A) 0
B) 1
C) 3
D) 4
Correct Answer: D
Solution :
\[C{{H}_{3}}-\underset{Br\,}{\mathop{\underset{|\,\,\,\,}{\overset{\overset{H}{\mathop{|\,\,}}\,}{\mathop{{{C}_{*}}}}}\,}}\,-\underset{Br\,}{\overset{H}{\mathop{\underset{|\,\,\,}{\overset{|\,\,}{\mathop{{{C}_{*}}}}}\,}}}\,-COOH\] Number of enantiomers = \[{{2}^{n}}\] (n = asymmetric Carbon atom) = \[{{2}^{2}}\] = 4.
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