A) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-COOH\]
B) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-COOH\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-COOH\]
D) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]
Correct Answer: C
Solution :
A carbon atom which is attached by four different group is called chiral centre. In optically active compound at least one chiral centre should be present. \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{{{C}^{*}}}}}\,-COOH\]
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