An organic compound 'A' on treatment with NH₃ gives 'B' which on heating gives 'C', 'C' on treatment with Br₂ in the presence of KOH produces ethylamine. Compound 'A' is  

CH₃CH₂COOH

The correct answer is option 4. CH₃CH₂COOH

Let us break down the problem and the reasoning behind identifying compound '\(A\)' as \(CH_3CH_2COOH\) (Propionic acid), based on the given transformations:

Compound '\(A\)' reacts with ammonia \((NH_3)\) to form '\(B\)'. This reaction typically involves the hydrogen of \(-OH\) of '\(A\)' being attacked by ammonia, leading to the formation of ammonium propanoate \((B)\). '\(B\)', formed from the reaction with \(NH_3\), is heated to give '\(C\)', i.e., Propanamide. '\(C\)', which results from the heating process, reacts with bromine \((Br_2)\) in the presence of potassium hydroxide \((KOH)\) to produce ethylamine \((CH_3CH_2NH_2)\). This reaction involves the conversion of the functional group present in 'C' to an amine group.

Therefore,\(CH_3CH_2COOH\) (Propionic acid) is identified as compound '\(A\)' based on its ability to undergo the specified reactions leading to the formation of ethylamine. This sequence of reactions is typical in organic chemistry transformations involving carboxylic acids and their derivatives.