• # question_answer The number of enantiomers of the compound $C{{H}_{3}}CHBrCHBrCOOH$ is [AIIMS 1997] A) 0 B) 1 C) 3 D) 4

$C{{H}_{3}}-\underset{Br\,}{\mathop{\underset{|\,\,\,\,}{\overset{\overset{H}{\mathop{|\,\,}}\,}{\mathop{{{C}_{*}}}}}\,}}\,-\underset{Br\,}{\overset{H}{\mathop{\underset{|\,\,\,}{\overset{|\,\,}{\mathop{{{C}_{*}}}}}\,}}}\,-COOH$ Number of enantiomers = ${{2}^{n}}$ (n = asymmetric Carbon atom) = ${{2}^{2}}$ = 4.