# Pólya's Enumeration formula and isomers

The hydrocarbon benzene has six carbon atoms arranged at the vertices of a regular hexagon, and six hydrogen atoms, with one bonded to each carbon atom.

I know that two molecules are said to be isomers if they are composed of the same number and types of atoms, but have different structure.

How many isomers may be obtained by replacing two of the hydrogen atoms with chlorine atoms, and two others with bromine atoms?

Show that exactly three isomers (ortho-dichlorobenzene, meta-di-chlorobenzene, and para-dichlorobenzene) may be constructed by replacing two of the hydrogen atoms of benzene with chlorine atoms.

So far I've ended up nowhere, I would appreciate the help.

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For the dichlorbenzene, observe that the two chlorine atoms may be adjacent or diagonally opposite or at distance 2.

The possible distributions of chlorine and bromine are (up to rotation and reflection)

• Cl Cl Br Br H H
• Cl Cl Br H Br H
• Cl Cl Br H H Br
• Cl Cl H Br Br H
• Cl Br Cl Br H H
• Cl Br Cl H Br H
• Cl H Cl Br Br H
• Cl H Cl Br H Br
• Cl Br Br Cl H H
• Cl Br H Cl Br H
• Cl Br H Cl H Br
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And how did you come down to this set? –  Max Nov 15 '12 at 18:13
Just "trial and error", starting with Cl, then filling up with Cl, Br, H in that order of preference, checking for repetitions. –  Hagen von Eitzen Nov 15 '12 at 18:15
Despite the "Polya" question's title, I like this answer. I think the problem is a good example of the fact that sometimes it's more work to apply general machinery than to just solve the problem with common sense and a little patience. Of course it takes some skill and/or experience to judge when trial and error is likely to be efficient and when you'd be better off with general machinery. –  Andreas Blass Jul 15 '13 at 16:15

This is just like what we did here. Draw the basic benzene hexagon and for now do not fill in the slots where the hydrogen atoms are attached. Now study the automorphisms of the benzene hexagon. You might think that for example a rotation that maps a carbon atom to its clockwise neighbor is an automorphism, but that would be wrong, because it does not take single and double bonds into account. Double bonds must be mapped to double bonds, and this simplifies things considerably.

It follows that the automorphism group $G$ of this compound is generated by a rotation by 120 degrees and three flips about the axes passing through the centers of opposite single-double bonds. It is not difficult to see that a series of rotations in fact simplifies to a single rotation, that two flips cancel each other, and that two flips with rotations between them yield a rotation. This means we have covered all of $G$.

Now the three flips have cycle structure $a_2^3$ and the two rotations, $a_3^2$, giving the cycle index $$Z(G) = \frac{1}{6} \left( a_1^6 + 3 a_2^3 + 2 a_3^2 \right).$$

The base generating function for the first question is $H + Cl + Br,$ which we substitute into the cycle index, getting $$1/6\, \left( H+{\it Cl}+{\it Br} \right) ^{6}+1/2\, \left( {H}^{2}+{{ \it Cl}}^{2}+{{\it Br}}^{2} \right) ^{3}+1/3\, \left( {H}^{3}+{{\it Cl }}^{3}+{{\it Br}}^{3} \right) ^{2}$$ It follows that the number of isomers with two hydrogen atoms, two chlorine atoms and two bromide atoms is 18 (extract the coefficient of $H^2 Cl^2 Br^2$).

I get four different isomers of dichlorobenzene (again extracting coefficients, this time for $H^4 Cl^2$). These are: 1, two chlorine atoms bonded to two carbon atoms with a double bond between them, 2, same, but with a single bond between the carbon atoms, 3, with one carbon atom between them and 4, with two carbon atoms between them (in number 3, the angle between the two chlorine atoms is 120 degrees and in number 4, 180 degrees). I am not a chemist and I looked this up. It turns out that 1-2 dichlorobenzene does not exist as the isomer with a single bond between the carbon atoms to which the chlorine atoms are attached. Why that is I don't know.

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Actually, because benzene is a resonance structure, all carbon-carbon double bonds should be treated the same. –  Andrew Salmon Jun 11 '13 at 2:20

For the problem of counting the number of distinct benzene rings with $2~$hydrogen, $2~$chlorine and $2~$bromine atoms attached, under the dihedral symmetry group and using Pólya's method (which is no great improvement on brute counting here), one can reason as follows. The $12$-element dihedral group has as conjugacy classes: the singleton identity, two rotations of order$~6$, two of order$~3$, a single rotation of order$~2$ (minus the identity), and two classes of $3~$reflections each.

We need to sum over all these symmetries the number of configurations (not of classes of configurations) invariant under the symmetry (so having at least that symmetry). For the identity just count all configurations, which are $\binom6{2,2,2}=90$ in number. All symmetries of order$~2$ happen to have at least $2$ orbits of length$~2$, for each of which the same atom has to be attached on their elements, and the remaining $2$ elements also necessarily have the same two (remaining) atoms attached. The gives $3!=6$ configurations for each one of those$~7$ symmetries, and $42$ in all. The other symmetries have some orbit longer than$~2$, which makes it impossible to have one of our configurations fixed by them. So adding up and dividing by the order of the group one obtains $\frac{90+42}{12}=11$ as answer. This agrees with the enumeration of Hagen von Eitzen.

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