'A' is a neutral organic compound (M.F. : C8H9ON). On treatment with aqueous Br2/HO , 'A' forms a compound 'B' which is soluble in dilute acid. 'B' on treatment with aqueous NaNO2/HCl (0-5 C) produces a compound 'C' which on treatment with CuCN/NaCN produces 'D'. Hydrolysis of 'D' produces 'E' which is also obtainable from the hydrolysis of 'A'. 'E' on treatment with acidified KMnO4 produces 'F'. 'F' contains two different types of hydrogen atoms. The structure of 'A' is
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Answer
(1) [figure]
Solution
Br2/HO converts a primary amide into a primary amine with one carbon atom less (Hoffmann bromamide degradation) A is a primary amide (-CONH2) and B is an amine (basic, soluble in dilute acid). Option (4) is a secondary amide ruled out.
For option (1), A = 4-methylbenzamide:
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B = 4-methylaniline, C = 4-methylbenzenediazonium chloride, D = 4-methylbenzonitrile, E = 4-methylbenzoic acid.
Hydrolysis of A gives the same acid E:
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F = benzene-1,4-dicarboxylic acid (terephthalic acid): all four ring H are equivalent and both –COOH H are equivalent two types of H.
From options (2) and (3), F would be benzene-1,3- and benzene-1,2-dicarboxylic acid, which have more than two types of H.
Answer: (1)