Q.75 A hydrocarbon 'A' (C4H8) on reaction with HCI gives a compound 'B,' (C4H9Cl), which on reaction with 1 mol of NH3 gives compound 'C,' (C4H11N). On reacting with NaNO2 and HCI followed by treatment with water, compound 'C' yields optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 mols of acetaldehyde. Identify compounds 'A' to 'D'. Explain the reactions involved.

(i) Addition of HCI to compound "A' shows that compound 'A' is an alkene. Compound B ' is C4H9Cl.
(ii) Compound E ' reacts with NH2, it forms amine 'C'.
(iii) C' gives diazonium salt with NaNO2/ HCI, which yields optically active alcohol. So, 'C' is an aliphatic amine.
(iv) A on ozonolysis produces 2 moles of CH3CHO. So. 'A is CH3 —CH= CH—CH3 (But-2-ene).