A) ethanamine
B) methanamine
C) methanamide
D) ethanenitrile
Correct Answer: B
Solution :
When acetamide is treated with NaBr or \[\text{B}{{\text{r}}_{\text{2}}}\text{/NaOH,}\]Hofmann degradation reaction takes place and a primary amine is obtained as: \[C{{H}_{3}}CON{{H}_{2}}+NaOBr\,or\,B{{r}_{2}}/NaOH\xrightarrow{{}}\] \[\underset{\begin{smallmatrix} \text{methyl}\,\,\text{amine} \\ \,\,\,\,\,\,\,\text{or} \\ \text{methanamine} \end{smallmatrix}}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{N}{{\text{H}}_{\text{2}}}}}\,\text{+2NaBr+NaHC}{{\text{O}}_{\text{3}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]
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