{CH3COOH}$ is ethanoic acid, and $\ce{CH3CH2COOH}$ is propanoic acid. If there are two carboxylic acid groups, the suffix changes to -dioic acid, e.g., $\ce{HOOC-COOH}$ is ethanedioc acid, and $\ce{HOOC-CH2-COOH}$ is propanedioic acid. (note that when the suffix begins with a consonant (the letter 'd' in this case), the last letter 'e' of the parent hydrocarbon name is not dropped) Start numbering from the $\ce{C}$ of the $\ce{-COOH}$ group. The $\ce{-COOH}$ group itself does not need a location number, as it is always at the end of the chain. If the $\ce{-COOH}$ group is bonded to a cyclic chain, the suffix -carboxylic acid is added to the name of the cyclic hydrocarbon. Numbering starts from the point of attachment of $\ce{-COOH}$ to the ring. If the $\ce{-COOH}$ group is bonded to a benzene ring, the parent name "benzoic acid is used. Numbering starts from the point of attachment of the $\ce{-COOH}$ group to the ring. Example $\PageIndex{1}$ What is the IUPAC name of the compound shown on the right? Solution The longest chain counting the $\ce{-COOH}$ group is three $\ce{C's}$ and a double bond, so the parent name is propene and replace the last letter 'e' with -oic acid, i.e., propenoic acid. There is a methyl group attached that becomes a prefix, i.e., methylpropenoic acid. A location number is needed for the methyl group and the double bond. Start numbering from the $\ce{C}$ of the $\ce{-COOH}$ group, double bond receives#2 and the methyl group receives #2. Answer: 2-methylprop-2-enoic acid The common name of 2-methylprop-2-enoic acid is methacrylic acid, and prop-2-enoic acid is acrylic acid, the monomers (the repeating units) in some polymers. Example $\PageIndex{2}$ What is the IUPAC name of the compound shown on the right? Solution The $\ce{-COOH}$ group is attached to a five $\ce{C}$ cyclic chain, so the name of the cyclic chain becomes the parent name: cyclopentane. Add the suffix -carboxylic acid to the parent name to indicate the carboxylic acid group attached to a cyclic chain. Answer: cyclopentanecarboxylic acid Example $\PageIndex{3}$ What is the IUPAC name of the compound shown on the right? Solution A carboxylic acid bonded to a benzene ring takes "benzoic acid" as the parent name. A $\ce{-OH}$ group takes the prefix "hydroxy" in the presence of a $\ce{-COOH}$ group, i.e., hydroxybenzoic acid. Start numbering from the point of attachment of the $\ce{-COOH}$ group: the $\ce{-OH}$ group receives #4. Answer: 4-hydroxybenzoic acid. Common names of carboxylic acids Common names of carboxylic acids are derived from the names of natural sources of these acids. Table 1 lists some of the common names of carboxylic acids. Condensed formula IUPACE name Common name Source of the common name $\ce{HCOOH}$ methanoic acid formic acid Latin: formica, ant $\ce{CH3COOH}$ ethanoic acid acetic acid Latin: acetum, vinegar $\ce{CH3CH2COOH}$ propanoic acid propionic acid Greek: propion, first fat (\ce{CH3(CH2)2COOH}\) butanoic acid butyric acid Latin: butyrum, butter (\ce{CH3(CH2)4COOH}\) hexanoic acid caproic acid Latin: caper, goat (\ce{CH3(CH2)14COOH}\) hexadecanoic acid palmitic acid Latin: palma, palm tree (\ce{CH3(CH2)16COOH}\) octadecanoic acid stearic acid Greek: stear, solid fat (\ce{CH3(CH2)18COOH}\) eicosanoic acid arachidic acid Greek: arachis, peanut The first syllable of the common names, e.g., form-, acet-, prop-, etc., are also used as the first syllable of the common names of related compounds. For example, $\ce{HCOH}$ is formaldehyde, $\ce{CH3COH}$ is acetaldehyde, etc. Physical properties of carboxylic acids The carboxylic acid group has a $\ce{C=O}$ and a $\ce{-OH}$ groups, i.e., an sp 2 - and an sp 3 hybridized $\ce{O}$ bonded to the same $\ce{C}$. Both $\ce{O's}$ have two lone pairs of electrons on them, i.e., $\ce{-\!\!{\overset{\overset{\huge\enspace\!{:O:}}\|\!\!\!\!\|\enspace}{C}}\!\!-\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{O}}-H}$. Lone pair of electrons are usually not shown except when needed, i.e., the carboxylic acid group is represented as $\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-OH}$ or as $\ce{-COOH}$. The carboxylic acid group has three polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$ and $\ce{\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$, resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}\|\!\!\|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{O}-\overset{\delta{+}}{H}}$. This is because $\ce{O}$ are more electronegative than $\ce{C}$ (3.3-2.6 = 0.7) and $\ce{H}$ (3.3-2.2 = 1.17), as shown in Figure $\PageIndex{1}$. It makes $\ce\overset{\delta{+}}{C}$ an electrophile, $\ce\overset{\delta{-}}{O}$ a nucleophile or a base, and $\ce\overset{\delta{+}}{H}$ an acid in reactivity. Due to the acid protons, carboxylic acids are also classified as organic acids. The polar $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$ and $\ce{\overset{\delta{-}}{O}{-}\overset{\delta{+}}{H}}$ bonds allow dipole-dipole interactions and hydrogen bonding in addition to the London dispersion forces. Carboxylic acids have stronger intermolecular forces, higher melting points, higher boiling points, and higher solubilities in water compared to alcohols and aldehydes of comparable molar mass due to more intermolecular forces, as compared in Table 2. Condensed formula IUPAC name Molar mass (g/mol) Boiling point (oC) Solubility in water $\ce{CH3(CH2)2COOH}$ Butanoic acid 88.1 163 Miscible $\ce{CH3(CH2)3CH2OH}$ Pentan-1-ol 88.1 137 2.3 g/100 mL $\ce{CH3(CH2)3CHO}$ Pentanal 86.1 103 Slightly soluble Two carboxylic acids can make two hydrogen bonds with each other, as illustrated in Figure $\PageIndex{2}$, behaving as a dimer with two times higher molecular mass. It explains their higher boiling points than alcohols of the same molar mass. Carboxylic acids of up to five $\ce{C's}$, i.e., methanoic acid, ethanoic acid, propanoic acid, butanoic acid, and pentanoic acid, are soluble in water. Hexanoic acid is slightly soluble, and higher acids are insoluble. Carboxylic acids have a sour taste because they are acids due to ionizable proton in their $\ce{-O-H}$ groups. For example, the sour taste of citrus fruits is due to citric acid, and the sour taste of vinegar is due to ethanoic acid. Oxidation is i) loss of electrons, ii) gain of $\ce{O}$, or loss of $\ce{H}$; and the reduction is the opposite of these. Oxidation and reduction happen together and are collectively called Redox reactions. Most chemical reactions in biological systems are redox reactions, e.g., photosynthesis is a reduction of $\ce{CO2}$ to convert solar energy into potential chemical energy, and digestion of food is the opposite, i.e., oxidation to release the energy for the activities of life. A major portion of organic compounds is hydrocarbon groups, gradually oxidized to alcohols, aldehydes or ketones, carboxylic acids, and finally, carbon dioxide and water, releasing energy. For example, methane ($\ce{CH4}$) oxidizes to methanol ($\ce{CH3OH}$), methanal ($\ce{CH2O}$), methanoic acid ($\ce{HCOOH}$), and finally to carbon dioxide ($\ce{CO2}$) that is exhaled, as shown below. The alcohols, aldehydes, ketones, and carboxylic acids also serve as intermediates for synthesizing compounds the body needs. Carboxylic acids commonly appear in the metabolic process. For example, glucose, a six $\ce{C}$ compound, is first converted to two pyruvic acid molecules. Under low oxygen conditions (anaerobic), pyruvic acid is reduced to lactic acid. In the presence of oxygen (aerobic), pyruvic acid releases a $\ce{CO2}$ and becomes a two $\ce{C}$ group that joins a four $\ce{C}$ compound oxalic acid to make a six $\ce{C}$ compound citric acid. Citric acid releases a $\ce{CO2}$ and becomes a five $\ce{C}$ compound $\alpha$-ketoglutaric acid, which releases another $\ce{CO2}$ and becomes four $\ce{C}$ compound succinic acid, as shown below. This process goes on through several intermediate carboxylic acids, and either all the $\ce{C's}$ of the starting compound convert to $\ce{CO2}$, or the intermediate is utilized to synthesize compounds needed by the body. , , and Note: The carboxylic acids are shown as neutral acids in the above example, but in the physiological conditions, they exist as anions, i.e., as a carboxylate group ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-O^{-}}$). These points will be discussed in a later chapter. Acid halides and acid anhydrides -the most reactive acid derivatives Nomenclature of acid halides Acid halides contain the ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-X}$) group, where $\ce{X}$ can be $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$. IUPAC name of the acid halide takes the name of the corresponding carboxylic acid with the suffix -oic acid replaced with -oil halide, as shown in the following examples. Acid chlorides and acid bromides are the most common. Nomenclature of acid anhydrides Acid anhydride contains two acyl $\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-}$ groups bonded to a common $\ce{O}$ atom, i.e., ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-O-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}-}}$) group. A carboxylic acid anhydride is derived by condensing two carboxylic acids by losing a $\ce{H2O}$ molecule. The acid anhydride is symmetric anhydride if both the acids are the same and mixed or asymmetric anhydride if two different acids are condensed to form the anhydride, Symmetric acid anhydrides are named using the name of the corresponding acid with the last word 'acid' replaced with 'anhydride', as shown in the following examples. Mixed or asymmetric anhydrides are named by listing the names of the two acids in alphabetic order without the last word 'acid', followed by the word 'anhydride', as shown in the following examples. Physical properties of acid halides and acid anhydrides The acid halides group has two polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{X}}$ resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}\|\!\!\|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{X}}$. The acid anhydrides have four polar bonds i.e., two $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and two $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$ resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}\|\!\!\|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{O}-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}\|\!\!\|\enspace}{\overset{\delta{+}}{C}}}-}$. This polarity of the group can be observed in the electrostatic potential maps of acid halide, acid anhydride, and an acid shown in Figure $\PageIndex{3}$ It is apparent from the comparison of the electrostatic potential maps shown in Figure $\PageIndex{3}$ that the carbonyl $\ce{C}$ is more bluish, i.e., higher $\delta{+}$ and stronger nucleophile, in the case of acid halide and acid anhydride than in carboxylic acid. The question is $\ce{Cl}$ in the acid halide is less electronegative than $\ce{O}$ in carboxylic acid, then why is the carbonyl $\ce{C}$ more $\delta{+}$ in the acid halide than in the acid? The answer is in the fact that the heteroatom not only draws the bonding electron away from the carbonyl $\ce{C}$, it also donates its lone pair through resonance that diminishes the $\delta{+}$ character on the carbonyl $\ce{C}$: The 2p-orbital of $\ce{O}$ and 2p-orbital of $\ce{C}$ of carboxylic acid are of similar size and overlap well for the reasonce to happen and diminish its $\delta{+}$ character. The 3p-orbital of $\ce{Cl}$ or 4p orbital of $\ce{Br}$ overlaps poorly with 2p-orbital of carbonyl $\ce{C}$ due to the size difference and does not diminish its $\delta{+}$ character. The (\ce{O}\) is shared between to $\ce{C=O}$ groups in the case of acid anhydride. So it diminish $\delta{+}$ character of $\ce{C=O}$ less than in carboxylic acids. Due to the highly nucleophilic carbonyl $\ce{C}$, the acid halides and acid anhydrides are very reactive and primarily used as reactive intermediates in chemical synthesis. Again due to their high reactivity, they can not survive in biological systems. Biochemical systems use carboxylic acid derivatives containing $\ce{S}$ or phosphate groups as reactive intermediates, which will be described later. Easters Esters have an acyl group (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-}$) bonded with an alkoxy group (easters ($\ce{-OR'}$), i.e., an easter ($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-OR'}$) group. Nomenclature of esters IUPAC name of an ester starts with the name of the alkyl group that is part of the alkoxy ($\ce{-OR'}$) group, followed by the name of the acid corresponding to the acyl (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-}$) group with the suffix -oic acid replaced with the suffix -oate, as shown in the following examples. Example $\PageIndex{1}$ What is the IUPAC name of the compound shown on the right? Solution Alkyl group in the alkoxy ($\ce{-OR'}$) group is four ($\ce{C's}$) i.e., butyl. The acyl group is three ($\ce{C's}$), i.e., propanoic acid; replace acid with -oate, i.e., propanoate. Answer: butyl propanoate Example $\PageIndex{2}$ What is the IUPAC of the compound shown on the right? Solution It is an easter where the alkyl group in the alkoxy ($\ce{-OR'}$) group is one ($\ce{C's}$), i.e., methyl. The acyl group contains benzene, i.e., benzoic acid, and replace the -ic acid with -oate, i.e., benzoate. Answer: methyl benzoate Physical properties of esters Esters group has a polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, and two polar $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{O}}$ groups, resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}\|\!\!\|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{O-R'}}$, as shown in Figure $\PageIndex{4}$. The reactivity of an ester's carbonyl $\ce{C}$ is almost the same as that of a carboxylic acid. Easters are pretty common in nature. Small esters are volatile and soluble in water, making them easier to smell and taste. The fragrances of many perfumes and flavors of several fruits are due to esters, as shown in Table 3. pentyl acetate pentyl methanoate ethyl heptanoate propyl acetate Banana Plum Grape Pear octyl acetate ethyl butanoate pentyl butanoate propyl pentanoate Orange Pineapple Apricot Apple Esters in fats Fats are esters of carboxylic acids that contain a long chain hydrocarbon, an alkane or alkene with cis double bonds, called fatty acids, and propane-1,2,3-triol also called glycerol, as shown in one example in Figure $\PageIndex{5}$. Aspirin -an ester in medical use Aspirin is an ester of salicylic acid found in a willow tree's bark. Salicylic acid reduces pain and fever, but it irritates the stomach lining. Aspirin, an ester of salicylic acid, overcomes this problem and is commonly used to reduce pain and fever and as an anti-inflammatory agent. Methyl salicylate is another ester of salicylic acid found in wintergreen oil and used as skin ointments to soothe sore muscles. 0 1 2 NaN NaN NaN Salicylic acid Aspirin Methyl salicylate Amides Amidess have an acyl group (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-}$) bonded with a nitrogen group ($\ce{-NRR'}$), i.e., an amide ($\ce{-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-NRR'}$) group, where $\ce{R}$ and $\ce{R'}$ may be $\ce{H}$ or a hydrocarbon group. Nomenclature of amides IUPAC name of an amide is the name of the acid corresponding to the acyl (($\ce{R-\!\!{\overset{\overset{\huge\enspace\!{O}}\|\!\!\|\enspace}{C}}\!\!-}$) group with the suffix -oic acid replaced with the word amide. If a hydrocarbon group is bonded to nitrogen, its name appears as a prefix preceded by N-. If two identical hydrocarbon groups are bonded to nitrogen, the group name is preceded by N, N-di. If two different hydrocarbon groups are bonded to nitrogen, the group names, each preceded by N-, are listed alphabetically as prefixes. Some examples are shown below. An amide group attached to a benzene ring takes the base name 'benzamide,' as shown in the following examples. Example $\PageIndex{1}$ What is the IUPAC name of the compound shown on the right? Solution There is an amide group on a four $\ce{C}$ chain, i.e., butanoic acid, which changes to butanamide. There is a methyl group on nitrogen that becomes the prefix N-methyl. Answer: N-methylbutanamide Example $\PageIndex{2}$ What is the IUPAC name of the compound shown on the right? Solution There are two amide groups, so it takes the suffix -diamide. The corresponding acid is a six $\ce{C}$ diacid named hexanedioic acid. Replace -oic acid with amide. Answer: hexanediamid Physical properties of amides Amide group has a polar bonds, i.e., $\ce{\overset{\delta{+}}{C}{=}\overset{\delta{-}}{O}}$, $\ce{\overset{\delta{+}}{C}{-}\overset{\delta{-}}{N}}$, and $\ce{\overset{\delta{-}}{N}{-}\overset{\delta{+}}{H}}$ groups, resulting in a polar group: $\ce{-\!\!{\overset{\overset{\huge\enspace\!{\overset{\Large{\delta{-}}}{O}}}\|\!\!\|\enspace}{\overset{\delta{+}}{C}}}\!\!-\overset{\delta{-}}{N}-R'R"}$, as shown in Figure $\PageIndex{5}$. Comparison of electrostatic potential maps in Figure $\PageIndex{5}$ shows that the carbonyl $\ce{C}$ of amide is less $\delta{+}$, i.e., less electrophilic than that of the corresponding carboxylic acid. This is because of two reasons i) $\ce{N}$ is less electronegative and draws electrons away less than $\ce{O}$ and ii) being less electronegative, $\ce{N}$ sends its lone pair of electrons more to the carbonyl $\ce{C}$ neutralizing its $\delta{+}$ by resonance than $\ce{O}$. Both of these factors make the carbonyl $\ce{C}$ less $\delta{+}$ and less electrophilic, which makes amides one of the lest reactive and the most stable carboxylic acid derivatives that are found commonly in nature. The resonance effect is illustrated in Figure $\PageIndex{6}$ below. Due to the resonance, the $\ce{C-N}$ bond has a significant double bond character. Since the lone pair of electrons of amide are occupied in the resonance, they are less available to protons of acids. Therefore, amides are much less basic than amines. Amides have $\ce{\overset{\delta{-}}{N}{-}\overset{\delta{+}}{H}}$ bonds that allow them to make hydrogen bonds with water molecules. Therefore, amides containing up to five $\ce{C's}$ are soluble in water due the hydrogen bonding. Those with larger alkyl groups, i.e., with more than five $\ce{C's}$ are slightly soluble or insoluble due to the hydrophobic character of the alkyl groups dominating over the hydrophilic nature of the amide group. Amides in nature and medicines Proteins are made of small repeat units, called amino acids, joined through amide groups, as illustrated in Figure $\PageIndex{7}$, which makes amide one of the most common groups present in nature. During the digestion of proteins, $\ce{N's}$ end up in urea excreted by kidneys. If kidneys malfunction, urea may build to a toxic level, resulting in uremia. Urea is also used as a fertilizer. Barbiturates derived from barbituric acid have sedative and hypnotic effects and are used in medicines for these effects. Barbiturate drugs include phenobarbital and pentobarbital. Phenacetin and acetaminophen are amides used in Tylenol as alternatives to reduce fever and pain but with little anti-inflammatory effect. The structures of these amides are shown below. 0 1 2 3 4 5 urea barbituric acid phenobarbital pentobarbital phenacetin acetaminophen cyclic amides are called lactams. A four-member lactam is a common feature in the structure of Penicillin and related synthetic antibiotics, as shown below. 0 1 2 3 NaN NaN NaN NaN A four-membered lactam Penicillin G Amoxicillin Cephalexin Nitriles Nitriles have cyano group that is a polar bond $\ce{-\overset{\delta{+}}{C}{≡}\overset{\delta{-}}{N}\!:}$ with a lone pair in one of the sp orbital of $\ce{N}$, as illustrated in Figure $\PageIndex{5}$. IUPAC name of nitriles is composed of the name of the hydrocarbon skeleton, including the $\ce{C}$ in the nitrile group with the suffix -nitrile. Another way of naming them is to take the name of the corresponding carboxylic acid but replace the suffix -oic acid with -onitrile, as shown in the following example. Nitriles are not very common in nature but are important as intermediates in synthesizing organic compounds. Phosphorous groups Phosphorous ($\ce{P}$) is in the same group with $\ce{N}$ in periodic table. Like $\ce{N}$ in ammonia ($\ce{{H}-\overset{\bullet\bullet}{\underset{\underset{\huge{H}} \|}{N}}\!-H}$), the $\ce{P}$ can have three bonds and a lone pair, as in phosphine ($\ce{{H}-\overset{\bullet\bullet}{\underset{\underset{\huge{H}} \|}{P}}\!-H}$) with eight valence electrons, i.e., octet complete. Unlike $\ce{N}$, the $\ce{P}$ is in fourth row in the periodic table and, like other elements of the fourth and higher row, can have more than eight valence electrons in its compounds, e.g., ten valence electrons in phosphoric acid ($\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-OH}$). Phosphorous groups are important in biological systems, e.g., they are part of DNA molecules, phospholipids in cell membranes, and energetic molecules like adenosine triphosphate that are used as energy currency in biochemical reactions. Phosphoric acid and phosphoric anhydrides Phosphoric acid or orthphosphosporic acid has three $\ce{-OH}$ groups and one $\ce{=O}$ bonded to a $\ce{P}$ atom, i.e., $\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-OH}$. Like carboxylic acid anhydride, which is two carboxylic acids joined through a common $\ce{O}$, two phosphoric acids joined through a common $\ce{O}$ is a diphosphoric acid or a pyrophosphoric acid, and three phosphoric acids condensed in this way is a triphosphoric acid. Their corresponding anions formed after ionizing the acidic protons from the $\ce{-OH}$ groups are called phosphates, and alkoxy ($\ce{-OR}$ replacing one or more $\ce{-OH}$ groups are phosphate esters, as shown below. 0 1 2 $\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-OH}$ Phosphoric acid or orthphosphosporic acid $\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-OH}$ diphosphoric acid or a pyrophosphoric acid $\ce{HO-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{OH}} \|}{P}}}\!\!\!\!-OH}$ triphosphoric acid $\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} \|}{P}}}\!\!\!\!-O^{-}}$ Phosphate ion or orthphosphate ion $\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} \|}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} \|}{P}}}\!\!\!\!-O^{-}}$ diphosphate ion or a pyrophosphate ion $\ce{^{-}O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} \|}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} \|}{P}}}\!\!\!\!-O-\!\!\!\!\!{\overset{\overset{\huge\enspace{O}}\|\!\!\|\enspace}{\underset{\underset{\huge\enspace\,{O^{-}}} \|}{P}}}\!\!\!\!-O^{-}}$ triphosphate ion Phosphoric esters Like carboxylic acid changes to an ester when its $\ce{-OH}$ group is replaced with an alkoxy ($\ce{-OR}$) group, phosphoric acid changes to mono phosphoric ester when one of its $\ce{-OH}$ groups are replaced with alkoxy ($\ce{-OR}$), to diester when two $\ce{-OH}$ groups are replaced with alkoxy ($\ce{-OR}$), and to triester when three $\ce{-OH}$ groups are replaced with alkoxy ($\ce{-OR}$). The phosphoric esters are named by listing the names of the alkyl parts of the alkoxy groups in alphabetic order, followed by the world phosphate. In more complex phosphodiesters, it is common practice to name the organic compound followed by the word 'phosphate' or prefixed phospho-, as shown in the following examples. 0 1 2 3 dimethyl phosphate ethyl methyl phosphate dihydroxyacetone phosphate adenosine triphosphate
Bookshelves/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/01%3A_Matter_energy_and_their_measurements/1.04%3A_Significant_Figures	Significant figures are related to errors associated with the measured numbers. It is important to understand the significant figures because when calculations are made using numbers with errors, the answer cannot have less error than the error in any original number. The answer needs correction for the significant figures. Precision and accuracy The measured numbers have two types of errors, systematic and random errors, that determine the accuracy and precision of the measured number, as illustrated in Fig. 1.4.1. Systematic errors Systematic errors are constant, i.e., they have the same value in every measurement. For example, meter rod is a little short or a little long than a meter, it will introduce a systematic error. Systematic errors usually happen due to inaccurate calibration of the measuring instrument. The systematic errors determine how much the measured value differs from the actual value. Random errors Random errors are the statistical variability of the measured number. Random errors vary from one observation to another. Random errors cancel out if many measurements are taken and averaged. Scientific measurements are usually taken at least in triplicate and averaged to minimize random errors. The random errors determine how close the repeat measured numbers are to each other. Accuracy Accuracy or trueness of the measurement is defined as how close the average value is to the actual value. The closer the average is to the actual value, the more accurate or true it is, as illustrated in Fig. 1.4.1. The trueness depends on systematic errors, i.e., less systematic error, more accurate the average. Precision Precision is defined as how close the individual measurements are to each other. The closer the individual values are to each other, the more precise the measurement is, irrespective of whether it is accurate or not, as illustrated in Fig. 1.4.1. Precision depends on random errors, i.e., more substantial random errors mean less precision. Exact and inexact number There are two types of numbers, count numbers that are exact and measured numbers that are inexact. If the value is a counted number, it is an exact number . That is, there is no error in it. For example, a purchase of one dozen oranges contains exactly 12 nos of oranges; it can not be 11.5 or 12.5. Inexact numbers and error range When a value is measured, it comes with an error of measurement. A measured number with an error is called an inexact number . For example, when the same one dozen oranges are purchased by mass, the balance may read it 1572.6 g, or 1573 g, or 1570 g, depending on whether the smallest digit that the balance displays is 0.1g, 1 g, or 10 g. Suppose the balance is accurate to 1 g and reports the mass 1573 g; the actual mass may be anywhere in the range of 1572.5g-to-1573.4g. The smallest measured digit, i.e., the number in one's place, in this case, is an estimated number associated with an error. By convention, the estimated digit has ±1 errors associated with it. For example, the above-mentioned measured numbers are reported in science as 1572. 6 g ± 0.1 g, 157 3 g ± 1 g, or 15 7 0 g ± 10 g, respectively. The estimated digits are shown in bold fonts in the examples. The smallest digit in the display of digital instruments is an estimated number. In measurement using instruments that do not have a digital display, the smallest digit marked on the instrument plus one digit less than the minimum marked digit is added to the reported value. The smallest reported digit is an estimated digit. For example, the length of the pencil in Fig 1.4.2 is reported as 17. 7 cm using the ruler on the bottom, where 17 includes the smallest digit marked on the ruler, and the last digit, i.e., 0.7 is an estimated digit. By convention, the error range in this value is shown as 17. 7 ± 0.1. The same length is 17.7 0 cm using the ruler on the top in Fig. 1.4.2l, where 17.7 includes the smallest digit marked on the ruler, and the last reported digit, i.e., 0, is an estimated digit. By convention, the error range in this value is shown as 17.7 0 ± 0.01. The estimated digits are marked in bold fonts. Rules to determine the significant figures in measured numbers Significant figures All the digits reported in the measured value, including the estimated digit, are significant figures (SF). For example, 1572.6 g, 1573 g, and 1570 g have significant figures of 5, 4, and 3, respectively. Caution Note that zero in the last reading 1570g is not significant; it is a placeholder zero that is needed to place the estimated digit 7 at tens place. It is crucial to find significant figures in measured numbers because, when they are used in calculations, the answer cannot have less error than the maximum error in any measured number used in the calculation. The rules to determine the significant numbers in a measured number are the following. All non-zero digits are significant, e.g., 1572 has 4 SFs. The zeros may or may not be significant. In the following examples, the zeros in bold fonts are nonsignificant. Zeros between non-zero digits are significant, e.g., 1305.6 has 5 SFs. Leading zeros are not significant, e.g., 0.0 134 has 3 SFs. Trailing zeros are not significant if there is no decimal point present, e.g., 157 0 has 3 SFs. Trailing zeros are significant if the decimal point is present, e.g., 1570. has 4 SFs because the decimal point is present. Similarly, 0.0 124 has 3 SFs, but 0.0 1240 has 4 SFs because the decimal point is present. Confusion arises when more than one trailing zeros and the decimal point is absent. For example, 1500 g has 2 SFs by convention, but if the balance was accurate to 10 g, one of the zero was an estimated digit and was significant. Converting the number to a scientific notation resolves this issue. The coefficient part of the scientific notation shows all the significant figures in the measurement. For example, the number 1500 g, if shown in scientific notation as 1.5 x 10 3 has 2 SFs, but the same number shown as 1.50 x 10 3 has 3 SFs. Rounding the calculated answer involving inexact numbers When inexact numbers are used in calculations, the answer needs to be rounded to an appropriate number of significant figures, determined by the following rules. Rules of rounding A number is rounded by keeping the larger digits equal to significant figures and dropping or replacing the remaining smaller digits with placeholder zeros. The placeholder zeros are in bold fonts in the following examples. For example, 13543 becomes 135 00 when rounded to three significant. If the largest digit dropped is 4 or less than 4, it is simply dropped. For example, 23145 becomes 231 00 , when rounded to three significant figures. If the largest digit dropped is 5 or more than 5, then the smallest digit retained is increased by one. For example, 13543 becomes 14 000 when rounded into two significant figures. Rules for determining the significant figures in a calculated answer In the following rules, the track of significant figures that dictate the significant figures in the answer is kept by using bold fonts. In addition and subtraction, the answer has the same number of decimal places as the number with the smallest number of decimal places in the original numbers. For example, 13. 2 + 12.252 = 25. 4 52 is rounded to 25. 5 to keep one decimal place. In multiplication and division, the answer has the same number of significant figures as the original number with the smallest number of significant figures. For example, 1.35 x 2.1 = 2.8 35 is rounded to 2.8. Note If mathematical operations are performed in a series of steps, keep track of the significant figures but do not round off intermediate answers. Carry as many digits as possible from the intermediate answers to the next calculation step. Round off the final answer following the above rules. For example, (13. 2 + 12.252) x (1.35 x 2.1 ) = 25.4 52 x 2.8 35 = 72 .15642 is round to 72 in agreement with 2.8 35 that should have been rounded to 2.8 in a one-step calculation. Rounding the intermediate answers will lead to incorrect final answer of 71 instead of more correct 72, i.e. (13.2 + 12.252) x (1.35 x 2.1) = 25.5 x 2.8 = 71 Caution Exact numbers have an unlimited number of significant figures, which means they do not restrict the significant figures in the calculated answer. Example $\PageIndex{1}$ If 12 oranges weigh 1572.6 g calculate the mass of 1 orange in grams? Solution \begin{equation} 1572.6 \mathrm{~g} / 12=131.05 \mathrm{~g}\nonumber \end{equation} Explanation: The answer has 5 SFs because 12 is a counted number and exact. The only inexact number in the calculation that dictates the significant figures in the solution is 1572.6, which has 5 SFs. Example $\PageIndex{2}$ One dozen oranges were were sold 11 times. Calculate the total oranges sold? Solution \begin{equation} 12 \times 11=132 \text { oranges }\nonumber \end{equation} Explanation: The answer is not rounded because both the numbers in the calculation are exact, so the answer is also exact with unlimited significant figures.
Courses/University_of_British_Columbia/UBC_Introductory_Chemistry/07%3A_Bonding_and_Intermolecular_Forces/7.02%3A_Lewis_Structures_of_Ionic_Compounds-_Electrons_Transferred	Learning Objectives State the octet rule. Define ionic bond . Draw Lewis structures for ionic compounds. In Section 4.7, we demonstrated that ions are formed by losing electrons to make cations, or by gaining electrons to form anions. The astute reader may have noticed something: many of the ions that form have eight electrons in their valence shell. Either atoms gain enough electrons to have eight electrons in the valence shell and become the appropriately charged anion, or they lose the electrons in their original valence shell; the lower shell, now the valence shell, has eight electrons in it, so the atom becomes positively charged. For whatever reason, having eight electrons in a valence shell is a particularly energetically stable arrangement of electrons. The octet rule explains the favorable trend of atoms having eight electrons in their valence shell. When atoms form compounds, the octet rule is not always satisfied for all atoms at all times, but it is a very good rule of thumb for understanding the kinds of bonding arrangements that atoms can make. It is not impossible to violate the octet rule. Consider sodium: in its elemental form, it has one valence electron and is stable. It is rather reactive, however, and does not require a lot of energy to remove that electron to make the Na + ion. We could remove another electron by adding even more energy to the ion, to make the Na 2 + ion. However, that requires much more energy than is normally available in chemical reactions, so sodium stops at a 1+ charge after losing a single electron. It turns out that the Na + ion has a complete octet in its new valence shell, the n = 2 shell, which satisfies the octet rule. The octet rule is a result of trends in energies and is useful in explaining why atoms form the ions that they do. Now consider an Na atom in the presence of a Cl atom. The two atoms have these Lewis electron dot diagrams and electron configurations: \[\mathbf{Na\, \cdot }\; \; \; \; \; \; \; \; \; \; \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :} \nonumber \] \[\left [ Ne \right ]3s^{1}\; \; \; \; \left [ Ne \right ]3s^{2}3p^{5} \nonumber \] For the Na atom to obtain an octet, it must lose an electron; for the Cl atom to gain an octet, it must gain an electron. An electron transfers from the Na atom to the Cl atom: \[\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :} \nonumber \] resulting in two ions—the Na + ion and the Cl − ion: \[\mathbf{Na}^{+}\; \; \; \; \; \; \; \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-} \nonumber \] \[\left [ Ne \right ]\; \; \; \; \; \left [ Ne \right ]3s^{2}3p^{6} \nonumber \] Both species now have complete octets, and the electron shells are energetically stable. From basic physics, we know that opposite charges attract. This is what happens to the Na + and Cl − ions: \[\mathbf{Na}^{+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\rightarrow Na^{+}Cl^{-}\; \; or\; \; NaCl \nonumber \] where we have written the final formula (the formula for sodium chloride) as per the convention for ionic compounds, without listing the charges explicitly. The attraction between oppositely charged ions is called an ionic bond, and it is one of the main types of chemical bonds in chemistry. Ionic bonds are caused by electrons transferring from one atom to another. In electron transfer, the number of electrons lost must equal the number of electrons gained. We saw this in the formation of NaCl. A similar process occurs between Mg atoms and O atoms, except in this case two electrons are transferred: The two ions each have octets as their valence shell, and the two oppositely charged particles attract, making an ionic bond: \[\mathbf{Mg\,}^{2+}\; + \; \left[\mathbf{:}\mathbf{\ddot{\underset{.\: .}O}}\mathbf{\: :}\right]^{2-}\; \; \; \; \; Mg^{2+}O^{2-}\; or\; MgO \nonumber \] Remember, in the final formula for the ionic compound, we do not write the charges on the ions. What about when an Na atom interacts with an O atom? The O atom needs two electrons to complete its valence octet, but the Na atom supplies only one electron: \[\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.}O}}\mathbf{\: :} \nonumber \] The O atom still does not have an octet of electrons. What we need is a second Na atom to donate a second electron to the O atom: These three ions attract each other to give an overall neutral-charged ionic compound, which we write as Na 2 O. The need for the number of electrons lost being equal to the number of electrons gained explains why ionic compounds have the ratio of cations to anions that they do. This is required by the law of conservation of matter as well. Example $\PageIndex{1}$: Synthesis of Calcium Chloride from Elements With arrows, illustrate the transfer of electrons to form calcium chloride from $Ca$ atoms and $Cl$ atoms. Solution A $Ca$ atom has two valence electrons, while a $Cl$ atom has seven electrons. A $Cl$ atom needs only one more to complete its octet, while $Ca$ atoms have two electrons to lose. Thus we need two $Cl$ atoms to accept the two electrons from one $Ca$ atom. The transfer process looks as follows: The oppositely charged ions attract each other to make CaCl 2 . Exercise $\PageIndex{1}$ With arrows, illustrate the transfer of electrons to form potassium sulfide from $K$ atoms and $S$ atoms. Answer Summary The tendency to form species that have eight electrons in the valence shell is called the octet rule. The attraction of oppositely charged ions caused by electron transfer is called an ionic bond. The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions.
Courses/University_of_Kentucky/UK%3A_General_Chemistry/21%3A_Nuclear_Chemistry/21.2%3A_Nuclear_Equations	Skills to Develop Identify common particles and energies involved in nuclear reactions Write and balance nuclear equations Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions . To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction. Types of Particles in Nuclear Reactions Many entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure $\PageIndex{1}$. Protons $ (\ce{^{1}_{1}p}$, also represented by the symbol $\ce{^1_1H})$ and neutrons $ (\ce{^1_0n})$ are the constituents of atomic nuclei, and have been described previously. Alpha particles $ (\ce{^4_2He}$, also represented by the symbol $\ce{^{4}_{2}\alpha})$ are high-energy helium nuclei. Beta particles $ (\ce{^{0}_{−1}\beta}$, also represented by the symbol $\ce{^0_{-1}e})$ are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons $ (\ce{^0_{+1}e}$, also represented by the symbol $\ce{^0_{+1}β})$ are positively charged electrons (“anti-electrons”). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4, so it is symbolized $\ce{^4_2He}$ . This works because, in general, the ion charge is not important in the balancing of nuclear equations. Figure $\PageIndex{1}$: Although many species are encountered in nuclear reactions, this table summarizes the names, symbols, representations, and descriptions of the most common of these. Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter , particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays (γ)—and other much smaller subnuclear particles, which are beyond the scope of this chapter—according to the mass-energy equivalence equation $E = mc^2$, seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created: \[\ce{^0_{−1}e + ^0_{+1}e } \rightarrow \gamma + \gamma \label{21.3.1}\] Gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays. Gamma rays are produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions. Balancing Nuclear Reactions A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways: The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products. The sum of the charges of the reactants equals the sum of the charges of the products. If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, was one of the two products. Example $\PageIndex{1}$ shows how we can identify a nuclide by balancing the nuclear reaction. Example $\PageIndex{1}$: Balancing Equations for Nuclear Reactions The reaction of an α particle with magnesium-25 $ (\ce{^{25}_{12}Mg})$ produces a proton and a nuclide of another element. Identify the new nuclide produced. Solution The nuclear reaction can be written as: \[\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X}\] where $\ce A$ is the mass number and $\ce Z$ is the atomic number of the new nuclide, $\ce X$. Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products: \[\mathrm{25+4=A+1}\] so \[ \mathrm{A=28}\] Similarly, the charges must balance, so: \[\mathrm{12+2=Z+1}\] so \[\mathrm{Z=13}\] Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{^{28}_{13}Al}$. Exercise $\PageIndex{1}$ The nuclide $\ce{^{125}_{53}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction? Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry: The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting α particles: \[\ce{^{212}_{84}Po⟶ ^{208}_{82}Pb + ^4_2He}\] The first nuclide to be prepared by artificial means was an isotope of oxygen, 17 O. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with α particles: \[\ce{^{14}_7N + ^4_2α⟶ ^{17}_8O + ^1_1H}\] James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with 12 C by the nuclear reaction between 9 Be and 4 He: \[\ce{^9_4Be + ^4_2He⟶ ^{12}_6C + ^1_0n}\] The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, $\ce{^2_1H}$ ) , by Emilio Segre and Carlo Perrier in 1937: \[ \ce{^2_1H + ^{97}_{42}Mo⟶2^1_0n + ^{97}_{43}Tc}\] The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was: \[ \ce{^{235}_{92}U + ^1_0n⟶ ^{87}_{35}Br + ^{146}_{57}La + 3^1_0n}\] Summary Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged. Glossary alpha particle (α or $\ce{^4_2He}$ or $\ce{^4_2α}$) high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons antimatter particles with the same mass but opposite properties (such as charge) of ordinary particles beta particle ( $β$ or $\ce{^0_{-1}e}$ or $\ce{^0_{-1}β}$ ) high-energy electron gamma ray (γ or $\ce{^0_0γ}$ ) short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality nuclear reaction change to a nucleus resulting in changes in the atomic number, mass number, or energy state positron ( $\ce{^0_{+1}β}$ or $\ce{^0_{+1}e}$) antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/21%3A_Resonance_and_Molecular_Orbital_Methods/21.08%3A_More_on_Stabilization_Energies	It was shown in Section 21-3 that benzene is $36$-$38 \: \text{kcal}$ more stable than the hypothetical molecule 1,3,5-cyclohexatriene on the basis of the differences between experimental heats of combustion, or hydrogenation, and heats calculated from bond energies. We call this energy difference the stabilization energy (SE) of benzene. We have associated most of this energy difference with $\pi$-electron delocalization, which is the delocalization energy (DE). The difference between SE and DE will be small only if our bond-energy tables are reliable and steric and strain effects are small. The problem with bond energies is that we use bond energies that neglect changes in bond strength caused by environment. Primary, secondary, tertiary, alkenic, and alkynic $\ce{C_H}$ bonds are assumed to have equal energies; $\ce{C-C}$ single bonds are assumed to be equal, regardless of whether the other bonds to the carbon atoms in question are single or multiple; and differences in energy between double bonds that are mono-, di-, tri-, or tetra-substituted are neglected, as are changes in bond energies associated with steric strain. Bond energies are strictly applicable to molecules in which the bonds are of the normal lengths. In the case of benzene, which has $\ce{C-C}$ bonds with lengths intermediate between normal single and double bonds, there seems to be no clear agreement as to how to take the bond distances into account in computing the delocalization energy. In spite of these uncertainties the stabilization energies seem to give a good qualitative idea of the importance of electron delocalization in organic molecules. Tables 21-1 and 21-2 give stabilization energies for several substances that are best represented as hybrid structures. Table 21-1: Stabilization Energies (or Approximate Delocalization Energies) from Heats of Formation of Some Aromatic Compounds Table 21-2: Stabilization Energies (SE) from Heats of Formation $\left( \Delta H^0 \right)$ of Some Conjugated Polyenes
Courses/Lansing_Community_College/LCC%3A_Chem_151_-_General_Chemistry_I/Text/10%3A_Molecular_Geometry_and_Bonding_Theories/10.08%3A_Second-Row_Diatomic_Molecules	Learning Objectives To apply Molecular Orbital Theory to the diatomic homonuclear molecule from the elements in the second period. If we combine the splitting schemes for the 2s and 2p orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the valence orbitals of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. We now describe examples of systems involving period 2 homonuclear diatomic molecules, such as N 2 , O 2 , and F 2 . four key points to remember when drawing molecular orbital diagrams: The number of molecular orbitals produced is the same as the number of atomic orbitals used to create them (the "law of conservation of orbitals" ). As the overlap between two atomic orbitals increases, the difference in energy between the resulting bonding and antibonding molecular orbitals increases. When two atomic orbitals combine to form a pair of molecular orbitals, the bonding molecular orbital is stabilized about as much as the antibonding molecular orbital is destabilized. The interaction between atomic orbitals is greatest when they have the same energy. We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F 2 . We use the diagram in part (a) in Figure $\PageIndex{1}$; the n = 1 orbitals (σ 1 s and σ 1 s * ) are located well below those of the n = 2 level and are not shown. As illustrated in the diagram, the σ 2 s and σ 2 s * molecular orbitals are much lower in energy than the molecular orbitals derived from the 2 p atomic orbitals because of the large difference in energy between the 2 s and 2 p atomic orbitals of fluorine. The lowest-energy molecular orbital derived from the three 2 p orbitals on each F is $ \sigma _{2p_{z}} $ and the next most stable are the two degenerate orbitals, $ \pi _{2p_{x}} $ and $ \pi _{2p_{y}} $ . For each bonding orbital in the diagram, there is an antibonding orbital, and the antibonding orbital is destabilized by about as much as the corresponding bonding orbital is stabilized. As a result, the $ \sigma ^{\star }_{2p_{z}} $ orbital is higher in energy than either of the degenerate $ \pi _{2p_{x}}^{\star } $ and $ \pi _{2p_{y}}^{\star } $ orbitals. We can now fill the orbitals, beginning with the one that is lowest in energy. Each fluorine has 7 valence electrons, so there are a total of 14 valence electrons in the F 2 molecule. Starting at the lowest energy level, the electrons are placed in the orbitals according to the Pauli principle and Hund’s rule. Two electrons each fill the σ 2 s and σ 2 s * orbitals, 2 fill the $ \sigma _{2p_{z}} $ orbital, 4 fill the two degenerate π orbitals, and 4 fill the two degenerate π * orbitals, for a total of 14 electrons. To determine what type of bonding the molecular orbital approach predicts F 2 to have, we must calculate the bond order. According to our diagram, there are 8 bonding electrons and 6 antibonding electrons, giving a bond order of (8 − 6) ÷ 2 = 1. Thus F 2 is predicted to have a stable F–F single bond, in agreement with experimental data. We now turn to a molecular orbital description of the bonding in O 2 . It so happens that the molecular orbital description of this molecule provided an explanation for a long-standing puzzle that could not be explained using other bonding models. To obtain the molecular orbital energy-level diagram for O 2 , we need to place 12 valence electrons (6 from each O atom) in the energy-level diagram shown in part (b) in Figure $\PageIndex{1}$. We again fill the orbitals according to Hund’s rule and the Pauli principle, beginning with the orbital that is lowest in energy. Two electrons each are needed to fill the σ 2 s and σ 2 s * orbitals, 2 more to fill the $ \sigma _{2p_{z}} $ orbital, and 4 to fill the degenerate $ \pi _{2p_{x}}^{\star } $ and $ \pi _{2p_{y}}^{\star } $ orbitals. According to Hund’s rule, the last 2 electrons must be placed in separate π * orbitals with their spins parallel, giving two unpaired electrons. This leads to a predicted bond order of (8 − 4) ÷ 2 = 2, which corresponds to a double bond, in agreement with experimental data (Table 4.5): the O–O bond length is 120.7 pm, and the bond energy is 498.4 kJ/mol at 298 K. None of the other bonding models can predict the presence of two unpaired electrons in O 2 . Chemists had long wondered why, unlike most other substances, liquid O 2 is attracted into a magnetic field. As shown in Figure $\PageIndex{2}$, it actually remains suspended between the poles of a magnet until the liquid boils away. The only way to explain this behavior was for O 2 to have unpaired electrons, making it paramagnetic, exactly as predicted by molecular orbital theory. This result was one of the earliest triumphs of molecular orbital theory over the other bonding approaches we have discussed. The magnetic properties of O 2 are not just a laboratory curiosity; they are absolutely crucial to the existence of life. Because Earth’s atmosphere contains 20% oxygen, all organic compounds, including those that compose our body tissues, should react rapidly with air to form H 2 O, CO 2 , and N 2 in an exothermic reaction. Fortunately for us, however, this reaction is very, very slow. The reason for the unexpected stability of organic compounds in an oxygen atmosphere is that virtually all organic compounds, as well as H 2 O, CO 2 , and N 2 , have only paired electrons, whereas oxygen has two unpaired electrons. Thus the reaction of O 2 with organic compounds to give H 2 O, CO 2 , and N 2 would require that at least one of the electrons on O 2 change its spin during the reaction. This would require a large input of energy, an obstacle that chemists call a spin barrier . Consequently, reactions of this type are usually exceedingly slow. If they were not so slow, all organic substances, including this book and you, would disappear in a puff of smoke! For period 2 diatomic molecules to the left of N 2 in the periodic table, a slightly different molecular orbital energy-level diagram is needed because the $ \sigma _{2p_{z}} $ molecular orbital is slightly higher in energy than the degenerate $ \pi ^{\star }_{np_{x}} $ and $ \pi ^{\star }_{np_{y}} $ orbitals. The difference in energy between the 2 s and 2 p atomic orbitals increases from Li 2 to F 2 due to increasing nuclear charge and poor screening of the 2 s electrons by electrons in the 2 p subshell. The bonding interaction between the 2 s orbital on one atom and the 2 pz orbital on the other is most important when the two orbitals have similar energies. This interaction decreases the energy of the σ 2 s orbital and increases the energy of the $ \sigma _{2p_{z}} $ orbital. Thus for Li 2 , Be 2 , B 2 , C 2 , and N 2 , the $ \sigma _{2p_{z}} $ orbital is higher in energy than the $ \sigma _{3p_{z}} $ orbitals, as shown in Figure $\PageIndex{3}$ Experimentally, it is found that the energy gap between the ns and np atomic orbitals increases as the nuclear charge increases (Figure $\PageIndex{3}$ ). Thus for example, the $ \sigma _{2p_{z}} $ molecular orbital is at a lower energy than the $ \pi _{2p_{x,y}} $ pair. Completing the diagram for N 2 in the same manner as demonstrated previously, we find that the 10 valence electrons result in 8 bonding electrons and 2 antibonding electrons, for a predicted bond order of 3, a triple bond. Experimental data show that the N–N bond is significantly shorter than the F–F bond (109.8 pm in N 2 versus 141.2 pm in F 2 ), and the bond energy is much greater for N 2 than for F 2 (945.3 kJ/mol versus 158.8 kJ/mol, respectively). Thus the N 2 bond is much shorter and stronger than the F 2 bond, consistent with what we would expect when comparing a triple bond with a single bond. Example $\PageIndex{3}$: Diatomic Sulfur Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in S 2 , a bright blue gas at high temperatures. Given: chemical species Asked for: molecular orbital energy-level diagram, bond order, and number of unpaired electrons Strategy: Write the valence electron configuration of sulfur and determine the type of molecular orbitals formed in S 2 . Predict the relative energies of the molecular orbitals based on how close in energy the valence atomic orbitals are to one another. Draw the molecular orbital energy-level diagram for this system and determine the total number of valence electrons in S 2 . Fill the molecular orbitals in order of increasing energy, being sure to obey the Pauli principle and Hund’s rule. Calculate the bond order and describe the bonding. Solution: A Sulfur has a [Ne]3 s 2 3 p 4 valence electron configuration. To create a molecular orbital energy-level diagram similar to those in Figure $\PageIndex{1}$ and Figure $\PageIndex{3}$, we need to know how close in energy the 3 s and 3 p atomic orbitals are because their energy separation will determine whether the $ \pi _{3p_{x,y}} $ or the $ \sigma _{3p_{z}} $ > molecular orbital is higher in energy. Because the ns – np energy gap increases as the nuclear charge increases (Figure $\PageIndex{3}$), the $ \sigma _{3p_{z}} $ molecular orbital will be lower in energy than the $ \pi _{3p_{x,y}} $ pair. B The molecular orbital energy-level diagram is as follows: Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. C Ten valence electrons are used to fill the orbitals through $ \pi _{3p_{x}} $ and $ \pi _{3p_{y}} $, leaving 2 electrons to occupy the degenerate $ \pi ^{\star }_{3p_{x}} $ and $ \pi ^{\star }_{3p_{y}} $ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of S 2 is $ \left ( \sigma _{3s} \right )^{2}\left ( \sigma ^{\star }_{3s} \right )^{2}\left ( \sigma _{3p_{z}} \right )^{2}\left ( \pi _{3p_{x,y}} \right )^{4}\left ( \pi _{3p ^{\star }_{x,y}} \right )^{2} $ with 2 unpaired electrons. The bond order is (8 − 4) ÷ 2 = 2, so we predict an S=S double bond. Exercise $\PageIndex{3}$: The Peroxide Ion Use a qualitative molecular orbital energy-level diagram to predict the electron configuration, the bond order, and the number of unpaired electrons in the peroxide ion (O 2 2 − ). Answer $ \left ( \sigma _{2s} \right )^{2}\left ( \sigma ^{\star }_{2s} \right )^{2}\left ( \sigma _{2p_{z}} \right )^{2}\left ( \pi _{2p_{x,y}} \right )^{4}\left ( \pi _{2p ^{\star }_{x,y}} \right )^{4} $ bond order of 1; no unpaired electrons Molecular Orbitals for Heteronuclear Diatomic Molecules Diatomic molecules with two different atoms are called heteronuclear diatomic molecules. When two nonidentical atoms interact to form a chemical bond, the interacting atomic orbitals do not have the same energy. If, for example, element B is more electronegative than element A (χ B > χ A ), the net result is a “skewed” molecular orbital energy-level diagram, such as the one shown for a hypothetical A–B molecule in Figure $\PageIndex{4}$. The atomic orbitals of element B are uniformly lower in energy than the corresponding atomic orbitals of element A because of the enhanced stability of the electrons in element B. The molecular orbitals are no longer symmetrical, and the energies of the bonding molecular orbitals are more similar to those of the atomic orbitals of B. Hence the electron density of bonding electrons is likely to be closer to the more electronegative atom. In this way, molecular orbital theory can describe a polar covalent bond. A molecular orbital energy-level diagram is always skewed toward the more electronegative atom. An Odd Number of Valence Electrons: NO Nitric oxide (NO) is an example of a heteronuclear diatomic molecule. The reaction of O 2 with N 2 at high temperatures in internal combustion engines forms nitric oxide, which undergoes a complex reaction with O 2 to produce NO 2 , which in turn is responsible for the brown color we associate with air pollution. Recently, however, nitric oxide has also been recognized to be a vital biological messenger involved in regulating blood pressure and long-term memory in mammals. Because NO has an odd number of valence electrons (5 from nitrogen and 6 from oxygen, for a total of 11), its bonding and properties cannot be successfully explained by either the Lewis electron-pair approach or valence bond theory. The molecular orbital energy-level diagram for NO (Figure $\PageIndex{13}$) shows that the general pattern is similar to that for the O 2 molecule (Figure $\PageIndex{11}$). Because 10 electrons are sufficient to fill all the bonding molecular orbitals derived from 2 p atomic orbitals, the 11th electron must occupy one of the degenerate π * orbitals. The predicted bond order for NO is therefore (8-3) ÷ 2 = 2 1/2 . Experimental data, showing an N–O bond length of 115 pm and N–O bond energy of 631 kJ/mol, are consistent with this description. These values lie between those of the N 2 and O 2 molecules, which have triple and double bonds, respectively. As we stated earlier, molecular orbital theory can therefore explain the bonding in molecules with an odd number of electrons, such as NO, whereas Lewis electron structures cannot. Note that electronic structure studies show the ground state configuration of $\ce{NO}$ to be $ \left ( \sigma _{2s} \right)^{2}\left ( \sigma ^{\star }_{2s} \right)^{2}\left ( \pi _{2p_{x,y}} \right)^{4} \left ( \sigma _{2p_{z}} \right)^{2} \left ( \pi _{2p ^{\star }_{x,y}} \right)^{1} $ in order of increasing energy. Hence, the $ \pi _{2p_{x,y}}$ orbitals are lower in energy than the $\sigma _{2p_{z}}$ orbital. This is because the $\ce{NO}$ molecule is near the transition of flipping energies levels observed in homonuclear diatomics where the sigma bond drops below the pi bond ( Figure $\PageIndex{11}$). Molecular orbital theory can also tell us something about the chemistry of $NO$. As indicated in the energy-level diagram in Figure $\PageIndex{13}$, NO has a single electron in a relatively high-energy molecular orbital. We might therefore expect it to have similar reactivity as alkali metals such as Li and Na with their single valence electrons. In fact, $NO$ is easily oxidized to the $NO^+$ cation, which is isoelectronic with $N_2$ and has a bond order of 3, corresponding to an N≡O triple bond. Nonbonding Molecular Orbitals Molecular orbital theory is also able to explain the presence of lone pairs of electrons. Consider, for example, the HCl molecule, whose Lewis electron structure has three lone pairs of electrons on the chlorine atom. Using the molecular orbital approach to describe the bonding in HCl, we can see from Figure $\PageIndex{6}$ that the 1 s orbital of atomic hydrogen is closest in energy to the 3 p orbitals of chlorine. Consequently, the filled Cl 3 s atomic orbital is not involved in bonding to any appreciable extent, and the only important interactions are those between the H 1 s and Cl 3 p orbitals. Of the three p orbitals, only one, designated as 3 p z , can interact with the H 1 s orbital. The 3 p x and 3 p y atomic orbitals have no net overlap with the 1 s orbital on hydrogen, so they are not involved in bonding. Because the energies of the Cl 3 s , 3 p x , and 3 p y orbitals do not change when HCl forms, they are called nonbonding molecular orbitals . A nonbonding molecular orbital occupied by a pair of electrons is the molecular orbital equivalent of a lone pair of electrons. By definition, electrons in nonbonding orbitals have no effect on bond order, so they are not counted in the calculation of bond order. Thus the predicted bond order of HCl is (2 − 0) ÷ 2 = 1. Because the σ bonding molecular orbital is closer in energy to the Cl 3 p z than to the H 1 s atomic orbital, the electrons in the σ orbital are concentrated closer to the chlorine atom than to hydrogen. A molecular orbital approach to bonding can therefore be used to describe the polarization of the H–Cl bond to give $ H^{\delta +} -- Cl^{\delta -} $. Electrons in nonbonding molecular orbitals have no effect on bond order. Example $\PageIndex{4}$: The Cyanide Ion Use a “skewed” molecular orbital energy-level diagram like the one in Figure $\PageIndex{4}$ to describe the bonding in the cyanide ion (CN − ). What is the bond order? Given: chemical species Asked for: “skewed” molecular orbital energy-level diagram, bonding description, and bond order Strategy: Calculate the total number of valence electrons in CN − . Then place these electrons in a molecular orbital energy-level diagram like Figure $\PageIndex{4}$ in order of increasing energy. Be sure to obey the Pauli principle and Hund’s rule while doing so. Calculate the bond order and describe the bonding in CN − . Solution: A The CN − ion has a total of 10 valence electrons: 4 from C, 5 from N, and 1 for the −1 charge. Placing these electrons in an energy-level diagram like Figure $\PageIndex{4}$ fills the five lowest-energy orbitals, as shown here: Because $\chi_N > \chi_C$, the atomic orbitals of N (on the right) are lower in energy than those of C. B The resulting valence electron configuration gives a predicted bond order of (8 − 2) ÷ 2 = 3, indicating that the CN − ion has a triple bond, analogous to that in N 2 . Exercise $\PageIndex{4}$: The Hypochlorite Ion Use a qualitative molecular orbital energy-level diagram to describe the bonding in the hypochlorite ion (OCl − ). What is the bond order? Answer All molecular orbitals except the highest-energy σ* are filled, giving a bond order of 1. Although the molecular orbital approach reveals a great deal about the bonding in a given molecule, the procedure quickly becomes computationally intensive for molecules of even moderate complexity. Furthermore, because the computed molecular orbitals extend over the entire molecule, they are often difficult to represent in a way that is easy to visualize. Therefore we do not use a pure molecular orbital approach to describe the bonding in molecules or ions with more than two atoms. Instead, we use a valence bond approach and a molecular orbital approach to explain, among other things, the concept of resonance, which cannot adequately be explained using other methods. Summary Molecular orbital energy-level diagrams for diatomic molecules can be created if the electron configuration of the parent atoms is known, following a few simple rules. Most important, the number of molecular orbitals in a molecule is the same as the number of atomic orbitals that interact. The difference between bonding and antibonding molecular orbital combinations is proportional to the overlap of the parent orbitals and decreases as the energy difference between the parent atomic orbitals increases. With such an approach, the electronic structures of virtually all commonly encountered homonuclear diatomic molecules , molecules with two identical atoms, can be understood. The molecular orbital approach correctly predicts that the O 2 molecule has two unpaired electrons and hence is attracted into a magnetic field. In contrast, most substances have only paired electrons. A similar procedure can be applied to molecules with two dissimilar atoms, called heteronuclear diatomic molecules , using a molecular orbital energy-level diagram that is skewed or tilted toward the more electronegative element. Molecular orbital theory is able to describe the bonding in a molecule with an odd number of electrons such as NO and even to predict something about its chemistry.
Courses/Duke_University/Textbook%3A_Modern_Applications_of_Chemistry_(Cox)/01%3A_Primer/1.01%3A_Unit_IV-_Electronic_Structure_and_Bonding/1.1.03%3A_Chemical_Bonding_I-_Basic_Concepts/1.1.3.08%3A_Bond_Order_and_Bond_Lengths	Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased. Introduction Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward. Bond Order Bond order is the number of bonding pairs of electrons between two atoms. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. To determine the bond order between two covalently bonded atoms, follow these steps: Draw the Lewis structure. Determine the type of bonds between the two atoms. 0: No bond 1: Single bond 2: double bond 3: triple bond If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have resonance bonding , the bond order does not need to be an integer. Example $\PageIndex{1}$: $CN^-$ Determine the bond order for cyanide, CN - . Solution 1) Draw the Lewis structure. 2) Determine the type of bond between the two atoms. Because there are 3 dashes, the bond is a triple bond. A triple bond corresponds to a bond order of 3. Example $\PageIndex{2}$: $H_2$ Determine the bond order for hydrogen gas, H 2 . Solution 1) Draw the Lewis structure. 2) Determine the type of bond between the two atoms. There is only one pair of shared electrons (or dash), indicating is a single bond, with a bond order of 1. Polyatomic molecules If there are more than two atoms in the molecule, follow these steps to determine the bond order: Draw the Lewis structure. Count the total number of bonds. Count the number of bond groups between individual atoms. Divide the number of bonds between atoms by the total number of bond groups in the molecule. Example $\PageIndex{3}$: $NO_3^-$ Determine the bond order for nitrate, $NO_3^-$. Solution 1) Draw the Lewis structure. 2) Count the total number of bonds. 4 The total number of bonds is 4. 3) Count the number of bond groups between individual atoms. 3 The number of bond groups between individual atoms is 3. 4) Divide the number of bonds between individual atoms by the total number of bonds. \[\dfrac{4}{3}= 1.33 \] The bond order is 1.33 Example $\PageIndex{4}$: $NO^+_2$ Determine the bond order for nitronium ion: $NO_2^+$. Solution 1) Draw the Lewis Structure. 2) Count the total number of bonds. 4 The total number of bonds is 4. 3) Count the number of bond groups between individual atoms. 2 The number of bond groups between atoms is 2. 4) Divide the bond groups between individual atoms by the total number of bonds. \[\frac{4}{2} = 2\] The bond order is 2. A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability. Trends in the Periodic Table Bond order increases across a period and decreases down a group. Bond Length Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond. To find the bond length, follow these steps: Draw the Lewis structure. Look up the chart below for the radii for the corresponding bond. Find the sum of the two radii. 4 Determine the carbon-to-chlorine bond length in CCl 4 . Solution Using Table A3 , a C single bond has a length of 75 picometers and that a Cl single bond has a length of 99 picometers. When added together, the bond length of a C-Cl bond is approximately 174 picometers. 2 Determine the carbon-oxygen bond length in CO 2 . Solution Using Table A3 , we see that a C double bond has a length of 67 picometers and that an O double bond has a length of 57 picometers. When added together, the bond length of a C=O bond is approximately 124 picometers. Trends in the Periodic Table Because the bond length is proportional to the atomic radius , the bond length trends in the periodic table follow the same trends as atomic radii: bond length decreases across a period and increases down a group. Problems What is the bond order of $O_2$? What is the bond order of $NO_3^-$? What is the carbon-nitrogen bond length in $HCN$? Is the carbon-to-oxygen bond length greater in $CO_2$ or $CO$? What is the nitrogen-fluoride bond length in $NF_3$? Solutions 1. First, write the Lewis structure for $O_2$. There is a double bond between the two oxygen atoms; therefore, the bond order of the molecule is 2. 2. The Lewis structure for NO 3 - is given below: To find the bond order of this molecule, take the average of the bond orders. N=O has a bond order of two, and both N-O bonds have a bond order of one. Adding these together and dividing by the number of bonds (3) reveals that the bond order of nitrate is 1.33. 3. To find the carbon-nitrogen bond length in HCN, draw the Lewis structure of HCN. The bond between carbon and nitrogen is a triple bond, and a triple bond between carbon and nitrogen has a bond length of approximately 60 + 54 =114 pm. 4. From the Lewis structures for CO 2 and CO, there is a double bond between the carbon and oxygen in CO 2 and a triple bond between the carbon and oxygen in CO. Referring to the table above, a double bond between carbon and oxygen has a bond length of approximately 67 + 57 = 124 pm and a triple bond between carbon and oxygen has a bond length of approximately 60 + 53 =113 pm. Therefore, the bond length is greater in CO 2 . Another method makes use of the fact that the more electron bonds between the atoms, the tighter the electrons are pulling the atoms together. Therefore, the bond length is greater in CO 2 . 5. To find the nitrogen-to-fluorine bond length in NF 3 , draw the Lewis structure. The bond between fluorine and nitrogen is a single bond. From the table above, a single bond between fluorine and nitrogen has a bond length of approximately 64 + 71 =135 pm. References Campbell, Neil A., Brad Williamson, and Robin J. Heyden. Biology: Exploring Life . Boston, Massachusetts: Pearson Prentice Hall, 2006. Petrucci, Ralph H., Harwood, William S., Herring, F. G., and Madura Jeffrey D. General Chemistry: Principles & Modern Applications . 9th Ed. New Jersey: Pearson Education, Inc., 2007. Print. Cordero, Beatriz, Verónica Gómez, Ana E. Platero-Prats, Marc Revés, Jorge Echeverría, Eduard Cremades, Flavia Barragán and Santiago Alvarez. Dalton's Transactions." Covalent radii revisited 2008: Pekka Pyykkö and Michiko Atsumi, Chem. Eur. J. Molecular Double-Bond Covalent Radii for Elements Li–E112 2009
Courses/University_of_British_Columbia/UBC_CHEM_154%3A_Chemistry_for_Engineering/02%3A_The_Periodic_Table_and_Atomic_Structure	Template:HideTOC In this chapter, we describe how electrons are arranged in atoms and how the spatial arrangements of electrons are related to their energies. We also explain how knowing the arrangement of electrons in an atom enables chemists to predict and explain the chemistry of an element. As you study the material presented in this chapter, you will discover how the shape of the periodic table reflects the electronic arrangements of elements. In this and subsequent chapters, we build on this information to explain why certain chemical changes occur and others do not. After reading this chapter, you will know enough about the theory of the electronic structure of atoms to explain what causes the characteristic colors of neon signs, how laser beams are created, and why gemstones and fireworks have such brilliant colors. In later chapters, we will develop the concepts introduced here to explain why the only compound formed by sodium and chlorine is NaCl, an ionic compound, whereas neon and argon do not form any stable compounds, and why carbon and hydrogen combine to form an almost endless array of covalent compounds, such as CH 4 , C 2 H 2 , C 2 H 4 , and C 2 H 6 . You will discover that knowing how to use the periodic table is the single most important skill you can acquire to understand the incredible chemical diversity of the elements.
Courses/University_of_Alberta_Augustana_Campus/AUCHE_250%3A_Organic_Chemistry_I/07%3A_Eliminations/7.03%3A_Sequence_Rules_-_The_EZ_Designation	Key Terms Make certain that you can define, and use in context, the key term below. sequence rules (Cahn-Ingold-Prelog rules) Study Notes The limitations of the cis/t rans system are illustrated in the examples given below. From your study of the IUPAC system, you should be able to identify this compound as 4-ethyl-3-methyl-3-heptene, but is it cis or trans ? At first you might say cis , because it appears that two ethyl groups appear on the same side of the double bond. However, the correct answer is trans . The rule is that the designation cis or trans must correspond to the configuration of the longest carbon chain. Tracing out the seven-carbon chain in the compound shown above, you change sides as you pass through the double bond: So, the full name for this compound is trans -4-ethyl-3methyl-3-heptene. The cis / trans system breaks down completely in a compound such as that shown below. The E / Z system, which is the subject of this section, is designed to accommodate such situations. In cases where two or more double bonds are present, you must be prepared to assign an E or Z designation to each of the double bonds. For example: Another use for these sequence rules will be part of the discussion of optical isomerism in Section 9.5. E/ Z nomenclature When each carbon in a double bond is attached to a hydrogen and and a non-hydrogen substituent, the geometric isomers can be identified by using the cis-trans nomenclature discussed in the previous section. However, when a double bond is attached to three or four non-hydrogen substituents there are some examples where cis-trans nomenclature is ineffective in describing the substituents orientation in geometric isomers. In these situations the rigorous IUPAC system for naming alkene isomers, called the E/Z system, is used. The E/Z system analyzes the two substituents attached to each carbon in the double bond and assigns each either a high or low priority. If the higher priority group on both carbons in the double bond the same side the alkene is said to have a Z isomer (from German zusammen = together). You could think of Z as Zame Zide to help memorize it. If the higher priority group on opposite sides the alkene has an E isomer (from German entgegen = opposite). Note, if both substituents an a double bond carbon are exactly the same there is no E/Z isomerizem is possible. Also, if E/Z isomerism is possible, interchanging the substituents attached on double-bond carbon converts one isomer to the other. Substituent priority for the E,Z system is assigned using the Cahn–Ingold–Prelog (CIP) sequence rules. These are the same rules used to assign R/S configurations to chiral centers in Section 5.5 . A brief overview of using CIP rules to determine alkene configuration is given here but CIP rules are discussed in greater detail in Section 5.5 . Note The priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system Rule 1) The "first point of difference" rule First, determine the two stubstitents on each double-bond carbon separately. Rank these substituents based on the atom which directly attached to the double-bond carbon. The substituent whose atom has a higher atomic number takes precedence over the substituent whose atom has a lower atomic number. Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins. To illustrate this, consider the molecule at the left. Is the double bond here E or Z ? At the left end of the double bond, Br > C. But the right end of the double bond requires a careful analysis. At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins". Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH 2 -O-CH 3 ). This means that the isomer shown is opposite = entgegen = E . And what is the name? The "name" feature of ChemSketch says it is (2 E )-2-(1-bromoethylidene)pent-3-ynyl methyl ether. Rule 2) If the first atom on both substituents are the identical, then proceed along both substituent chains until the first point of difference is determined. An easy example which shows the necessity of the E/Z system is the alkene, 1-bromo-2-chloro-2-fluoro-1-iodoethene, which has four different substituents attached to the double bond. The figure below shows that there are two distinctly different geometric isomers for this molecule neither of which can be named using the cis-trans system. Consider the left hand structure. On the double bond carbon on the left, the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at carbon on the right. The attached atoms are Cl and F, with Cl having the higher atomic number and the higher priority. When considering the relative positions of the higher priority groups, the higher priority group is "down" on the left double bond carbon and "down" at right double bond carbon. Since the two higher priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E). Example 7.5.1: Butene 0 1 cis-2-butene (Z)-2-butene trans-2-butene (E)-2-butene The Figure above shows the two isomers of 2-butene. You should recognize them as cis and trans . Let's analyze them to see whether they are E or Z . Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is ( Z )-2-butene. Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on opposite sides of the double bond, they are entgegen = opposite. Therefore, this is ( E )-2-butene. E/ Z will work – even when cis / trans fails In simple cases, such as 2-butene, Z corresponds to cis and E to trans . However, that is not a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the E / Z system is that it will always work. In contrast, the cis / trans system breaks down with many ambiguous cases. Example 7.5.2 The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene. 0 1 (Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene (E)-1-bromo-2-chloro-2-fluoro-1-iodoethene It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as cis or trans . On the other hand, the E / Z system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the ( Z ) isomer. Similarly, the right hand structure is ( E ). E/ Z will work, but may not agree with cis / trans Consider the molecule shown at the left. This is 2-bromo-2-butene -- ignoring the geometric isomerism for now. Cis or trans ? This molecule is clearly cis . The two methyl groups are on the same side. More rigorously, the "parent chain" is cis . E or Z ? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group -- because the other group is -H. On the right, the methyl is the low priority group -- because the other group is -Br. That is, the high priority groups are -CH 3 (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = E . Note This example should convince you that cis and Z are not synonyms. Cis/ trans and E / Z are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine cis / trans or E / Z separately, as needed. Multiple double bonds If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z . Example 7.5.3 The configuration at the left hand double bond is E ; at the right hand double bond it is Z . Thus this compound is (1 E ,4 Z )-1,5-dichloro-1,4-hexadiene. The double-bond rule in determining priorities Example 7.5.4 Consider the compound below This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end. What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH 2 -CH 3 (an ethyl group) and -CH=CH 2 (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z -isomer; the compound is ( Z )-1-chloro-2-ethyl-1,3-butadiene. Example 7.5.5 The configuration about double bonds is undoubtedly best specified by the cis / trans notation when there is no ambiguity involved. Unfortunately, many compounds cannot be described adequately by the cis / trans system. Consider, for example, configurational isomers of 1-fluoro-1-chloro-2-bromo-2-iodo-ethene, 9 and 10. There is no obvious way in which the cis / trans system can be used: A system that is easy to use and which is based on the sequence rules already described for the R , S system works as follows: An order of precedence is established for the two atoms or groups attached to each end of the double bond according to the sequence rules of Section 19-6. When these rules are applied to 1-fluoro- 1-chloro-2-bromo-2- iodoethene, the priority sequence is: at carbon atom 1, C1 > F at carbon atom 2, I > Br Examination of the two configurations shows that the two priority groups- one on each end- are either on the same side of the double bond or on opposite sides: 0 1 priority groups on opposite sides (E) configuration priority groups on same side (Z) configuration The Z isomer is designated as the isomer in which the top priority groups are on the same side ( Z is taken from the German word zusammen- together). The E isomer has these groups on opposite sides ( E , German for entgegen across). Two further examples show how the nomenclature is used: 0 1 (Z)-1-chloro-1-butene (1Z,3E)-1,3-butadiene-1,4-d2 Exercises
Courses/BethuneCookman_University/BCU%3A_CH_332_Physical_Chemistry_II/Text/14%3A_Nuclear_Magnetic_Resonance_Spectroscopy	Sorry, but all current instructors at UCD use the commercial textbook for this course. Nuclear magnetic resonance (NMR) is a physical phenomenon in which nuclei in a magnetic field absorb and re-emit electromagnetic radiation. This energy is at a specific resonance frequency which depends on the strength of the magnetic field and the magnetic properties of the isotope of the atoms. Many scientific techniques exploit NMR phenomena to study molecular physics, crystals, and non-crystalline materials through nuclear magnetic resonance spectroscopy. NMR is also routinely used in advanced medical imaging techniques, such as in magnetic resonance imaging (MRI).
Courses/University_of_South_Carolina__Upstate/CHEM_U109%3A_Chemistry_of_Living_Things_-_Mueller/01%3A_Chemistry_Matter_and_Measurement/1.5%3A_Expressing_Numbers%3A_Significant_Figures	Skills to Develop Identify the number of significant figures in a reported value. Use significant figures correctly in arithmetical operations. Scientists have established certain conventions for communicating the degree of precision of a measurement. Imagine, for example, that you are using a meterstick to measure the width of a table. The centimeters (cm) are marked off, telling you how many centimeters wide the table is. Many metersticks also have millimeters (mm) marked off, so we can measure the table to the nearest millimeter. But most metersticks do not have any finer measurements indicated, so you cannot report the table’s width any more exactly than to the nearest millimeter. All you can do is estimate the next decimal place in the measurement (Figure $\PageIndex{1}$). Figure $\PageIndex{1}$ : Measuring an Object to the Correct Number of Digits. How many digits should be reported for the length of this object? The concept of significant figures takes this limitation into account.The significant figures of a measured quantity are defined as all the digits known with certainty and the first uncertain, or estimated, digit. It makes no sense to report any digits after the first uncertain one, so it is the last digit reported in a measurement. Zeros are used when needed to place the significant figures in their correct positions. Thus, zeros may not be significant figures. “ Sig figs ” is a common abbreviation for significant figures. Your instructor might even write " s f " on a graded assignment as a reminder about significant figures. For example, if a table is measured and reported as being 1,357 mm wide, the number 1,357 has four significant figures. The 1 (thousands), the 3 (hundreds), and the 5 (tens) are certain; the 7 (units) is assumed to have been estimated. It would make no sense to report such a measurement as 1,357.0 or 1,357.00 because that would suggest the measuring instrument was able to determine the width to the nearest tenth or hundredth of a millimeter, when in fact it shows only tens of millimeters and the units have to be estimated. On the other hand, if a measurement is reported as 150 mm, the 1 (hundreds) and the 5 (tens) are known to be significant, but how do we know whether the zero is or is not significant? The measuring instrument could have had marks indicating every 10 mm or marks indicating every 1 mm. Is the zero an estimate, or is the 5 an estimate and the zero a placeholder? The rules for deciding which digits in a measurement are significant are as follows: All nonzero digits are significant. In 1,357, all the digits are significant. Captive (or embedded) zeros , which are zeros between significant digits, are significant. In 405, all the digits are significant. Leading zeros , which are zeros at the beginning of a decimal number less than 1, are not significant. In 0.000458, the first four digits are leading zeros and are not significant. The zeros serve only to put the digits 4, 5, and 8 in the correct positions. This number has three significant figures. Trailing zeros , which are zeros at the end of a number, are significant only if the number has a decimal point. Thus, in 1,500, the two trailing zeros are not significant because the number is written without a decimal point; the number has two significant figures. However, in 1,500.00, all six digits are significant because the number has a decimal point. Example $\PageIndex{1}$ How many significant digits does each number have? 6,798,000 6,000,798 6,000,798.00 0.0006798 SOLUTION four (by rules 1 and 4) seven (by rules 1 and 2) nine (by rules 1, 2, and 4) four (by rules 1 and 3) Combining Numbers It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division. For addition or subtraction, the rule is to stack all the numbers with their decimal points aligned and then limit the answer’s significant figures to the rightmost column for which all the numbers have significant figures. Consider the following: The arrow points to the rightmost column in which all the numbers have significant figures—in this case, the tenths place. Therefore, we will limit our final answer to the tenths place. Is our final answer therefore 1,459.0? No, because when we drop digits from the end of a number, we also have to round the number. Notice that the second dropped digit, in the hundredths place, is 8. This suggests that the answer is actually closer to 1,459.1 than it is to 1,459.0, so we need to round up to 1,459.1. The rules in rounding are simple: If the first dropped digit is 5 or higher, round up. If the first dropped digit is lower than 5, do not round up. For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows: The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up. Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 10 2 , whereas 450.0 has four significant figures and would be written as 4.500 × 10 2 . In scientific notation, all significant figures are listed explicitly. Example $\PageIndex{2}$ Write the answer for each expression using scientific notation with the appropriate number of significant figures. 23.096 × 90.300 125 × 9.000 1,027 + 610 + 363.06 14.84 x 0.55 – 8.02 SOLUTION The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the hundredths place) is greater than 5, we round up to 2,085.6, which in scientific notation is 2.0856 × 10 3 . The calculator gives 1,125 as the answer, but we limit it to three significant figures and convert into scientific notation: 1.13 × 10 3 . (1130 is also technically three significant figures also, but scientific notation is more clear.) The calculator gives 2,000.06 as the answer, but because 610 has its farthest-right significant figure in the tens column, our answer must be limited to the tens position: 2.00 × 10 3 . (In 200 0.06 the underlined one's, tenth's, and hundredth's places are not significant.) This involves two different rules, one for the multiplication and one for the subtraction, so the sig. fig.s must be evaluated after each step. Here the digits that are not significant will be underlined, then the answer will be rounded at the end. Algebra rules dictate that the multiplication be done before subtraction. There are four significant figures in 14.84 and two significant figures in 0.55, so the answer is limited to two sig.figs. 14.84 x 0.55 = 8.1 62 When subtracting we limit by decimal places, not total significant figures. The result of the multiplication, 8.1 62 , is only known to one spot to the right of the decimal place. (14.84 x 0.55) – 8.02 = (8.1 62 ) – 8.02 = 0.1 42 = 0.1 The final answer, 0.1, should only be listed to one decimal place. NOTE ON ROUNDING. It is most accurate to round your answer at the very end (while still keeping track of the significant figures at each step), like done by underlining the digits that are not significant in Part d above. While slightly less accurate, it is acceptable to round at each step if it helps you keep track of significant figures as you go. Notice that the answer to Part d changes slightly here: d. (14.84 x 0.55) – 8.02 = (8.1 62 ) – 8.02 ≈ (8.2) – 8.02 = 0.1 8 ≈ 0.2 The "≈" symbol means "approximately equal to" and is used here when rounding early. Concept Review Exercises Explain why the concept of significant figures is important in numerical calculations. State the rules for determining the significant figures in a measurement. When do you round a number up, and when do you not round a number up? Answers Significant figures represent all the known digits of a measurement plus the first estimated one. All nonzero digits are significant; zeros between nonzero digits are significant; zeros at the end of a nondecimal number or the beginning of a decimal number are not significant; zeros at the end of a decimal number are significant. Round up only if the first digit dropped is 5 or higher. Key Takeaways Significant figures properly report the number of measured and estimated digits in a measurement. There are rules for applying significant figures in calculations. Contributors Anonymous
Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Appendix_I%3A_Index_of_enzymatic_reactions_by_pathway	While Organic Chemistry With a Biological Emphasis is organized, like most sophomore-level organic chemistry texts, around a structural and mechanistic framework, students of biochemistry may often want to clarify the mechanism of an enzymatic reaction which they encounter when studying the central metabolic pathways. An excellent resource for this purpose is John McMurry's The Organic Chemistry of Biological Pathways (Roberts and Company, 2005), but also helpful will be this index of enzymatic reactions organized by pathway, with links to sections/problems in this text where the reaction mechanism is addressed. Sections/problems listed with an asterisk (*) do not discuss the exact reaction indicated, but do discuss a closely related reaction. EC numbers are provided whenever possible, with links to the corresponding page in the BRENDA database of enzymes. Clicking on the 'reaction flask' icon on a BRENDA page brings up the reaction diagram. Pathways Amino acid biosynthesis Amino acid catabolism Citric Acid Cycle Fatty acid metabolism Glycolysis, Gluconeogenesis, Fermentation Isoprenoid biosynthesis Nucleoside biosynthesis Nucleotide catabolism Pentose Phosphate Pathway, Calvin Cycle Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/08%3A_Alkenes_-_Reactions_and_Synthesis/8.10%3A_Radical_Additions_to_Alkenes_-__Chain-Growth_Polymers	In our brief introduction to radical reactions in Section 6.7, we said that radicals can add to $\ce{C=C}$ bonds, taking one electron from the double bond and leaving one behind to yield a new radical. Let’s now look at the process in more detail, focusing on the industrial synthesis of alkene polymers. A polymer is a large—sometimes very large—molecule, built up by repetitive joining together of many smaller molecules, called monomers . Nature makes wide use of biological polymers. Cellulose, for instance, is a polymer built of repeating glucose monomer units; proteins are polymers built of repeating amino acid monomers; and nucleic acids are polymers built of repeating nucleotide monomers. Synthetic polymers, such as polyethylene, are much simpler chemically than biopolymers, but there is still a great diversity to their structures and properties, depending on the identity of the monomers and on the reaction conditions used for polymerization. The simplest synthetic polymers are those that result when an alkene is treated with a small amount of a suitable catalyst. Ethylene, for example, yields polyethylene, an enormous alkane that may have a molecular weight up to 6 million u and may contain as many as 200,000 monomer units. Worldwide production of polyethylene is approximately 88 million tons per year. Polyethylene and other simple alkene polymers are called chain-growth polymers because they are formed in a chain-reaction process in which an initiator adds to a carbon–carbon double bond to yield a reactive intermediate. The intermediate then reacts with a second molecule of monomer to yield a new intermediate, which reacts with a third monomer unit, and so on. Historically, ethylene polymerization was carried out at high pressure (1000–3000 atm) and high temperature (100–250 °C) in the presence of a radical initiator such as benzoyl peroxide. Like many radical reactions, the mechanism of ethylene polymerization occurs in three steps: initiation, propagation, and termination: Initiation: The polymerization reaction is initiated when a few radicals are generated on heating a small amount of benzoyl peroxide catalyst to break the weak O−O bond. The initially formed benzoyloxy radical loses $\ce{CO2}$ and gives a phenyl radical (Ph·), which adds to the $\ce{C=C}$ bond of ethylene to start the polymerization process. One electron from the ethylene double bond pairs up with the odd electron on the phenyl radical to form a new C−C bond, and the other electron remains on carbon. Propagation: Polymerization occurs when the carbon radical formed in the initiation step adds to another ethylene molecule to yield another radical. Repetition of the process for hundreds or thousands of times builds the polymer chain. Termination: The chain process is eventually ended by a reaction that consumes the radical. The combination of two growing chains is one possible chain-terminating reaction. \[2 \mathrm{R}-\mathrm{CH}_2 \mathrm{CH}_2 \cdot \longrightarrow \mathrm{R}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{R} \nonumber\] Ethylene is not unique in its ability to form a polymer. Many substituted ethylenes, called vinyl monomers, also undergo polymerization to yield polymers with substituent groups regularly spaced on alternating carbon atoms along the chain. Propylene, for example, yields polypropylene, and styrene yields polystyrene. When an unsymmetrically substituted vinyl monomer such as propylene or styrene is polymerized, the radical addition steps can take place at either end of the double bond to yield either a primary radical intermediate (RCH 2 · ) or a secondary radical (R 2 CH · ). Just as in electrophilic addition reactions, however, we find that only the more highly substituted, secondary radical is formed. Figure $\PageIndex{1}$ shows some commercially important alkene polymers, their uses, and the monomers from which they are made. Monomer Formula Trade or common name of polymer Uses Ethylene $\ce{H2C═CH2}$ Polyethylene Packaging, bottles Propene (propylene) $\ce{H2C═CHCH3}$ Polypropylene Moldings, rope, carpets Chloroethylene (vinyl chloride) $\ce{H2C═CHCl}$ Poly(vinyl chloride) Insulation, films, pipes Styrene $\ce{H2C═CHC6H5}$ Polystyrene Foam, moldings Tetrafluoroethylene $\ce{F2C═CF2}$ Teflon Gaskets, nonstick coatings Acrylonitrile $\ce{H2C═CHCN}$ Orlon, Acrilan Fibers Methyl methacrylate NaN Plexiglas, Lucite Paint, sheets, moldings Vinyl acetate $\ce{H2C═CHOCOCH3}$ Poly(vinyl acetate) Paint, adhesives, foams Worked Example $\PageIndex{1}$: Predicting the Structure of a Polymer Show the structure of poly(vinyl chloride), a polymer made from H 2 C=CHCl, by drawing several repeating units. Strategy Mentally break the carbon–carbon double bond in the monomer unit, and form single bonds by connecting numerous units together. Solution The general structure of poly(vinyl chloride) is Exercise $\PageIndex{1}$ Show the monomer units you would use to prepare the following polymers: Answer H 2 C ═ CHOCH 3 ClCH ═ CHCl Exercise $\PageIndex{2}$ One of the chain-termination steps that sometimes occurs to interrupt polymerization is the following reaction between two radicals. Propose a mechanism for the reaction, using fishhook arrows to indicate electron flow. Answer
Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/04%3A_Evaluating_Analytical_Data/4.03%3A_Propagation_of_Uncertainty	Suppose we dispense 20 mL of a reagent using the Class A 10-mL pipet whose calibration information is given in Table 4.2.8 . If the volume and uncertainty for one use of the pipet is 9.992 ± 0.006 mL, what is the volume and uncertainty if we use the pipet twice? As a first guess, we might simply add together the volume and the maximum uncertainty for each delivery; thus (9.992 mL + 9.992 mL) ± (0.006 mL + 0.006 mL) = 19.984 ± 0.012 mL It is easy to appreciate that combining uncertainties in this way overestimates the total uncertainty. Adding the uncertainty for the first delivery to that of the second delivery assumes that with each use the indeterminate error is in the same direction and is as large as possible. At the other extreme, we might assume that the uncertainty for one delivery is positive and the other is negative. If we subtract the maximum uncertainties for each delivery, (9.992 mL + 9.992 mL) ± (0.006 mL – 0.006 mL) = 19.984 ± 0.000 mL we clearly underestimate the total uncertainty. So what is the total uncertainty? From the discussion above, we reasonably expect that the total uncertainty is greater than ±0.000 mL and that it is less than ±0.012 mL. To estimate the uncertainty we use a mathematical technique known as the propagation of uncertainty. Our treatment of the propagation of uncertainty is based on a few simple rules. A Few Symbols A propagation of uncertainty allows us to estimate the uncertainty in a result from the uncertainties in the measurements used to calculate that result. For the equations in this section we represent the result with the symbol R , and we represent the measurements with the symbols A , B , and C . The corresponding uncertainties are u R , u A , u B , and u C . We can define the uncertainties for A , B , and C using standard deviations, ranges, or tolerances (or any other measure of uncertainty), as long as we use the same form for all measurements. The requirement that we express each uncertainty in the same way is a critically important point. Suppose you have a range for one measurement, such as a pipet’s tolerance, and standard deviations for the other measurements. All is not lost. There are ways to convert a range to an estimate of the standard deviation. See Appendix 2 for more details. Uncertainty When Adding or Subtracting When we add or subtract measurements we propagate their absolute uncertainties. For example, if the result is given by the equation \[R = A + B - C \nonumber\] the the absolute uncertainty in R is \[u_R = \sqrt{u_A^2 + u_B^2 + u_C^2} \label{4.1}\] If we dispense 20 mL using a 10-mL Class A pipet, what is the total volume dispensed and what is the uncertainty in this volume? First, complete the calculation using the manufacturer’s tolerance of 10.00 mL±0.02 mL, and then using the calibration data from Table 4.2.8 . Solution To calculate the total volume we add the volumes for each use of the pipet. When using the manufacturer’s values, the total volume is \[V = 10.00 \text{ mL} + 10.00 \text{ mL} = 20.00 \text{ mL} \nonumber\] and when using the calibration data, the total volume is \[V = 9.992 \text{ mL} + 9.992 \text{ mL} = 19.984 \text{ mL} \nonumber\] Using the pipet’s tolerance as an estimate of its uncertainty gives the uncertainty in the total volume as \[u_R = \sqrt{(0.02)^2 + (0.02)^2} = 0.028 \text{ mL} = 0.028 \text{ mL} \nonumber\] and using the standard deviation for the data in Table 4.2.8 gives an uncertainty of \[u_R = \sqrt{(0.006)^2 + (0.006)^2} = 0.0085 \text{ mL} \nonumber\] Rounding the volumes to four significant figures gives 20.00 mL ± 0.03 mL when we use the tolerance values, and 19.98 ± 0.01 mL when we use the calibration data. Uncertainty When Multiplying or Dividing When we multiple or divide measurements we propagate their relative uncertainties. For example, if the result is given by the equation \[R = \frac {A \times B} {C} \nonumber\] then the relative uncertainty in R is \[\frac {u_R} {R}= \sqrt{\left( \frac {u_A} {A} \right)^2 + \left( \frac {u_B} {B} \right)^2 + \left( \frac {u_C} {C} \right)^2} \label{4.2}\] The quantity of charge, Q , in coulombs that passes through an electrical circuit is \[Q = i \times t \nonumber\] where i is the current in amperes and t is the time in seconds. When a current of 0.15 A ± 0.01 A passes through the circuit for 120 s ± 1 s, what is the total charge and its uncertainty? Solution The total charge is \[Q = (0.15 \text{ A}) \times (120 \text{ s}) = 18 \text{ C} \nonumber\] Since charge is the product of current and time, the relative uncertainty in the charge is \[\frac {u_R} {R} = \sqrt{\left( \frac {0.01} {0.15} \right)^2 + \left( \frac {1} {120} \right)^2} = 0.0672 \nonumber\] and the charge’s absolute uncertainty is \[u_R = R \times 0.0672 = (18 \text{ C}) \times (0.0672) = 1.2 \text{ C} \nonumber\] Thus, we report the total charge as 18 C ± 1 C. Uncertainty for Mixed Operations Many chemical calculations involve a combination of adding and subtracting, and of multiply and dividing. As shown in the following example, we can calculate the uncertainty by separately treating each operation using Equation \ref{4.1} and Equation \ref{4.2} as needed. For a concentration technique, the relationship between the signal and the an analyte’s concentration is \[S_{total} = k_A C_A + S_{mb} \nonumber\] What is the analyte’s concentration, C A , and its uncertainty if S total is 24.37 ± 0.02, S mb is 0.96 ± 0.02, and k A is $0.186 \pm 0.003 \text{ ppm}^{-1}$? Solution Rearranging the equation and solving for C A \[C_A = \frac {S_{total} - S_{mb}} {k_A} = \frac {24.37 - 0.96} {0.186 \text{ ppm}^{-1}} = \frac {23.41} {0.186 \text{ ppm}^{-1}} = 125.9 \text{ ppm} \nonumber\] gives the analyte’s concentration as 126 ppm. To estimate the uncertainty in C A , we first use Equation \ref{4.1} to determine the uncertainty for the numerator. \[u_R = \sqrt{(0.02)^2 + (0.02)^2} = 0.028 \nonumber\] The numerator, therefore, is 23.41 ± 0.028. To complete the calculation we use Equation \ref{4.2} to estimate the relative uncertainty in C A . \[\frac {u_R} {R} = \sqrt{\left( \frac {0.028} {23.41} \right)^2 + \left( \frac {0.003} {0.186} \right)^2} = 0.0162 \nonumber\] The absolute uncertainty in the analyte’s concentration is \[u_R = (125.9 \text{ ppm}) \times (0.0162) = 2.0 \text{ ppm} \nonumber\] Thus, we report the analyte’s concentration as 126 ppm ± 2 ppm. To prepare a standard solution of Cu 2 + you obtain a piece of copper from a spool of wire. The spool’s initial weight is 74.2991 g and its final weight is 73.3216 g. You place the sample of wire in a 500-mL volumetric flask, dissolve it in 10 mL of HNO 3 , and dilute to volume. Next, you pipet a 1 mL portion to a 250-mL volumetric flask and dilute to volume. What is the final concentration of Cu 2 + in mg/L, and its uncertainty? Assume that the uncertainty in the balance is ±0.1 mg and that you are using Class A glassware. Answer The first step is to determine the concentration of Cu 2 + in the final solution. The mass of copper is \[74.2991 \text{ g} - 73.3216 \text{ g} = 0.9775 \text{ g Cu} \nonumber\] The 10 mL of HNO 3 used to dissolve the copper does not factor into our calculation. The concentration of Cu 2 + is \[\frac {0.9775 \text{ g Cu}} {0.5000 \text{ L}} \times \frac {1.000 \text{ mL}} {250.0 \text{ mL}} \times \frac {1000 \text{ mg}} {\text{g}} = 7.820 \text{ mg } \ce{Cu^{2+}} \text{/L} \nonumber\] Having found the concentration of Cu 2 + , we continue with the propagation of uncertainty. The absolute uncertainty in the mass of Cu wire is \[u_\text{g Cu} = \sqrt{(0.0001)^2 + (0.0001)^2} = 0.00014 \text{ g} \nonumber\] The relative uncertainty in the concentration of Cu 2 + is \[\frac {u_\text{mg/L}} {7.820 \text{ mg/L}} = \sqrt{\left( \frac {0.00014} {0.9775} \right)^2 + \left( \frac {0.20} {500.0} \right)^2 + \left( \frac {0.006} {1.000} \right)^2 + \left( \frac {0.12} {250.0} \right)^2} = 0.00603 \nonumber\] Solving for u mg /L gives the uncertainty as 0.0472. The concentration and uncertainty for Cu 2 + is 7.820 mg/L ± 0.047 mg/L. Uncertainty for Other Mathematical Functions Many other mathematical operations are common in analytical chemistry, including the use of powers, roots, and logarithms. Table 4.3.1 provides equations for propagating uncertainty for some of these function where A and B are independent measurements and where k is a constant whose value has no uncertainty. Function $u_R$ Function.1 $u_R$.1 $R = kA$ $u_R = ku_A$ $R = \ln (A)$ $u_R = \frac {u_A} {A}$ $R = A + B$ $u_R = \sqrt{u_A^2 + u_B^2}$ $R = \log (A)$ $u_R = 0.4343 \times \frac {u_A} {A}$ $R = A - B$ $u_R = \sqrt{u_A^2 + u_B^2}$ $R = e^A$ $\frac {u_R} {R} = u_A$ $R = A \times B$ $\frac {u_R} {R} = \sqrt{\left( \frac {u_A} {A} \right)^2 +\left( \frac {u_B} {B} \right)^2}$ $R = 10^A$ $\frac {u_R} {R} = 2.303 \times u_A$ $R = \frac {A} {B}$ $\frac {u_R} {R} = \sqrt{\left( \frac {u_A} {A} \right)^2 +\left( \frac {u_B} {B} \right)^2}$ $R = A^k$ $\frac {u_R} {R} = k \times \frac {u_A} {A}$ If the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, what is the [H + ] and its uncertainty? Solution The concentration of H + is \[[\ce{H+}] = 10^{-\text{pH}} = 10^{-3.72} = 1.91 \times 10^{-4} \text{ M} \nonumber\] or $1.9 \times 10^{-4}$ M to two significant figures. From Table 4.3.1 the relative uncertainty in [H + ] is \[\frac {u_R} {R} = 2.303 \times u_A = 2.303 \times 0.03 = 0.069 \nonumber\] The uncertainty in the concentration, therefore, is \[(1.91 \times 10^{-4} \text{ M}) \times (0.069) = 1.3 \times 10^{-5} \text{ M} \nonumber\] We report the [H + ] as $1.9 (\pm 0.1) \times 10^{-4}$ M, which is equivalent to $1.9 \times 10^{-4} \text{ M } \pm 0.1 \times 10^{-4} \text{ M}$. A solution of copper ions is blue because it absorbs yellow and orange light. Absorbance, A , is defined as \[A = - \log T = - \log \left( \frac {P} {P_\text{o}} \right) \nonumber\] where, T is the transmittance, P o is the power of radiation as emitted from the light source and P is its power after it passes through the solution. What is the absorbance if P o is $3.80 \times 10^2$ and P is $1.50 \times 10^2$? If the uncertainty in measuring P o and P is 15, what is the uncertainty in the absorbance? Answer The first step is to calculate the absorbance, which is \[A = - \log T = -\log \frac {P} {P_\text{o}} = - \log \frac {1.50 \times 10^2} {3.80 \times 10^2} = 0.4037 \approx 0.404 \nonumber\] Having found the absorbance, we continue with the propagation of uncertainty. First, we find the uncertainty for the ratio P / P o , which is the transmittance, T. \[\frac {u_{T}} {T} = \sqrt{\left( \frac {15} {3.80 \times 10^2} \right)^2 + \left( \frac {15} {1.50 \times 10^2} \right)^2 } = 0.1075 \nonumber\] Finally, from Table 4.3.1 the uncertainty in the absorbance is \[u_A = 0.4343 \times \frac {u_{T}} {T} = (0.4343) \times (0.1075) = 4.669 \times 10^{-2} \nonumber\] The absorbance and uncertainty is 0.40 ± 0.05 absorbance units. Is Calculating Uncertainty Actually Useful? Given the effort it takes to calculate uncertainty, it is worth asking whether such calculations are useful. The short answer is, yes. Let’s consider three examples of how we can use a propagation of uncertainty to help guide the development of an analytical method. One reason to complete a propagation of uncertainty is that we can compare our estimate of the uncertainty to that obtained experimentally. For example, to determine the mass of a penny we measure its mass twice—once to tare the balance at 0.000 g and once to measure the penny’s mass. If the uncertainty in each measurement of mass is ±0.001 g, then we estimate the total uncertainty in the penny’s mass as \[u_R = \sqrt{(0.001)^2 + (0.001)^2} = 0.0014 \text{ g} \nonumber\] If we measure a single penny’s mass several times and obtain a standard deviation of ±0.050 g, then we have evidence that the measurement process is out of control. Knowing this, we can identify and correct the problem. We also can use a propagation of uncertainty to help us decide how to improve an analytical method’s uncertainty. In Example 4.3.3 , for instance, we calculated an analyte’s concentration as 126 ppm ± 2 ppm, which is a percent uncertainty of 1.6%. Suppose we want to decrease the percent uncertainty to no more than 0.8%. How might we accomplish this? Looking back at the calculation, we see that the concentration’s relative uncertainty is determined by the relative uncertainty in the measured signal (corrected for the reagent blank) \[\frac {0.028} {23.41} = 0.0012 \text{ or } 0.12\% \nonumber\] and the relative uncertainty in the method’s sensitivity, k A , \[\frac {0.003 \text{ ppm}^{-1}} {0.186 \text{ ppm}^{-1}} = 0.016 \text{ or } 1.6\% \nonumber\] Of these two terms, the uncertainty in the method’s sensitivity dominates the overall uncertainty. Improving the signal’s uncertainty will not improve the overall uncertainty of the analysis. To achieve an overall uncertainty of 0.8% we must improve the uncertainty in k A to ±0.0015 ppm –1 . Verify that an uncertainty of ±0.0015 ppm –1 for k A is the correct result. Answer An uncertainty of 0.8% is a relative uncertainty in the concentration of 0.008; thus, letting u be the uncertainty in k A \[0.008 = \sqrt{\left( \frac {0.028} {23.41} \right)^2 + \left( \frac {u} {0.186} \right)^2} \nonumber\] Squaring both sides of the equation gives \[6.4 \times 10^{-5} = \left( \frac {0.028} {23.41} \right)^2 + \left( \frac {u} {0.186} \right)^2 \nonumber\] Solving for the uncertainty in k A gives its value as $1.47 \times 10^{-3}$ or ±0.0015 ppm –1 . Finally, we can use a propagation of uncertainty to determine which of several procedures provides the smallest uncertainty. When we dilute a stock solution usually there are several combinations of volumetric glassware that will give the same final concentration. For instance, we can dilute a stock solution by a factor of 10 using a 10-mL pipet and a 100-mL volumetric flask, or using a 25-mL pipet and a 250-mL volumetric flask. We also can accomplish the same dilution in two steps using a 50-mL pipet and 100-mL volumetric flask for the first dilution, and a 10-mL pipet and a 50-mL volumetric flask for the second dilution. The overall uncertainty in the final concentration—and, therefore, the best option for the dilution—depends on the uncertainty of the volumetric pipets and volumetric flasks. As shown in the following example, we can use the tolerance values for volumetric glassware to determine the optimum dilution strategy [Lam, R. B.; Isenhour, T. L. Anal. Chem. 1980 , 52 , 1158–1161]. Which of the following methods for preparing a 0.0010 M solution from a 1.0 M stock solution provides the smallest overall uncertainty? (a) A one-step dilution that uses a 1-mL pipet and a 1000-mL volumetric flask. (b) A two-step dilution that uses a 20-mL pipet and a 1000-mL volumetric flask for the first dilution, and a 25-mL pipet and a 500-mL volumetric flask for the second dilution. Solution The dilution calculations for case (a) and case (b) are \[\text{case (a): 1.0 M } \times \frac {1.000 \text { mL}} {1000.0 \text { mL}} = 0.0010 \text{ M} \nonumber\] \[\text{case (b): 1.0 M } \times \frac {20.00 \text { mL}} {1000.0 \text { mL}} \times \frac {25.00 \text{ mL}} {500.0 \text{mL}} = 0.0010 \text{ M} \nonumber\] Using tolerance values from Table 4.2.1 , the relative uncertainty for case (a) is \[u_R = \sqrt{\left( \frac {0.006} {1.000} \right)^2 + \left( \frac {0.3} {1000.0} \right)^2} = 0.006 \nonumber\] and for case (b) the relative uncertainty is \[u_R = \sqrt{\left( \frac {0.03} {20.00} \right)^2 + \left( \frac {0.3} {1000} \right)^2 + \left( \frac {0.03} {25.00} \right)^2 + \left( \frac {0.2} {500.0} \right)^2} = 0.002 \nonumber\] Since the relative uncertainty for case (b) is less than that for case (a), the two-step dilution provides the smallest overall uncertainty. Of course we must balance the smaller uncertainty for case (b) against the increased opportunity for introducing a determinate error when making two dilutions instead of just one dilution, as in case (a).
Bookshelves/Biological_Chemistry/Chemistry_of_Cooking_(Rodriguez-Velazquez)/04%3A_Sugar/4.01%3A_Sugar_Chemistry_(ADD_US)	Chemically, sugar consists of carbon (C), oxygen (O), and hydrogen (H) atoms, and is classified as a carbohydrate . There are three main groups of sugars, classified according to the way the atoms are arranged together in the molecular structure. These groups are the following: Monosaccharides or simple sugars. Dextrose (glucose) is the major monosaccharide. Others are l evulose or fructose (found in honey and many fruits), and galactose, which is a milk sugar. Such sugars do not readily crystallize. (Mono means one, indicating that the sugar consists of only one molecule.) Disaccharides or complex sugars. Sucrose (common sugar) is the primary example of a disaccharide. Maltose, found in cereals, and lactose , found in milk, are others. Polysaccharides . Examples are starches, dextrins, and cellulose. Bakers are not concerned with polysaccharides but rather with the monosaccharides and disaccharides. The latter two both sweeten, but they cannot be used interchangeably because they have different effects on the end product. These differences are touched on later in the book. Sugar Names It is helpful to understand some of the conventions of the names of different sugars. Note that sugar names often end in “ose”: sucrose, dextrose, maltose, lactose, etc. Sucrose is the chemical name for sugar that comes from the cane and beet sugar plants. Note that glucose is the chemical name for a particular type of sugar. What is sometimes confusing is that glucose occurs naturally, as a sugar molecule in substances such as honey, but it is also produced industrially from the maize plant (corn). The Canadian Food and Drug Regulations (FDR) govern the following definitions: Sugars: All monosaccharides and disaccharides. Used for nutrition labelling purposes. Sweetening agent: Any food for which a standard is provided in Division 18 of the Food and Drug Regulation, or any combination of these. Includes sugar (sucrose), sugar syrups, and molasses derived from sugar cane or sugar beet, dextrose, glucose and syrups, honey and lactose. Excludes sweeteners considered to be food additives. Sweetening ingredient: Any sugar, invert sugar, honey, dextrose, glucose, or glucose solids, or any combination of these in dry or liquid form. Designed for sweetening fruits, vegetables, and their products and substitutes. Maple syrup: The syrup obtained by the concentration of maple sap or by the dilution or solution of a maple product, other than maple sap, in potable water. Sweetener: Any food additive listed as a sweetener. Includes both sugar alcohols and high intensity- sweeteners such as acesulfame-potassium, aspartame, and sucralose. Sugar alcohols: Food additives that may be used as sweeteners. Includes isomalt, lactitol, maltitol, maltitol syrup, mannitol, sorbitol, sorbitol syrup, xylitol, and erythritol.
Courses/Palomar_College/PC%3A_CHEM100_-_Fundamentals_of_Chemistry/13%3A_Solutions/13.07%3A_Solution_Dilution	Learning Objectives Explain how concentrations can be changed in the lab Understand how stock solutions are used in the laboratory We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms - dilute and concentrated . A dilute solution is one in which there is a relatively small amount of solute dissolved in the solution. A concentrated solution contains a relatively large amount of solute. These two terms do not provide any quantitative information (actual numbers) - but they are often useful in comparing solutions in a more general sense. These terms also do not tell us whether or not the solution is saturated or unsaturated, or whether the solution is "strong" or "weak". These last two terms will have special meanings when we discuss acids and bases, so be careful not to confuse these terms. Stock Solutions It is often necessary to have a solution whose concentration is very precisely known. Solutions containing a precise mass of solute in a precise volume of solution are called stock (or standard) solutions. To prepare a standard solution a piece of lab equipment called a volumetric flask should be used. These flasks range in size from 10 mL to 2000 mL are are carefully calibrated to a single volume. On the narrow stem is a calibration mark . The precise mass of solute is dissolved in a bit of the solvent and this is added to the flask. Then enough solvent is added to the flask until the level reaches the calibration mark. Often it is convenient to prepare a series of solutions of known concentrations by first preparing a single stock solution as described in the previous section. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration. Dilutions of Stock (or Standard) Solutions Imagine we have a salt water solution with a certain concentration. That means we have a certain amount of salt (a certain mass or a certain number of moles) dissolved in a certain volume of solution. Next we willl dilute this solution - we do that by adding more water, not more salt: $\rightarrow$ Before Dilution After Dilution The molarity of solution 1 is \[ M_1 = \dfrac{\text{moles}_1}{\text{liter}_1}\] and the molarity of solution 2 is \[ M_1 = \dfrac{\text{moles}_1}{\text{liter}_1}\] rearrange the equations to find moles: \[ \text{moles}_1 = M_1 \text{liter}_1 \] and \[ \text{moles}_2 = M_2 \text{liter}_2 \] What stayed the same and what changed between the two solutions? By adding more water, we changed the volume of the solution. Doing so also changed it's concentration. However, the number of moles of solute did not change. So, \[moles_1 = moles_2\] Therefore \[ \boxed{M_1V_1= M_2V_2 } \label{diluteEq}\] where $M_1$ and $M_2$ are the concentrations of the original and diluted solutions and $V_1$ and $V_2$ are the volumes of the two solutions Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand the above relationship, the calculations are easy to do. Suppose that you have $100. \: \text{mL}$ of a $2.0 \: \text{M}$ solution of $\ce{HCl}$. You dilute the solution by adding enough water to make the solution volume $500. \: \text{mL}$. The new molarity can easily be calculated by using the above equation and solving for $M_2$. \[M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}\] The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Another common dilution problem involves deciding how much of a highly concentrated solution is required to make a desired quantity of solution of lesser concentration. The highly concentrated solution is typically referred to as the stock solution. Example $\PageIndex{1}$: Diluting NITRIC ACID Nitric acid $\left( \ce{HNO_3} \right)$ is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is $16 \: \text{M}$. How much of the stock solution of nitric acid needs to be used to make $8.00 \: \text{L}$ of a $0.50 \: \text{M}$ solution? Solution Steps for Problem Solving Unnamed: 1 Identify the "given"information and what the problem is asking you to "find." Given: M1, Stock $\ce{HNO_3} = 16 \: \text{M}$ $V_2 = 8.00 \: \text{L}$ $M_2 = 0.50 \: \text{M}$ Find: Volume stock $\ce{HNO_3} \left( V_1 \right) = ? \: \text{L}$ List other known quantities none Plan the problem First, rearrange the equation algebraically to solve for $V_1$. \[V_1 = \frac{M_2 \times V_2}{M_1}\] Calculate and cancel units Now substitute the known quantities into the equation and solve. \[V_1 = \frac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}\] Think about your result. $250 \: \text{mL}$ of the stock $\ce{HNO_3}$ needs to be diluted with water to a final volume of $8.00 \: \text{L}$. The dilution is by a factor of 32 to go from $16 \: \text{M}$ to $0.5 \: \text{M}$. Exercise $\PageIndex{1}$ A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution? Answer 135.4 mL Diluting and Mixing Solutions 0 1 How to Dilute a Solution by CarolinaBiological NaN Note The calculated volume will have the same dimensions as the input volume, and dimensional analysis tells us that in this case we don't need to convert to liters, since L cancels when we divide M (mol/L) by M (mol/L).
Courses/University_of_South_Carolina__Upstate/CHEM_U109%3A_Chemistry_of_Living_Things_-_Mueller/14%3A_Amino_Acids_Proteins_and_Enzymes/14.06_Enzyme_Action	Skills to Develop To describe the interaction between an enzyme and its substrate. Enzyme-catalyzed reactions occur in at least two steps. In the first step, an enzyme molecule (E) and the substrate molecule or molecules (S) collide and react to form an intermediate compound called the enzyme-substrate (E–S) complex . (This step is reversible because the complex can break apart into the original substrate or substrates and the free enzyme.) Once the E–S complex forms, the enzyme is able to catalyze the formation of product (P), which is then released from the enzyme surface: \[S + E \rightarrow E–S \tag{$\PageIndex{1}$}\] \[E–S \rightarrow P + E \tag{$\PageIndex{2}$}\] Hydrogen bonding and other electrostatic interactions hold the enzyme and substrate together in the complex. The structural features or functional groups on the enzyme that participate in these interactions are located in a cleft or pocket on the enzyme surface. This pocket, where the enzyme combines with the substrate and transforms the substrate to product is called the active site of the enzyme (Figure $\PageIndex{1}$). Figure $\PageIndex{1}$: Substrate Binding to the Active Site of an Enzyme. The enzyme dihydrofolate reductase is shown with one of its substrates: NADP + (a) unbound and (b) bound. The NADP + (shown in red) binds to a pocket that is complementary to it in shape and ionic properties. The active site of an enzyme possesses a unique conformation (including correctly positioned bonding groups) that is complementary to the structure of the substrate, so that the enzyme and substrate molecules fit together in much the same manner as a key fits into a tumbler lock. In fact, an early model describing the formation of the enzyme-substrate complex was called the lock-and-key model (Figure $\PageIndex{2}$). This model portrayed the enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. Figure $\PageIndex{2}$: The Lock-and-Key Model of Enzyme Action. (a) Because the substrate and the active site of the enzyme have complementary structures and bonding groups, they fit together as a key fits a lock. (b) The catalytic reaction occurs while the two are bonded together in the enzyme-substrate complex. Working out the precise three-dimensional structures of numerous enzymes has enabled chemists to refine the original lock-and-key model of enzyme actions. They discovered that the binding of a substrate often leads to a large conformational change in the enzyme, as well as to changes in the structure of the substrate or substrates. The current theory, known as the induced-fit model , says that enzymes can undergo a change in conformation when they bind substrate molecules, and the active site has a shape complementary to that of the substrate only after the substrate is bound, as shown for hexokinase in Figure $\PageIndex{3}$. After catalysis, the enzyme resumes its original structure. Figure $\PageIndex{3}$: The Induced-Fit Model of Enzyme Action. (a) The enzyme hexokinase without its substrate (glucose, shown in red) is bound to the active site. (b) The enzyme conformation changes dramatically when the substrate binds to it, resulting in additional interactions between hexokinase and glucose. The structural changes that occur when an enzyme and a substrate join together bring specific parts of a substrate into alignment with specific parts of the enzyme’s active site. Amino acid side chains in or near the binding site can then act as acid or base catalysts, provide binding sites for the transfer of functional groups from one substrate to another or aid in the rearrangement of a substrate. The participating amino acids, which are usually widely separated in the primary sequence of the protein, are brought close together in the active site as a result of the folding and bending of the polypeptide chain or chains when the protein acquires its tertiary and quaternary structure. Binding to enzymes brings reactants close to each other and aligns them properly, which has the same effect as increasing the concentration of the reacting compounds. Example $\PageIndex{1}$ What type of interaction would occur between an OH group present on a substrate molecule and a functional group in the active site of an enzyme? Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified. SOLUTION An OH group would most likely engage in hydrogen bonding with an appropriate functional group present in the active site of an enzyme. Any of the polar amino acid side chains would be able to engage in hydrogen bonding with an OH group. One example would be asparagine, which has an amide functional group. Example $\PageIndex{2}$ What type of interaction would occur between an COO − group present on a substrate molecule and a functional group in the active site of an enzyme? Suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you just identified. SOLUTION Because the carboxyl group has a negative charge, it would likely engage in ion-ion interactions (ionic bonding). Any of the positive amino acid side chains: arginine, histidine & lysine. One characteristic that distinguishes an enzyme from all other types of catalysts is its substrate specificity . An inorganic acid such as sulfuric acid can be used to increase the reaction rates of many different reactions, such as the hydrolysis of disaccharides, polysaccharides, lipids, and proteins, with complete impartiality. In contrast, enzymes are much more specific. Some enzymes act on a single substrate, while other enzymes act on any of a group of related molecules containing a similar functional group or chemical bond. Some enzymes even distinguish between D- and L-stereoisomers, binding one stereoisomer but not the other. Urease, for example, is an enzyme that catalyzes the hydrolysis of a single substrate—urea—but not the closely related compounds methyl urea, thiourea, or biuret. The enzyme carboxypeptidase, on the other hand, is far less specific. It catalyzes the removal of nearly any amino acid from the carboxyl end of any peptide or protein. Enzyme specificity results from the uniqueness of the active site in each different enzyme because of the identity, charge, and spatial orientation of the functional groups located there. It regulates cell chemistry so that the proper reactions occur in the proper place at the proper time. Clearly, it is crucial to the proper functioning of the living cell. Summary A substrate binds to a specific region on an enzyme known as the active site, where the substrate can be converted to product. The substrate binds to the enzyme primarily through hydrogen bonding and other electrostatic interactions. The induced-fit model says that an enzyme can undergo a conformational change when binding a substrate. Enzymes exhibit varying degrees of substrate specificity. Concept Review Exercises Distinguish between the lock-and-key model and induced-fit model of enzyme action. Answers The lock-and-key model portrays an enzyme as conformationally rigid and able to bond only to substrates that exactly fit the active site. The induced fit model portrays the enzyme structure as more flexible and is complementary to the substrate only after the substrate is bound. Exercises What type of interaction would occur between each group present on a substrate molecule and a functional group of the active site in an enzyme? COOH NH 3 + OH CH(CH 3 ) 2 For each functional group in Exercise 1, suggest an amino acid whose side chain might be in the active site of an enzyme and form the type of interaction you identified. Answers hydrogen bonding ionic bonding hydrogen bonding dispersion forces 2. The amino acid has a polar side chain capable of engaging in hydrogen bonding; serine (answers will vary). The amino acid has a negatively charged side chain; aspartic acid (answers will vary). The amino acid has a polar side chain capable of engaging in hydrogen bonding; asparagine (answers will vary). The amino acid has a nonpolar side chain; isoleucine (answers will vary).
Courses/Duke_University/CHEM_310L%3A_Physical_Chemistry_I_Laboratory/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/03%3A_Estimating_and_Reporting_Experimental_Error/3.05%3A_Rejection_of_Outliers_(Q-test)	In a set of collected data, there may be one or more values that deviate markedly from the trend of the others. You should not reject a piece of data that seems suspect until you perform a Q-test. In a set of data of three to ten measurements, the Q-test can be performed on the suspect data point: \[Q_{exp} \equiv \frac{\mid(\text { suspected outlier's value })-(\text {value closest to it }) \mid}{\text { (highest value })-(\text { lowest value })}\] Compare the value of $Q_{exp}$ from the equation above to the critical value of $Q_r$. If $Q_{exp}$ is greater or equal to $Q_r$ then the suspect value can be rejected. If $Q_{exp}$ is less than $Q_r$,the value should be retained in the data set. The following table provides critical values for $Q(\alpha, n)$, where $\alpha$ is the probability of incorrectly rejecting the suspected outlier and $n$ is the number of samples in the data set. There are several versions of the Q -Test, each of which calculates a value for Q ij where i is the number of suspected outliers on one end of the data set and j is the number of suspected outliers on the opposite end of the data set. The critical values for Q here are for a single outlier, Q 10 , where \[Q_\text{exp} = Q_{10} = \frac{\mid(\text { suspected outlier's value })-(\text {value closest to it }) \mid}{\text { (highest value })-(\text { lowest value })} \nonumber \] The suspected outlier is rejected if Q exp is greater than $Q(\alpha, n)$. For additional information consult Rorabacher, D. B. “Statistical Treatment for Rejection of Deviant Values: Critical Values of Dixon’s ‘ Q ’ Parameter and Related Subrange Ratios at the 95% confidence Level,” Anal. Chem. 1991 , 63 , 139–146. 0 1 2 3 4 5 $\frac {\alpha \ce{->}} {n \ce{ v }}$ 0.1 (90% confidence limit) 0.05 (95% confidence limit) 0.04 (96% confidence limit) 0.02 (98% confidence limit) 0.01 (99% confidence limit) 3 0.941 0.970 0.976 0.988 0.994 4 0.765 0.829 0.846 0.889 0.926 5 0.642 0.710 0.729 0.780 0.821 6 0.560 0.625 0.644 0.698 0.740 7 0.507 0.568 0.586 0.637 0.680 8 0.468 0.526 0.543 0.590 0.634 9 0.437 0.493 0.510 0.555 0.598 10 0.412 0.466 0.483 0.527 0.568 (see also http://chem-net.blogspot.com/2013/01...atistical.html )
Courses/National_Yang_Ming_Chiao_Tung_University/Chemical_Principles_for_Medical_Students/02%3A_Periodic_Properties_of_the_Elements/2.03%3A_Sizes_of_Atoms_and_Ions	Learning Objectives To understand periodic trends in atomic radii. To predict relative ionic sizes within an isoelectronic series. Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes; however, some atoms are larger or smaller than others, and this influences their chemistry. In this section, we discuss how atomic and ion “sizes” are defined and obtained. Atomic Radii Recall that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure $\PageIndex{1}$ which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom. Figure $\PageIndex{1}$ also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell ( n = 1), it shows only a single peak. In contrast, neon, with filled n = 1 and 2 principal shells, has two peaks. Argon, with filled n = 1, 2, and 3 principal shells, has three peaks. The peak for the filled n = 1 shell occurs at successively shorter distances for neon ( Z = 10) and argon ( Z = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1 s 2 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of n . Consequently, the two electrons in the n = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the n = 1 shell also decreases tremendously (the filled 1 s orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled n = 2 shell in argon is located closer to the nucleus and has a lower energy than the n = 2 shell in neon. Figure $\PageIndex{1}$ illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl 2 molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a covalent atomic radius ($r_{cov}$), which is half the distance between the nuclei of two like atoms joined by a covalent bond in the same molecule, of 99 pm or 0.99 Å (Figure $\PageIndex{2a}$). Atomic radii are often measured in angstroms (Å), a non-SI unit: 1 Å = 1 × 10 −10 m = 100 pm. In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds. A similar approach for measuring the size of ions is discussed later in this section. Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the metallic atomic radius ($r_{met}$) is defined as half the distance between the nuclei of two adjacent metal atoms in the solid (Figure $\PageIndex{2b}$). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the van der Waals atomic radius ($r_{vdW}$), which is half the internuclear distance between two nonbonded atoms in the solid (Figure $\PageIndex{2c}$). This is somewhat difficult for helium which does not form a solid at any temperature. An atom such as chlorine has both a covalent radius (the distance between the two atoms in a $\ce{Cl2}$ molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, $\ce{Cl2(s)}$ at low temperatures). These radii are generally not the same (Figure $\PageIndex{2d}$). Periodic Trends in Atomic Radii Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions. Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure $\PageIndex{3}$). In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure $\PageIndex{4}$). Trends in atomic size result from differences in the effective nuclear charges ($Z_{eff}$) experienced by electrons in the outermost orbitals of the elements. For all elements except H, the effective nuclear charge is always less than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius. Atomic radii decrease from left to right across a row and increase from top to bottom down a column. The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. All have a filled 1 s 2 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2 s and 2 p orbitals, electrons in the same principal shell are not very effective at shielding one another from the nuclear charge . Thus the single 2 s electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1 s 2 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of Z eff = +1.26 for Li.) In contrast, the two 2 s electrons in beryllium do not shield each other very well, although the filled 1 s 2 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2 s electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2 s and 2 p orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size (Figure $\PageIndex{5}$). The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number n is not constant. As we saw in Chapter 2, the size of the orbitals increases as n increases, provided the nuclear charge remains the same . In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to ns orbitals with increasing values of n . However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number n increases from 2 to 6, but the nuclear charge increases from +3 to +55! As a consequence the radii of the lower electron orbitals in Cesium are much smaller than those in lithium and the electrons in those orbitals experience a much larger force of attraction to the nucleus. That force depends on the effective nuclear charge experienced by the the inner electrons. If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6 s 1 valence electron configuration, is much larger than lithium, with a 2 s 1 valence electron configuration. The effective nuclear charge changes relatively little for electrons in the outermost, or valence shell, from lithium to cesium because electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge . Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 shells, abbreviated as [Xe]5 s 2 4 d 10 5 p 6 , which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge. Not all Electrons shield equally Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. Example $\PageIndex{1}$ On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon. Given: three elements Asked for: arrange in order of increasing atomic radius Strategy: Identify the location of the elements in the periodic table. Determine the relative sizes of elements located in the same column from their principal quantum number n . Then determine the order of elements in the same row from their effective nuclear charges. If the elements are not in the same column or row, use pairwise comparisons. List the elements in order of increasing atomic radius. Solution: A These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater. B Combining the two inequalities gives the overall order: C < Si < Al. Exercise $\PageIndex{1}$ On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur. Answer O < S < P < K Atomic Radius: Atomic Radius, YouTube(opens in new window) [youtu.be] Ionic Radii and Isoelectronic Series An ion is formed when either one or more electrons are removed from a neutral atom to form a positive ion (cation) or when additional electrons attach themselves to neutral atoms to form a negative one (anion). The designations cation or anion come from the early experiments with electricity which found that positively charged particles were attracted to the negative pole of a battery, the cathode, while negatively charged ones were attracted to the positive pole, the anode. Ionic compounds consist of regular repeating arrays of alternating positively charged cations and negatively charges anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it is possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radius (the radius of a cation or anion) of one or both. As illustrated in Figure $\PageIndex{6}$, the internuclear distance corresponds to the sum of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na + ion is essentially the same in NaCl and Na 2 S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed. A comparison of ionic radii with atomic radii (Figure $\PageIndex{7}$) shows that a cation, having lost an electron, is always smaller than its parent neutral atom, and an anion, having gained an electron, is always larger than the parent neutral atom . When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases and the ion shrinks (compare Li at 167 pm with Li + at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe 2 + at 78 pm with Fe 3 + at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases and the ion expands (compare F at 42 pm with F − at 133 pm). Cations are always smaller than the neutral atom and anions are always larger. Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na − ion, allowing comparison of its size with that of the far more familiar Na + ion, which is found in many compounds. The radius of sodium in each of its three known oxidation states is given in Table $\PageIndex{1}$. All three species have a nuclear charge of +11, but they contain 10 (Na + ), 11 (Na 0 ), and 12 (Na − ) electrons. The Na + ion is significantly smaller than the neutral Na atom because the 3 s 1 electron has been removed to give a closed shell with n = 2. The Na − ion is larger than the parent Na atom because the additional electron produces a 3 s 2 valence electron configuration, while the nuclear charge remains the same. Unnamed: 0 Na+ Na0 Na− Electron Configuration 1s22s22p6 1s22s22p63s1 1s22s22p63s2 Radius (pm) 102 154* 202† The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253. The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253. The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253. The metallic radius measured for Na(s). †Source: M. J. Wagner and J. L. Dye, “Alkalides, Electrides, and Expanded Metals,” Annual Review of Materials Science 23 (1993): 225–253. Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of n lie at successively greater distances from the nucleus. Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an isoelectronic series . For example, the isoelectronic series of species with the neon closed-shell configuration (1 s 2 2 s 2 2 p 6 ) is shown in Table $\PageIndex{3}$. The sizes of the ions in this series decrease smoothly from N 3− to Al 3 + . All six of the ions contain 10 electrons in the 1 s , 2 s , and 2 p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al 3 + ) is the smallest, and the ion with the smallest nuclear charge (N 3− ) is the largest. The neon atom in this isoelectronic series is not listed in Table $\PageIndex{3}$, because neon forms no covalent or ionic compounds and hence its radius is difficult to measure. Ion Radius (pm) Atomic Number N3− 146 7 O2− 140 8 F− 133 9 Na+ 98 11 Mg2+ 79 12 Al3+ 57 13 Example $\PageIndex{2}$ Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl − , K + , S 2− , and Se 2 − . Given: four ions Asked for: order by increasing radius Strategy: Determine which ions form an isoelectronic series. Of those ions, predict their relative sizes based on their nuclear charges. For ions that do not form an isoelectronic series, locate their positions in the periodic table. Determine the relative sizes of the ions based on their principal quantum numbers n and their locations within a row. Solution: A We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K + , Cl − , and S 2− form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K + has the greatest nuclear charge ( Z = 19), its radius is smallest, and S 2− with Z = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se 2 − ion to be even larger than S 2− . B The order must therefore be K + < Cl − < S 2− < Se 2 − . Exercise $\PageIndex{2}$ Based on their positions in the periodic table, arrange these ions in order of increasing size: Br − , Ca 2 + , Rb + , and Sr 2 + . Answer Ca 2 + < Sr 2 + < Rb + < Br − Summary Ionic radii share the same vertical trend as atomic radii, but the horizontal trends differ due to differences in ionic charges. A variety of methods have been established to measure the size of a single atom or ion. The covalent atomic radius ( r cov ) is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the metallic atomic radius ( r met ) is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The van der Waals radius ( r vdW ) of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The ionic radii of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series , shows a clear correlation between increasing nuclear charge and decreasing size.
Courses/University_of_WisconsinStevens_Point/CHEM_101%3A_Basic_Chemistry_(D'Acchioli)/08%3A_Acids_and_Bases/8.04%3A_Strong_and_Weak_Acids_and_Bases	Learning Objectives Define a strong and a weak acid and base. Recognize an acid or a base as strong or weak. Strong and Weak Acids Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way; consider, for example, HCl(aq). When HCl is dissolved in H 2 O, it completely dissociates into H + (aq) and Cl − (aq) ions; all the HCl molecules become ions: \[HCl\overset{100\%}{\rightarrow}H^{+}(aq)+Cl^{-}(aq) \nonumber \] Any acid that dissociates 100% into ions is called a strong acid . If it does not dissociate 100%, it is a weak acid . HC 2 H 3 O 2 is an example of a weak acid: \[HC_{2}H_{3}O_{2}\overset{\sim 5\%}{\longrightarrow}H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq) \nonumber \] Because this reaction does not go 100% to completion, it is more appropriate to write it as a reversible reaction : \[HC_{2}H_{3}O_{2}\rightleftharpoons H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq) \nonumber \] As it turns out, there are very few strong acids, which are given in Table $\PageIndex{1}$. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid. Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. Acids Bases HCl LiOH HBr NaOH HI KOH HNO3 RbOH H2SO4 CsOH HClO3 Mg(OH)2 HClO4 Ca(OH)2 NaN Sr(OH)2 NaN Ba(OH)2 Strong and Weak Bases The issue is similar with bases: a strong base is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a weak base . There are very few strong bases (Table $\PageIndex{1}$); any base not listed is a weak base. All strong bases are OH – compounds. So a base based on some other mechanism, such as NH 3 (which does not contain OH − ions as part of its formula), will be a weak base. Example $\PageIndex{1}$: Identifying Strong and Weak Acids and Bases Identify each acid or base as strong or weak. HCl Mg(OH) 2 C 5 H 5 N Solution Because HCl is listed in Table $\PageIndex{1}$, it is a strong acid. Because Mg(OH) 2 is listed in Table $\PageIndex{1}$, it is a strong base. The nitrogen in C 5 H 5 N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base. Exercise $\PageIndex{1}$ Identify each acid or base as strong or weak. $\ce{RbOH}$ $\ce{HNO_2}$ Answer a strong base Answer b weak acid Example $\PageIndex{2}$: Characterizing Base Ionization Write the balanced chemical equation for the dissociation of Ca(OH) 2 and indicate whether it proceeds 100% to products or not. Solution This is an ionic compound of Ca 2 + ions and OH − ions. When an ionic compound dissolves, it separates into its constituent ions: \[\ce{Ca(OH)2 → Ca^{2+}(aq) + 2OH^{−}(aq)} \nonumber \] Because Ca(OH) 2 is listed in Table $\PageIndex{1}$, this reaction proceeds 100% to products. Exercise $\PageIndex{2}$ Write the balanced chemical equation for the dissociation of hydrazoic acid (HN 3 ) and indicate whether it proceeds 100% to products or not. Answer a The reaction is as follows: \[\ce{HN3 → H^{+}(aq) + N3^{−}(aq)} \nonumber \] It does not proceed 100% to products because hydrazoic acid is not a strong acid. Summary Strong acids and bases are 100% ionized in aqueous solution. Weak acids and bases are less than 100% ionized in aqueous solution.
Courses/Georgia_Southern_University/CHEM_1152%3A_Survey_of_Chemistry_II_(Osborne)/09%3A_Nucleic_Acids/9.01%3A_Nucleotides	Learning Objectives Identify the components of nucleosides and nucleotides. Identify structural differences between the nitrogenous bases. To identify the different molecules that combine to form nucleotides. Differentiate between the components in DNA and RNA. Demonstrate naming nucleosides and nucleotides. Nucleic acids are molecules that store and replicate information for cellular growth and reproduction. The two types of nucleic acids are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) . The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides . Structural Components of Nucleotides The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 10 9 nucleotides. Nucleotides can be further broken down to phosphoric acid (H 3 PO 4 ), a pentose sugar (a sugar with five carbon atoms), and a nitrogenous base (a base containing nitrogen atoms). \[\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar}\] If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide , and the resulting nucleic acid is ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide , and the nucleic acid is DNA. Figure $\PageIndex{1}$: Structures of pentose sugar found in nucleic acids: ribose (left) and 2-deoxyribose (right). The nitrogenous bases found in nucleotides are classified as pyrimidines or purines . Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure $\PageIndex{2}$). The formation of a glycosidic bond between C1′ of the pentose sugar and N1 of the pyrimidine base or N9 of the purine base joins the pentose sugar to the nitrogenous base. In the formation of this bond, a molecule of water is removed. Table $\PageIndex{1}$ summarizes the similarities and differences in the composition of nucleotides in DNA and RNA. The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring. Composition DNA RNA purine bases adenine and guanine adenine and guanine pyrimidine bases cytosine and thymine cytosine and uracil pentose sugar 2-deoxyribose ribose inorganic acid phosphoric acid (H3PO4) H3PO4 Nucleosides A nucleoside is produced during the condensation reaction between the pentose and nitrogenous base. Nucleosides are named according to the identity of the nitrogenous base, by changing the ending of the name. If the base is a purine, the -ine ending is replaced - osine . If the base is a pyrimidine, the -ine or -il ending is replaced with - idine . If the nucleotide represents a deoxyribonucleotide, the word deoxy- is added to the front of the name. Using these guidelines, the name of the nucleoside shown in Figure $\PageIndex{3}$ is deteremined to be adenosine. Naming Nucleotides Nucleotides are named by adding monophosphate (sometimes shown as 5’-monophosphate) to the end of the name of the corresponding nucleoside. In addition to the full name, abbreviations can be used to indicate the composition of the nucleotide. If deoxyribose is present, a lower case d is used. The names (full and abbreviated) and structures of the major ribonucleotides and one of the deoxyribonucleotides are given in Figure $\PageIndex{4}$. Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure $\PageIndex{5}$, have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD + ), and coenzyme A, contain adenine nucleotides as structural components. Example $\PageIndex{1}$ Classify each compound as a pentose sugar, a purine, or a pyrimidine. Indicate whether it can exist in DNA, RNA, or both. adenine guanine deoxyribose Solution purine, it can exist in both DNA and RNA purine, it can exist in both DNA and RNA pentose sugar, it can only exist in DNA Exercise $\PageIndex{1}$ Classify each compound as a pentose sugar, a purine, or a pyrimidine. Indicate whether it can exist in DNA, RNA, or both. thymine ribose cytosine Summary Nucleosides are composed of a pentose sugar (ribose or deoxyribose) and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Nucleosides containing a purine base end with osine and those containing a pyrimidine end with idine . If the deoxyribose sugar is present, deoxy is added to the front of the name. Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose. Nucleotides are named by adding monophosphate to the end of the nucleoside name.
Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.11%3A_Solution_Concentrations/3.11.03%3A_Environment-_Determining_Water_Purity_via_Biological_Oxygen_Demand	Prior to studying this section, you may wish to revise your solution calculations and your molar calculations. Aquatic life is dependent upon gases dissolved in the water that they live in. Gases, such as carbon dioxide (CO 2 ) will be required by aquatic plants for photosynthesis, whereas most aquatic plants and animals require oxygen for anaerobic respiration. Microorganisms will also require oxygen as they go about decomposing organic matter. One important indicator of water quality is the dissolved oxygen content of the water. At a pressure of 1atm and a temperature of 20 o C, the maximum solubility of oxygen is about 9ppm, which equates to 0.009 g dm -3 . Oxygen, being a non-polar molecule, has a low solubility in water, which is a polar solvent. The polar water molecule induces a dipole moment on the oxygen molecule and the two molecules are now weakly attracted. The diagram below shows the polar water molecule on the left inducing a dipole moment on the non-polar oxygen molecule leading to a weak force of attraction; To ensure a balanced and diverse aquatic community, the oxygen content should not fall below 6ppm, although some species of fish can survive in environments with oxygen contents as low as 3ppm. Bacteria are able to survive in water with even lower levels of oxygen. When organic matter decomposes aerobically in water, the bacteria responsible for this process use up some of the dissolved oxygen present in the water. The amount of oxygen required by the bacteria to decompose this organic matter is defined as biological oxygen demand (BOD). This is often measured in a fixed volume of water over a fixed period of time such as 5 days. If water has a high BOD without the means of replenishing the used oxygen, then very soon it will not be able to support aquatic life. There is a high risk of this happening in bodies of water that are still and do not have much mechanical mixing, e.g. lakes and ponds, whereas fast flowing rivers are able to replenish this oxygen via the mechanical action of its flow. So what factors could cause an increase in BOD? If excessive biodegradable materials find their way into water, this will lead to an increase in decomposing bacteria and hence an increase in BOD. Possible sources of this material include sewerage and industrial wastes from food processing or paper mills. BOD of water can also increase due to the addition of nutrients such as nitrates and phosphates that can be found in fertilizers or laundry detergents. This causes an increase in algae growth, which eventually dies, and decays. As plant growth becomes excessive, the volume of dead and decaying organic material increases rapidly. This decay requires O 2 , leading to the depletion of O 2 in the water. We can use BOD values as an indicator of water quality. BOD/ppm Quality of water < 1 Almost pure water 5 Doubtful purity 10 Unacceptable quality 100 - 400 Waste from untreated sewerage 100 - 10000 Waste from meat processing The BOD of a sample of water can be tested using a redox titration called the Winkler method. The principle of the Winkler method is as follows; oxygen in a water sample is made to oxidize iodide ions into iodine. The amount of iodine produced is determined by titrating with a standard thiosulphate solution. The amount of oxygen present in the original sample of water can be determined from the titer. The reactions are summarized as follows: 1) 2 Mn 2 + (aq) + 4 OH - (aq) + O 2(aq) → 2 MnO 2 (s) + 2 H 2 O (l) 2) MnO 2 (s) + 4 H + (aq) + 2 I - (aq) → Mn 2 + (aq) + I 2(aq) + 2 H 2 O (l) 3) I 2(aq) + 2 S 2 O 3 2 - (aq) → S 4 O 6 2 - (aq) + 2 I - (aq) Using the steps outlined above, please calculate the following. A 500 cm 3 sample of water was saturated with oxygen and left for 5 days. The final oxygen content was measured using the sequence of reactions highlighted above. It was found that 5.00cm 3 of a 0.0500 mol dm -3 solution of Na 2 S 2 O 3 (aq) was required to react with the iodine produced. a) Calculate how many moles of Na 2 S 2 O 3 (aq) reacted with the iodine in reaction (3) b) Deduce how many moles of iodine had been produced in reaction (2). c) Deduce how many moles of MnO 2 (s) had been produced in reaction (1). d) Deduce how many moles of O 2(g) were present in the water. e) Calculate the solubility of oxygen in the water in g dm -3 . f) Assume the maximum solubility of the water is 0.009 g dm -3 and deduce the BOD of the water sample. Solution a) Amount of Na 2 S 2 O 3 (aq) = 5.00 x 0.0500/1000 = 2.5 x 10 -4 moles. b) Amount of I 2(aq) = ½ (2.50 x 10 -4 moles) = 1.25 x 10 -4 moles. c) Amount of MnO 2 (s) = 1.25 x 10 -4 moles. d) Amount of O 2(g) = ½ (1.25 x 10 -4 ) moles = 6.25 x 10 -5 moles. e) Amount of O2(g) in 1dm3 = 1.25 x 10-4 moles Mass in 1dm3 = 0.004 g dm-3 f) Oxygen used by bacteria (BOD) = 0.009 g dm -3 – 0.004 g dm -3 = 0.005 g dm -3
Courses/Indiana_Tech/EWC%3A_CHEM_1000_-_Introductory_Chemistry_(Budhi)/04%3A_Atoms_and_Elements/4.9%3A_Atomic_Mass%3A_The_Average_Mass_of_an_Element%E2%80%99s_Atoms	Learning Objectives Explain what is meant by the atomic mass of an element. Calculate the atomic mass of an element from the masses and relative percentages of the isotopes of the element. In chemistry we very rarely deal with only one isotope of an element. We use a mixture of the isotopes of an element in chemical reactions and other aspects of chemistry, because all of the isotopes of an element react in the same manner. That means that we rarely need to worry about the mass of a specific isotope, but instead we need to know the average mass of the atoms of an element. Using the masses of the different isotopes and how abundant each isotope is, we can find the average mass of the atoms of an element. The atomic mass of an element is the weighted average mass of the atoms in a naturally occurring sample of the element. Atomic mass is typically reported in atomic mass units. Calculating Atomic Mass You can calculate the atomic mass (or average mass) of an element provided you know the relative abundance (the fraction of an element that is a given isotope), the element's naturally occurring isotopes, and the masses of those different isotopes. We can calculate this by the following equation: \[\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \nonumber \] Look carefully to see how this equation is used in the following examples. Example $\PageIndex{1}$: Boron Isotopes Boron has two naturally occurring isotopes. In a sample of boron, $20\%$ of the atoms are $\ce{B}$-10, which is an isotope of boron with 5 neutrons and mass of $10 \: \text{amu}$. The other $80\%$ of the atoms are $\ce{B}$-11, which is an isotope of boron with 6 neutrons and a mass of $11 \: \text{amu}$. What is the atomic mass of boron? Solution Boron has two isotopes. We will use the equation: \[\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots \nonumber \] Isotope 1: $\%_1 = 0.20$ (Write all percentages as decimals), $\text{mass}_1 = 10$ Isotope 2: $\%_2 = 0.80$, $\text{mass}_2 = 11$ Substitute these into the equation, and we get: \[\text{Atomic mass} = \left( 0.20 \right) \left( 10 \right) + \left( 0.80 \right) \left( 11 \right)\nonumber \] \[\text{Atomic mass} = 10.8 \: \text{amu}\nonumber \] The mass of an average boron atom, and thus boron's atomic mass, is $10.8 \: \text{amu}$. Example $\PageIndex{2}$: Neon Isotopes Neon has three naturally occurring isotopes. In a sample of neon, $90.92\%$ of the atoms are $\ce{Ne}$-20, which is an isotope of neon with 10 neutrons and a mass of $19.99 \: \text{amu}$. Another $0.3\%$ of the atoms are $\ce{Ne}$-21, which is an isotope of neon with 11 neutrons and a mass of $20.99 \: \text{amu}$. The final $8.85\%$ of the atoms are $\ce{Ne}$-22, which is an isotope of neon with 12 neutrons and a mass of $21.99 \: \text{amu}$. What is the atomic mass of neon? Solution Neon has three isotopes. We will use the equation: \[\text{Atomic mass} = \left( \%_1 \right) \left( \text{mass}_1 \right) + \left( \%_2 \right) \left( \text{mass}_2 \right) + \cdots\nonumber \] Isotope 1: $\%_1 = 0.9092$ (write all percentages as decimals), $\text{mass}_1 = 19.99$ Isotope 2: $\%_2 = 0.003$, $\text{mass}_2 = 20.99$ Isotope 3: $\%_3 = 0.0885$, $\text{mass}_3 = 21.99$ Substitute these into the equation, and we get: \[\text{Atomic mass} = \left( 0.9092 \right) \left( 19.99 \right) + \left( 0.003 \right) \left( 20.99 \right) + \left( 0.0885 \right) \left( 21.99 \right)\nonumber \] \[\text{Atomic mass} = 20.17 \: \text{amu}\nonumber \] The mass of an average neon atom is $20.17 \: \text{amu}$ The periodic table gives the atomic mass of each element. The atomic mass is a number that usually appears below the element's symbol in each square. Notice that the atomic mass of boron (symbol $\ce{B}$) is 10.8, which is what we calculated in Example $\PageIndex{1}$, and the atomic mass of neon (symbol $\ce{Ne}$) is 20.8, which is what we calculated in Example $\PageIndex{2}$. Take time to notice that not all periodic tables have the atomic number above the element's symbol and the mass number below it. If you are ever confused, remember that the atomic number should always be the smaller of the two and will be a whole number, while the atomic mass should always be the larger of the two and will be a decimal number. Exercise $\PageIndex{1}$ Chlorine has two naturally occurring isotopes. In a sample of chlorine, $75.77\%$ of the atoms are $\ce{Cl}$-35, with a mass of $34.97 \: \text{amu}$. Another $24.23\%$ of the atoms are $\ce{Cl}$-37, with a mass of $36.97 \: \text{amu}$. What is the atomic mass of chlorine? Answer 35.45 amu Summary An element's atomic mass is the weighted average of the masses of the isotopes of an element An element's atomic mass can be calculated provided the relative abundance of the element's naturally occurring isotopes and the masses of those isotopes are known. The periodic table is a convenient way to summarize information about the different elements. In addition to the element's symbol, most periodic tables will also contain the element's atomic number and the element's atomic mass.
Bookshelves/Organic_Chemistry/Understanding_Organic_Chemistry_Through_Computation_(Boaz_and_Pearce)/04%3A_Measuring_Equilibrium_on_Cyclohexane_Chair_Structures/4.02%3A_Background	As was discussed in the lecture portion of the course, substituted cyclohexane rings exist as chair structures that can place the substituent in either the axial or equatorial position. As shown in Figure 1, the equilibrium generally favors the conformer that places the substituent in the equatorial position over the one that places the substituent in the axial position. One measure of how much a substituent causes the equatorial conformer to be favored is known as an A value. An A value measures the difference in energy ($\Delta G^{\circ}$) in kcal/mol for a monosubstituted cyclohexane ring in the equilibrium as written in Figure 1. The bigger in magnitude an A value is, the greater the substituent will cause the equilibrium to favor the equatorial conformer. The difference in energy between conformers can also be related to the equilibrium constant ($K_{eq}$) by equation $\PageIndex{1}$ . $R$ represents the ideal gas constant [1.987 cal/(mol*K)] and $T$ represents the temperature in Kelvin, and ln is the natural logarithm. This means if we can calculate the change in energy between two cyclohexane conformers, it will be possible to estimate the equilibrium constant of the process. By extension you can then estimate the proportion of each conformer (axial and equatorial) at equilibrium. \[G_{ax}-G_{eq}=\Delta G^{\circ}=-R \operatorname{Tln}\left(K_{e q}\right)\]
Courses/Bellingham_Technical_College/CHEM_110%3A_Bellingham_Technical_College/02%3A_Module_2/2.01%3A_Prelude_to_Atoms_Molecules_and_Ions	Although not an SI unit, the angstrom (Å) is a useful unit of length. It is one ten-billionth of a meter, or 10 −10 m. Why is it a useful unit? The ultimate particles that compose all matter are about 10 −10 m in size, or about 1 Å. This makes the angstrom a natural—though not approved—unit for describing these particles. The angstrom unit is named after Anders Jonas Ångström, a nineteenth-century Swedish physicist. Ångström's research dealt with light being emitted by glowing objects, including the sun. Ångström studied the brightness of the different colors of light that the sun emitted and was able to deduce that the sun is composed of the same kinds of matter that are present on the earth. By extension, we now know that all matter throughout the universe is similar to the matter that exists on our own planet. Anders Jonas Ångstrom, a Swedish physicist, studied the light coming from the sun. His contributions to science were sufficient to have a tiny unit of length named after him, the angstrom, which is one ten-billionth of a meter. Source: Photo of the sun courtesy of NASA's Solar Dynamics Observatory.
Courses/University_of_Illinois_Chicago/NATS_106%3A_Chemical_and_Biological_Systems_(UIC)/01%3A_Elements_and_the_Periodic_Table/1.06%3A_Elements_and_Atomic_Number	Learning Objectives Use the atomic number and mass number to describe an atom. The modern atomic theory states that atoms of one element are the same, while atoms of different elements are different. What makes atoms of different elements different? The fundamental characteristic that all atoms of the same element share is the number of protons . All atoms of hydrogen have one and only one proton in the nucleus; all atoms of iron have 26 protons in the nucleus. This number of protons is so important to the identity of an atom that it is called the atomic number ( Z ). The number of protons in an atom is the atomic number of the element. Thus, hydrogen has an atomic number of 1, while iron has an atomic number of 26. Each element has its own characteristic atomic number. Atoms are electrically neutral , meaning that the overall electric charge is zero. This is because the number of protons (positive charge) equals the number of electrons (negative charge). Therefore, the atomic number also provides the number of electrons. For example, helium has Z = 2, which tells us that there are 2 protons in the nucleus and 2 electrons outside of the nucleus. Sometimes atoms will gain or lose electrons resulting in a difference in the number of protons and electrons, which means the charge is no longer zero. Atoms that have a charge are called ions and will be discussed further in later chapters. \[atomic\ number = Z = p^+ = e^{–1}\] As we learned previously, protons and neutrons, which are found in the nucleus of an atom, each have a mass of ~1 amu. Because an electron has negligible mass relative to that of a proton or a neutron, the majority of an atom's mass is in the nucleus. The mass number ( A ) is defined as the total number of protons ($p^+$) and neutrons ($n$) in an atom: \[mass\ number = A = p^+ + n\] Atoms of the same element always have the same number of protons, same Z , but often have different numbers of neutrons, therefore, different mass numbers. These atoms are called isotopes and will be discussed in more detail in the next chapter. Example $\PageIndex{1}$ The most common carbon atoms have six protons and six neutrons in their nuclei. What are the atomic number and the mass number of these carbon atoms? An isotope of uranium has an atomic number of 92 and a mass number of 235. What are the number of protons and neutrons in the nucleus of this atom? Solution If a carbon atom has six protons in its nucleus, its atomic number is 6. If it also has six neutrons in the nucleus, then the mass number is 6 + 6, or 12. If the atomic number of uranium is 92, then that is the number of protons in the nucleus. Because the mass number is 235, then the number of neutrons in the nucleus is 235 − 92, or 143. Exercise $\PageIndex{1}$ The number of protons in the nucleus of a tin atom is 50, while the number of neutrons in the nucleus is 68. What are the atomic number and the mass number of this isotope? Answer Atomic number = 50, mass number = 118 Example $\PageIndex{2}$: What is the symbol for an isotope of uranium that has an atomic number of 92 and a mass number of 235? How many protons and neutrons are in $\ce{_{26}^{56}Fe}$ Solution The symbol for this isotope is $\ce{_{92}^{235}U}$ This iron atom has 26 protons and 56 − 26 = 30 neutrons. Exercise $\PageIndex{2}$ How many protons are in $\ce{_{11}^{23} Na}$ Answer 11 protons Key Takeaways Atoms themselves are composed of protons, neutrons, and electrons. Each element has its own atomic number, which is equal to the number of protons in its nucleus. Isotopes of an element contain different numbers of neutrons. Elements are represented by an atomic symbol. Contributors Lisa Sharpe Elles, University of Kansas
Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/General_Chemistry_Labs/Online_Chemistry_Lab_Manual/Chem_11_Experiments/17%3A_VSEPR_Theory_and_Shapes_of_Molecules_(Experiment)	To build a variety of molecules and ions using molecular model kits. To draw Lewis structures (both projection and perspective drawings) for each of these molecules and ions. To determine the hybridization of the central atoms, the number and types of bonds, the geometries, and the polarities of the molecules and ions. Lewis Structures A Lewis Structure is a representation of covalent molecules (or polyatomic ions) where all the valence electrons are shown distributed about the bonded atoms as either shared electron pairs (bond pairs) or unshared electron pairs (lone pairs). A shared pair of electrons is represented as a short line (a single bond). Sometimes atoms can share two pairs of electrons, represented by two short lines (a double bond). Atoms can even share three pairs of electrons, represented by three short lines (a triple bond). Pairs of dots are used to represent lone pair electrons. Examples are shown for the molecules $ce{SF2}$ and $\ce{CH2O}$ below. Please review (in your text or notes) the rules for drawing Lewis structures before performing this exercise. This includes rules for structures which obey the octet rule, as well as those which involve expanded or reduced octets. Resonance Structures Resonance refers to bonding in molecules or ions that cannot be correctly represented by a single Lewis structure. These structures are often equivalent, meaning that they contain the same number of bonds at different locations. The molecule $\ce{SO2}$ (shown above) has two such resonance forms. Resonance structures can also be non-equivalent, in which case they will have different numbers and/or locations of bonds. Note that any valid resonance structure of a molecule can be used to determine its shape and polarity. VSEPR Theory The VSEPR ( V alence S hell E lectron P air R epulsion) model is used to predict the geometry of molecules based on the number of effective electron pairs around a central atom. The main postulate for the VSEPR theory is that the geometrical structure around a given atom is principally determined by minimizing the repulsion between effective electron pairs. Both the molecular geometry and the polarity of individual bonds then determine whether the molecule is polar or not. Before determining the shape of a molecule, the Lewis structure must be properly drawn. The shape of a molecule is then determined by the number of areas of electron density (or, number of effective electron pairs) around a central atom. Areas of electrons density include: Lone pairs of electrons: these electrons tend to take more space than the bonded pairs in space leading into somewhat distorted structures. Bonds (single, double and triple bonds count as one (1) area of electron density or one effective electron pair). Before performing this exercise, please review (in your text or notes) the various geometries and bond angles that can be produced by different numbers of effective electron pairs around the central atom. Molecular Polarity A polar bond is one in which the electron cloud is closer to the nucleus of one atom (the more electronegative one) than the other (the less electronegative one). Knowledge of both the bond polarities and the shape are required in the determination of the molecule’s overall polarity (dipole moment). A polar molecule is one that shows an imbalance in its electron distribution. When placed in an electric field, these molecules tend to align themselves with the electric field. Some molecules have polar bonds but no dipole moment. This happens when the bonds in a molecule are arranged in a way in which polarities cancel each other out. One way this occurs is when molecules have all identical bonds and there is no lone pair on the central atom (for example, $\ce{CO2}$). Molecules that do not fit both these criteria may be polar or not depending on how atoms are bonded and the electron pairs arranged around the central atom (for example, $\ce{XeCl2F2}$ shown below). Procedure Materials and Equipment Molecular model kits from the stockroom Instructions Complete the following for each of the molecules and ions on your Report form: Draw Lewis structures, including all resonance structures if applicable (1). Build models and then draw perspective structures (2) that accurately represent bond angles and molecular shapes. The molecular model kits contain different colored balls and different size stick connectors. Three-dimensional models will be constructed from these balls and sticks. The stick connectors represent bonds. Use the short rigid sticks for single bonds. The long flexible sticks must be used to create double (2 sticks) and triple (3 sticks) bonds. The different colored balls represent different atoms. There is a key on the inside of the lid of the model kit which indicates which colors correspond to which atoms. There are, however, some exceptions to this. For example, the kit indicates that the green balls with just one hole are to be used for the halogens. But if a halogen (such as $\ce{Cl}$) appears in a molecule as a central atom with an expanded octet, you would need to use a ball that has 5 or 6 holes to build the model (brown or silver ball). Another example is oxygen. The kit indicates that the red balls with two holes should be be used for oxygen. However, if an oxygen atom in a compound requires more than two bonds, the red balls cannot be used. In this case you would substitute a blue ball for oxygen. The 3-D models will serve as a visual guide to help you with your perspective structures. Use the following guidelines to draw them correctly: For bonds lying in the plane of the paper, use a regular solid line. For bonds that project down into the paper away from you, use a hatched wedge-shaped line ( ). For bonds that project up out of the paper towards you, use a solid wedge-shaped line ( ). An example showing both the Lewis structure and perspective representation of$\ce{CH4}$ is provided below. Determine the number of atoms bonded to the central atom (or, number of $\sigma $-bonds) (3). Determine the number of lone electron pairs on the central atom (4). Predict the electronic geometry using all areas of electron density (or, effective electron pairs) and the ideal bond angles associated with this geometry (5). Predict the actual geometry of the molecule or ion (6). Determine the hybridization of the central atom (7). Determine the polarity of the molecule (8). Use an arrow to show the direction of electron density for polar molecules on the perspective drawing. Please be sure to return all balls and stick connectors to the model kit when finished. Pre-laboratory Assignment: VSEPR Theory and Shapes of Molecules Complete the following table Chemical Species $\ce{KrF2}$ $\ce{PH4^+}$ $\ce{TeCl6}$ $\ce{ClBr3}$ $\ce{H2S}$ Lewis Structure NaN NaN NaN NaN NaN Perspective Drawing NaN NaN NaN NaN NaN Number of atoms bonded to central atom NaN NaN NaN NaN NaN Number of lone pairs on central atom NaN NaN NaN NaN NaN Electronic geometry NaN NaN NaN NaN NaN Molecular Geometry NaN NaN NaN NaN NaN Polarity NaN NaN NaN NaN NaN Lab Report: VSEPR Theory and Shapes of Molecules $\ce{HCN}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{CH3OH}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{SeF6}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{NO2^-}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{AsF5}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{XeF2}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{TeF5^-}$: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{H2CO}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{SF4}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{XeF4}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{PO4^{3-}}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{BrF3}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{NH3}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: $\ce{CH3NH2}$ 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: Molecule: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: Molecule: 0 1 2 3 1. Lewis Structure 1. Lewis Structure 2. Perspective drawing 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs on the central atom 4. Number of non-bonding electron pairs on the central atom 5. Electronic geometry: 6. Molecular geometry with ideal bond angles 7. Hybridization of central atom 7. Hybridization of central atom 8. Polarity: Questions For each one of the molecules with lone pairs of electrons on the central atom that is non polar, give an explanation why they have no dipole moment. Find all molecules (or ions) with resonance structures and draw them in the box below: 0 1 2 3 1 NaN NaN NaN 2 NaN NaN NaN 3 NaN NaN NaN Summary of Types of Shapes Areas of electron density Areas of electron density.1 Areas of electron density.2 Molecular geometry Example of species Polarity of the example Number atoms bonded to cenral atom ($\sigma$-bonds) Number of lone pairs Electronic geometry NaN NaN NaN 2 0 NaN NaN NaN NaN 3 0 NaN NaN NaN NaN 2 1 NaN NaN NaN NaN 4 0 NaN NaN NaN NaN 3 1 NaN NaN NaN NaN 2 2 NaN NaN NaN NaN 5 0 NaN NaN NaN NaN 4 1 NaN NaN NaN NaN 3 2 NaN NaN NaN NaN 2 3 NaN NaN NaN NaN 6 0 NaN NaN NaN NaN 5 1 NaN NaN NaN NaN 4 2 NaN NaN NaN NaN
Courses/Saint_Marys_College_Notre_Dame_IN/CHEM_431%3A_Inorganic_Chemistry_(Haas)/CHEM_431_Readings/09%3A_Acid-Base_and_Donor-Acceptor_Chemistry/9.04%3A_Lewis_Concept_and_Frontier_Orbitals/9.4.07%3A_Bulky_groups_weaken_the_strength_of_Lewis_acids_and_bases_because_they_introduce_steric_strain_into_the_resulting_acid-base_adduct.	Steric effects can influence the ability of a Lewis acid or base to form adducts by introducing: front strain (F-strain) whereby bulky groups make it difficult for the Lewis acid and Lewis base centers to approach and interact. back strain (B-strain) associated with steric interactions that do not directly impede the Lewis acid and base centers from interacting but instead occur as the acid and base rearrange upon adduct formation. For instance, when trivalent boron compounds form adducts with amines, the boron center changes from a more open trigonal geometry to a more hindered tetrahedral one: internal strain (I-strain) is also associated with the geometry changes incident on adduct formation. However, while B-strain involves direct steric clashes that occur on adduct formation, I-strain is the strain involved in deforming bond and torsional angles away from more stable local geometries. Thus it is more important for Lewis base centers embedded in rings or clusters. References 1. Alder, R. W., Strain effects on amine basicities. Chemical Reviews 1989, 89 (5), 1215-1223.
Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/21%3A_Resonance_and_Molecular_Orbital_Methods/21.04%3A_The_Benzene_Problem	We already have alluded to the difficulties encountered in the interpretation of the structure of benzene in Sections 1-1G and 6-5 . Our task here is to see what new insight the VB and MO treatments can give us about benzene, but first we will indicate those properties of benzene that are difficult to explain on the basis of simple structure theory. Some of the Unusual Properties of Benzene From x-ray diffraction and spectroscopic measurements, benzene is known to be a planar molecule with six carbons $1.390$ Å apart in a hexagonal ring, $5$. Six hydrogen atoms, one associated with each carbon, are located $1.09$ Å from those carbons. All $\ce{H-C-C}$ and $\ce{C-C-C}$ bond angles are $120^\text{o}$: The 1,3,5-cyclohexatriene structure, $6$, proposed for benzene in 1866 by Kekule, has alternating single and double bonds around the ring, which would be predicted to have bond lengths of $1.48$ Å and $1.34$ Å, respectively (see Table 2-1): The knowledge that the bond lengths are equal in the ring in benzene is a point against the Kekule formulation, but a more convincing argument is available from a comparison of the chemistry of benzene with that of 1,3,5-hexatriene, $7$: Benzene also is more stable by about $36$-$38 \: \text{kcal mol}^{-1}$ than anticipated for the 1,3,5-cyclohexatriene structure. You will recall from earlier discussions that the heat of combustion of one mole of benzene is $38 \: \text{kcal}$ less than calculated for cyclohexatriene (see Section 6-5A ). Also, the heat of hydrogenation of benzene is only $49.8 \: \text{kcal mol}^{1}$, which is $36 \: \text{kcal}$ less than expected for 1,3,5-cyclohexatriene; this estimate is based on the assumption that the heat of hydrogenation of 1,3,5-cyclohexatriene (with three double bonds) would be three times that of cyclohexane ($28.5 \: \text{kcal mol}^{-1}$, for one double bond), or $3 \times 28.2 = 85.5 \: \text{kcal mol}^{-1}$. The extra stability of benzene relative to the hypothetical 1,3,5-cyclohexatriene can be called its stabilization energy . Most (but not all) of this stabilization may be ascribed to resonance or electron delocalization. The Atomic-Orbital Model of Benzene In Section 6-5 an atomic model of benzene was discussed in some detail. Each carbon in the ring was considered to form three coplanar $sp^2$-hybrid $\sigma$ bonds at $120^\text{o}$ angles. These carbon-carbon and carbon-hydrogen $\sigma$ bonds use three of the four valence electrons of each carbon. The remaining six carbon electrons are in parallel $p$ orbitals, one on each of the six carbons. Each of the $\pi$ electrons can be regarded as being paired with its immediate neighbors all around the ring, as shown by $8$: As mentioned in Section 21-2B , delocalization of the electrons over all six centers in benzene should give a more stable electron distribution than any structure in which the electrons are localized in pairs between adjacent carbons (as in the classical 1,3,5-cyclohexatriene structure). The simple MO and VB treatments of benzene begin with the same atomic-orbital model and each treats benzene as a six-electron $\pi$-bonding problem. The assumption is that the $\sigma$ bonds of benzene should not be very much different from those of ethene and may be regarded as independent of the $\pi$ system. The Molecular Orbital Method for Benzene Extension of the ideas of Section 21-2 for the MO treatment of an electron-pair bond between two nuclei to the $\pi$ bonding in benzene is fairly straightforward. What is very important to understand is that there must be more than one molecular orbital for the $\pi$ electrons because there are six $\pi$ electrons, and the Pauli principle does not allow more than two paired electrons to occupy a given orbital. In fact, combination (or mixing) os the six $2p$ orbitals of benzene, shown in $8$, gives six $\pi$ molecular orbitals. Without exception, the number of molecular orbitals obtained by mixing is always the same as the number of atomic orbitals mixed . The details of the mathematics of the mixing process to give an optimum set of molecular orbitals will not be described here,$^1$ but the results are shown in Figure 21-5. Of the six predicted molecular orbitals, three are bonding and three are antibonding. The six $\pi$ electrons are assigned to the three bonding orbitals in pairs and are calculated to have a total $\pi$-electron energy of $6 \alpha + 8 \beta$. The calculation that leads to the results shown in Figure 21-5 is not very sophisticated. It is based on the assumption that the $\pi$ bonding between each carbon and its immediate neighbors is equal all around the ring and that bonding involving carbons more than $2$ Å apart is unimportant. What happens if we use the MO method to calculate the $\pi$-electron energy of classical 1,3,5-cyclohexatriene? The procedure is exactly as for benzene, except that we decree that each carbon $p$ orbital forms a $\pi$ bond with only one of its neighboring $p$ orbitals. The results are shown in Figure 21-6. The $\pi$-electron energy turns out to be three times that of ethene, or $6 \alpha + 6 \beta$ compared to $6 \alpha + 8 \beta$ for benzene. The calculated delocalization energy for benzene is the difference between these quantities, or $\left( 6 \alpha + 8 \beta \right) - \left( 6 \alpha + 6 \beta \right) = 2 \beta$. That is to say, the calculated delocalization energy is the difference between the energy of benzene with full $\pi$ bonding and the energy of 1,3,5-cyclohexatriene with alternating single and double bonds. If the electron delocalization energy $\left( 2 \beta \right)$ is equal to the stabilization energy $\left( 38 \: \text{kcal mol}^{-1} \right)$, then $\beta = 19 \: \text{kcal mol}^{-1}$. Whether this is a valid method for determining $\beta$ has been a matter of dispute for many years. Irrespective of this, the results of the calculations do account for the fact that benzene is more stable than would be expected for 1,3,5-cyclohexatriene. However, do the results also account for the low reactivity toward the various reagents in Figure 214, such as those that donate $\ce{Br}^\oplus$ to double bonds (see Section 10-3A )? To settle this question, we have calculated the changes in $\pi$-electron energy that occur in each of the following reactions: This means calculating the $\pi$-electron energies of all four entities and assuming that the differences in the $\sigma$-bond energies cancel between the two reactions. The result of this rather simple calculation is that attack of $\ce{Br}^\oplus$ on benzene is thermodynamically less favorable than on 1,3,5-cyclohexatriene by about $\beta$. If $\beta$ is $19 \: \text{kcal mol}^{-1}$, this is clearly a sizable energy difference, and we can conclude that the simple MO method does indeed account for the fact that benzene is attacked by $\ce{Br}^\oplus$ far less readily than is 1,3,5-cyclohexatriene. The Valence Bond Method for Benzene Extension of the basic ideas of the VB treatment described in Section 21-2 to the atomic-orbital model of benzene is straightforward. We can write VB structures that represent pairing schemes of electrons in the atomic orbitals as shown in $9$ through $13$: Pairing schemes $9$ and $10$ correspond to Kekule's structures, whereas $11$, $12$, and $13$ are called "Dewar structures" because J. Dewar suggested, in 1869, that benzene might have a structure such as $14$: The electrons are paired in the configurations represented by $11$, $12$, and $13$, but these pairing schemes are not as energetically favorable as $9$ and $10$. The reason is that the two electrons paired according to the dashed lines in $11$, $12$, and $13$ are on nuclei separated by $2.8$ Å, which is too far apart for effective bonding. The dashed lines between the distant carbons in $11$, $12$, and $13$ are significant only in that they define a pairing scheme. Such lines sometimes are said to represent "formal bonds". We hope that it is clear from what we have said here and previously that the electron-pairing schemes $9$ through $13$ do not separately have physical reality or independent existence ; indeed, the energy of the actual molecule is less than any one of the contributing structures. The double-headed arrow between the structures is used to indicate that they represent different electron-pairing schemes for a molecule and not different forms of the molecule in equilibrium with one another. When we use the resonance method in a qualitative way, we consider that the contribution of each of the several structures is to be weighted in some way that accords with the degree of bonding each would have, if it were to represent an actual molecule with the specified geometry. Thus the Kekule-type electron-pairing schemes, $9$ and $10$, are to be taken as contributing equally and predominantly to the hybrid structure of benzene - equally because they are energetically equivalent, and predominantly because they can contribute much more to the overall bonding than $11$, $12$, and $13$. In using the resonance method, we assume that all the resonance structures contributing to a given resonance hybrid have exactly the same spatial arrangements of the nuclei but different pairing schemes for the electrons. Therefore $11$, $12$, and $13$ are not to be confused with bicyclo[2.2.0]-2,5-hexadiene, $15$, because $15$ is a known (albeit not very stable) molecule with different atom positions and therefore vastly different bond angles and bond lengths from benzene: The electron-pairing schemes $9$ and $10$ represent the electron pairing that $15$ would have if it were grossly distorted, with each carbon at the corner of a regular hexagon and a formal bond in place of a carbon-carbon single bond. Thus $9$ and $10$ would not contribute in a significant way to the resonance hybrid of $15$. Clearly, it is inconvenient and tedious to write the structures of the contributing forms to show the structure of a resonance hybrid. A shorthand notation is therefore desirable. Frequently, dashed rather than full lines are used where the bonding electrons are expected to be delocalized over several atoms. For benzene, $16a$ or $16b$ is quite appropriate: Unfortunately, although these are clear and explicit renderings, they are tedious to draw. As a result, many authors use (as we will most often) a single Kekule structure to represent benzene, with the understanding that all the $\ce{C-C}$ bonds are equivalent. Other authors choose to represent benzene as a hexagon with an inscribed circle: This is a simple notation for benzene, but is quite uninformative and even can be actively misleading with some aromatic ring systems, and thus should be used with this limitation in mind. In calculations of the resonance energy of benzene, the five electronic configurations (valence-bond structures $9$ through $13$) are combined mathematically to give five hybrid states, and of these the lowest-energy state is assumed to correspond to the normal state of the molecule. Thus benzene is considered by this approach to be a resonance hybrid of the valence-bond structures $9$ through $13$. In this simple treatment, $9$ and $10$ are calculated to contribute about $80\%$ and $11$, $12$, and $13$ about $20\%$ to the hybrid. The actual numerical VB calculations, which are much more difficult to carry through than the corresponding MO calculations, give an energy of $Q + 2.61 J$ for benzene and $Q + 1.50 J$ for classical 1,3,5-cyclohexatriene.$^3$ The resonance or delocalization energy then is $\left( Q + 2.61 J \right) - \left( Q + 1.50 J \right) = 1.11 J$, which makes $J \sim 35 \: \text{kcal mol}^{-1}$ if the resonance energy is taken to be equal to the $38 \: \text{kcal}$ value obtained for the stabilization energy. If one carries through a simple VB calculation of the relative energy change associated with attack of $\ce{Br}^\oplus$ on benzene as compared to 1,3,5-cyclohexatriene, the value obtained is $0.63 J$, which corresponds to $22 \: \text{kcal}$. This is in excellent agreement with the $19 \: \text{kcal}$ value obtained by the MO method ( Section 21-3C ). $^1$There are many excellent books that cover this subject in great detail; however, the simplest introductory work is J. D. Roberts, Molecular Orbital Calculations , W. A. Benjamin, Inc., Menlo Park, Calif., 1961. $^2$ Note that in $9$ and also in $10$, we show a particular way of pairing the electrons. However, just as $1 \leftrightarrow 2$, and $4a \leftrightarrow 4b$, we also must consider other sets that represent exchanges of electrons across the dashed lines of $9$ and also of $10$. $^3$$Q$ and $J$ are negative VB energy parameters that correspond roughly to the MO parameters $\alpha$ and $\beta$.
Courses/Lumen_Learning/Book%3A_General_College_Chemistry_I_(Lumen)/01%3A_Main_Body/1.06%3A_Measurement_Uncertainty_Accuracy_and_Precision	Learning Objectives By the end of this section, you will be able to: Define accuracy and precision Distinguish between exact and uncertain numbers Correctly represent uncertainty in measured quantities using significant figures Identify the number of significant figures in a given quantity Apply proper significant figure and rounding rules to calculated quantities Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number . If we count eggs in a carton, we know exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used. Significant Figures in Measurement The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid. Figure 1. To measure the volume of liquid in this graduated cylinder, you must mentally subdivide the distance between the 21 and 22 mL marks into tenths of a milliliter, and then make a reading (estimate) at the bottom of the meniscus. Refer to the illustration in Figure 1. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL. This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If we weigh the quarter on a more sensitive balance, we may find that its mass is 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty , which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits . Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values. Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them. Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point. Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located. The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407 × 10 −3 ; then the number 8.32407 contains all of the significant figures, and 10 −3 locates the decimal point. The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 × 10 3 (two significant figures), 1.30 × 10 3 (three significant figures, if the tens place was measured), or 1.300 × 10 3 (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant. Summary of Significant Figures Rules: Nonzero digits are significant. Leading zeros are not significant. Captive zeros are significant. Trailing zeros are significant if there is a decimal point written anywhere in the number. If there is no decimal point written in the number, trailing zeros are assumed not to be significant. When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 3.17 × 10 8 people. Significant Figures in Calculations A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers: When we add or subtract numbers , we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction). When we multiply or divide numbers , we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division). When we have to apply multiple rules within a calculation (must add or subtract numbers AND multiply or divide numbers in the same calculation), follow the order of operations and apply each rule to determine significant figures before moving on to the next operation. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is greater than or equal to 5, we “round up” and increase the retained digit by 1. The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures: 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5) 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is lesser than 5) Let’s work through these rules with a few examples. Example 1: Rounding Numbers Round the following to the indicated number of significant figures: 31.67 (to two significant figures) 8.1649 (to three significant figures) 0.90275 (to four significant figures) [reveal-answer q=”347710″]Show Answer[/reveal-answer] [hidden-answer a=”347710″] 31.67 rounds “up” to 32 (the dropped digit, 6, is greater than or equal to 5) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5) 0.90275 rounds “up” to 0.9028 (the dropped digit, 5, and is greater than or equal to 5) [/hidden-answer] Check Your Learning Round the following to the indicated number of significant figures: 0.424 (to two significant figures) 0.0038661 (to three significant figures) 421.25 (to four significant figures) 28,683.5 (to five significant figures) [reveal-answer q=”776155″]Show Answer[/reveal-answer] [hidden-answer a=”776155″] 0.42 0.00387 421.3 28,684 [/hidden-answer] Example 2: Addition and Subtraction with Significant Figures Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction). Add 1.0023 g and 4.383 g. Subtract 421.23 g from 486 g. [reveal-answer q=”998865″]Show Answer[/reveal-answer] [hidden-answer a=”998865″] Answer is 5.385 g (round to the thousandths place; three decimal places) Answer is 65 g (round to the ones place; no decimal places) [/hidden-answer] Check Your Learning Add 2.334 mL and 0.31 mL. Subtract 55.8752 m from 56.533 m. [reveal-answer q=”409370″]Show Answer[/reveal-answer] [hidden-answer a=”409370″] 2.64 mL 0.658 m [/hidden-answer] Example 3: Multiplication and Division with Significant Figures Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division). Multiply 0.6238 cm by 6.6 cm. Divide 421.23 g by 486 mL. [reveal-answer q=”481696″]Show Answer[/reveal-answer] [hidden-answer a=”481696″] [/hidden-answer] Check Your Learning Multiply 2.334 cm and 0.320 cm. Divide 55.8752 m by 56.53 s. [reveal-answer q=”155520″]Show Answer[/reveal-answer] [hidden-answer a=”155520″] 0.747 cm 2 0.9884 m/s [/hidden-answer] In the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation. Example 4: Calculation with Significant Figures One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters. [reveal-answer q=”519671″]Show Answer[/reveal-answer] [hidden-answer a=”519671″] [/hidden-answer] Check Your Learning What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm 3 ? [reveal-answer q=”129357″]Show Answer[/reveal-answer] [hidden-answer a=”129357″]1.034 g/mL[/hidden-answer] Example 5: Experimental Determination of Density Using Water Displacement A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown. Use these values to determine the density of this piece of rebar. Rebar is mostly iron. Does your result in number 1 support this statement? How? [reveal-answer q=”565576″]Show Answer[/reveal-answer] [hidden-answer a=”565576″] The volume of the piece of rebar is equal to the volume of the water displaced: (rounded to the nearest 0.1 mL, per the rule for addition and subtraction) The density is the mass-to-volume ratio: (rounded to two significant figures, per the rule for multiplication and division) The density of iron is 7.9 g/cm 3 , very close to that of rebar, which lends some support to the fact that rebar is mostly iron. [/hidden-answer] Check Your Learning An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown. Use these values to determine the density of this material. Do you have any reasonable guesses as to the identity of this material? Explain your reasoning. [reveal-answer q=”76927″]Show Answer[/reveal-answer] [hidden-answer a=”76927″] 19 g/cm 3 It is likely gold; the right appearance for gold and very close to the density given for gold in Measurements . [/hidden-answer] Accuracy and Precision Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure 2). Figure 2. (a) These arrows are close to both the bull’s eye and one another, so they are both accurate and precise. (b) These arrows are close to one another but not on target, so they are precise but not accurate. (c) These arrows are neither on target nor close to one another, so they are neither accurate nor precise. Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table 1. Table 1. Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers Dispenser #1 Dispenser #2 Dispenser #3 283.3 298.3 296.1 284.1 294.2 295.9 283.9 296.0 296.1 284.0 297.8 296.0 284.1 293.9 296.1 Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL). Key Concepts and Summary Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the measurement. The uncertainty of a calculated value depends on the uncertainties in the values used in the calculation and is reflected in how the value is rounded. Measured values can be accurate (close to the true value) and/or precise (showing little variation when measured repeatedly). Exercises Express each of the following numbers in scientific notation with correct significant figures: 711.0 0.239 90743 134.2 0.05499 10000.0 0.000000738592 Express each of the following numbers in exponential notation with correct significant figures: 704 0.03344 547.9 22086 1000.00 0.0000000651 0.007157 Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty: the number of eggs in a basket the mass of a dozen eggs the number of gallons of gasoline necessary to fill an automobile gas tank the number of cm in 2 m the mass of a textbook the time required to drive from San Francisco to Kansas City at an average speed of 53 mi/h Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty: the number of seconds in an hour the number of pages in this book the number of grams in your weight the number of grams in 3 kilograms the volume of water you drink in one day the distance from San Francisco to Kansas City How many significant figures are contained in each of the following measurements? 38.7 g 2 × 10 18 m 3,486,002 kg 9.74150 × 10 −4 J 0.0613 cm 3 17.0 kg 0.01400 g/mL How many significant figures are contained in each of the following measurements? 53 cm 2.05 × 10 8 m 86,002 J 9.740 × 10 4 m/s 10.0613 m 3 0.17 g/mL 0.88400 s The following quantities were reported on the labels of commercial products. Determine the number of significant figures in each. 0.0055 g active ingredients 12 tablets 3% hydrogen peroxide 5.5 ounces 473 mL 1.75% bismuth 0.001% phosphoric acid 99.80% inert ingredients Round off each of the following numbers to two significant figures: 0.436 9.000 27.2 135 1.497 × 10 −3 0.445 Round off each of the following numbers to two significant figures: 517 86.3 6.382 × 10 3 5.0008 22.497 0.885 Perform the following calculations and report each answer with the correct number of significant figures. 628 × 342 (5.63 × 10 2 ) × (7.4 × 10 3 ) 8119 × 0.000023 14.98 + 27,340 + 84.7593 42.7 + 0.259 Perform the following calculations and report each answer with the correct number of significant figures. 62.8 × 34 0.147 + 0.0066 + 0.012 38 × 95 × 1.792 15 – 0.15 – 0.6155 140 + 7.68 + 0.014 28.7 – 0.0483 Consider the results of the archery contest shown in this figure. Which archer is most precise? Which archer is most accurate? Who is both least precise and least accurate? Classify the following sets of measurements as accurate, precise, both, or neither. Checking for consistency in the weight of chocolate chip cookies: 17.27g, 13.05g, 19.46g, 16.92g Testing the volume of a batch of 25-mL pipettes: 27.02 mL, 26.99 mL, 26.97 mL, 27.01 mL Determining the purity of gold: 99.9999%, 99.9998%, 99.9998%, 99.9999% [reveal-answer q=”369115″]Show Selected Answers[/reveal-answer] [hidden-answer a=”369115″] 2. (a) 7.04 × 10 2 ; (b) 3.344 × 10 −2 ; (c) 5.479 × 10 2 ; (d) 2.2086 × 10 4 ; (e) 1.00000 × 10 3 ; (f) 6.51 × 10 −8 ; (g) 7.157 × 10 −3 4. (a) exact; (b) exact; (c) uncertain; (d) exact; (e) uncertain; (f) uncertain 6. (a) two; (b) three; (c) five; (d) four; (e) six; (f) two; (g) five 8. (a) 0.44; (b) 9.0; (c) 27; (d) 140; (e) 1.5 × 10 −3 ; (f) 0.45 10. (a) 2.15 × 10 5 ; (b) 4.2 × 10 6 ; (c) 2.08; (d) 0.19; (e) 27,440; (f) 43.0 12. (a) Archer X; (b) Archer W; (c) Archer Y [/hidden-answer] Glossary accuracy: how closely a measurement aligns with a correct value exact number: number derived by counting or by definition precision: how closely a measurement matches the same measurement when repeated rounding: procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation significant figures: (also, significant digits) all of the measured digits in a determination, including the uncertain last digit uncertainty: estimate of amount by which measurement differs from true value CC licensed content, Shared previously Chemistry. Provided by : OpenStax College. Located at : http://openstaxcollege.org . License : CC BY: Attribution . License Terms : Download for free at https://openstaxcollege.org/textbooks/chemistry/get
Courses/Martin_Luther_College/Organic_Chemistry_-_MLC/03%3A_Alcohols_Ethers_Thiols_Sulfides_and_Amines/3.01%3A_Alcohols_and_Phenols/3.1.05%3A_Alcohols_from_Carbonyl_Compounds-_Reduction	Objectives After completing this section, you should be able to determine whether a given reaction should be classified as an oxidation or a reduction. write an equation to represent the reduction of an aldehyde or ketone using sodium borohydride or lithium aluminum hydride. discuss the relative advantages and disadvantages of using sodium borohydride or lithium aluminum hydride to reduce aldehydes or ketones to alcohols. identify the product formed from the reduction of a given aldehyde or ketone. identify the aldehyde or ketone that should be used to produce a given alcohol in a reduction reaction. identify the best reagent to carry out the reduction of a given aldehyde or ketone. write an equation to represent the reduction of an ester or a carboxylic acid to an alcohol. identify the product formed from the reduction of a given ester or carboxylic acid. identify the esters or carboxylic acids that could be reduced to form a given alcohol. Key Terms Make certain that you can define, and use in context, the key terms below. (organic) oxidation (organic) reduction Study Notes In your course in first‑year general chemistry, you probably discussed oxidation‑reduction reactions in terms of the transfer of electrons and changes in oxidation numbers (oxidation states). In organic chemistry, it is often more convenient to regard reduction as the gain of hydrogen or loss of oxygen, and oxidation as the gain of oxygen or the loss of hydrogen. There is no contradiction in using these various definitions. For example, when hydrogen is added across the double bond of ethene to reduce it to ethane, the oxidation number of the doubly bonded carbon atoms decreases from −II to −III. Similarly, when 2‑propanol is oxidized to acetone hydrogen is removed from the compound and the oxidation number of the central carbon atom increases from 0 to +II. If necessary, review the concept of oxidation number . Reduction of Aldehydes and Ketones Like carbon, hydrogen can be used as a nucleophile if it is bonded to a metal in such a way that the electron density balance favors the hydrogen side. A hydrogen atom that carries a net negative charge and bears a pair of unshared electrons is called a hydride ion ( - :H). How much of the negative charge density resides on hydrogen depends on the difference in electronegativity between hydrogen and the metal it’s bonded to. The most common sources of the hydride nucleophile are lithium aluminum hydride (LiAlH 4 ) and sodium borohydride (NaBH 4 ). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH 4 is more polar, thereby, making LiAlH 4 a stronger reducing agent. Addition of a hydride anion (H: - ) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols. These reactions will be discussed in more detail in Chapter 19 . Predicting the Product of a Hydride Addition to a Carbonyl During the reduction the C=O double bond in the reactant and forms a C-O single bond in the product. The breaking of the C=O double bond allows for the formation of two single bonds in the product. One will be attached to the oxygen and one to the carbon which originally in the carbonyl. Both of these single bonds will be attached to an "H" in the product formed. Example $\PageIndex{1}$ Mechanism Both NaBH 4 and LiAlH 4 act as if they were a source of the hydride anion nucleophile. They hydride anion undergoes nucleophilic addition to the carbonyl carbon to form a C-H single bond and forming a tetrahedral alkoxide ion intermediate. The alkoxide ion is subsequently converted to an alcohol by reaction with a proton source. In the LiAlH 4 reduction, the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the borohydride reduction the hydroxylic solvent system achieves this hydrolysis automatically. The lithium, sodium, boron and aluminum end up as soluble inorganic salts. Note! The reaction and the corresponding mechanism of hydride reductions of carbonyls is fairly complicated. The following mechanism has been simplified for easier understanding.. 1) Nucleophilic attack to form a tetrahedral alkoxide intermediate 2) Protonation to form an alcohol In metal hydride reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH 4 and NaBH 4 are both capable of reducing aldehydes and ketones to the corresponding alcohol. Biological Reduction Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role. NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Instead of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD + . However, NAD + is stabilized by the fact that its nicotinamide ring is aromatic; it was not aromatic in NADH. CONADH2.png Reduction of Carboxylic Acids and Esters Carboxylic acids and esters can be converted to 1 o alcohols using Lithium aluminum hydride (LiAlH 4 ). Note that NaBH 4 is not strong enough to convert carboxylic acids or esters to alcohols. Notice that during these reaction two hydrogen atoms are added the to carbonyl carbon whereas only one hydrogen atom was added during aldehyde and ketone reductions. These reactions will be discussed in more detail in Chapter 21 . Example $\PageIndex{2}$ Worked Example How would you prepare the following molecules using a hydride reduction? a) b) Answer Analysis: The first step is to note if the target molecule is a 1 o or 2 o alcohol. Primary alcohols can be made by the hydride reduction of an aldehyde, carboxylic acid, or ester while secondary alcohols are made by the reduction of ketones. A 3 o alcohol cannot be made by a hydride reduction. Also, it is important to remember than aldehydes and ketones can be reduced by both NaBH 4 and LiAlH 4 whereas carboxylic acids and ester can only be reduced by LiAlH 4 . a) Because the target molecule is a secondary alcohol the starting material must be a ketone. b) Because the target molecule is a primary alcohol the starting material could be an aldehyde, a carboxylic acid, or an ester. Exercises Exercise $\PageIndex{1}$ 1) Give the aldehyde, ketone, or carboxylic acid (there can be multiple answers) that could be reduced to form the following alcohols. a) b) c) d) 2) Given the following alcohol, draw the structure from which it could be derived using only NaBH 4 a) b) c) d) Answer 1) a) b) c) d) 2) Note, NaBH 4 is only a strong enough reducing agent to reduce ketones and aldehydes. a) b) Exercise $\PageIndex{2}$ Give the aldehyde, ketone, or carboxylic acid (there can be multiple answers) that could be reduced to form the following alcohols. Answer Exercise $\PageIndex{3}$ Given the following alcohol, draw the structure from which it could be derived using only NaBH 4 . Answer Note: NaBH 4 is only a strong enough reducing agent to reduce ketones and aldehydes.

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