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Consider the following reactions, (i) NaNO2/HCl, 0-5°C (ii) β-naphthol/NaOH [P] → Colored Solid [P] → Br2/H2O → C7H6NBr3 The compound [P] is:

1. (1) A benzene ring with NH2 at position 1 and CH3 at position 2 (o-toluidine)
2. (2) A benzene ring with NH2 at position 1 and CH3 at position 3 (m-toluidine)
3. (3) A benzene ring with NH2 at position 1 and CH3 at position 4 (p-toluidine)
4. (4) A benzene ring with NHCH3 at position 1 and CH3 at position 2 (N-methyl-o-toluidine)

Correct answer: (2) A benzene ring with NH2 at position 1 and CH3 at position 3 (m-toluidine)

Solution

P must be a toluidine (C7H9N): it couples with diazotised P + beta-naphthol to give an azo dye (colored solid). Bromination giving C7H6NBr3 needs three positions ortho/para to NH2 free, which only m-toluidine (NH2 at 1, CH3 at 3) provides (2,4,6-tribromo). So P is m-toluidine.

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