A) \[C{{H}_{3}}CHN+2{{H}_{2}}O\xrightarrow{{}}C{{H}_{3}}COO+N{{H}_{3}}\]
B) \[C{{H}_{3}}CN+4H\xrightarrow{Na/ErOH}C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]
C) \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/NaOH}C{{H}_{3}}N{{H}_{2}}\]
D) \[C{{H}_{3}}COCl+C{{H}_{3}}OH\xrightarrow{{}}\]\[C{{H}_{3}}COOC{{H}_{3}}+HCl\]
Correct Answer: C
Solution :
Key Idea: In Hofmann bromamide reaction acid amide reacts with\[B{{r}_{2}}/C{{l}_{2}}\]in\[NaOH/KOH\]to produce amine having one carbon atom less than amide. So the following is Hofmann bromamide reaction. \[\underset{\text{ethanamide}}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,\xrightarrow{B{{r}_{2}}/NaOH}\underset{\text{methanamine}}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\]
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