A) \[BrC{{H}_{2}}\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}Br\]
B) \[{{C}_{2}}{{H}_{5}}C{{H}_{2}}\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}Br\]
C) \[{{C}_{3}}{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}Br\]
D) \[HOOC\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,CHCOOH\]
E) \[OHC-CH(OH)-C{{H}_{2}}OH\]
Correct Answer: A
Solution :
A chiral carbon atom is that whose all the four valencies are satisfied by different atoms or groups. In the alternate (a), two valencies are satisfied by the same group i. e., \[-C{{H}_{2}}Br\] So, the compound without a chiral carbon atom is \[\begin{align} & C{{H}_{3}} \\ & | \\ & BrC{{H}_{2}}CHC{{H}_{2}}Br \\ \end{align}\]
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