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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: (4) A benzene ring with NHCH3 at position 1 and CH3 at position 2 (N-methyl-o-toluidine)

Solution

The correct option is N-methyl-o-toluidine because the reactions indicate the presence of an amino group (NH) and a methyl group (CH3) on the benzene ring, specifically at the ortho position, which aligns with the formation of a colored solid in the presence of NaNO2 and HCl, followed by bromination.

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