A) \[C{{H}_{3}}COOH\]
B) \[C{{H}_{3}}N{{H}_{2}}\]
C) \[C{{H}_{3}}Br\]
D) \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\]
Correct Answer: B
Solution :
Key Idea: This is Hofmann bromamide reaction. In this reaction acid amide reacts with bromine and KOH (or \[NaOH\]) to form amine having one carbon atom less than acid amide \[\underset{acetamide}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,+B{{r}_{2}}+4KOH\to \underset{\underset{methyl\text{ }amine}{\mathop{(X)}}\,}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\] \[+2KBr+{{K}_{2}}C{{O}_{3}}+2{{H}_{2}}O\] \[\therefore \] The principle product X in the reaction is methyl amine.
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