A) \[C{{H}_{2}}=CHCOC{{H}_{2}}C{{H}_{2}}OH\]
B) \[C{{H}_{2}}=CH-\underset{OH}{\mathop{\underset{|}{\mathop{CH}}\,}}\,C{{H}_{2}}CHO\]
C) \[C{{H}_{2}}=CHCOC{{H}_{2}}CHO\]
D) \[C{{H}_{2}}=CHCOC{{H}_{2}}COOH\]
Correct Answer: A
Solution :
\[Mn{{O}_{2}}\]oxidises allylic and benzylic alcohols. \[\underset{allylic}{\mathop{\underset{\nearrow }{\mathop{C{{H}_{2}}=CH}}\,}}\,\underset{OH}{\mathop{\underset{|}{\mathop{CH}}\,}}\,\underset{This\,is\,not\,affected}{\mathop{\underset{\nearrow }{\mathop{C{{H}_{2}}C{{H}_{2}}}}\,}}\,OH\xrightarrow{Mn{{O}_{4}}}\] \[C{{H}_{2}}=CHCOC{{H}_{2}}C{{H}_{2}}OH\]
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