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The hydrocarbon naphthalene has ten carbon atoms arranged in a double hexagon, and eight hydrogen atoms attached at each of the corners of the hexagons. Naphthol is obtained by replacing one of the hydrogen atoms of naphthalene with a hydroxyl group (OH). How many isomers of naphthol are there?

So far I considered N={1,2,3,4,5,6,7,8} the set of Hydrogen and C={hydrogen, hydroxil} the set of types. Considering hydrogen H and hydroxil D, I got, following an example, that $ZGv(H+D,..,H^8+D^8)$. No idea how to get the number of isomers out of this. The example I followed was at this link at page 40.

Is it the same thing as the cyclobutane from 6.6 on page 40 or does it come differently? It has two types so it should be the same?

Tetramethylnaphthalene is obtained by replacing four of the hydro- gen atoms of naphthalene with methyl groups (CH3). How many isomers of tetramethylnaphthalene are there?

This is another one that implies the use of the same formula, is it solved on the same principle like the one above? Appreciate the help.

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1 Answer 1

First draw your compound leaving all hydrogen atoms blank for the moment. These will be our slots, to be filled with a hydrogen atom or a hydroxyl group. The base generating function is $1+z$, with $z$ marking the hydroxyl group.

The slots are being permuted by the automorphism group $G$ of the compound, which is very simple in structure. It contains four permutations: the identity, the horizontal flip, the vertical flip, and a rotation by 180 degrees. The cycle index of $G$ can then be read off the diagram and it is $$ Z(G) = \frac{1}{4} \left( a_1^8 + 3 a_2^4 \right).$$

Now do the usual substitution to obtain the orbit enumerator, which is $$1+2\,z+10\,{z}^{2}+14\,{z}^{3}+22\,{z}^{4}+14\,{z}^{5}+10\,{z}^{6}+2\, {z}^{7}+{z}^{8}$$

It follows that there are two isomers.

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Where did you get that 1/4(a1^8+3a2^4) ? and where do I substitute? Thank you for the help so far and why the 1+Z? –  Max Nov 15 '12 at 17:24
    
The cycle index can be read off the diagram with a bit of practice. There is a mechanical procedure, as well, if you prefer that. Mark all 8 slots with a number from one to eight then apply one of the four permutations from $G$. This gives you the table representation of the permutation (it shows what element each element gets mapped to). Factor that into cycles and represent cycles of length $k$ by $a_k$ to get the cycle index. –  Marko Riedel Nov 15 '12 at 17:42
    
The base generating function includes $z^0 = 1$ for hydrogen compounds (zero because we're not counting those) and $z^1$ for hydroxyl groups, which we are counting. You could have used $H$ and $D$ as well, it's the same thing. –  Marko Riedel Nov 15 '12 at 17:44
    
I understand. There's this another that I need some help with as well. math.stackexchange.com/questions/238116/… –  Max Nov 15 '12 at 17:51
    
I forgot, it's not 100% precise to say that we apply a permutation from $G$. We should say "an automorphism" from $G$, which in turn corresponds to a permutation of the eight slots. If we were only placing hydroxyl groups on the four left and right carbon atoms the automorphism would still be the same, but not the permutation, which would represent four rather than eight slots. –  Marko Riedel Nov 15 '12 at 17:53

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