A) ethanamine
B) ethyl alcohol
C) methanamine
D) bromoform
Correct Answer: C
Solution :
Amides, on reaction with \[\text{B}{{\text{r}}_{\text{2}}}\text{/NaOH,}\]gives one carbon less amines. This is Hofmann-bromamide reaction. \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/NaOH}C{{H}_{3}}N{{H}_{2}}\]You need to login to perform this action.
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