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.
When hydrocarbon ‘A’ reacts with HCl to give a compound ‘B’ with a chloride moiety, it means that A is an alkene and –Cl from HCl is added across the double bond to form ‘B’.
When ‘B’ reacts with NH3, it forms an amine with the –Cl being substituted with –NH2. Compound ‘C’ is thus C4H11N or C4H9NH2.
When amine ‘C’ reacts with NaNO2 and HCl followed by treatment with water forms a diazonium salt and later gives an alcohol. The product ‘D’ is an optically active alcohol, so the amine ‘C’ is aliphatic in nature.
Ozonolysis of ‘A’ gives 2 moles of CH3CHO or acetaldehyde. Ozonolysis of alkenes gives aldehydes or ketones or a mixture depending on the substitution pattern of the alkene. The products – two moles of acetaldehyde mean the compound ‘A’ is But-2-ene or CH3—CH=CH—CH3.
The reactions are given as follows.
