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Reaction of propanamide with ethanolic sodium hydroxide and bromine will give
A
Aniline
B
Ethylamine
C
Methylamine
D
Propylamine
Detailed Solution
In the Hofmann bromamide degradation reaction, an amide $RCONH_2$ reacts with $Br_2$ and alcoholic $NaOH$ to give a primary amine with one carbon less than the starting amide: $CH_3CH_2CONH_2$ (propanamide) $\rightarrow CH_3CH_2NH_2$ (ethylamine).