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How many stereoisomers does this molecule have ?
$CH_3CH=CHCH_2CHBrCH_3$
$CH_3CH=CHCH_2CHBrCH_3$
A
8
B
2
C
4
D
6
Detailed Solution
The given compound is $CH_3-CH=CH-CH_2-\overset{*}{C}HBr-CH_3$.
The double bond has two different groups on each carbon ($CH_3$ and $H$ on one carbon; $H$ and $CH_2CHBrCH_3$ on the other), so geometrical isomerism (cis and trans forms) is possible.
The carbon carrying Br is attached to four different groups ($H$, $Br$, $CH_3$ and $CH_2CH=CHCH_3$), so it is an asymmetric (chiral) carbon and optical isomerism is also possible.
Number of optical isomers $= 2^n = 2^1 = 2$ (here $n$ = number of asymmetric carbons).
Each geometrical form (cis and trans) exists as a pair of enantiomers ($R$ and $S$).
Total number of stereoisomers $= 2 \times 2 = 4$ (cis-R, cis-S, trans-R, trans-S).
The double bond has two different groups on each carbon ($CH_3$ and $H$ on one carbon; $H$ and $CH_2CHBrCH_3$ on the other), so geometrical isomerism (cis and trans forms) is possible.
The carbon carrying Br is attached to four different groups ($H$, $Br$, $CH_3$ and $CH_2CH=CHCH_3$), so it is an asymmetric (chiral) carbon and optical isomerism is also possible.
Number of optical isomers $= 2^n = 2^1 = 2$ (here $n$ = number of asymmetric carbons).
Each geometrical form (cis and trans) exists as a pair of enantiomers ($R$ and $S$).
Total number of stereoisomers $= 2 \times 2 = 4$ (cis-R, cis-S, trans-R, trans-S).