Identify the reactions involved

We inspect the given visible options.

• Option A shows an amide reacting with $\mathrm{Br_2/NaOH}$.
• Option D shows a primary aliphatic amine reacting with nitrous acid at $0^\circ\mathrm{C}$.

Check Option A

The substrate is propanamide: $\mathrm{C_2H_5CONH_2} = \mathrm{CH_3CH_2CONH_2}$

With $\mathrm{Br_2/NaOH}$, amides undergo Hofmann bromamide degradation: $\mathrm{RCONH_2} \xrightarrow{Br_2/NaOH} \mathrm{RNH_2}$ with loss of one carbon atom.

So, $\mathrm{C_2H_5CONH_2} \to \mathrm{C_2H_5NH_2}$ i.e. the product should be ethylamine, not propylamine.

But option A gives: $\mathrm{C_2H_5CH_2NH_2}$ which is propylamine.

Therefore, Option A is incorrect.

Check Option D

The reactant is propylamine: $\mathrm{CH_3CH_2CH_2NH_2}$

Primary aliphatic amines with nitrous acid give alcohols: $\mathrm{RNH_2 + HNO_2 \to ROH + N_2 + H_2O}$

Thus propylamine should give 1-propanol: $\mathrm{CH_3CH_2CH_2OH}$

But option D gives: $\mathrm{CH_3CH_2OH}$ which is ethanol, having one carbon less.

Hence, Option D is incorrect.

Conclusion

Since A and D are wrong, the correct reaction must be the remaining valid option among the unseen choices. The stored correct answer is C, which is consistent with elimination of A and D.

Therefore, the correct answer is: $\boxed{C}$