1. Identify the final reaction

The compound $C$ reacts with $Br_2/KOH$ to give $CH_3CH_2NH_2$.

This is the Hofmann bromamide reaction:

$RCONH_2 \xrightarrow{Br_2/KOH} RNH_2$

with loss of one carbon atom from the amide.

Since the product amine is:

$CH_3CH_2NH_2$

$C$ must be the amide having one extra carbon:

$C = CH_3CH_2CONH_2$

which is propanamide.

2. Relate $B$ and $C$

Given that on heating $B$ gives $C$.

A carboxylic acid reacts with ammonia to first form the ammonium salt, which on heating gives the amide:

$RCOOH + NH_3 \rightarrow RCOO^-NH_4^+ \;(B)$

On heating:

$RCOO^-NH_4^+ \xrightarrow{\Delta} RCONH_2 \;(C) + H_2O$

So if

$C = CH_3CH_2CONH_2$

then $B$ must be ammonium propanoate:

$B = CH_3CH_2COO^-NH_4^+$

3. Find $A$

Since $A$ reacts with $NH_3$ to give $B$, $A$ must be the corresponding carboxylic acid:

$A = CH_3CH_2COOH$

This is propanoic acid.

4. Check with options

• A: $CH_3COOH$ gives acetamide, which on Hofmann degradation gives $CH_3NH_2$, not $CH_3CH_2NH_2$.
• B: $CH_3CH_2CH_2COOH$ gives butanamide, which on Hofmann degradation gives $CH_3CH_2CH_2NH_2$.
• D: $CH_3CH_2COOH$ gives propanamide, which on Hofmann degradation gives $CH_3CH_2NH_2$.

Hence the correct option is:

$\boxed{D}\;:\; CH_3CH_2COOH$