Question:
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.
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.
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Aqueous KOH favours substitution to form alcohols, while alcoholic KOH favours elimination to form alkenes.
Updated On: Jun 15, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Write the structure of 1-chlorobutane.
1-chlorobutane is
\[ \mathrm{CH_3CH_2CH_2CH_2Cl} \]
Step 2: Reaction with aqueous KOH.
Aqueous KOH provides \(\mathrm{OH^-}\) ions and favours nucleophilic substitution.
So, chlorine is replaced by \(\mathrm{OH}\).
\[ \mathrm{CH_3CH_2CH_2CH_2Cl+KOH(aq)\rightarrow CH_3CH_2CH_2CH_2OH+KCl} \]
Hence,
\[ P=\text{Butan-1-ol} \]
Step 3: Reaction with alcoholic KOH.
Alcoholic KOH favours elimination reaction.
So, HCl is removed from 1-chlorobutane to form an alkene.
\[ \mathrm{CH_3CH_2CH_2CH_2Cl+KOH(alc)\rightarrow CH_3CH_2CH=CH_2+KCl+H_2O} \]
Hence,
\[ Q=\text{But-1-ene} \]
Step 4: Final conclusion.
Therefore,
\[ \boxed{P=\text{Butan-1-ol},\quad Q=\text{But-1-ene}} \]
1-chlorobutane is
\[ \mathrm{CH_3CH_2CH_2CH_2Cl} \]
Step 2: Reaction with aqueous KOH.
Aqueous KOH provides \(\mathrm{OH^-}\) ions and favours nucleophilic substitution.
So, chlorine is replaced by \(\mathrm{OH}\).
\[ \mathrm{CH_3CH_2CH_2CH_2Cl+KOH(aq)\rightarrow CH_3CH_2CH_2CH_2OH+KCl} \]
Hence,
\[ P=\text{Butan-1-ol} \]
Step 3: Reaction with alcoholic KOH.
Alcoholic KOH favours elimination reaction.
So, HCl is removed from 1-chlorobutane to form an alkene.
\[ \mathrm{CH_3CH_2CH_2CH_2Cl+KOH(alc)\rightarrow CH_3CH_2CH=CH_2+KCl+H_2O} \]
Hence,
\[ Q=\text{But-1-ene} \]
Step 4: Final conclusion.
Therefore,
\[ \boxed{P=\text{Butan-1-ol},\quad Q=\text{But-1-ene}} \]
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