question_answer

An organic compound A upon reacting with \[~N{{H}_{3}}\] gives B. On heating B gives C. C in presence of KOH reacts with \[B{{r}_{2}}\] to give \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}.\] A is:

A) \[C{{H}_{3}}COOH\]

B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COOH\]

C) \[C{{H}_{3}}-\underset{\underset{C{{H}_{3}}}{\mathop{|}}\,}{\mathop{CH}}\,-COOH\]

D) \[C{{H}_{3}}C{{H}_{2}}COOH\]

Correct Answer: D

Solution :

[d]

Reaction (III) is a Hofmann bromamide reaction formation of \[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\] is possible only from a compound \[C{{H}_{3}}C{{H}_{2}}CO{{O}^{-}}NH_{4}^{+}\][b] in (II) reaction further propanic acid \[\left( C{{H}_{3}}C{{H}_{2}}COOH \right)\] on reaction with \[N{{H}_{3}}\] produce \[N{{H}_{3}}C{{H}_{2}}CO{{O}^{-}}NH_{4}^{-}\] (reaction I) hence the reaction will be