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How many geometrical isomers will exist for compound of this type which has cycloalkene having carbon atoms greater or equal to 8 in the middle of chain?

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I know that due to ring in middle of chain geometrical isomerism exist and also due to cycloalkene having carbon atoms greater or equal to 8. But I am confused in a case when a cycloalkene exist in middle of the chain. Whether we will consider geometrical isomers due to cycloalkene or due to both, i.e. the ring and also due to cycloalkene case?

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Geometrical isomers refers to isomerism in which the connectivity of atoms is preserved, and changes refer only to the spatial arrangement of the atoms. In other words, it only allows transformations around certain features that can be arranged in different ways in three dimensions (and which aren't just conformers that will easily transform into each other due to molecular motion).

I mention this because your reference to a "cycloalkene [...] having [8 or more] carbon atoms" suggest that you are looking for constitutional isomers (i.e. compounds with the same atoms but different connectivity) and not just geometrical isomers (i.e. compounds with the same connectivity but different spatial arrangement).

If we consider geometrical isomers of the compound you gave, 1) the cycloalkene ring, 2) the side chains and 3) the substitution sites in the ring are fixed (as changing them would produce constitutional isomers), so we are left with isomerism due to cis/trans double bonds.

You have three double bonds in the side chains that will produce $2^3=8$ isomers. Cyclooctenes are in principle large enough rings that their trans form can be isolated, so this leads to $16$ isomers; you can figure them out from your posted example (which is cis in the cycloalkene and trans in the side chains) by changing double bonds.

If you want to figure out constitutional isomers, that's a far more complex task; you'll have to account for ring size, number of side chains, ring substitution pattern, branching, position of insaturations and so on; with a hydrocarbon of this size, the number of possibilities will be huge.

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  • $\begingroup$ could you please explain that- if we just have the ring(without any unsaturation in the ring) then also we will have 16 geometrical isomers i.e. due to the different spatial arrangement of side branches.If we have unsaturation in the ring then won't be there 2× more (i.e 16×2=32) isomers due to the different spatial arrangement of atoms linked with double bonded carbon of the ring? $\endgroup$ – Anirudh May 10 '18 at 15:14
  • $\begingroup$ @Anirudh ,The question is about geometrical isomers. You are also thinking of Optical isomers (Spatial arrangement of groups around an atom not double bond) which can also be possible. If it is asked to find total no. of stereoisomers, we then also have to think about those. But here in this context, those are not considered. $\endgroup$ – Soumik Das May 11 '18 at 3:48
  • $\begingroup$ @SoumikDas thank you for your comment .If we have just a ring in the middle of chain then also there will be 16 geometrical isomers(as disubstituted cycloalkane also show CIS-trans isomerism) and also when a cycloalkene(having more of equal to 8 carbon) is at the end of chain then also there will be 16 geometrical isomers .The above case is the combination of two therefore we should consider both(that is my actual doubt)Is my consideration wrong? $\endgroup$ – Anirudh May 11 '18 at 5:09
  • $\begingroup$ As @SoumikDas said, if you consider asymmetric carbons in the cycle, you find four optical isomers for each geometrical isomer (not just cis / trans isomerism, as the two possible cis arrangements are enantiomers, and the same holds for the two possible trans arrangements); so, if we enlarge our scope to consider these, we can find 64 possible stereoisomers, 32 pairs of enantiomers. $\endgroup$ – user41033 May 11 '18 at 9:24

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