Give the structures of A, B and C in the following reactions: (i) CH₃CH₂I →[NaCN] A →[OH⁻] B →[NaOH+Br₂] C (ii) C₆H₅N₂Cl →[CuCN] A →[H₂O/H⁺] B →[NH₃/Δ] C (iii) CH₃CH₂Br →[KCN] A →[LiAlH₄] B →[HNO₂/0°C] C (iv) C₆H₅NO₂ →[Fe/HCl] A →[NaNO₂+HCl] B →[H₂O/H⁺] C (v) CH₃COOH →[NH₃/Δ] A →[NaOBr] B →[NaNO₂/HCl] C (vi) C₆H₅NO₂ →[Fe/HCl] A →[HNO₂/273K] B →[C₆H₅OH] C
A = CH₃CH₂CN (Propanenitrile), B = CH₃CH₂CONH₂ (Propanamide), C = CH₃CH₂NH₂ (Ethylamine)
A = C₆H₅CN (Benzonitrile), B = C₆H₅COOH (Benzoic acid), C = C₆H₅CONH₂ (Benzamide)
A = CH₃CH₂CN (Propanenitrile), B = CH₃CH₂CH₂NH₂ (Propan-1-amine), C = CH₃CH₂CH₂OH (Propan-1-ol)
A = C₆H₅NH₂ (Aniline), B = C₆H₅N₂Cl (Benzene diazonium chloride), C = C₆H₅OH (Phenol)
A = CH₃CONH₂ (Acetamide), B = CH₃NH₂ (Methylamine), C = CH₃OH (Methanol)
A = C₆H₅NH₂ (Aniline), B = C₆H₅N₂Cl (Benzene diazonium chloride), C = C₆H₅-N=N-C₆H₄OH (p-Hydroxyazobenzene)
Explanation
This question tests understanding of amine chemistry reactions. In (i), NaCN substitutes the halide forming nitrile A, which hydrolyzes to amide B, then undergoes Hofmann bromamide degradation to amine C. In (ii), Sandmeyer reaction converts diazonium salt to benzonitrile A, hydrolysis gives benzoic acid B, and reaction with ammonia yields benzamide C. In (iii), KCN gives nitrile A, LiAlH₄ reduces it to amine B, and nitrous acid converts primary amine to alcohol C. In (iv), nitrobenzene reduces to aniline A, diazotization gives diazonium salt B, and hydrolysis yields phenol C. In (v), acetic acid with ammonia gives acetamide A, Hofmann degradation yields methylamine B, and nitrous acid gives methanol C. In (vi), aniline A undergoes diazotization to B, which couples with phenol to give azo compound C.