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Compound (A) with molecular formula C_(6)H_(6)O gives violet color with neutral FeCl_(3). (A) reacts with CHCl_(3) and NaOH gives two isomers (B) and ( C) with molecular formula C_(7)H_(6)O_(2). Compound (A) reacts with ammonial at 473 K in the presence of ZnCl_(2) gives compound (D) with molecular formula C_(6)H_(7)N. Compound (D) undergoes carbylamine test. Identify (A), (B),(C ) and (D) and explain the reactions. |
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Answer» Solution :(i) COMPOUND (A) with molecular formula `C_(6)H_(6)O` gives violet colour with neutral `FeCl_(3)`. (ii) (A) reacts with `CHCl_(3)` and NaOH gives two isomers (B) and( C) with molecular formula `C_(7)H_(6)O_(2)`.  (iii) Compound (A) reacts with ammonia at 473 K in the presence of `ZnCl_(2)` gives compound (D) with molecular formula `C_(6)H_(7)N`.  (iv) Compound (D) undergoes carbylamine test. `C_(6)H_(5)NH_(2) + CHCl_(3) + 3KOH overset(Delta)to C_(6)H_(5)NC + 3KCl + 3H_(2)O` 
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