In a set of reactions propanoic acid yielded a compound D.${\text{CH}}_3{\text{CH}}_2\text{COOH}\stackrel{{\text{SOCl}}_2}{\to }\text{B}\stackrel{{\text{NH}}_3}{\to }\text{C}\underset{\text{KOH}}{\overset{{\text{Br}}_2}{\to }}\text{D}$ The structure of D would be :

Propanoic acid reacts with ${\text{SOCl}}_2$ to give propanoyl chloride (B). Treatment with ${\text{NH}}_3$ forms propanamide (C). Finally, the Hoffmann bromamide degradation reaction with ${\text{Br}}_2/\text{KOH}$ removes the carbonyl carbon to yield a primary amine with one carbon less, which is ethanamine (${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$).