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A compound X shows the following properties: (i) it liberates CO2 gas on treatment with NaHCO3; (ii) it forms an azo dye when diazotized and coupled with beta-naphthol; (iii) it gives a white precipitate of tribromoaniline on treatment with bromine water. Identify compound X.

1. Aniline
2. p-Aminobenzenesulphonic acid (NH2 and SO3H para to each other)
3. m-Aminobenzenesulphonic acid (NH2 and SO3H meta to each other)
4. m-Aminobenzoic acid (NH2 and COOH meta to each other)

Correct answer: m-Aminobenzoic acid (NH2 and COOH meta to each other)

Solution

Aniline lacks -COOH so fails test (i). Sulphanilic acids have -SO3H not -COOH, so also fail test (i). m-Aminobenzoic acid has both -COOH (liberates CO2 with NaHCO3) and -NH2 (forms azo dye and tribromoaniline precipitate), satisfying all three observations.

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