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) Addition of HCl to compound A shows that compound 'B' is `C_(4)H_(9)Cl`.
(ii) Compound 'B' reacts with `NH_(2)`. It forms amine 'C'.
(iii) 'C' gives diazonium salt with `NaNO_(2)//HCl` , which yields an optically active alcohol. So, 'C' is aliphatic amine.
(iv) 'A' on ozonolysis produces 2 moles of `CH_(3)CHO`. So, 'A' is `CH_(3)-CH =CH-CH_(3)` (But-2-ene)