Q. 363.7( 9 Votes )
An organic compou
The given compound A is C6H5CONH2, an amide which undergoes Hoffmann Bromamide synthesis which gives the compound B, primary amine C6H5NH2 aniline. Aniline reacts with NaNO2 and HCl at 0°C to give compound C Benzene Diazonium Chloride, a diazo salt. Benzene diazonium chloride reacts with ethanol, CH3CH2OH to give a compound D hydrocarbon Benzene by elimination of N2+Cl- and replacement with H. The given compound B is aniline because only aniline reacts with Br2 water to give a white precipitate of compound E, 2,4,6-tribromophenylamine.
The reactions are as follows.
a(i) Aniline is treated with NaNO2 and HCl at 0°C to form benzene diazonium chloride. This salt on treatment with HBF4 or Fluroboric acid gives Fluorobenzene and has toxic gas BF3 as by-product.
(ii)Benzamide is converted to Benzylamine by Hoffmann bromamide synthesis, where an amide is treated with Br2 water and aqueous or ethanolic NaOH to give benzylamine, a primary amine.
(iii)Ethanamine, which is a primary amine, can be converted to N,N-Diethylethanamine which is a tertiary amine by ammonolysis of alkyl halides. An alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group. The reaction is carried out in a sealed tube at 100C. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt. Ethanamine, C2H5—NH2, the primary amine is further treated with ethyl halide, C2H5X, leading to the substitution of the –H groups on ethanamine with the ethyl groups on the halide, first forming secondary amine N-ethylethanamine (C2H5)2NH, and then tertiary amine N, N-Diethylethanamine (C2H5)3N.
C2H5—NH2 + C2H5X → (C2H5)2NH + C2H5X → (C2H5)3N
(b)(i) CH3CH2CN, ethyl nitrile, on partial hydrolysis with mild heat with NaOH or KOH gives compound A, ethyl amide, CH3CH2CONH2.
CH3CH2CN + NaOH/KOH → CH3CONH2
Ethyl amide on treatment with NaOH and Br2 water undergoes Hoffmann Bromamide degradation to form compound B, a primary amine, ethyl amine, or CH3CH2NH2.
CH3CONH2 + Br2 + NaOH → CH3CH2NH2.
(ii) CH3CH2Br, ethyl bromide, when treated with KCN, undergoes nucleophilic substitution reaction to form CH3CH2CN. CH3CH2CN on treatment with LiAlH4, a strong reducing agent, forms compound A, CH3CH2NH2, ethyl amine, primary amine. CH3CH2NH2 on treatment with nitrous acid at 0C, forms compound B which is a diazo salt, ethyldiazonium ion or ethyldiazonium chloride, CH3CH2N2+Cl-
Rate this question :
(a) Write the reaChemistry - Board Papers