# Number of geometrical isomers when a cyclic ring is involved along with axis of symmetry

This is an question from my Chemistry study material and it asks to find the number of Geometrical Isomers of the following compound in the picture.

Basically the method/formula given in my book to find number of Geometrical Isomers of a compound is:

Using this formula I put $n=3$.Morever since both ends are same I get $p=(3+1)/2=2$.So final answer according to me is $2^{3-1}+2^{2-1}=4+2=6$.

But the answer given is $2^3=8$.

Where did I go wrong?

• Didn't read the formula, but 6 looks right to me. See, the ends are either both trans, or both cis, or cis and trans (no matter which is which, since they are otherwise identical), which makes 3 options, and on top of that they are either on the same side of the cycle or not, which multiplies it by 2. – Ivan Neretin Aug 10 '16 at 6:13
• Yeah right.Even I thought the same.In all probability the given answer is wrong.BTW i wonder why don't you write your comments as answers.They are elaborate enough to be an answer! :-) @IvanNeretin – user14857 Aug 10 '16 at 6:15
• Besides, I was wrong on this one, as explained by vapid. Indeed, if we consider enantiomers, we have 8 all right. – Ivan Neretin Aug 10 '16 at 7:40
• @IvanNeretin I think he is wrong.He considered enantiomers which are optical isomers and not geometrical. – user14857 Aug 10 '16 at 7:42
• Are they? I never was able to remember correctly. What are geometrical isomers, then? Everything with similar connectivity, yet different, excluding optical isomers? – Ivan Neretin Aug 10 '16 at 7:44

The answer in your textbook is correct. Here are all possible isomers: I added a symmetry axis to the right, so you can see that these compounds are mirror images and are not superimposable. The formula you used for calculations is too complicated. The way to calculate it is simple: there are four stereogenic centers so: $$2^4=16$$There is one axis of symmetry so:$$16/2=8$$Have more faith in your books:)