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Step-by-Step Solution
Step 1: Identify the sequence of reactions
We are told that an organic compound A (a carboxylic acid) first reacts with $NH_3$ to give B, which then on heating forms C. When C is treated with $Br_2$ and $KOH$, the final product is $CH_3CH_2NH_2$ (ethylamine). This series of steps indicates the formation of an amide followed by a Hoffmann bromamide (or Hoffmann rearrangement) reaction that reduces the carbon chain by one.
Step 2: Determine the structure of B and C
B formation: A carboxylic acid $RCOOH$ reacts with $NH_3$ to form its ammonium salt, which upon mild heating can directly convert to the amide $RCONH_2$.
C formation: On stronger heating, the ammonium salt or intermediate converts to the amide (let us call it C). Thus, $B \rightarrow C$ implies $RCONH_2$ is formed.
Step 3: Apply the Hoffmann bromamide reaction
When an amide $RCONH_2$ is treated with $KOH$ and $Br_2$, the Hoffmann bromamide reaction occurs, converting $RCONH_2$ into a primary amine $RNH_2$ with one less carbon atom in the chain.
Step 4: Match the final product with the correct starting acid
The final amine given is $CH_3CH_2NH_2$ (ethylamine). To arrive at ethylamine via Hoffmann rearrangement, the original amide must have had three carbons in its chain (since the process removes one carbon):
If C is $CH_3CH_2CONH_2$ (propionamide), then after Hoffmann bromamide reaction:
$CH_3CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2NH_2$
Hence, the carboxylic acid A must be $CH_3CH_2COOH$ (propanoic acid), which is option 4.
Step 5: Confirm the answer
Therefore, the organic compound A is propionic acid ($CH_3CH_2COOH$).
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