Question

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

Answer 1

Addition of HCl has occurred on ‘A’. This implies ‘A’ is an alkene.

Cl in compound ‘B’ is substituted by NH₂ to give ‘C’.

‘C’ gives a diazonium salt with NaNO₂/HCl that liberates N₂ to give optically active alcohol. This means that ‘C’ is an aliphatic amine. Number of carbon atoms in amine is same as in compound ‘A’.

Since products of ozonolysis of compound ‘A’ are CH₃ — CH = O

and O = CH—CH₃. The compound ‘A’ is CH₃—CH= CH—CH₃

On the basis of structure of ‘A’ reactions can be explained as follows :