Number of stereoisomers of a polyene

Find the number of stereoisomers of the following compound:


[![4-(buta-1,3-dien-1-yl)-1,2-dimethyl-5-(penta-1,2,3-trien-1-yl)cyclohex-1-ene](https://i.stack.imgur.com/2q0u2.png)](https://i.stack.imgur.com/2q0u2.png)


I think the answer is 16:


* The double bond in the ring cannot show geometrical isomerism.
* The first double bond on the top substituent can show *cis-trans* isomerism and the one on the end will not show *cis-trans* isomerism because of two same groups on one carbon.
* The substituent on the bottom that is: $\ce{-CH=C=C=CHMe}$ can also show *cis-trans* isomerism.
* Then, there are two chiral centres and the number of optical isomers is $2^2=4$. The total number of stereoisomers is 16.


But the answer given is 4. Am I correct or not?




A picture is worth a thousand words so I've redrawn your compound and have numbered the areas to be considered.


[![enter image description here](https://i.stack.imgur.com/vs97s.png)](https://i.stack.imgur.com/vs97s.png)


* Double bond "1" is not capable of displaying *cis-trans*
stereoisomers since there are two identical substituents (hydrogens)
on one end of the double bond.
* Double bond "2" is capable of displaying *cis* and *trans* stereoisomers since there are two different substituents on each end of the double bond.
* Double bond "3" can only exist as a *cis* isomer due to the geometric constraints imposed by the 6-membered ring.
* Double bond "4" is a 1,2,3-triene (allene would be a 1,2-diene); such trienes display the same stereochemical *cis-trans* possibilities as a double bond.
* The substituents on the two cyclohexene carbons contained in circle "5" also have *cis-trans* arrangements; these are further illustrated in the cyclohexene drawings pictured below.
* Finally, each of the carbons contained in circle "5" are chiral (they each have 4 different substituents) and can exist as R and S stereoisomers.


Considering only the circles that can have stereoisomers (2, 4 and 5) we can have the following spatial arrangements:


* 2-*cis*, 4-*cis*, 5-*trans*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*trans*, 4-*cis*, 5-*trans*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*cis*, 4-*trans*, 5-*trans*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*trans*, 4-*trans*, 5-*trans*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*cis*, 4-*cis*, 5-*cis*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*trans*, 4-*cis*, 5-*cis*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*cis*, 4-*trans*, 5-*cis*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
* 2-*trans*, 4-*trans*, 5-*cis*, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.


Therefore, there are a total of 16 stereoisomers for your molecule, 8 diastereomers, each of which can exist as an enantiomeric pair.


**Thanks to Loong** for pointing out that the *cis* and *trans* geometry in circle "5" limits the R and S possibilities at the two chiral carbon atoms in circle "5". That is, these carbons in (for example) the *cis* isomer cannot be R,S R,R S,R and S,S, - in the *cis* isomer they can only be either R,S and S,R **or** R,R and S,S.


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