Propanoic acid gives a series of reactions as given below. $CH_3CH_2COOH \xrightarrow{SOCl_2} B \xrightarrow{NH_3} C \xrightarrow{Br_2/KOH} D$ The structure of D would be:

The given sequence of reactions is as follows:

1. Formation of Acid Chloride (B): Propanoic acid ($CH_3CH_2COOH$) reacts with thionyl chloride ($SOCl_2$) to form propanoyl chloride ($B$). $CH_3CH_2COOH + SOCl_2 \longrightarrow CH_3CH_2COCl (B) + SO_2 + HCl$

2. Formation of Amide (C): Propanoyl chloride ($B$) reacts with ammonia ($NH_3$) to undergo nucleophilic acyl substitution, forming propanamide ($C$). $CH_3CH_2COCl + 2NH_3 \longrightarrow CH_3CH_2CONH_2 (C) + NH_4Cl$

3. Hofmann Bromamide Degradation (D): Propanamide ($C$) reacts with bromine ($Br_2$) in the presence of a strong base like potassium hydroxide ($KOH$). This is the Hofmann bromamide degradation reaction, which converts a primary amide to a primary amine with one carbon atom less than the original amide. $CH_3CH_2CONH_2 + Br_2 + 4KOH \longrightarrow CH_3CH_2NH_2 (D) + K_2CO_3 + 2KBr + 2H_2O$

Therefore, the final product D is ethanamine ($CH_3CH_2NH_2$).