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प्रश्न
An amide ‘A’ with molecular formula \[\ce{C7H7ON}\] undergoes Hoffmann Bromamide degradation reaction to give amine ‘B’. 'B’ on treatment with nitrous acid at 273-278 K form ‘C’ and on treatment with chloroform and ethanolic potassium hydroxide forms 'D’. ‘C’ on treatment, with ethanol gives ‘E’. Identify ‘A’, ‘B’, ‘C’ ‘D’ and ‘E' and write the sequence of chemical equations.
उत्तर
\[\ce{\underset{A}{C6H5CONH2} + Br2 + 4NaOH -> C6H5NH2 + Na2CO3 + 2NaBr + 2H2O}\]
\[\ce{\underset{B}{C6H5NH2} + HNO2 + HCl -> C6H5N^2+Cl- + 2H2O}\]
\[\ce{\underset{C}{C6H5N^2+Cl-} + CHCl3 + 3KOH -> \underset{D}{C6H5NC} + 3KCl + 3H2O}\]
\[\ce{C6H5N^2+Cl- + C2H5OH -> \underset{E}{C6H6} + N2 + CH3CHO}\]
A = \[\ce{C6H5CONH2}\] Benzamide
B = \[\ce{C6H5NH2}\] Aniline.
C = \[\ce{C6H5N^2+Cl-}\] Benzenediazonium Chloride
D = \[\ce{C6H5NC}\] phenyl isocyanide
E = \[\ce{C6H6}\] Benzene
