Identify the reaction

Propanamide is an amide: $\mathrm{CH_3CH_2CONH_2}$

Reaction with $\mathrm{Br_2/KOH(aq)}$ is the Hofmann bromamide reaction (also called Hofmann rearrangement).

Key rule of Hofmann bromamide reaction

In this reaction, a primary amide $\mathrm{RCONH_2}$ gives a primary amine $\mathrm{RNH_2}$ with one carbon less than the original amide.

General reaction: $\mathrm{RCONH_2 \xrightarrow{Br_2/KOH} RNH_2}$

Apply to propanamide

For propanamide, $\mathrm{R = CH_3CH_2}$

So, $\mathrm{CH_3CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2NH_2}$

The product $\mathrm{CH_3CH_2NH_2}$ is ethylamine.

Check options

• A: Propanenitrile $\rightarrow$ incorrect, nitrile is not formed in Hofmann bromamide reaction.
• B: Propylamine $\rightarrow$ incorrect, this would have the same number of carbons as the amide; Hofmann reaction reduces carbon count by 1.
• C: Ethylnitrile $\rightarrow$ incorrect.
• D: Ethylamine $\rightarrow$ correct.

Final answer

$\boxed{\text{D: Ethylamine}}$

Comparison with stored correct answer

Stored correct answer is D, which matches the derived answer.