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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

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?

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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

  • 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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    $\begingroup$ The absolute configurations R or S for the two chiral carbon atoms in circle “5” implicitly include the relative configuration cis or trans in circle “5”. That leaves us with only 16 individual stereoisomers. $\endgroup$ – Loong Feb 21 '16 at 18:45
  • $\begingroup$ I missed that, thanks for pointing it out. I'll correct my answer. $\endgroup$ – ron Feb 21 '16 at 19:01

protected by orthocresol Nov 25 '17 at 15:23

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