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How will you convert C_(2)H_(5) - Br into :(i) C_(2)H_(6)(ii) C_(2)H_(5)-O - C_(2)H_(5) (iii) C_(2)H_(5) - CN

Answer»

Solution :(i) `C_(2)H_(5)Br+ 2[H] overset(Zn-Hg+ "Conc.HCl")hArr C_(6)H_(6)+HBr`
(ii)`C_(2)H_(5)Br+ Na^(+)O^(-)C_(2)H_(5) to C_(2)H_(5)-O-C_(2)H_(5)+NABR`
(III) `C_(2)H_(5)Br+ underset("(ALC)")(KCN) to C_(2)H_(5)CN+ KBr`


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