A hydrocarbon 'A' `(C_(4)H_(8))` on reaction HCl gives a compound 'B', `(C_(4)H_(9)Cl)` which on reaction with 1 mol of `NH_(3)` gives compounds 'C' `(C_(4)H_(11)N)`. On reacting with `NaNO_(2)` and HCl followed by treatment with water compound 'C' yields an optically active alcohol, 'D'. Ozonolysis of 'A' given 2mols of acetyldehyde. Identify compound 'A' to 'D'. Explain the reaction involved.

Explanation

(i) M.F. of compound A `(C_4H_8)` corresponds to the general formula `C_nH_(2n)` where n = 4, therefore, A is an alkene.
Since azonolysis of hydrocarbon .A. `(C_4H_8)` gives two moles of acetaldehyde, therefore, hydrocarbon .A. must be a symmetrical alkene, i.e., but-2-ene.

(ii) Since but-2-ene reacts with HCl to form compound .B. (`C_4H_9Cl`), therefore, .B. must be 2-chlorobutane.
(iii) Since compound (B), i.e., 2-chlorobutane reacts with one mole of `NH_3` to give compound (C), therefore, compound (C) must be a `1^@` amine, i.e., butan - 2- amine.

(iv) Since compound .C. reacts with `NaNO_2//HCl (HNO_2)` to give an alcohol .D.. therefore, compound .D. must be butan - 2- ol.