For the following compound, find
- Total number of stereoisomers
- Number of optically active stereoisomers
- 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.
- Total number of stereoisomers = 4
- Number of optically active stereoisomers = 4
- Total number of fractions on fractional distillation of all stereoisomers = 2
Are my approach and my answers right?