A) \[B{{r}_{2}}\] in aqueous KOH
B) \[B{{r}_{2}}\] in alcoholic KOH
C) \[C{{l}_{2}}\] in sodium
D) Sodium in ether
Correct Answer: A
Solution :
: Only treatment of amide with \[B{{r}_{2}}\] in aqueous solution of \[Na\] or \[KOH\] will give an amine with lesser no. of carbon atoms than in the reactant. \[\underset{Amide}{\mathop{R-CON{{H}_{2}}}}\,\xrightarrow[\begin{smallmatrix} Hofmann\,bromamide\,or \\ \text{degradation} \end{smallmatrix}]{B{{r}_{2}}/KOH,\,\Delta }\underset{A\min e}{\mathop{R-N{{H}_{2}}}}\,\]
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