Chapter 4: The Interaction of Light and Matter

Major Concept Area: The Atomic/Molecular View of Matter

• Atoms and molecules undergo a variety of motions having characteristic energies.
• Light interacts (couples) with these motions and provides information about them.

The Quantized Atom. As we saw in Chapter 2, a very interesting fact about atoms has been known since the middle of the 19th century. When atoms are irradiated with white light (light consisting of photons of all wavelengths), they absorb only certain definite frequencies of the light, not all frequencies. Atoms of different elements absorb different frequencies. How do we interpret this? We have seen that the frequency of light is a measure of its energy. Thus atoms absorb only light of certain energies. This implies that the atom is only allowed to have certain energies. We now know that the electric field associated with the light photon interacts with the electrons of the atom. When a photon is absorbed by an atom, the energy of the photon is used to cause an electron of the atom to make a transition ("jump") to a higher energy state within the atom. The photon that was absorbed disappears. Only certain energy states are allowed to the electrons. That is why only photons of certain frequencies (energies) can be absorbed. The photon energy is exactly the same as the energy difference between 2 allowed electron states. We write this as

This equation has been tremendously important in the development of our understanding of the detailed structures of atoms and molecules. Virtually everything we know about the atom, its internal structure, and the structures of molecules containing several atoms, is a result of studying the interaction of light with the atom or molecule. For you to have a qualitative understanding of this interaction is therefore very important.

4-1 The Absorption of Light by Atoms and Molecules. We begin this section by stating some very important ideas, which will then be illustrated in subsequent discussion. The ideas are these:

• Light consists of oscillating electric and magnetic fields.
• Because nuclei and electrons are charged particles, their motions in atoms and molecules generate oscillating electric fields.
• An atom or molecule can absorb energy from light if the frequency of the light oscillation and the frequency of the electron or molecular "transition motion" match. Unless these frequencies match, light absorption cannot occur. The "transition motion" frequency is related to the frequencies of motion in the higher and lower energy states (Equation 4-1-1).
• By measuring the frequencies of light absorbed by an atom or molecule, we can determine the frequencies of the various transition motions that the atom or molecule can have. Thus we can use light absorption to probe the dynamics of atoms and molecules!

These ideas are the basis for the techniques of spectroscopy, the study of the interaction of light and matter. The word means "measurement of spectrum," where the spectrum of a substance is the array of frequencies of light absorbed by its atoms or molecules.

You should already be familiar with the idea of energy absorption through matching frequencies. This is exactly the way in which you transfer energy to a child on a swing. You push at a frequency that matches that of the swing. Energy is transferred successfully, making the swing go to a higher energy state (higher amplitude), only if the push is made at the right time. If the frequency of pushing does not match the frequency of swinging, very little if any energy is transferred. In fact, the swing may actually lose energy if you try to push it while it is moving back toward you. This idea operates in the same way when light and matter interact. It is complicated only by the fact that, in contrast to the swing, molecules can undergo more than one type of motion.

Molecular Motions and Energy. There are many ways in which a molecule can move, and each motion contributes to the energy of the molecule. Molecular motion is quantized -- that is, each motion can occur with only certain energies. A molecule may gain or lose energy in one of these motions by absorbing or emitting a photon of light. The photon has energy that matches the allowed energy gain or loss associated with the motion. By measuring the energies of the photons absorbed by the molecule, we obtain its spectrum. The spectrum provides information about the motions of the molecule. We begin with an expression for the energy of a molecule, written as a sum of the energies of the various motions possible to it. Each motion will then be described, along with the region of the electromagnetic spectrum with which it interacts. The total energy of a molecule may be written as follows:

We now examine each term in this equation.

4-2 Nuclear Magnetic Resonance--E_spin. The nuclei of some (but not all) atoms behave as if they were rotating, or spinning, about an axis passed through them, much as a top spins about its central axis. This is quite easily visualized in terms of the particle nature of the nucleus; it is less easily visualized in terms of the wave picture, which by now we know is another face of nature that is manifested in the atomic/molecular realm. Spin of nuclei is similar to electron spin, discussed in Chapter 2. Examples of atoms whose nuclei spin are hydrogen, fluorine, the 13 isotope of carbon, and the 15 isotope of nitrogen. There are many others that will not be listed here. The focus in this chapter will be on the spin of the hydrogen nucleus, because hydrogen atoms occur very frequently in molecules and are therefore very useful in determination of structure. When placed in a strong magnetic field, the spinning motion of the hydrogen atom gives rise to two energy levels. These are shown in an energy level diagram in Figure 4-1.

The spin can be promoted from the lower to the higher level by interacting with and absorbing energy from electromagnetic radiation in the radio frequency (RF) region of the spectrum. A typical radio photon has a frequency of 10⁸ s^-1, a wavelength of 3 meters (about ten feet), and an energy of 6.6 x 10^-26 Joules. The separation between the two spin energy levels is a sensitive function of the molecular environment of the hydrogen atom in a molecule. Thus hydrogen atoms in different environments absorb photons of different energies. By plotting the amount of energy absorbed by the spinning nuclei versus the frequency of the RF radiation applied to the molecule, we obtain the nuclear magnetic resonance (NMR) spectrum of the molecule. This spectrum provides information about the chemical environment of the spinning hydrogen nucleus, and can be used to deduce the atomic bonding patterns in the molecule. NMR has now been developed to such a level of sophistication that it can be used to determine the 3-dimensional structures of huge protein molecules in solution. Recently, nuclear magnetic resonance spectroscopy has found medical applications, in which context it is known as magnetic resonance imaging (MRI). MRI utilizes the spinning motions of atomic nuclei to provide a map of the internal structure of human body tissue.

We will now discuss the nature of NMR spectra in a systematic way, beginning with spectra of simple molecules, and gradually escalating in complexity. As we proceed through the discussion, we will introduce some of the jargon used by practicing chemists when discussing NMR. As a first bit of jargon, we will henceforth refer to hydrogen atoms as protons. This is appropriate, because it is the nucleus of the hydrogen atom that is spinning, and this nucleus indeed consists of a single proton. We begin by considering the NMR spectrum resulting from the molecule shown in Figure 4-2a. The molecule has a single proton on the right (on carbon 1), and a pair of protons on the left (on carbon 3). Carbon atoms 1 and 3 are separated by a carbon atom attached to two chlorine atoms. The single proton on the right is attached to a carbon atom that is attached to 2 chlorine atoms. In contrast, each proton on the left is attached to a carbon atom that is attached to one chlorine atom and a second hydrogen atom. The environments of the protons on carbon 1 and carbon 3 are clearly different. The energy level spacing of the two types of protons should thus be different, and two signals, one for each type of proton, should appear in the NMR spectrum. The spectrum is shown in Figure 4-3a.

Several aspects of this spectrum are important. First, the single sharp signal at exactly 0 ppm is not due to the subject molecule, but is instead due to a reference compound named tetramethylsilane, or TMS. The structure of TMS is shown in Figure 4-2d. By general agreement, the positions of NMR signals are reported relative to the position of the signal for TMS. Second, the unit used to measure resonance position, ppm (parts per million), is not a unit of energy. It has no units at all. It may be regarded as a measure of the shift in position between the TMS resonance and a particular resonance in the compound of interest. It is usually called the chemical shift. The chemical shift indicates the chemical environment in which the spinning nucleus is found, as we will see shortly. Ignoring the TMS signal, then, which is present in all NMR spectra, we see that there are indeed two signals. Interestingly, the sizes, or intensities, of the signals are different. The right signal appears to be twice as tall as the one on the left. This suggests that the right signal is due to the pair of protons on carbon 3, and the left signal to the single proton on carbon 1. It also appears that there is a relationship between the position of a signal in the spectrum and the number of chlorine atoms in the immediate environment of the proton(s) giving rise to the signal. We summarize our observations of the simple spectrum in Figure 4-3a in terms of three rules:

• Rule 1: Each unique proton environment generates one signal.
• Rule 2: The intensity (size) of an NMR signal gives a measure of the number of equivalent atoms responsible for the signal. A two-proton signal is twice as large as a one-proton signal.
• Rule 3: The greater the number of electronegative atoms near a proton, the further to the left of TMS the proton signal will be.

Rule 3 requires some comment. In Figure 4-3a, the signal due to the hydrogen atom on carbon 1 is shifted further away from TMS than is the signal of the hydrogens on carbon 3. In NMR jargon, the left signal is said to be further downfield (meaning further to the left) than the right signal. The reason for this is that the carbon-1 proton is close to two electronegative chlorine atoms. These pull electrons away from carbon 1, which in turn pulls them away from the hydrogen atom. The less electron density a hydrogen atom has around it, the further downfield its NMR resonance occurs. This type of consideration helps us to assign signals to particular hydrogen atoms in molecules.

Figure 4-2b shows a molecule that appears simpler than the molecule in Figure 4-2a, but is in fact somewhat more complex in NMR terms. The molecule in Figure 4-2b can be obtained from that in Figure 4-2a by removing the central carbon atom, bringing the proton-bearing carbons together. Clearly the carbon 1 proton is still in a different environment than are the pair of carbon 2 protons, so we might expect the NMR spectrum of this new molecule to be very similar to that of the three-carbon molecule. Specifically, we might expect two signals, one twice as intense as the other, with the less intense one downfield. The spectrum is shown in Figure 4-3b. Although there are two signals, in roughly the expected locations, the signals appear to be split into subsignals. The upfield signal now consists of two, and the downfield signal of three, closely-spaced subsignals. Chemists call the upfield signal a doublet, and the downfield signal a triplet, to indicate the number of subsignals. Apparently, there is a complication showing up here that is not present in the spectrum in Figure 4-3a. As we will see, this complication is what makes NMR so useful for determination of structure. The splitting of signals seen in Figure 4-3b occurs because protons on neighboring atoms "feel" each other's spins. A spinning nucleus is, in fact, a tiny magnet. When it is close enough to another proton, the second proton feels the small magnetic field of its neighbor as a tiny perturbation on the large magnetic field applied to the sample. The effect of this proton-proton interaction is to cause small changes in the energy level spacings of the proton that show up as the splittings that we see in Figure 4-3b. Interestingly, the number of subsignals into which a signal is split appears to be one greater than the number of neighboring protons. We can now add a couple of new rules to our list:

• Rule 4: Protons on neighboring atoms interact with each other, and split each other's signals.
• Rule 5: The number of subsignals in a multiplet is one more than the number of neighbor protons.

Clearly the splitting of signals due to neighboring atoms (called spin-spin splitting in the jargon of the chemist) is of potentially great value in assigning signals and in determining structure; it tells us the number of protons on the atom next door to the atom bearing the protons being observed.

To test our rules, we consider the molecule in Figure 4-2c. Again we see only two different proton environments here. The two carbon-1 protons are in one environment; the three carbon-2 protons are in a different environment. We expect two signals, one of total intensity two and the other of total intensity three. The signal of intensity two should be further downfield because the carbon 1 protons have an electronegative chlorine atom in their neighborhood. We expect the downfield signal to be split to a quartet due to interaction with the three protons on carbon 2; and the upfield signal to be split to a triplet via interaction with the two protons on carbon 1. The spectrum is in Figure 4-3c, and appears much as we expected it to. The NMR rules seem to work fine.

We consider one final example, the NMR spectrum of isobutyraldehyde, which will reveal a couple of other useful subtleties of NMR spectra. The molecule of interest and its NMR spectrum are shown in Figure 4-4a and Figure 4-4b. The protons labelled "a" in the figure experience the same environment (this is clear from a model of the molecule), and are expected to give one signal. This signal should be the furthest upfield because these protons are furthest removed from the electonegative oxygen atom. Because there is a single proton, "b", on the neighboring carbon atom, the signal for protons "a" should be a doublet with total relative intensity 6. This is clearly seen in the spectrum. Proton "c" should give the furthest downfield signal because it is attached to an oxygen-bearing carbon atom. This, too, is expected to be a doublet due to interaction with proton "b". Although the splitting of the proton "c" signal is small, it is discernible in Figure 4-4b. Note that the signal from "c" is very far downfield, much more so than the signals influenced by chlorine atoms in previous examples. This results from the very high electronegativity of the oxygen atom, which substantially depletes the electron density of the attached carbon atom. Finally, we examine the signal for proton "b". This is our first encounter with a proton that interacts with two other sets of protons, and it is important to understand how the splitting works in a situation like this. Figure 4-4c shows a magnified view of the signal for proton "b", which occurs at a chemical shift of 6.4 in 4-4b. The signal shows a total of 14 lines, which is many more than we have encountered in a multiplet up to this point. The explanation for all of these lines runs as follows. First, the signal for "b" is split to 7 lines by the interaction with the 6 "a"-type protons on neighboring carbon atoms. Then, each component of the 7-line pattern is split to two lines by the interaction with proton "c". The total is 14 lines. In terms of this interpretation, Figure 4-4c clearly shows a 7-line pattern, each component of which is split into two lines. This pattern is called a doublet of septets, to indicate that it results from interaction of "b" with two distinctly different types of protons. On the basis of the isobutyraldehyde example, we are in a position to state two additional rules.

• Rule 6: A proton on an oxygen-bearing carbon is shifted very far downfield.
• Rule 7: A proton with "n" neighboring protons on one side and "m" on the other will give a signal split into (n+1)(m+1) components.

Example 4-1. The structure of propionaldehyde, C₃H₆O, is shown in Figure 4-5a. Predict what the NMR spectrum of propionaldehyde should look like.

Solution. The spectrum should show 3 signals of relative intensities 3 to 2 to 1. The signal of intensity 3, due to the hydrogen atoms labelled "a" in the figure, will be furthest to the right because these atoms are far removed from the oxygen atom. The signal should be split into 3 components from the effect of the spins of the two H atoms on the middle carbon. The signal of intensity 1, due to hydrogen "c", is furthest left because this hydrogen atom is near the oxygen atom. This signal also will be split into 3 components due to the effect of the spinning motions of the two H atoms on the middle carbon atom. Finally, the signal due to hydrogens "b" will appear at an intermediate position, with relative intensity 2. It will be split to two components by hydrogen "c"; and each of these will be split to 4 components by hydrogens "a". This signal will consist of 8 components altogether. The spectrum is shown in Figure 4-5b.

Chemists have invented a shorthand system for presenting the information in an NMR spectrum without having to have the spectrum itself (i.e., the picture) available. Each signal in the spectrum is described by stating, first, its multiplicity (the amount of splitting); second, the chemical shift in parts per million downfield from TMS; and third, the relative intensity of the signal, in terms of number of identical protons giving rise to the signal. The propionaldehyde spectrum is described as follows in this shorthand system:

The Greek letter, d, symbolizes the chemical shift in units of parts per million. This brief description is sufficient to allow us to visualize and, if desired, even draw the spectrum, and to interpret it in terms of structure.

In this section, we have considered a number of NMR spectra and from them have developed a simple set of rules for their interpretation and prediction. It is important to realize that, although these rules are very useful in understanding the NMR spectra of relatively simple molecules, the spectra of many molecules involve complexities that our simple rules are unable to handle. Nonetheless, we have laid the groundwork for an understanding of NMR spectra that can be expanded in more advanced books.

Table 4-1 gives chemical shifts characteristic of a variety of proton environments.

Spin-spin Splitting--A Closer Look. To this point, we have discussed the most basic aspects of nuclear magnetic resonance spectroscopy. We have acknowledged the existence of spin-spin splitting, but have not said much about its molecular origins, and have not explained the origin of the number of subsignals in a multiplet. In this section we will pursue these matters in a bit more depth. As a basis for discussion, consider a molecule in which there are three protons on carbon 1 and two protons on the neighboring carbon atom 2. Our rules tell us that the signal due to carbon-1 protons will be split to a triplet, and that the signal due to carbon-2 protons will be split to a quartet. Why a triplet? Why a quartet? Why do the subsignals of the triplet have relative intensities 1:2:1, while those of the quartet have relative intensities 1:3:3:1? What is the significance of the magnitude of the splitting between the components of a multiplet?

Focus first on the splitting of the signal due to the carbon-1 protons. The spinning protons on carbon 2 behave like tiny magnets. The magnetic field is generated by the spinning negative charge, which is much like an electric current. The direction of the magnetic field generated by a carbon-2 proton depends on its direction of spin (clockwise or counterclockwise). Either direction is equally probable for each of the three carbon-2 protons. Given this, it is easy to appreciate that there are 3 possible combinations of spin directions for the 2 protons on carbon 2. Both can be spinning clockwise, in which case their magnetic fields add together and reinforce the large laboratory magnetic field. One can be spinning clockwise and the other counterclockwise, in which case their magnetic fields cancel. This arrangement can occur in two different ways because either proton 1 or proton 2 can have clockwise spin. The arrangement of one spin "up" (cw) and the other "down" (ccw) is thus twice as probable as the arrangement with both spins "up". Finally, both protons can have spin "down", in which case their magnetic fields add together but oppose the applied laboratory field. The spin arrangements of the carbon-2 protons are perhaps better appreciated when presented pictorially. We conclude that the carbon-1 protons will feel three slightly different magnetic fields, one slightly larger than, one equal to, and one slightly smaller than the applied field. The middle possibility is twice as probable as the other two.

Now we are in a position to appreciate why the signal due to the carbon-1 protons is split to 3 subsignals of relative intensities 1:2:1. This is a simple consequence of the fact that the carbon-2 protons create 3 slightly different magnetic fields that can be experienced by the carbon-1 protons. One signal arises from each of these magnetic fields, with intensity proportional to the probability of the spin arrangement giving rise to the magnetic field perturbation caused by protons on carbon 2.

In similar fashion, we can understand why the signal due to carbon-2 protons is split to 4 subsignals. The 3 protons on carbon 1 can have their spins aligned in 4 different ways (all "up" (one possibility), two "up" one "down" (3 possibilities), one "up" two "down" (3 possibilities), and all "down" (again only one possibility). The carbon-2 protons will experience 4 slightly different magnetic fields with probabilities 1:3:3:1, so will be split to a quartet whose components have intensities reflecting these probabilities. The number of components in a multiplet and their relative intensities are thus seen to have a physically plausible explanation.

In general, as we have seen, "n" neighboring protons cause a splitting into "n+1" components. For reasons that we need not pursue, the relative intensities of the components resulting from splitting by any number of equivalent neighboring protons can be readily computed from Pascal's Triangle, in which a particular number is calculated as the sum of the numbers to the left and right of it in the row above.

We conclude our examination of spin-spin splitting by considering the magnitude of the separation between two adjacent signals within a multiplet. We can base the initial part of the discussion on the same molecule used above, in which there are three protons (call them protons A) on carbon 1 and two protons (protons B) on the neighboring carbon atom 2 (later we will add carbon 3, bearing one proton, D). The magnetic interaction between protons A and B is called spin-spin coupling. Protons A are said to couple with (meaning that they interact with) protons B. The strength of the interaction determines the magnitude of the splitting between adjacent components of a multiplet. The stronger the interaction, the larger the splitting. Further, spin-spin coupling between the two sets of protons A and B obeys a version of Newton's third law, that for every action there is an equal and opposite reaction. If protons A interact with protons B with a certain strength, splitting the B signal by a certain amount, then protons B interact with protons A with exactly the same strength, splitting the A signal by exactly the same amount. The size of the splitting within a multiplet is called the coupling constant, J, and is usually measured in frequency units. We will not be concerned with measuring coupling constants, so do not worry at this point about the units. The importance of coupling constants for us will be this: the signals for two sets of interacting (coupled) protons show exactly the same coupling constant. Thus it is often possible by looking at the NMR spectrum to decide which signals in the spectrum are coupled. This is an additional aid in building the structure of a molecule from an analysis of its NMR spectrum.

To show how this works, we complicate our test molecule somewhat by adding a third carbon to which is attached a single proton, D. The structure is then as shown. We expect the following coupling interactions within this structure:

• A with B only; the signal for A should be a 1:2:1 triplet
  
• D with B only; the signal for D should be a 1:2:1 triplet
  
• B with A and also with D; the signal for B should be a 1:3:3:1 quartet as a result of coupling with A. Each component of the quartet should be split to a doublet as a result of coupling with D.
  

Suppose that the interaction between A and B is stronger than that between B and D. For discussion sake, we will assume that B and D couple with only 1/3 the strength of A and B. Thus if the coupling constant for interaction between A and B is 12 s^-1, then that for interaction between B and D is 4 s^-1. Then we can make the following additional statements about the appearance of the NMR spectrum:

• The splitting within the multiplet due to protons A (12 s^-1) will be 3 times bigger than the splitting within the multiplet due to protons D.
• The B multiplet will consist of 4 pairs of signals. The members of a pair will be separated by 4 s^-1; the corresponding members of two adjacent pairs will be separated by 12 s^-1.

The spectrum of our test molecule might look as shown, where we have assumed chemical shifts of 1 ppm, 2 ppm, and 4 ppm for the A, B, and D protons, respectively.

4-3 Infrared Spectroscopy--E_vibration. Infrared spectroscopy allows us to examine the vibrational motions of molecules. We will discuss these motions as one of several types of motion that the molecule as a whole can undergo.

E_translation, E_rotation, E_vibration. The nuclear spinning motion discussed above involves only one atom of a molecule. The three motions to be discussed now -- translation, rotation, and vibration -- are motions of the molecule as a whole in space, involving concerted movements of all of the atoms. For illustration, we use a very simple molecule consisting of only 2 atoms. Such molecules are called diatomic molecules. Examples are O₂, H₂, HCl, and CO. A generic diatomic molecule is pictured in Figure 4-6a.

First, the entire molecule may move through space in an arbitrary direction and with a particular velocity. This is called translational motion and with it we associate the translational kinetic energy of the molecule:

The translational velocity of the molecule has components along each axis of a Cartesian coordinate system, so

The total translational KE of the molecule is made up of three parts, each representing the kinetic energy of the molecule along one reference direction. The molecule has 3 translational degrees of freedom (i.e., independent directions in which it can translate), one corresponding to each Cartesian axis. A molecule travelling through space at a temperature of 300 K has translational kinetic energy of 6.2 x 10^-21 Joules. However, the spacing between the allowed translational energies for an HCl molecule in a container 20 cm in length is on the order of only 10^-40 J. This is such a small energy that transitions between translational kinetic energy levels are difficult to observe. At the present time, no spectroscopy is done involving these energy levels.

Second, the molecule may rotate about an internal axis. For example, rotation of a diatomic molecule is diagrammed in Figure 4-6b. Like translation, a molecular rotation may be resolved into three mutually perpendicular components. The rotational kinetic energy of the molecule is

where I_x, I_y, and I_z are the moments of inertia (the moment of inertia is the rotational equivalent of mass) about the x, y, and z axes, and w_x, w_y, w_z are the angular velocities (in units of radians/s) about these axes. There are 3 rotational degrees of freedom, one for each Cartesian axis. An exception arises in the case of a linear molecule, for which one of the three axes is coincident with the molecular axis. The molecule has no rotational energy about this axis, because the corresponding moment of inertia is zero (i.e., no mass lies off the axis). Linear molecules have only two rotational degrees of freedom.

Rotation is a type of oscillation, in which the number of rotations per second is the frequency. If the molecule has associated with it a permanent electric field, then rotation of the molecule about some axis causes an oscillation in the electric field. If this oscillation matches in frequency the oscillation in the electric field of an impinging photon, the molecule can absorb the photon and go to a higher energy rotational state. Rotations of molecules at room temperature typically occur with energy on the order of 10^-23 J. Photons with frequencies near 10¹¹ s^-1 (wavelength, 0.03 m), in the microwave region of the electromagnetic spectrum, are required to interact with such motions. The rotational (or microwave) spectrum of a molecule gives information about the moments of inertia, from which can be extracted the bond lengths and bond angles in the molecule. These properties of molecules were introduced in Chapter 3. Interpretation of rotational spectra is complex, and we will not be concerned with it at this time.

Finally, the molecule may vibrate. Chemically bonded atoms are like masses connected by a spring (the bond). The bonds can be stretched and compressed somewhat, and the bond angles can vary within certain limits. Motions of the molecule in which bonds are compressed and/or stretched, or in which bond angles get larger then smaller, but in which the center of mass of the molecule does not move through space, are called vibrational motions. A diatomic molecule vibrates by repeated stretching and contraction of the bond joining the two atoms, as shown in Figure 4-6c. The molecule has one vibrational degree of freedom in addition to 3 translational and 2 rotational degrees of freedom.

For a polyatomic molecule the number of vibrational degrees of freedom (called vibrational modes) is the total number of degrees of freedom minus those of translation and rotation. The total is 3N, where N = the number of atoms in the molecule. This is true because each atom may independently move in any of three directions and therefore has 3 degrees of freedom available to it; the molecule has a total of 3N. The number of vibrational degrees of freedom is 3N-6 for a non-linear polyatomic molecule, and 3N-5 for a linear polyatomic molecule. The water molecule, with N = 3, has 3*3-6 = 3 vibrational degrees of freedom. The three vibrations are shown in Figure 4-6d.

Vibrational motions occur with characteristic frequencies (just as does the motion of the playground swing). In addition, the vibration causes a change in the distribution of the electrons in the molecule, which may give rise to an oscillating electric field. If the frequency of impinging light matches the frequency of the vibrational motion, the molecule may absorb light photons (energy) and move to a vibrational state in which the amplitude of vibration is increased. By measuring the frequencies of light absorbed, we may observe how rapidly the molecular vibrations occur! Since the frequencies of vibrations are directly related to the strengths of the chemical bonds (i.e., to the stiffness of the spring), vibrational frequencies give us information about chemical bond strengths. Typically, molecules vibrate at between 10¹³ and 10¹⁴ times per second. Photons with frequency 10¹⁴ s^-1 (wavelength 3 * 10^-6 m) have energy 6.6 x 10^-20 J. Vibrational motions therefore have frequencies that are about 10³ times larger than the frequencies of molecular rotation and a million times larger than nuclear spin frequencies. Higher energy photons in the infrared region of the spectrum are required to excite the vibrations.

Infrared (IR) spectroscopy deals with transitions between vibrational energy levels in molecules; it is therefore also called vibrational spectroscopy. Because IR spectra are easily acquired and provide much structural information, we will discuss them in some detail. An IR spectrum is a plot of the energy of the infrared radiation (expressed either in micrometers (1 mm = 10^-6 m) or wavenumbers, the reciprocal of wavelength in centimeters) versus the percent of light transmitted by the compound, and is displayed as shown in Figure 4-7. Within the energy range between 200 and 4000 cm^-1, the vibrational spectrum of the molecule appears as a series of absorption bands of variable intensity. The absorption bands are the downward-facing spikes, or peaks. Note that when absorption occurs, the percent of light transmitted by the sample decreases from near 100 to various smaller values. Each absorption band in the spectrum corresponds to a vibrational transition within the molecule, and gives a measure of the frequency at which the vibration occurs. For the water molecule, with three vibrational degrees of freedom, there are three sets of energy levels within which transitions may occur. These are shown in Figure 4-8, with the associated vibrations. The spacing between energy levels depends upon the nature of the vibration. Each spacing requires a photon of different energy to cause the transition, so infrared photons of three different energies are absorbed by H₂O. The frequencies of the photons (and of the vibrations) are 3600, 3500, and 1650 cm^-1, respectively, for the asymmetric stretch, the symmetric stretch, and the bend.

All molecules vibrate, but not all vibrations interact with electromagnetic radiation. For a vibrational motion to absorb infrared electromagnetic radiation, it must produce a change in the dipole moment of the molecule. (Dipole moments were discussed in Chapter 3. Recall that the dipole moment is a small electric field within the molecule, resulting from its structure.) HCl, for example, with a center of positive charge located near the H atom and a center of negative charge located near the Cl atom, has a dipole moment. Moreover, the magnitude of the dipole moment changes as the bond stretches, so this vibration is able to absorb IR radiation and is said to be IR active. HCl thus exhibits an IR spectrum. The N₂ molecule, on the other hand, has no dipole moment. Furthermore, stretching the NN triple bond does not change the dipole moment. Consequently, the molecular vibration is infrared inactive (cannot directly absorb IR radiation). There are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that change the value of the dipole moment from 0 to some non-zero value. Consider the CO₂ molecule in Figure 4-9a. This molecule has no permanent dipole moment, since the individual bond dipoles exactly cancel. However, when it undergoes a bending vibration (shown in Figure 4-9b) its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is therefore IR active. All three vibrations of the water molecule change the dipole moment and are IR active. Three bands appear in the IR spectrum. The requirement that a vibration must cause a change in the dipole moment in order for a molecule to absorb radiation is another situation in which an oscillating electric field in the molecule must match that in the radiation. Oscillation of the dipole moment generates such an oscillating electric field.

In addition to the number of distinct vibrations expected for a molecule, we can anticipate their locations in the spectrum. For a diatomic molecule, A-B, the wavenumber, n, of the infrared radiation absorbed by vibration of the molecule is given by

k_AB is the force constant for the A-B bond and measures the bond strength; m_AB is called the reduced mass; and M_A and M_B are the masses of the atoms. This equation indicates that the heavier the atoms involved in the bond, the lower the absorption frequency, given a constant bond strength. Care is required in applying this equation, however. For the series HF, HCl, HBr, and HI, the IR absorptions occur at, respectively, 3958, 2885, 2559, and 2230 cm^-1. It is tempting to conclude that this is due to the increase in mass of the halogen atom. Calculation of m for each molecule reveals that the reduced mass actually changes very little (try it). The change in frequency is instead due to the decrease in bond strengh along the series. Equation 4-3-4 suggests that, because the frequency of a vibration depends on the identities of the two bonded atoms and the strength of the bond between them, a particular type of bond should always absorb at more or less the same location in the infrared spectrum. This turns out to be true. Thus, for example, all molecules containing C-H bonds show absorptions in the region of 3000 cm^-1, corresponding to the stretching motions of these bonds.

To develop a feel for the manner in which structural changes in the framework of a molecule are reflected in the infrared spectrum, we will consider the IR spectra of a series of relatively simple molecules based on a chain of 5 carbon atoms. The first member of the series has formula C₅H₁₂, and is called pentane. The infrared spectrum of pentane is in Figure 4-10a. The spectrum is relatively uncomplicated. There is a group of absorption bands in the region 2800-3000 cm^-1 that correspond to C-H stretching vibrations; a pair of bands in the 1350-1500 cm^-1 region, due to C-H bending vibrations; and a series of very weak bands in the region between 700 and 1400 cm^-1, due to various complex motions of the molecule that involve C-C stretches, C-C bends, and more complicated C-H motions. The most intense absorption in the spectrum is clearly that due to C-H stretching, because the largest dipole moment change arises from these motions.

The IR spectrum of 1-chloropentane, C₅H₁₁Cl, is shown in Figure 4-10b. The number preceding the name of this compound indicates that the chlorine atom is attached to the first carbon atom in the chain. The spectrum is remarkably like that of pentane, with two obvious changes. First, the bands in the 700-1400 region, although they show the same pattern as in this region of the pentane spectrum, are more intense than those for pentane. This is a result of the chlorine atom, which makes the molecule polar, resulting in a larger change in dipole moment during many of the vibrations. Thus intensity of absorption is increased. Second, there is a strong band at 650 cm^-1 that is absent in the pentane spectrum. This band must be associated with the C-Cl stretching motion.

Next, we consider the spectrum of 1-pentanol, C₅H₁₁OH, shown in Figure 4-10c. Again, the bands in the 2800-3000 and 1350-1500 regions appear much as in pentane; and those in the 700-1400 region again show a similar pattern, but with enhanced intensity due to the presence of the electronegative OH group, and with a strong band near 1050. The new features of the spectrum are the intense, broad band at 3300-3400 cm^-1, and the very broad feature in the 600-700 cm^-1 region. The high energy 3300-3400 band corresponds to stretching of the O-H bond. A band in this region is characteristic of molecules containing -OH. The broad low energy feature is also a result of the presence of OH. The band is extremely broad because the O-H group is involved in hydrogen bonding with O-H groups of neighboring molecules. In most cases, molecules containing O-H groups are involved in hydrogen bonding to some extent, and produce a broad O-H stretching band in the IR spectrum. Hydrogen bonding, one of several types of intermolecular force, will be discussed in Chapter 6.

Finally, we examine the spectrum for the pentane derivative, CH₃-C(=O)-CH₂-CH₂-CH₃, which is called 2-pentanone (the -one ending tells us that the molecule contains the C=O unit). The spectrum is in 4-10d. The striking feature of this spectrum is the intense band at 1700 cm^-1, corresponding to the stretching vibration of the C=O group. All molecules containing the C=O group have a band in the region 1680-1750 cm^-1 in the infrared spectrum. The presence of such a band is an indication of this structural feature.

This brief discussion of the IR spectra of pentane derivatives should give you an appreciation of the relationship between structure and the positions of absorption bands in the IR spectrum. Let's look at two further examples.

Example 4-2. The structure and infrared spectrum of acetone, C₃H₆O, are shown in Figure 4-11. What structural information can be determined from this spectrum?

Solution. The bands just above 2800 cm^-1 are due to C-H stretching motions, which occur at high frequency because the H atom has such small mass. Any compound containing C-H bonds will have an absorption band in this region. The intense absorption band at 1730 cm^-1 is due to the stretch of the C=O bond. C=O bonds invariably absorb in the region 1750-1680 cm^-1, independent of the other groups bonded to carbon. The appearance of a band in this region in the spectrum of an unknown compound is diagnostic of a carbon-oxygen double bond. The bands at energy lower than 1500 cm^-1 arise from bending motions of C-H bonds and stretching and bending motions of C-C bonds. As a general rule, it is difficult to assign bands in this region to particular bonds in the molecule.

Example 4-3. The structure and infrared spectrum of t-butanol, (CH₃)₃COH, are shown in Figures 4-12a and 4-12b. What structural information is available from the spectrum?

Solution. The most prominent feature of the spectrum is the strong, broad band at about 3400 cm^-1. This is due to the stretching vibration of the O-H bond. The band is strong (i.e. intense) because there is a substantial change in dipole moment when the O-H bond stretches. It is broad as a result of hydrogen bonding, which is the interaction between the hydrogen atom of the O-H group of one molecule with the oxygen atom of the O-H group of another. Hydrogen bonding, an example of an intermolecular force, will be discussed more fully in Chapter 6. Many molecules containing an OH group exhibit a similar vibrational band in the region 3200-3600 cm^-1. Next, the intense series of bands between 2800 and 3000 cm^-1 is due to stretching motions of the various C-H bonds in the molecule. That they all occur at 3000 or less indicates that the hydrogen atoms are attached to carbon atoms that form single bonds only (rather than, say, double bonds). Finally, the intense, somewhat broad band at 1200 cm^-1 is due to the stretching vibration of the C-O single bond. The remaining bands cannot be assigned to particular bond types; instead, they are due to complex, interdependent vibrational motions of the molecule. However, the pattern of bands in the region below 1400 cm^-1, usually called the fingerprint region, is unique to a particular molecule, and can be used to identify it.

Tables 4-2 and 4-3 present characteristic frequencies for stretching and bending of a variety of bond types.

Functional Group Analysis and Fingerprinting. An organic functional group is a small group of atoms, bonded together in a characteristic way, that occurs in a large variety of compounds. Examples are C=O (carbonyl), C(=O)OH (carboxyl), and NH₂ (amine). The functional group is often emphasized in writing the formula of a compound containing it. Thus R-COOH represents any molecule containing the carboxyl functional group. R is a generic symbol for an organic "radical", consisting of primarily carbon and hydrogen atoms bonded together, and in which one carbon atom is capable of forming one additional bond. Organic functional groups differ from one another both in the strength of the bond(s) and in the masses of the atoms involved. For instance, the O-H and C=O functional groups contain atoms of different masses connected by bonds of different strengths. According to equation 4-3-4, we therefore expect the O-H and C=O groups to absorb IR radiation at different positions in the spectrum. The presence of a strong, broad band between 3200 and 3400 cm^-1 indicates the presence of an O-H group in the molecule, while the presence of a strong band around 1700 cm^-1 confirms the presence of a C=O group.

For organic molecules, the infrared spectrum can be divided into three regions. Absorptions between 4000 and 1300 cm^-1 are primarily due to specific functional groups and bond types. Those between 1300 and 900 cm^-1, the fingerprint region, arise from more complex vibrations in the molecules; and those between 900 and 650 cm^-1 indicate benzene rings in the molecule. Particularly important regions are indicated in Table 4-4.

We will briefly discuss the characteristic infrared absorption frequencies of a variety of common organic functional groups. While studying these, realize that the infrared spectrum of even a simple compound can be very complex. Some absorption bands are weak while others are strong. Some are close enough together to coincide or overlap. Interactions such as hydrogen bonding (Chapter 6) cause absorption bands to shift and broaden. Accordingly, do not expect to be able to identify every absorption band in a spectrum. Rather, look for the presence or absence of absorption bands in their characteristic spectral regions to confirm the presence or absence of particular functional groups.

Alkanes. Alkanes are hydrocarbons containing only single bonds. The most prominent bands in infrared spectra of alkanes and from alkane portions of more complicated organic compounds are due to C-H stretching and bending. The symmetric and asymmetric C-H stretching frequencies are in the regions 3000 to 2900 and 2900 to 2800 cm^-1 respectively, and are usually weak. Because most organic compounds contain several -CH₃, CH₂, and/or -CH groupings, there is often overlap of absorption bands in this region. Thus the presence or absence of bands is taken to indicate the presence or absence of C-H bonds in the molecule. The major bending modes of CH₂ and CH₃ groups appear at 1470-1420 and 1380-1340 cm^-1. The exact positions depend upon the nature of adjacent atoms. Interpretation is usually complicated by the presence of several bands of this type, or additional bands from other sources. The (usually) large number of C-C bonds in an organic molecule makes the C-C stretching vibrations in the 1300-1100 cm^-1 region uninterpretable in most cases.

Alkenes. Alkenes are hydrocarbons containing at least one C=C double bond. When H is attached to a double-bonded C, C-H stretching bands generally appear in the region 3100-3000 cm^-1. Alkanes and alkenes can thus be differentiated. The C=C stretching frequency occurs in the 1675-1600 cm^-1 region and varies with the atoms attached to the double-bonded carbons. Table 4-5 shows specific alkene bending absorptions.

Alkynes (or acetylenes) Alkynes are hydrocarbons containing a carbon-carbon triple bond, C(3)C). The C-H stretching vibration of terminal acetylenes (meaning that the last carbon in a chain is triple bonded to its neighboring carbon) generally appears at 3300 cm^-1 as a strong sharp band. The CC stretching band is found at 2150-2100 cm^-1 if the alkyne is monosubstituted (i.e., R-C(3)C-H) and at 2270-2150 cm^-1 if disubstituted (i.e., R-C(3)C-R'). These absorptions are usually weak.

Aromatic (benzene) Rings. Aromatic C-H stretching absorption appears in the region 3100-3000 cm^-1 and is readily distinguished from alkane, alkene, or alkyne C-H, allowing a reliable determination of the types of carbon-bound hydrogen in the molecule. The positions of aromatic C-H bending bands at 900-690 cm^-1 depend upon the substitution pattern of the benzene ring as indicated in Table 4-6. In the absence of other interfering absorptions, these strong, usually sharp bands can be used to distinguish postitional isomers of substituted benzenes. Sharp bands at ~1600 and ~1500 cm^-1 are characteristic of all benzene compounds; a band at 1580 cm^-1 appears when the ring is conjugated with a substituent. (The word conjugated means the substituent has a double bond one bond removed from the benzene ring).

Alcohols. Alcohols are compounds containing the O-H functional group. The characteristic infrared band due to the O-H stretching vibration appears at 3650-3600 cm^-1 in dilute solution. With increasing concentration in solution, or in pure liquids or solids, intermolecular hydrogen bonding broadens the band and shifts its position to lower wavenumber (3500-3200 cm^-1). The infrared spectrum of pure t-butyl alcohol in Figure 4-12b clearly shows the broad O-H stretching band at 3440 cm^-1. In addition to the O-H stretching vibration, alcohols exhibit O-H bending and C-O stretching transitions at 1500-1300 cm^-1 and 1220-1000 cm^-1, respectively. Although C-O stretching bands occur in a spectral region where there are usually many other bands, they are easy to identify because they are intense (strong). They are often coupled to (i.e., move in concert with) C-C absorption bands and therefore exhibit splitting. The exact location of the C-O bands depends on the degree of branching of the carbon atom that is attached to oxygen. Thus, as Table 4-7 reveals, the position of the C-O absorption band can allow distinction between primary, secondary, and tertiary alcohols. In a primary alcohol, the OH group is attached to a carbon that is in turn attached to only one other carbon atom. In secondary and tertiary alcohols, the OH group is located on a carbon attached to 2 and 3 other carbons, respectively.

Ethers. Ethers are compounds containing the C-O-C functional group. The asymmetric C-O stretching absorption of ethers appears in the region 1280-1050 cm^-1. As in alcohols, the position of this strong band is dependent on the nature of the attached groups.

Aldehydes and Ketones. The carbonyl, (C=O) stretching absorption is particularly important because it is present in a variety of functional groups including aldehydes, ketones, and carboxylic acids and their derivatives. Table 4-8 shows the basic structures of compounds containing C=O.

The C=O stretching frequencies of aldehydes in which R is an alkane are observed at 1735-1710 cm^-1. Although the position of the carbonyl absorption does not allow distinction between aldehydes and ketones, the former are recognizable by the C-H stretching vibration that appears as two bands in the 2850-2700 cm^-1 region. Double bonds in conjugation with the carbonyl group lower the wavenumber by 25 to 50 cm^-1. Thus, aromatic aldehydes (benzene-C(=O)H) generally absorb at 1700-1690 cm^-1 and diaromatic ketones at 1670-1660 cm^-1. Intramolecular hydrogen bonding to the carbonyl oxygen also lowers the wavenumber by 25 to 50 cm^-1.

Carboxylic Acids. Carboxylic acids contain the C(=O)OH functional group. The carbonyl group of acids in which R is not aromatic appears at 1730-1700 cm^-1 and is shifted to 1720-1680 cm^-1 when the C=O double bond is conjugated with a C=C double bond. The most characteristic absorption of carboxylic acids is a broad band from 3300-2500 cm^-1 due to hydrogen bonded O-H stretching. The C-H stretching vibrations appear as small bands on top of this band.

Amines. Amines are compounds containing the N atom singly bonded to R or H. Primary amines have formula NRH₂, while secondary and tertiary amines have formula NR₂H and NR₃ respectively. Primary and secondary amines show N-H stretching vibrations in the 3500-3300 cm^-1 region. Primary amines generally have two bands approximately 70 cm^-1 apart due to asymmetric and symmetric stretching modes. Secondary amines show only one band. Inter- or intramolecular hydrogen bonding broadens the absorptions and lowers the frequency. In general the intensities of N-H bands are less than of O-H bands. The N-H bending and C-N stretching absorptions are not as strong as the corresponding alcohol bands and occur at approximately 100 cm^-1 higher frequencies. NH₂ groups give an additional broad band at 900-700 cm^-1 due to bending.

Nitriles. Nitriles are compounds containing the C(3)N group. A sharp, unusually strong absorption at 2260-2220 cm^-1 due to CN stretching is characteristic of nitriles.

To complete the section on IR spectroscopy, we make one broad and extremely useful generalization. Every compound, organic or inorganic, has a unique infrared spectrum. This spectrum serves as a fingerprint for the compound. Thus if a compound prepared in the laboratory is thought to be a particular known substance, the fastest way to determine whether or not it is is to obtain the IR spectra of the compound and the known substance. If the spectra are identical (band-for-band match), then so are the substances.

4-4 Electron Motion in Molecules--E_electrons. The electrons can change energy states within the molecule. When they do this, they change location with respect to the atomic nuclei. Motions of electrons within the molecule are associated with frequencies that are typically in the range 10¹⁴ to 10¹⁵ s^-1. When this frequency matches the frequency of light impinging on the molecule, the light can be absorbed, and an electron moves to a higher energy state. The frequencies of light absorbed give us information about the motions of the electrons in the molecule. Photons with frequency 10¹⁵ s^-1 have energy of 6.6 x 10^-19 J, and fall in the visible and ultraviolet regions of the electromagnetic spectrum. Spectroscopy involving changes in electron motions is therefore commonly called UV-Visible spectroscopy. It is also called electronic absorption spectroscopy (EAS).

The wavelengths of light absorbed and the extent of absorption by a substance can be measured using an instrument called an electronic absorption spectrometer. A typical spectrometer covers the wavelength range between 1100 and 190 nm and thus ranges from the near IR region (1100-750 nm) to the near UV region (400-190 nm) ("near" means bordering on the visible region). To measure the UV-Vis spectrum of a substance requires the preparation of a solution of the substance in an appropriate solvent. The spectrum is a plot produced by the spectrometer of the amount of light absorbed (the absorbance) as a function of the wavelength of the light. The spectrum of potassium permanganate, KMnO₄, dissolved in water is shown in Figure 4-13. Note that the absorption does not consist of sharp lines, as in NMR, or even as relatively narrow signals, as in IR. Instead, absorption usually occurs over a range of wavelengths. Each absorption region is therefore called an absorption band, to indicate that it covers a band, or range, of wavelengths, rather than occurring at just a single wavelength.

The amount of light absorbed by the substance -- the absorbance -- at a particular wavelength is directly proportional to 1) the length of solution through which the light beam passes; and 2) the concentration of substance (number of molecules or moles per unit volume) in the solution. This is written in shorthand form in equation 4-4-1.

where A = absorbance (amount of light absorbed); l = the length of solution through which the light passes, in cm; M = the molarity (number of moles of substance per liter) of the solution; and e = the proportionality constant relating A to l and M. e is called the molar absorptivity of the substance at the particular wavelength at which A is measured. The most important aspect of this equation for us at this point is this: the amount of light absorbed is directly proportional to the amount of substance per unit volume of solution. This is called Beer's Law. The molar absorptivity is characteristic of the substance and the wavelength. Values of e for a substance are usually quoted for the wavelengths of maximum absorbance, l_max -- that is, the tops of the absorption bands. The value of e gives a measure of the effectiveness of a molecule of substance at capturing a photon of light. Values can range from as low as 0.01 to as high as 10⁵ L/mole-cm. Note that substitution of 1000 cm³ for 1 L reduces the units of e to cm²/mole. This shows that e measures the effective cross-sectional area of the substance with respect to capturing light. Since the e values of a substance are characteristic of it, we may use the amount of light absorbed to measure the concentration of the substance in solution.

Example 4-4. The spectrum of a mixture of KMnO₄ and K₂Cr₂O₇ dissolved in water has A = 0.266 at 565 nm and A = 0.790 at 350 nm. For KMnO₄, e = 1.13 * 10³ at l = 350 nm and 1.27 * 10³ at 565 nm. For K₂Cr₂O₇, e = 2.84 * 10³ at l = 350 nm and essentially zero at 565 nm. Calculate the concentrations of KMnO₄ and K₂Cr₂O₇ in the solution.

Solution. Since dichromate does not absorb at 565 nm, the absorbance at 565 is due entirely to permanganate. It's concentration can be calculated from Beer's Law:

The absorbance at 350 nm is a sum of contributions from the two species; however, the permanganate contribution can be calculated and subtracted, allowing the dichromate concentration to be obtained:

Example 4-5. The complex ion, FeL(CH₃CN)(CO)²⁺ has the structure shown in Figure 4-14a. L represents the cyclic structure consisting of carbon, hydrogen, and nitrogen atoms. The complex decomposes when dissolved in acetonitrile, CH₃CN, according to the following equation

The course of this reaction can be monitored by measuring the absorbance of the solution at 556 nm, a wavelength where FeL(CH₃CN)₂²⁺ absorbs but FeL(CH₃CN)(CO)²⁺ does not. The following absorbance data are obtained at various times after the reaction begins:

The initial concentration of FeL(CH₃CN)(CO) was 0.694 * 10^-4 M. The value of e for FeL(CH₃CN)₂²⁺ at 556 nm is 9650 M^-1cm^-1.

1. Plot [FeL(CH₃CN)(CO)²⁺] vs time.
2. What percentage of the reactant remains after 45 minutes?
3. What is the average rate of the reaction over the first 15 minutes?

Solution. 1. The concentration of FeL(CH₃CN)(CO)²⁺ at each time must be calculated from the measured absorbance, the molar absorptivity of the product, FeL(CH₃CN)₂²⁺, and the known initial concentration of FeL(CH₃CN)(CO)²⁺. For example, at t = 3 minutes, the measured absorbance is 0.121. The corresponding [FeL(CH₃CN)₂²⁺] is

The reaction stoichiometry indicates that 1 mole of product is obtained from 1 mole of reactant. It follows that [FeL(CH₃CN)(CO)²⁺] = (6.94 - 1.25) * 10^-5 = 5.69 * 10^-5 M. An analogous calculation can be performed at each time in the table. This enables a plot of [FeL(CH₃CN)(CO)²⁺] versus time to be constructed, as shown in Figure 4-14b.

2. Find the concentration of FeL(CH₃CN)(CO)²⁺ from the curve at t = 45 minutes. Divide this by the initial concentration of reactant to find the fraction remaining at 45 minutes. Then multiply by 100.

When t = 45 min, the curve indicates that [FeL(CH₃CN)(CO)²⁺] has fallen to 5.7 * 10^-6 M. The fraction of reactant remaining is thus

3. Divide the change in [FeL(CH₃CN)(CO)²⁺] that occurs over the first 15 minutes by the elapsed time, 15 minutes. This gives the average change in concentration per minute over this time interval:

Example 4-6. The electronic absorption spectrum of a water solution of CrCl₃ (chromium(III) chloride) is shown in Figure 4-15a. The absorption bands, resulting from the Cr³⁺ ion, are in the visible region of the spectrum, and are responsible for the violet color of the solution. What are the energies of the most strongly absorbed photons? Plot the electronic energy levels of the Cr³⁺ ion on an energy level diagram.

Solution. The most strongly absorbed photons have energies corresponding to the wavelengths at the tops of the absorption bands. The energies, read directly from the absorption maxima in the spectrum, are at 16500 and 23000 cm^-1. These are in units of reciprocal wavelength and must be converted to energy units using the relationship

To construct the energy level diagram, we must realize that the observed absorptions correspond to transitions of an electron in the Cr³⁺ ion from the lowest energy state to two higher energy states, one corresponding to each absorption band. The photon energies calculated above give us the gaps between the energy levels. The energy level diagram should thus show three states. The middle state should be 3.28 * 10^-19J above the lowest one. The highest state should be 4.57 * 10^-19 J above the lowest one. The diagram is shown in Figure 4-15b.

The Correspondence of Color and Spectrum. Many substances are colored, resulting from absorption of light in the visible region of the spectrum. In addition to making these substances visually interesting, color is an important physical property, and is used with such quantities as melting point and boiling point in describing the substance. It is important to have a qualitative understanding of the relationship between the color of light absorbed by a substance and the color of light that we perceive when we look at the substance. These colors are complementary. We see the light that is NOT absorbed by the colored substance. The color that we see is that of the light that is not absorbed. Although this statement may contradict what you have previously thought, it nonetheless makes sense. Photons that are absorbed by a colored substance cannot reach the eye to be observed. The photons reaching the eye are those that are NOT absorbed.

White light is made up of the three primary colors, red (R), green (G), and blue (B). The relationships between and combinations of these three colors are best shown in the form of the color wheel in Figure 4-16a. The wheel is set up with the primary colors at 12, 4, and 8 of a clock face. The complements of the primaries are shown directly opposite the corresponding primary. Thus cyan, the complement of red, is at 6 oclock. Cyan, C, is a combination of equal amounts of B and G, which occur to either side of it. Similarly, magenta (R + B) occurs opposite its complement, green; and yellow (R + G) appears opposite its compement, blue. A substance that absorbs light of a particular color appears to have the color opposite the absorbed color on the wheel. Thus, if red is absorbed, cyan is seen. Figure 4-16b shows the correspondence between color and wavelength of light, which may be seen by viewing white light through a simple device called a spectroscope. With the information in Figure 4-16a and b, the color of a substance may be understood in terms of its electronic absorption spectrum. Thus if a substance has an absorption band at 550 nm, in the green region of the spectrum, it absorbs green photons, transmits red and blue, and should appear to us to be magenta (R + B).

Relationship of Structure to the Electronic Absorption Spectrum. It is important to have some idea of the types of molecular electronic structural features that are likely to give rise to absorption bands in the near-IR (1100-750 nm), visible (750-400 nm), near-UV (400-190 nm), and far UV (shorter wavelength than 190 nm) regions of the electromagnetic spectrum. We will make some broad statements for which you can expect many exceptions. The purpose of the statements is to enable you to make reasonable predictions about electronic transitions in molecules. In making predictions, of course, we run the risk of being wrong. But an incorrect prediction based on sound reasoning is far preferable to no prediction (and no reasoning) at all. You are thus encouraged to be fearless in making predictions.

The near-IR and Visible regions (1100-400 nm). Several structural features are notable for generating light absorption in these regions. First, compounds containing transition metal ions (ions of the metals found in the d block of the periodic table) show an interesting rainbow of colors. These ions are known for their ability to form complexes, compounds in which up to six molecules or ions are covalently bonded to the transition metal ion. In this context, the molecules or ions are called ligands. In theory, any molecule or ion with at least one lone (nonbonding) electron pair can serve as a ligand by using its electron pair to bond to a transition metal ion. In practice, molecules containing oxygen (2 lone pairs), nitrogen (1 lone pair), sulfur (2 lone pairs), phosphorus (1 lone pair), or halogen (3 lone pairs) frequently serve as ligands. Table 4-9 gives formulas and colors for a collection of complexes containing the Co³⁺ ion.

Without going into too much detail, the colors of transition metal ions are a consequence of the partial filling of the d valence orbitals with electrons. When ligands bind to the transition metal, the d orbitals are caused to have different energies, so that the d electrons can undergo light-promoted transitions between them. The energy spacing between the d orbitals is influenced by the chemical identity of the ligands; thus different ligands coax different colors from the same metal ion. These so-called d-d transitions are typically quite weak, with molar absorptivities in the range 0.1-100 M^-1cm^-1

Second, organic molecules with extended conjugation (alternating single and double bonds) often absorb visible light. As we will see in Chapter 13, acid-base indicators have structures of this type. Extended conjugation normally leads to quite intense color (strong absorbance, large molar absorptivity), in contrast to transition metal complexes, whose colors are relatively pale. The porphyrin ring, which occurs in many biologically-important molecules, has benzene-like resonance involving its double bonds. It absorbs intensely in the visible region. Electronic absorption spectral data for a number of compounds of this type are presented in Table 4-10.

Third, species (molecules or ions) in which an atom in a high oxidation state is bonded to species with lone pairs of electrons (such as oxide ions and halide ions) are often colored, due to the transition of an electron from one of the lone pairs to the electron-deficient high-oxidation-state atom. Such transitions are called charge transfers, because they essentially transfer negative charge (an electron) from a region of negative to a region of positive charge. The permanganate ion, MnO₄^-, is a good example of a species whose color is due to a charge transfer transition. The anion is intensely purple both in the solid state and in aqueous solution. Table 4-11 gives spectral data for charge transfer absorptions in a number of species.

Near UV region (400-190 nm). Many organic molecules with double bonds, particularly those in which one of the double bonded atoms has at least one lone pair, absorb radiation in the near UV region. Examples include acetone, with a carbon-oxygen double bond and 2 lone pairs on oxygen; benzene, with 3 conjugated double bonds; acetonitrile, with a carbon-nitrogen triple bond and a lone pair on nitrogen; and SO₄^2- (an inorganic ion), with sulfur-oxygen double bonds and lone pairs on oxygen. Table 4-12 presents absorption data for a number of common molecules of this type.

Far UV region (< 190 nm). Molecules containing strong sigma bonds and no lone pairs tend to absorb in this region. The only promotable electrons are those forming the single bonds, and they must receive a large amount of energy in order to undergo a transition to the lowest energy empty orbital of the molecule. Thus molecules such as ethane, C₂H₆ and other hydrocarbons have no absorptions at wavelengths longer than 190 nm. Strongly single bonded molecules with lone pairs on electronegative atoms also tend to absorb only at very short wavelength (high energy). Thus water, HF, NH₃, and the like have no absorptions in the visible or near-UV regions of the spectrum. Table 4-13 shows some typical far-UV data.

The foregoing discussion should give you an idea of the importance and utility of spectroscopy. Molecular structures, discussed in Chapter 3 from the standpoint of theory, have been elucidated largely via spectroscopic methods.

Supplement 1: ¹³C Nuclear Magnetic Resonance Spectroscopy.

The characteristics of ¹³C NMR spectra are fundamentally similar to those of proton spectra. That is, a signal is obtained for each structurally different type of carbon atom; the positions of the signals are measured in ppm with respect to the ¹³C signal for TMS; the signals tend to be further downfield for C atoms attached to electronegative atoms like Cl and O; and it is possible to observe spin-splitting of the ¹³C resonances by protons that are directly attached to the observed carbon. However, there are some differences in detail between ¹H and ¹³C spectra. First, the chemical shift range exhibited by carbon is more than 10 times larger than that of protons. Thus although most proton resonances occur within 10 ppm downfield from the TMS resonance, carbon resonances can range more than 100 ppm from the TMS ¹³C signal. Second, although there is still one signal from each different type of carbon atom, the intensities of the signals do not provide information about the number of carbon atoms generating the signal. Recall that this was one of the very useful aspects of proton NMR; it is not useful for carbon. The reasons for this are complex and we need not bother with them. Third, we expect the signal from a particular type of carbon atom to be split into subsignals by protons attached to the carbons. Indeed such splitting can be observed, and it follows the usual rules, but most often chemists choose NOT to observe it. Instead, they set the NMR instrument to electronically decouple the proton spins, so that only a single line is seen for each chemically distinct type of carbon atom. Fourth, ¹³C spectra are often obtained by dissolving the compound of interest in deuterated chloroform, CDCl₃. The carbon atom of the solvent gives rise to an NMR signal that is split into three subsignals due to coupling with the deuterium (heavy hydrogen) atom. This triplet of signals always falls at about 77 ppm downfield from the TMS signal. It should be ignored when interpreting ¹³C NMR spectra.

The ¹³C NMR spectrum for chloropentane is shown in Figure 4S-1. The spectrum has the following features:

• It shows a TMS signal at 0 ppm, a solvent signal at 77 ppm, and 4 additional signals, all singlets.
• The signals all fall within 80 ppm of the TMS signal.
• The 4 singlets have different intensities (heights).

The spectrum tells us immediately that there are 4 distinct types of carbon atom in the molecule, although it does not tell us how many of each. Figure 4S-2 shows several possible structures (isomers) for chloropentane. Of those shown the observed spectrum could only arise from options D, E, and G, all of which have 4 carbon atoms in structurally different environments. Without further information, we cannot say which isomer we have. However, suppose that we then re-run the spectrum, this time allowing the attached protons to couple and split the ¹³C signals, to obtain the result in Figure 4S-3. The small letters in the spectrum indicate the extent of splitting of each signal: s for singlet, d for doublet, and so on. Of structures D, E, and G, only D is consistent with the observed proton couplings. It should produce quartets from carbons 1 and 4, each coupling with 3 protons; a triplet from carbon 2 coupling with 2 protons, and a singlet from carbon 3 coupling with no protons. Structure E is expected to produce quartets from carbons 1 and 4 and doublets from carbons 2 and 3, while G should show a quartet from carbons 1, 2 triplets from carbons 3 and 4, and a doublet from carbon 2. Let's look at a couple more examples.

Example 4S-1. The structure and proton NMR of isobutyraldehyde are shown in Figures 4-4a and 4-4b, respectively. Predict the appearance of the ¹³C NMR spectrum of this molecule.

Solution. The structure shows 3 types of carbon atom, so 3 signals are expected. If coupling to protons is allowed, the signal from the CH₃ carbons should be a quartet, that from the carbonyl carbon should be a singlet, and the final carbon should produce a doublet. The signal from the carbonyl carbon should be quite far downfield due to the proximity of oxygen.

Example 4S-2. A compound of formula C₆H₁₂ produces the ¹³C spectrum shown in Figure 4S-4a. What is the molecular structure of the compound?

Solution. Possible atom arrangements corresponding to the molecular formula are shown in Figure 4S-4b. Clearly there are quite a few of them! The structures are expected to produce the following numbers of ¹³C signals:

Because 4 signals are observed in the spectrum the structure must be M, which is called methylcyclopentane. If coupling with attached hydrogen atoms were allowed, we would expect to see a quartet, a doublet, and 2 triplets. You might want to predict the splitting pattern that would be observed for each of the other possible structures.

It may have occurred to you to wonder whether coupling of one ¹³C nucleus with another can occur, and whether it is observable in the NMR spectra. In fact, it is not observable; perhaps we should be thankful for this, as it would greatly complicate things! We should spend a bit of time to understand why this type of coupling is not seen. As said above, the natural abundance of ¹³C is about 1% (actually 1.11%, but we use the simpler number because calculations are easier). This means that 1 carbon atom in a hundred has mass number 13. We can use this fact to determine the probabilities that varying numbers of ¹³C atoms will be found in a molecule with, say, 4 carbon atoms. Let's begin with the probability that a particular molecule will have NO ¹³C atoms. Here is the question we are asking: "What is the probability that carbon 1 is ¹²C AND that carbon 2 is ¹²C AND that carbon 3 is ¹²C AND that carbon 4 is ¹²C?" In probability mathematics, AND means multiply the probabilities for the events that it links. Thus the probability that we seek is (99/100)*(99/100)*(99/100)*(99/100) = (0.99)⁴ = 0.96. So of 100 of these molecules, 96 will have no carbon-13 atoms in them! The remaining 4 will have between 1 and 4 carbon 13 atoms. Next, let's calculate the probability that a molecule contains four carbon-13 atoms by calculating (1/100)⁴ = 1*10^-8! This is one atom in 100 million, such a small number that we can ignore this possibility. Much more likely is that a molecule will contain a single carbon-13 atom; this situation can be shown to have probability of about 0.039. The probability that a molecule will contain 2 carbon-13 atoms next to each other, a situation required for spin-spin coupling, is 2.9*10^-4. Only 3 molecules in 10000 qualify! Thus the intensity of the NMR signal from the molecules containing one carbon-13 atom is expected to be more than 100 times greater than the intensity from molecules with two neighboring carbon-13 atoms; the latter is so weak that it will not be seen. Only coupling with attached protons is significant in carbon NMR.

The modern chemist, armed with the techniques of NMR, IR, and UV-visible spectroscopy, is able to discern the detailed structure of just about any molecule of interest. Even large biological molecules yield to some of these techniques. We can be confident that in years to come, science will develop yet other ways to exploit the interaction of light and matter to reveal nature's details at the molecular level.

Supplement 2: An Intuitive Development of Beer's Law. It is possible to arrive at Beer's Law using a simple macroscopic physical system consisting of an inclined plane perforated with holes at a certain density, and divided into four sections of equal length. The plane is pictured in Figure 4S-5. Suppose that 10000 identical small steel spheres are released at the top of the plane, and that we observe that 9000 arrive at the end of the top section, having avoided falling into one of the holes. Based on this information, how many spheres will arrive at the bottom of the plane? Two approaches seem reasonable on the surface. The first is to assume that the same number of spheres will fall into holes in each section. If 1000 fell in the first section, then a total of 4000 will fall along the entire length of the plane, leaving 6000 to reach the bottom. The second approach is to assume that the number of spheres that fall into holes in a given section is proportional to the number that enter the section; i.e., that the same fraction of spheres will disappear in each section. By this reasoning, we predict that 8100 spheres will survive the first 2 sections (1/10 of 9000 fall into holes in the second section), 7290 will survive through 3 sections (1/10 of 8100 fall in holes in the third section), and 6561 will reach the bottom. This second line of reasoning proves correct: the number of balls that survive a section is proportional to the number of balls that start into the section. We write this in equation form.

This equation states that the number of spheres arriving at mark n is proportional to the number of spheres that leave the preceding mark, n-1. k is a proportionality constant that must have a value between 0 and 1; it has the value 0.9 for our example above. We can also express the equation in terms of the change in the number of balls:

Note that because k < 1, the quantity k-1 is negative, consistent with the fact that balls are being lost and therefore that the change in the number of balls defined in equation 4S-2 is also negative. We have therefore replaced k-1 with -K, where K is a new, positive, constant.

Here are two questions that we now consider: How does the value of K depend on the length of the interval, l? How does the value of K depend on the density of holes, r? Intuitively, we expect that a longer interval will result in more balls being lost. Thus K increases as l increases. Also, we expect more balls to be lost if the density of holes is increased. So K increases as r increases. In equation form, this becomes

where z is a constant representing the number of balls lost per unit density of spheres per unit length of section. Substituting 4S-3 into 4S-2 gives

Assuming an interval of infinitesimal length will result in an infinitesimal loss of spheres:

Separating the variable, N, from the variable, l, gives

Finally, integration provides

Of course this equation can be expressed in exponential form as N/N_o = 10^-e*r*l. In words, this equation states that the fraction of balls surviving an interval decreases exponentially with the length of the interval and the density of holes in the interval. Let's now make a direct analogy between our inclined plane and a solution of a substance that absorbs light at some wavelength. We think of l as the length of solution through which the light beam must pass; r as the density (concentration) of absorbing molecules in the solution; the small spheres as photons of light; and N/N_o as the fraction of light transmitted by the solution:

We have replaced the density of holes with the concentration of the solution in moles absorbing substance per liter, and the number of spheres with the intensity, I, of light. The intensity is simply the number of photons per second arriving at a particular location, so is a close analog of N. We now define a quantity called the absorbance, A, as A = e*M*l so that equation 4S-9 becomes

A represents the amount of light absorbed (the number of spheres lost). Note that when all light is transmitted by a sample, I_o/I = 1 and A = 0. When 1/10 of the light is transmitted, I_o/I = 10 and A = 1. Thus A goes up as the fraction transmitted goes down. The equation A = e*M*l is, of course, Beer's Law.

Applications

4-10. Predict the general appearance of the NMR spectrum of methyl ethyl ketone from its structure.

4-11. Nuclear Magnetic Resonance spectroscopy was used to establish the structure of the compound diborane, B₂H₆. The spectrum of diborane is shown. Which structure in the figure fits with the NMR spectrum?

4-12. The NMR spectrum and structure of benzene are shown. Explain how the spectrum is related to molecular structure.

4-13. The NMR spectrum and structure of ethylbenzene are shown. Interpret the spectrum in terms of molecular structure.

4-14. The NMR spectrum and structure of benzaldehyde are shown. Explain how the spectrum is related to molecular structure.

4-15. The NMR spectrum and structure of styrene are shown. Interpret the spectrum in terms of molecular structure.

4-16. The NMR spectrum and structure of 3-chloro-2-chloromethyl-1-propene are shown. Interpret the spectrum in terms of molecular structure.

4-17. The NMR spectrum and structure of 1,2-dichloropropane are shown. Interpret the spectrum in terms of molecular structure.

4-18. The IR spectrum and structure of acetaldehyde are shown. Explain how the spectrum is related to molecular structure.

4-19. Predict the locations of absorption bands in the IR spectrum of acetonitrile, CH₃CN.

4-20. The structure of vinyl bromide is shown. The IR spectrum of vinyl bromide shows absorption bands at 3080, 1840, 1600, 1360, 1250, 900-1000, and 600 cm^-1. Explain how the spectrum is related to molecular structure.

4-21. Develop a Lewis structure for each molecule and predict the appearance of the ¹H NMR spectrum and the IR spectrum.

4-25. Propose a structure consistent with the formula and NMR data

4-26. Propose a structure consistent with the formula and with NMR and IR data

4-27. Propose a structure consistent with the formula and with NMR and IR data.

4-28. Propose a structure consistent with the formula and NMR data.

4-29. Propose a structure consistent with the formula and with NMR and IR data.

4-30. Propose a structure consistent with the formula and NMR data.

4-31. Propose a structure consistent with the formula and with NMR and IR data.

4-33. Propose a structure consistent with the formula and NMR data.

4-34. Propose a structure consistent with the formula and with NMR and IR data.

4-35. Propose a structure consistent with the formula and the NMR data.

4-37. Predict the appearance of the NMR spectrum for the molecule shown.

4-38. The IR spectrum of acetonitrile, CH₃CN, is shown. Interpret the spectrum in terms of structure.

4-39. The IR spectrum of benzaldehyde, C₆H₅CHO, is shown. Interpret the spectrum in terms of structure.

4-40. Predict the major features of the IR spectrum of benzonitrile, C₆H₅CN.

4-41. Predict the major features of the IR spectrum of benzoic acid, C₆H₅COOH.

4-42. The IR spectrum and structure of sodium N,N-diethyldithiocarbamate trihydrate are shown. The IR absorptions are due to the dithiocarbamate anion and to the water of hydration. Interpret the spectrum in terms of structure.

4-43. The IR spectrum of imidazole, C₃H₄N₂, is shown. Interpret the spectrum in terms of structure.

4-44. The IR spectrum of dichloromethane, CH₂Cl₂, is shown. Interpret the spectrum in terms of structure.

4-45. The IR spectrum of dimethylsulfoxide (DMSO), (CH₃)₂SO, is shown. Interpret the spectrum in terms of structure.

4-46. The IR spectrum of the amino acid, glycine, NH₂CH₂COOH, is shown. Interpret the spectrum in terms of structure.

4-47. The electronic absorption spectrum of a 0.10 M water solution of NiCl₂ is shown. Determine the molar absorptivity at the wavelengths of maximum absorption. (If you need a reminder about the meaning of 0.10 M, see Chapter 1.)

4-48. From the electronic absorption spectrum shown in Application 4-47, determine the frequencies of the most strongly absorbed photons.

4-49. A solution containing both MnO₄^- and Cr₂O₇^2- has an absorbance of 0.896 at 350 nm, and 0.210 at 565 nm. Calculate the concentrations in moles per liter of MnO₄^- and Cr₂O₇^2- in the solution (see Example 4-3).

4-50. Develop a Lewis structure for each molecule and predict whether the substance should have electronic absorption bands in the visible (750-400 nm), near UV (400-190 nm) and/or far UV regions of the electromagnetic spectrum:

4-51. Develop a Lewis structure for each molecule and predict whether the substance should have electronic absorption bands in the visible (750-400 nm), near UV (400-190 nm) and/or far UV regions of the electromagnetic spectrum:

4-52. The electronic absorption spectrum for a 6 * 10^-5 M solution of the anion, Fe(CN)₅(L)^2-, where L is the N-methylpyrazinium cation (shown in the figure), is shown. Determine the molar absorptivity of the ion at its wavelengths of maximum absorbance. Do you think that the absorption at 660 nm is due to d-d transitions? To charge transfer? Why?

4-53. The electronic absorption spectrum of a 0.0062 M aqueous solution of the transition metal complex, Cr(EDTA) (see figure), shows an absorption band in the visible region of the spectrum. Do you think the band arises from a d-d transition? From charge transfer? Support your answer.

4-54. A substance has absorption bands at 520 nm and 290 nm, and absorbs strongly at wavelengths less than 250 nm. The substance is most likely to be:

4-55. A substance has absorption bands at 360 and 270 nm, and absorbs strongly below 250 nm. The substance is most likely to be

4-56. A substance has no absorption bands in the wavelength range from 1100 to 200 nm. The substance is most likely to be

4-57. A substance has an absorption band at 520 nm. The substance is most likely to be

4-58. A substance has an absorption band at 350 nm, and absorbs strongly at wavelengths below 300 nm. The substance is most likely to be

4-59. An aqueous solution is prepared containing the cation, Cr(H₂O)₄Cl₂⁺. Over a period of several hours, the color of the solution gradually changes to violet. The solution remains violet indefinitely. Suggest a possible explanation for this observation.

4-60. A solution of I₂ in CHCl₃ is prepared, and its spectrum run. Successive small volumes of dimethylsulfoxide (DMSO), C₂H₆SO, are then added, and the spectrum obtained after each addition. Explain the spectral series obtained.