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An organic compound (A) contains 69.42% C, 5.78% H and 11.57% N. Its vapour density is 60.5. It evolves ammonia when boiled with caustic potash. On heating with P_(2)O_(5), it gives a compound (B) containing 81.55%C, 4.85% H and 13.59% N. On reduction with Na and alcohol, (B) forms a base which reacts with HNO_(3) giving off nitrogen and yielding and alcohol (C ). The alcohol can be oxidised to benzoic acid. Assign structural formulae to (A) , (B) and (C ) |
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Answer» Solution :For compound (A): Moles of `C: H: N: O= (69.42)/(12): (5.78)/(1): (11.57)/(14): (13.23)/(16)` `=5.66: 5.78: 0.82: 0.82` `=7:7:1:1` Empirical FORMULA of (A) is `C_(7)H_(7)NO` (121) `therefore` molecular formula of (A) is also `C_(7)H_(7)NO` `(("mol.wt"=2 xxV.D),(=2 xx 60.5= 121))` For compound (B): Moles of `C:H: N= (81.55)/(12): (4.85)/(1): (13.59)/(14)` `=6.8 : 4.85: 0.97` `=7:5:1` `therefore` empirical formula of (B) is `C_(7)H_(5)N` Molecular formula of (B) will also be `C_(7)H_(5)N` as it is derived from `C_(7)H_(7)NO` (A) by dehydration with `P_(2)O_(5)` As (A) evolves ammonia when BOILED with caustic potash, it must be an amide `(C_(6)H_(5).CO.NH_(2))`. Thus, (B) and 9C ) can also be named from the given REACTION sequence, `underset(underset("Benzamide")((A)))(C_(6)H_(5)CONH_(2)) underset(-H_(2)O)overset(P_(2)O_(5))rarr underset((B))(C_(6)H_(5)CN) underset(Na//alc)overset("Red")rarr underset(darrHNO_(2))(C_(6)H_(5)CH_(2)NH_(2)) underset("Benzoic acid")(C_(6)H_(5)COOH) overset((O))larr underset((C ))(C_(6)H_(5)CH_(2)OH) + H_(2)O + N_(2)` The structure may be represented as 
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