Give the structures of A and B in the following sequences:
(a) CH₃COOH →NH₃, Δ→ A →NaOBr→ B
(b) C₆H₅NO₂ →Fe/HCl→ A →NaNO₂ + HCl, 0–5 °C→ B
(c) C₆H₅N₂⁺Cl⁻ →CuCN, Δ→ A →H₂O/H⁺→ B
(a) CH₃COOH →NH₃, Δ→ A →NaOBr→ B
(b) C₆H₅NO₂ →Fe/HCl→ A →NaNO₂ + HCl, 0–5 °C→ B
(c) C₆H₅N₂⁺Cl⁻ →CuCN, Δ→ A →H₂O/H⁺→ B
Answer: (a) A = CH₃CONH₂, B = CH₃NH₂. (b) A = C₆H₅NH₂, B = C₆H₅N₂⁺Cl⁻. (c) A = C₆H₅CN, B = C₆H₅COOH.
Solution
(a) A: acetamide; B: methanamine (NaOBr = Hofmann conditions).
(b) A: aniline; B: benzenediazonium chloride.
(c) A: benzonitrile (Sandmeyer); B: benzoic acid (hydrolysis).
(b) A: aniline; B: benzenediazonium chloride.
(c) A: benzonitrile (Sandmeyer); B: benzoic acid (hydrolysis).
CBSE Class 12 Board · 2020 · Subjective