More organic chemistry
Let’s see how the same ideas can be used to study other molecules. Consider a molecule like butadiene $(1,3)$—it is drawn in Fig. 15–9 according to the usual valence bond picture. We can play the same game with the extra four electrons corresponding to the two double bonds. If we remove four electrons, we have four carbon atoms in a line. You already know how to solve a line. You say, “Oh no, I only know how to solve an infinite line.” But the solutions for the infinite line also include the ones for a finite line. Watch. Let $N$ be the number of atoms on the line and number them from $1$ to $N$ as shown in Fig. 15–10. In writing the equations for the amplitude at position $1$ you would not have a term feeding from position $0$. Similarly, the equation for position $N$ would differ from the one that we used for an infinite line because there would be nothing feeding from position $N+1$. But suppose that we can obtain a solution for the infinite line which has the following property: the amplitude to be at atom $0$ is zero and the amplitude to be at atom $(N+1)$ is also zero. Then the set of equations for all the locations from $1$ to $N$ on the finite line are also satisfied. You might think no such solution exists for the infinite line because our solutions all looked like $e^{ikx_n}$ which has the same absolute value of the amplitude everywhere. But you will remember that the energy depends only on the absolute value of $k$, so that another solution, which is equally legitimate for the same energy, would be $e^{-ikx_n}$. And the same is true of any superposition of these two solutions. By subtracting them we can get the solution $\sin kx_n$, which satisfies the requirement that the amplitude be zero at $x=0$. It still corresponds to the energy $(E_0-2A\cos kb)$. Now by a suitable choice for the value of $k$ we can also make the amplitude zero at $x_{N+1}$. This requires that $(N+1)kb$ be a multiple of $\pi$, or that \begin{equation} \label{Eq:III:15:29} kb=\frac{\pi}{(N+1)}\,s, \end{equation}  where $s$ is an integer from $1$ to $N$. (We take only positive $k$’s because each solution contains $+k$ and $-k$; changing the sign of $k$ gives the same state all over again.) For the butadiene molecule, $N=4$, so there are four states with \begin{equation} \label{Eq:III:15:30} kb=\pi/5,\quad 2\pi/5,\quad 3\pi/5,\quad \text{and}\quad 4\pi/5. \end{equation}  We can represent the energy levels using a circle diagram similar to the one for benzene. This time we use a semicircle divided into five equal parts as shown in Fig. 15–11. The point at the bottom corresponds to $s=0$, which gives no state at all. The same is true of the point at the top, which corresponds to $s=N+1$. The remaining $4$ points give us four allowed energies. There are four stationary states, which is what we expect having started with four base states. In the circle diagram, the angular intervals are $\pi/5$ or $36$ degrees. The lowest energy comes out $(E_0-1.618A)$. (Ah, what wonders mathematics holds; the golden mean of the Greeks5  gives us the lowest energy state of the butadiene molecule according to this theory!) Now we can calculate the energy of the butadiene molecule when we put in four electrons. With four electrons, we fill up the lowest two levels, each with two electrons of opposite spin. The total energy is  \begin{equation} \label{Eq:III:15:31} E=2(E_0-1.618A)+2(E_0-0.618A)= 4(E_0-A)-0.472A. \end{equation} 
 \begin{align} E&=2(E_0-1.618A)+2(E_0-0.618A)\notag\\[1ex] \label{Eq:III:15:31} &=4(E_0-A)-0.472A. \end{align}  This result seems reasonable. The energy is a little lower than for two simple double bonds, but the binding is not so strong as in benzene. Anyway this is the way the chemist analyzes some organic molecules. The chemist can use not only the energies but the probability amplitudes as well. Knowing the amplitudes for each state, and which states are occupied, he can tell the probability of finding an electron anywhere in the molecule. Those places where the electrons are more likely to be are apt to be reactive in chemical substitutions which require that an electron be shared with some other group of atoms. The other sites are more likely to be reactive in those substitutions which have a tendency to yield an extra electron to the system. The same ideas we have been using can give us some understanding of a molecule even as complicated as chlorophyll, one version of which is shown in Fig. 15–12. Notice that the double and single bonds we have drawn with heavy lines form a long closed ring with twenty intervals. The extra electrons of the double bonds can run around this ring. Using the independent particle method we can get a whole set of energy levels. There are strong absorption lines from transitions between these levels which lie in the visible part of the spectrum, and give this molecule its strong color. Similar complicated molecules such as the xanthophylls, which make leaves turn red, can be studied in the same way. There is one more idea which emerges from the application of this kind of theory in organic chemistry. It is probably the most successful or, at least in a certain sense, the most accurate. This idea has to do with the question: In what situations does one get a particularly strong chemical binding? The answer is very interesting. Take the example, first, of benzene, and imagine the sequence of events that occurs as we start with the six-times ionized molecule and add more and more electrons. We would then be thinking of various benzene ions—negative or positive. Suppose we plot the energy of the ion (or neutral molecule) as a function of the number of electrons. If we take $E_0=0$ (since we don’t know what it is), we get the curve shown in Fig. 15–13. For the first two electrons the slope of the function is a straight line. For each successive group the slope increases, and there is a discontinuity in slope between the groups of electrons. The slope changes when one has just finished filling a set of levels which all have the same energy and must move up to the next higher set of levels for the next electron. The actual energy of the benzene ion is really quite different from the curve of Fig. 15–13 because of the interactions of the electrons and because of electrostatic energies we have been neglecting. These corrections will, however, vary with $n$ in a rather smooth way. Even if we were to make all these corrections, the resulting energy curve would still have kinks at those values of $n$ which just fill up a particular energy level.  Now consider a very smooth curve that fits the points on the average like the one drawn in Fig. 15–14. We can say that the points above this curve have “higher-than-normal” energies, and the points below the curve have “lower-than-normal” energies. We would, in general, expect that those configurations with a lower-than-normal energy would have an above average stability—chemically speaking. Notice that the configurations farther below the curve always occur at the end of one of the straight line segments—namely when there are enough electrons to fill up an “energy shell,” as it is called. This is the very accurate prediction of the theory. Molecules—or ions—are particularly stable (in comparison with other similar configurations) when the available electrons just fill up an energy shell. This theory has explained and predicted some very peculiar chemical facts. To take a very simple example, consider a ring of three. It’s almost unbelievable that the chemist can make a ring of three and have it stable, but it has been done. The energy circle for three electrons is shown in Fig. 15–15. Now if you put two electrons in the lower state, you have only two of the three electrons that you require. The third electron must be put in at a much higher level. By our argument this molecule should not be particularly stable, whereas the two-electron structure should be stable. It does turn out, in fact, that the neutral molecule of triphenyl cyclopropenyl is very hard to make, but that the positive ion shown in Fig. 15–16 is relatively easy to make. The ring of three is never really easy because there is always a large stress when the bonds in an organic molecule make an equilateral triangle. To make a stable compound at all, the structure must be stabilized in some way. Anyway if you add three benzene rings on the corners, the positive ion can be made. (The reason for this requirement of added benzene rings is not really understood.)  In a similar way the five-sided ring can also be analyzed. If you draw the energy diagram, you can see in a qualitative way that the six-electron structure should be an especially stable structure, so that such a molecule should be most stable as a negative ion. Now the five-ring is well known and easy to make and always acts as a negative ion. Similarly, you can easily verify that a ring of $4$ or $8$ is not very interesting, but that a ring of $14$ or $10$—like a ring of $6$—should be especially stable as a neutral object.