The number of optical isomers of the compound | vedclass

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(D) The given compound is $CH_{3}CHBrCHBrCOOH$. It contains two chiral carbon atoms $(n=2)$.Since the molecule is unsymmetrical (the two ends are diff

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$4$

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(D) The given compound is $CH_{3}CHBrCHBrCOOH$. It contains two chiral carbon atoms $(n=2)$.Since the molecule is unsymmetrical (the two ends are different: $-CH_{3}$ and $-COOH$),it cannot be divided into two equal halves.For an unsymmetrical molecule,the number of optical isomers is given by $2^{n}$,where $n$ is the number of chiral centers.Here,$n=2$,so the number of optical isomers $= 2^{2} = 4$.These $4$ isomers consist of $2$ pairs of enantiomers,and there are no meso forms $(m=0)$.Thus,the total number of optical isomers is $4$.

The number of optical isomers of the compound $CH_{3}CHBrCHBrCOOH$ is

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