This question involves identifying an organic compound through a series of reactions. Let's analyze the reaction sequence step by step:

Step 1: Compound A reacts with NH₃ to give B. This suggests that A is a carboxylic acid, and B is the corresponding ammonium salt.

General reaction: $\mathrm{RCOOH}+\mathrm{NH}{}_3\to \mathrm{RCOO}{}^{-}{\mathrm{NH}}^4{}^{+}$

Step 2: Heating B gives C. When ammonium carboxylate salts are heated, they undergo dehydration to form amides.

General reaction: $\mathrm{RCOO}{}^{-}{\mathrm{NH}}^4{}^{+}\to \mathrm{RCONH}{}_2+H_{2}O$

So C is an amide: RCONH₂

Step 3: C (amide) in presence of KOH reacts with Br₂ to give CH₃CH₂NH₂ (ethylamine). This is the Hofmann bromamide reaction, which converts amides to primary amines with one less carbon atom.

General reaction: $\mathrm{RCONH}{}_2+\mathrm{Br}{}_2+4\mathrm{KOH}\to \mathrm{RNH}{}_2+2\mathrm{KBr}+K_2\mathrm{CO}{}_3+2H_{2}O$

Since the final product is CH₃CH₂NH₂ (ethylamine), the amide C must be CH₃CH₂CONH₂ (propanamide). Therefore, the original carboxylic acid A must be CH₃CH₂COOH (propanoic acid).

Final Answer: A is CH₃CH₂COOH

Related Topics

Carboxylic Acids: Organic compounds containing the carboxyl functional group (-COOH). They react with bases like ammonia to form salts.

Amides: Derivatives of carboxylic acids where the -OH group is replaced by -NH₂. They can be prepared by heating ammonium salts of carboxylic acids.

Hofmann Bromamide Reaction: An important reaction where amides are converted to primary amines with one less carbon atom using bromine and potassium hydroxide.

Important Formulae

Formation of ammonium salt: R-COOH + NH₃ → R-COO^-NH₄⁺

Formation of amide: R-COO^-NH₄⁺ → R-CONH₂ + H₂O (on heating)

Hofmann degradation: R-CONH₂ + Br₂ + 4KOH → R-NH₂ + 2KBr + K₂CO₃ + 2H₂O