For the following compound, find

  1. Total number of stereoisomers
  2. Number of optically active stereoisomers
  3. Total number of fractions on fractional distillation of all stereoisomers


There is a chiral carbon marked with "*" in the compound. The compound is thus chiral and is optically active. The configuration at the chirality centre can either be R or S. There is also a double bond in the compound and cis-trans (geometrical isomerism) is possible here. So using basic permutations there are totally $2\times2=4$ stereoisomers (optical and geometrical).

Since there is a chirality centre the compound exists in right-handed and left handed systems and thus is optically active. Thus all the 4 stereoisomers are optically active.

Out of R-cis, R-trans, S-cis, S-trans isomers there are two pairs of diastereomers - R-cis, S-trans and R-trans, S-cis - so there are 2 fractions obtained due to fractional distillation.

Final Answer

  1. Total number of stereoisomers = 4
  2. Number of optically active stereoisomers = 4
  3. Total number of fractions on fractional distillation of all stereoisomers = 2

Are my approach and my answers right?


1 Answer 1


The approach is right, and your answers are right, but there are a couple of things to look at in your explanation.

First, the presence of a chiral centre does not in itself make a compound chiral.

It is in this case, but a molecule is chiral if, and only if, it is not superimposable on its mirror image. You could have a molecule with more than one chiral centre, which was superimposable on its mirror image, and was therefore not chiral.

Second, a chiral molecule is not necessarily optically active. Its specific rotation may be zero (or at least unmeasurable), in which case it is said to exhibit cryptochirality.

Thirdly, the R- and S- form of each of the geometric isomers are enantiomers (ie non-superimposable mirror images) of one another, not diastereomers. Diastereomers occur when a molecule has (for instance) two chiral centres and rather than inverting both (giving the enantiomer, its mirror image) you only invert one, which is actually a different compound, with different physical and chemical properties.

Personally I think using the term stereoisomer to refer to the cis and trans versions is misleading, even if strictly geometric isomers are a subset of stereoisomers. I'd prefer to say that the R- and S- forms of each geometric isomer are enantiomers of one another, and leave it at that, rather than lumping all four together as stereoisomers. But you don't get to choose the wording of the question, I guess.


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