Amines Part 3

Concept:The reaction sequence involves electrophilic addition of HCl to an alkene, nucleophilic substitution with ammonia, and diazotization followed by hydrolysis to convert an amine to an alcohol.Explanation:Step 1: Hydrocarbon P ($\mathrm{C_4H_8}$) is an alkene. Addition of HCl yields an alkyl halide Q ($\mathrm{C_4H_9Cl}$). Since Q is optically active, it must have a chiral carbon. This occurs when the alkene is unsymmetrical such that the product has a chiral centre. Hence, P is but-1-ene ($\mathrm{CH_2=CH-CH_2-CH_3}$) and Q is 2-chlorobutane ($\mathrm{CH_3-CHCl-CH_2-CH_3}$).Step 2: Q reacts with one mole of ammonia in a nucleophilic substitution reaction. The chlorine atom is replaced by an amino group, giving R ($\mathrm{C_4H_{11}N}$) which is 2-aminobutane ($\mathrm{CH_3-CH(NH_2)-CH_2-CH_3}$).Step 3: R undergoes diazotization with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ to form a diazonium salt. This salt is then hydrolyzed with water, replacing the diazonium group with a hydroxyl group. The product S ($\mathrm{C_4H_{10}O}$) is butan-2-ol ($\mathrm{CH_3-CH(OH)-CH_2-CH_3}$).Answer:The correct option is the one showing P as but-1-ene, Q as 2-chlorobutane, R as 2-aminobutane, and S as butan-2-ol.