InterviewSolution
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InterviewSolution
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A colourless substance 'A' (C_(6)H_(7)N) is sparingly soluble in water and gives a water soluble compound 'B' on treating with mineral acid. On reacting with CHCl_(3) and alcoholic potash 'A' produces an obnoxious smell due to the formation of compound 'C'. reaction of 'A' with benzenesulphonyl chloride gives compound 'D' which is soluble in alkali. with NaNO_(2) and HCl, 'A'forms compound 'E' which reacs with phenol in alkaline medium to give an orange dye 'F'. identify compounds 'A' to 'F'. |
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Answer» Solution :(i) Compound `A(C_(6)H_(7)N)` is sparingly soluble in water and gives a WAER soluble substnace (B), on treatment with mineralacid, therefore, compound (A) must be an amine. (ii) Since compound (A) with m.p. `C_(6)H_(7)N` reacts with `NaNO_(2)//HCl` to form a compound (E) which reacs with phenol in ALKALINE medium to give an ORANGE dye (F), therefore, compound (A) must be a aromatic `1^(@)` amine, i.e., aniline, `C_(6)H_(5)NH_(2)`. if A is aniline, then E must be benzenediazonium chloride and the orange dye (F) must be p-hydroxyazobenzene.  (iii) Since compound (A), i.e., anililne reacts `CHCl_(3)` in presence of alcoholic KOH to form a compound (C) having obnoxious smell, therefore, compound (C) must be benzene isonitrole.  (iv) Since compound (A), i.e., aniline reacts with benzenesulphonyl chloride to form compound (D) with is soluble in alkali, therefore, (D) must be N-phenylbenzenesulphonamide.  (v) Since, compound (A), i.e., aniline reacts with mineral acids (say HCl) to form water soluble compound (B), therefore, (B) must be anilinium chloride. 
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