The number of enantiomers of the compound C{H | vedclass

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Question

(d) $C{H_3} - \mathop {\mathop {{C_ * }}\limits_{|\,\,\,\,}^{\mathop {|\,\,}\limits^H } }\limits_{Br\,} - \mathop {\mathop {{C_ * }}\limits_{|\,\,\,}^

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) $C{H_3} - \mathop {\mathop {{C_ * }}\limits_{|\,\,\,\,}^{\mathop {|\,\,}\limits^H } }\limits_{Br\,} - \mathop {\mathop {{C_ * }}\limits_{|\,\,\,}^{|\,\,} }\limits_{Br\,}^H - COOH$

Number of enantiomers = ${2^n}$ ($n =$ asymmetric Carbon atom) = ${2^2}$ = $4$.

The number of enantiomers of the compound $C{H_3}CHBrCHBrCOOH$ is

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