A) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,COOH\]
B) \[\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}COOH\]
C) \[C{{H}_{2}}=CHCOOH\]
D) \[\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,COOH\]
Correct Answer: C
Solution :
\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[\begin{smallmatrix} \operatorname{Re}d\,(HVZ \\ reduction) \end{smallmatrix}]{C{{l}_{2}}}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{CH}}\,COOH\] \[\xrightarrow[-KCl,-{{H}_{2}}O]{Alc\,KOH}\underset{\alpha ,\beta -unsaturated\text{ }acid}{\mathop{C{{H}_{2}}=CHCOOH}}\,\]
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