The benzene molecule
Although quantum mechanics provides the basic laws that determine the structures of molecules, these laws can be applied exactly only to the most simple compounds. The chemists have, therefore, worked out various approximate methods for calculating some of the properties of complicated molecules. We would now like to show you how the independent particle approximation is used by the organic chemists. We begin with the benzene molecule. We discussed the benzene molecule from another point of view in Chapter 10. There we took an approximate picture of the molecule as a two-state system, with the two base states shown in Fig. 15–3. There is a ring of six carbons with a hydrogen bonded to the carbon at each location. With the conventional picture of valence bonds it is necessary to assume double bonds between half of the carbon atoms, and in the lowest energy condition there are the two possibilities shown in the figure. There are also other, higher-energy states. When we discussed benzene in Chapter 10, we just took the two states and forgot all the rest. We found that the ground-state energy of the molecule was not the energy of one of the states in the figure, but was lower than that by an amount proportional to the amplitude to flip from one of these states to the other. Now we’re going to look at the same molecule from a completely different point of view—using a different kind of approximation. The two points of view will give us different answers, but if we improve either approximation it should lead to the truth, a valid description of benzene. However, if we don’t bother to improve them, which is of course the usual situation, then you should not be surprised if the two descriptions do not agree exactly. We shall at least show that also with the new point-of-view the lowest energy of the benzene molecule is lower than either of the three-bond structures of Fig. 15–3. Now we want to use the following picture. Suppose we imagine the six carbon atoms of a benzene molecule connected only by single bonds as in Fig. 15–4. We have removed six electrons—since a bond stands for a pair of electrons—so we have a six-times ionized benzene molecule. Now we will consider what happens when we put back the six electrons one at a time, imagining that each one can run freely around the ring. We assume also that all the bonds shown in Fig. 15–4 are satisfied, and don’t need to be considered further. What happens when we put one electron back into the molecular ion? It might, of course, be located in any one of the six positions around the ring—corresponding to six base states. It would also have a certain amplitude, say $A$, to go from one position to the next. If we analyze the stationary states, there would be certain possible energy levels. That’s only for one electron. Next put a second electron in. And now we make the most ridiculous approximation that you can think of—that what one electron does is not affected by what the other is doing. Of course they really will interact; they repel each other through the Coulomb force, and furthermore when they are both at the same site, they must have considerably different energy than twice the energy for one being there. Certainly the approximation of independent particles is not legitimate when there are only six sites—particularly when we want to put in six electrons. Nevertheless the organic chemists have been able to learn a lot by making this kind of an approximation. Before we work out the benzene molecule in detail, let’s consider a simpler example—the ethylene molecule which contains just two carbon atoms with two hydrogen atoms on either side as shown in Fig. 15–5. This molecule has one “extra” bond involving two electrons between the two carbon atoms. Now remove one of these electrons; what do we have? We can look at it as a two-state system—the remaining electron can be at one carbon or the other. We can analyze it as a two-state system. The possible energies for the single electron are either $(E_0-A)$ or $(E_0+A)$, as shown in Fig. 15–6.  Now add the second electron. Good, if we have two electrons, we can put the first one in the lower state and the second one in the upper. Not quite; we forgot something. Each one of the states is really double. When we say there’s a possible state with the energy $(E_0-A)$, there are really two. Two electrons can go into the same state if one has its spin up and the other, its spin down. (No more can be put in because of the exclusion principle.) So there really are two possible states of energy $(E_0-A)$. We can draw a diagram, as in Fig. 15–7, which indicates both the energy levels and their occupancy. In the condition of lowest energy both electrons will be in the lowest state with their spins opposite. The energy of the extra bond in the ethylene molecule therefore is $2(E_0-A)$ if we neglect the interaction between the two electrons. Let’s go back to the benzene. Each of the two states of Fig. 15–3 has three double bonds. Each of these is just like the bond in ethylene, and contributes $2(E_0-A)$ to the energy if $E_0$ is now the energy to put an electron on a site in benzene and $A$ is the amplitude to flip to the next site. So the energy should be roughly $6(E_0-A)$. But when we studied benzene before, we got that the energy was lower than the energy of the structure with three extra bonds. Let’s see if the energy for benzene comes out lower than three bonds from our new point of view. We start with the six-times ionized benzene ring and add one electron. Now we have a six-state system. We haven’t solved such a system yet, but we know what to do. We can write six equations in the six amplitudes, and so on. But let’s save some work—by noticing that we’ve already solved the problem, when we worked out the problem of an electron on an infinite line of atoms. Of course, the benzene is not an infinite line, it has $6$ atomic sites in a circle. But imagine that we open out the circle to a line, and number the atoms along the line from $1$ to $6$. In an infinite line the next location would be $7$, but if we insist that this location be identical with number $1$ and so on, the situation will be just like the benzene ring. In other words we can take the solution for an infinite line with an added requirement that the solution must be periodic with a cycle six atoms long. From Chapter 13 the electron on a line has states of definite energy when the amplitude at each site is $e^{ikx_n}=e^{ikbn}$. For each $k$ the energy is \begin{equation} \label{Eq:III:15:25} E=E_0-2A\cos kb. \end{equation}  We want to use now only those solutions which repeat every $6$ atoms. Let’s do first the general case for a ring of $N$ atoms. If the solution is to have a period of $N$ atomic spacing, $e^{ikbN}$ must be unity; or $kbN$ must be a multiple of $2\pi$. Taking $s$ to represent any integer, our condition is that \begin{equation} \label{Eq:III:15:26} kbN=2\pi s. \end{equation}  We have seen before that there is no meaning to taking $k$’s outside the range $\pm\pi/b$. This means that we get all possible states by taking values of $s$ in the range $\pm N/2$. We find then that for an $N$-atom ring there are $N$ definite energy states3 and they have wave numbers $k_s$ given by \begin{equation} \label{Eq:III:15:27} k_s=\frac{2\pi}{Nb}\,s. \end{equation}  Each state has the energy (15.25). We have a line spectrum of possible energy levels. The spectrum for benzene ($N=6$) is shown in Fig. 15–8(b). (The numbers in parentheses indicate the number of different states with the same energy.) There’s a nice way to visualize the six energy levels, as we have shown in part (a) of the figure. Imagine a circle centered on a level with $E_0$, and with a radius of $2A$. If we start at the bottom and mark off six equal arcs (at angles from the bottom point of $k_sb=2\pi s/N$, or $2\pi s/6$ for benzene), then the vertical heights of the points on the circle are the solutions of Eq. (15.25). The six points represent the six possible states. The lowest energy level is at $(E_0-2A)$; there are two states with the same energy $(E_0-A)$, and so on.4 These are possible states for one electron. If we have more than one electron, two—with opposite spins—can go into each state. For the benzene molecule we have to put in six electrons. For the ground state they will go into the lowest possible energy states—two at $s=0$, two at $s=+1$, and two at $s=-1$. According to the independent particle approximation the energy of the ground state is \begin{align} E_{\text{ground}} &=2(E_0-2A)+4(E_0-A)\notag\\[1ex] \label{Eq:III:15:28}  &=6E_0-8A. \end{align} The energy is indeed less than that of three separate double bonds—by the amount $2A$. By comparing the energy of benzene to the energy of ethylene it is possible to determine $A$. It comes out to be $0.8$ electron volt, or, in the units the chemists like, $18$ kilocalories per mole. We can use this description to calculate or understand other properties of benzene. For example, using Fig. 15–8 we can discuss the excitation of benzene by light. What would happen if we tried to excite one of the electrons? It could move up to one of the empty higher states. The lowest energy of excitation would be a transition from the highest filled level to the lowest empty level. That takes the energy $2A$. Benzene will absorb light of frequency $\nu$ when $h\nu=2A$. There will also be absorption of photons with the energies $3A$ and $4A$. Needless to say, the absorption spectrum of benzene has been measured and the pattern of spectral lines is more or less correct except that the lowest transition occurs in the ultraviolet; and to fit the data one would have to choose a value of $A$ between $1.4$ and $2.4$ electron volts. That is, the numerical value of $A$ is two or three times larger than is predicted from the chemical binding energy. What the chemist does in situations like this is to analyze many molecules of a similar kind and get some empirical rules. He learns, for example: For calculating binding energy use such and such a value of $A$, but for getting the absorption spectrum approximately right use another value of $A$. You may feel that this sounds a little absurd. It is not very satisfactory from the point of view of a physicist who is trying to understand nature from first principles. But the problem of the chemist is different. He must try to guess ahead of time what is going to happen with molecules that haven’t been made yet, or which aren’t understood completely. What he needs is a series of empirical rules; it doesn’t make much difference where they come from. So he uses the theory in quite a different way than the physicist. He takes equations that have some shadow of the truth in them, but then he must alter the constants in them—making empirical corrections. In the case of benzene, the principal reason for the inconsistency is our assumption that the electrons are independent—the theory we started with is really not legitimate. Nevertheless, it has some shadow of the truth because its results seem to be going in the right direction. With such equations plus some empirical rules—including various exceptions—the organic chemist makes his way through the morass of complicated things he chooses to study. (Don’t forget that the reason a physicist can really calculate from first principles is that he chooses only simple problems. He never solves a problem with $42$ or even $6$ electrons in it. So far, he has been able to calculate reasonably accurately only the hydrogen atom and the helium atom.)