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 |
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