# 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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## 2 Answers

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

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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