A) \[HO-\underset{H}{\mathop{\underset{|}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-COOH\]
B) \[C{{H}_{3}}-\underset{Cl}{\mathop{\underset{|}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-COOH\]
C) \[C{{H}_{3}}-\underset{OH\,\,}{\mathop{\underset{|\,\,\,\,\,\,}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|\,\,\,\,\,}{\mathop{C-\,}}\,}}\,}}\,}}\,COOH\]
D) \[C{{H}_{3}}-\underset{Cl\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|\,\,\,\,\,\,}{\mathop{C-\,}}\,}}\,}}\,}}\,COOH\]
Correct Answer: B
Solution :
\[C{{H}_{3}}-\overset{H}{\mathop{\overset{|*}{\mathop{\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,\,}}\,}}\,-COOH\] In this structure chiral carbon atom is present since it is optical active.
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