A) \[C{{H}_{3}}CON{{H}_{2}}\]
B) \[{{C}_{2}}{{H}_{5}}NC\]
C) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\]
D) \[C{{H}_{3}}N{{H}_{2}}\]
Correct Answer: D
Solution :
Key Idea: This is Hofmann bromamide reaction. In this reaction amide react with \[\text{B}{{\text{r}}_{\text{2}}}\]and NaOH to form amine having one carbon less than parent amide. \[\underset{\text{ethanamide}}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,+4NaOH+B{{r}_{2}}\xrightarrow{{}}\] \[C{{H}_{3}}N{{H}_{2}}+{{H}_{2}}O+4NaBr+C{{O}_{2}}\] \[\therefore \]Methyl amine is formed during reaction.
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