A) \[C{{H}_{3}}C{{H}_{2}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HCOOH\]
B) \[C{{H}_{3}}\underset{OMe}{\mathop{\underset{|}{\mathop{C}}\,}}\,HCOOH\]
C) \[C{{H}_{3}}\underset{C{{H}_{2}}OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HCOOH\]
D) \[C{{H}_{3}}\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,HC{{H}_{2}}COOH\]
Correct Answer: C
Solution :
\[X\xrightarrow{NaHC{{O}_{3}}}C{{O}_{2}}\] Hence, compound X has\[~-COOH\] group. \[\underset{\begin{smallmatrix} \text{Chiral compound} \\ \text{optically active} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{2}}OH}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{\underset{|}{\mathop{C*}}\,}}\,}}\,}}\,-COOH}}\,\xrightarrow{LiAl{{H}_{4}}}\underset{\text{Achiral compound}}{\mathop{C{{H}_{3}}-\underset{C{{H}_{2}}OH}{\mathop{\underset{|}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-C{{H}_{2}}OH}}\,\]You need to login to perform this action.
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