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An organic compound A, when treated with NH3, forms B. On heating, B gives C, and C on treatment with Br2 in the presence of KOH yields ethylamine. The compound A is:

1. CH3COOH
2. CH3CH2CH2COOH
3. CH3CH2CH2COOH
4. CH3CH2COOH

Correct answer: CH3CH2COOH

Solution

A + NH3 -> ammonium salt (B); heating -> amide (C); C + Br2/KOH (Hofmann bromamide) gives an amine with one fewer carbon. To get ethylamine (2 C) the amide must be propanamide, so A = propanoic acid, CH3CH2COOH.

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