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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.

enter image description here

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

enter image description here

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?

Related: How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?

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    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 06:13
  • 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 –  Aug 10 '16 at 06:15
  • 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 07:40
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    @IvanNeretin I think he is wrong.He considered enantiomers which are optical isomers and not geometrical. –  Aug 10 '16 at 07:42
  • 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 07:44
  • Yes you are right.See this http://www.avogadro.co.uk/organic/isomer.gif @IvanNeretin –  Aug 10 '16 at 07:45
  • Wait, I got lost in this thread, so deleting my previous comment. The issue is that this molecule is symmetric, so you lose a factor of two. Consider the case of 2,4,6-octatetraene. There's 3 double bonds, but there are only 6 isomers because the two ends are the same. If you consider 2,4,6-nonatetraene, then you'll have the full 8 isomers. – Zhe Oct 27 '16 at 21:36

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The answer in your textbook is correct. Here are all possible isomers: enter image description here 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:)

EDIT: Because my answer has raised an interesting discussion, I will try to present my point of view. First of all, the cis/trans isomerism is a descriptive way to represent a stereochemistry of a molecule, but its scope is limited. It is useful for teaching organic chemistry, but not for a systematic nomenclature (especially for a ring chirality). I think that the main source of confusion here is understanding the difference between a local and global (absolute) stereochemistry. In the above picture the controversy is whether the four enantiomers in the two rightmost columns can be treated as geometrical isomers. Let's simplify the problem and consider 1,2-dimethylcyclopentane: dimethylcyclopentane

The situation is clear for the first two isomers (a and b) - these are geometrical isomers, no doubts about it. What about b and c? We see that in both isomers the methyl groups are in the trans orientation, so one may conclude that they are not geometrical isomers. Period. However, cis/trans notation describes only the local geometry (E/Z notation obviously cannot be used in this example). Isomers b and c have the same relative orientation of methyl groups, but in a global scope these are two different molecules, and thus two different geometrical isomers. Therefore, 1,2-dimethylcyclopentane has one cis and two trans isomers. This may be counterintuitive at first, because there are no such situations in case of double bonds, where the cis/trans notation is more frequently used, but in rings we have an additional layer of complexity. I am aware that some people may disagree with this explanation, but as far as I know there are no official (i.e. IUPAC) directives to resolve this controversy.

vapid
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    You considered optical isomers(enantiomers) but the question asks for geometrical isomers only. –  Aug 10 '16 at 07:42
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    @Sanchayan Dutta So are you saying that the two compounds in the first column are not geometrical isomers? That makes no sense. – vapid Aug 10 '16 at 07:48
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    Tell me you definition of geometrical isomers and optical isomers first. –  Aug 10 '16 at 07:49
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    This is not a case of different definitions of geometrical and optical isomerism - we all agree in this matter. So if you exclude optical isomerism there are only 3 geometrical isomers - don't you agree? Still, makes no sense. – vapid Aug 10 '16 at 07:57
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    No there are 6(exclude the enantiomers on the right) as the ring too shows Geometrical isomerism. –  Aug 10 '16 at 07:58
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    @Sanchayan Dutta Now explain me what is the difference between geometrical isomerism in the ring and optical isomerism. – vapid Aug 10 '16 at 08:13
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    No difference.However the rightmost two cannot be considered GI in any way as they have been taken before. –  Aug 10 '16 at 08:35
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    @SanchayanDutta I disagree. The author of your textbook also seems to disagree. I am stopping this discussion and letting others voice their opinion. – vapid Aug 10 '16 at 08:49
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    You are both right and both wrong in the sense that there is no accepted definition of geometric isomers. It seems that IUPAC discourages its usage . @vapid maybe you can edit this in your answer or make more research on it first – K_P Oct 28 '16 at 08:15
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    @K_P I agree with you that the problem is controversial. I will edit the question in the near future to show my point of view, today I am a bit busy. – vapid Oct 28 '16 at 10:28
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    @K_P The link is broken – Soham Jan 04 '19 at 17:11