When 1-chlorobutane is treated with aqueous KOH it gives \(P\). However, when it is treated with alcoholic KOH it gives \(Q\). Identify the products \(P\) and \(Q\) respectively.
Show Hint
Aqueous KOH favours substitution to form alcohols, while alcoholic KOH favours elimination to form alkenes.
Step 1: Write the substrate. 1-chlorobutane is $CH_3CH_2CH_2CH_2Cl$, a primary alkyl halide. Step 2: Recall the role of aqueous KOH. Aqueous KOH provides hydroxide ions in water and favours nucleophilic substitution, where $OH^-$ replaces the chlorine. Step 3: Find product P. \[ CH_3CH_2CH_2CH_2Cl + KOH(aq) \rightarrow CH_3CH_2CH_2CH_2OH + KCl \] so $P$ is butan-1-ol. Step 4: Recall the role of alcoholic KOH. Alcoholic KOH provides a stronger base and favours elimination, removing $HCl$ to create a carbon to carbon double bond. Step 5: Find product Q. \[ CH_3CH_2CH_2CH_2Cl + KOH(alc) \rightarrow CH_3CH_2CH=CH_2 + KCl + H_2O \] so $Q$ is but-1-ene. Step 6: State both products. Therefore $P$ is butan-1-ol and $Q$ is but-1-ene, matching option 4. \[ \boxed{P=\text{Butan-1-ol},\ Q=\text{But-1-ene}} \]
