An organic compound A upon reacting with $$NH_3$$ gives B. On heating, B gives C. C in presence of KOH reacts with $$Br_2$$ to give $$CH_3CH_2NH_2$$. A is :

The final step involves reacting C with $$Br_2/KOH$$ to produce $$CH_3CH_2NH_2$$ (ethylamine). This is the characteristic Hofmann Bromamide Degradation reaction, which converts an amide into a primary amine with one fewer carbon atom. 

Since the product is $$CH_3CH_2NH_2$$ (2 carbons), the starting amide C must have 3 carbons. Therefore, C is Propanamide: $$CH_3CH_2CONH_2$$.

The problem states that heating B gives C ($$CH_3CH_2CONH_2$$). Amides are typically formed by heating ammonium salts of carboxylic acids. Thus, B is the ammonium salt: $$CH_3CH_2COONH_4$$ (Ammonium propionate).

The compound A reacts with $$NH_3$$ to give B ($$CH_3CH_2COONH_4$$). Carboxylic acids react with ammonia to form ammonium salts. To form a 3-carbon ammonium salt, the starting acid A must be Propanoic acid. 

A = $$CH_3CH_2COOH$$.

Complete Reaction Sequence:

1. $$CH_3CH_2COOH \xrightarrow{NH_3} CH_3CH_2COONH_4$$  (Compound A to B)
2. $$CH_3CH_2COONH_4 \xrightarrow{\Delta} CH_3CH_2CONH_2 + H_2O$$ (Compound B to C)
3. $$CH_3CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2NH_2$$ (Hofmann Bromamide Degradation)