Question

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.

Answer 1

\[\ce{\underset{A}{C6H5CONH2} + Br2 + 4NaOH -> C6H5NH2 + Na2CO3 + 2NaBr + 2H2O}\]

\[\ce{\underset{B}{C6H5NH2} + HNO2 + HCl -> C6H5N^2+Cl- + 2H2O}\]

\[\ce{\underset{C}{C6H5N^2+Cl-} + CHCl3 + 3KOH -> \underset{D}{C6H5NC} + 3KCl + 3H2O}\]

\[\ce{C6H5N^2+Cl- + C2H5OH -> \underset{E}{C6H6} + N2 + CH3CHO}\]

A = \[\ce{C6H5CONH2}\] Benzamide

B = \[\ce{C6H5NH2}\] Aniline.

C = \[\ce{C6H5N^2+Cl-}\] Benzenediazonium Chloride

D = \[\ce{C6H5NC}\] phenyl isocyanide

E = \[\ce{C6H6}\] Benzene