Question:
Identify the product obtained when 2-chlorobutane is heated with aqueous solution of potassium hydroxide.
Identify the product obtained when 2-chlorobutane is heated with aqueous solution of potassium hydroxide.
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Aqueous KOH causes substitution; alcoholic KOH causes elimination.
Updated On: Oct 1, 2026
- But - 1 - ene
- But - 2 - ene
- Butan - 1 - ol
- Butan - 2 - ol
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
With aqueous KOH, the hydroxide ion acts as a nucleophile and replaces halogen (substitution). With alcoholic KOH it acts as a base and removes HX (elimination).
Step 2: Reaction:
\[ \text{CH}_3\text{CHClCH}_2\text{CH}_3 + \text{KOH(aq)} \xrightarrow{\Delta} \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 + \text{KCl} \]
Chlorine on C-2 is replaced by OH on C-2, giving butan-2-ol. The alkenes (but-1-ene and but-2-ene) would need alcoholic KOH, and butan-1-ol would need the halogen on C-1.
Final Answer:
The product is butan-2-ol, option (D). \[ \boxed{\text{Butan-2-ol}} \]
With aqueous KOH, the hydroxide ion acts as a nucleophile and replaces halogen (substitution). With alcoholic KOH it acts as a base and removes HX (elimination).
Step 2: Reaction:
\[ \text{CH}_3\text{CHClCH}_2\text{CH}_3 + \text{KOH(aq)} \xrightarrow{\Delta} \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 + \text{KCl} \]
Chlorine on C-2 is replaced by OH on C-2, giving butan-2-ol. The alkenes (but-1-ene and but-2-ene) would need alcoholic KOH, and butan-1-ol would need the halogen on C-1.
Final Answer:
The product is butan-2-ol, option (D). \[ \boxed{\text{Butan-2-ol}} \]
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