A) \[C{{H}_{3}}COCl\]
B) \[C{{H}_{3}}CON{{H}_{2}}\]
C) \[C{{H}_{3}}COOH\]
D) \[C{{H}_{3}}C{{H}_{2}}COOH\]
Correct Answer: D
Solution :
\[C{{H}_{3}}C{{H}_{2}}l\xrightarrow[-KI]{KCN}C{{H}_{3}}C{{H}_{2}}CN\xrightarrow[{}]{{{H}_{2}}{{O}^{+}}/{{H}_{2}}O}\] \[C{{H}_{3}}C{{H}_{2}}CHO\xrightarrow[{}]{{{H}_{3}}{{O}^{+}}/{{H}_{2}}O}\underset{propanoic\,acid}{\mathop{C{{H}_{3}}C{{H}_{2}}COOH}}\,\]
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