Question:
Identify the product obtained when 1-Chlorobutane is reacted with alc. KOH.
Identify the product obtained when 1-Chlorobutane is reacted with alc. KOH.
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Alcoholic KOH causes dehydrohalogenation, giving an alkene.
Updated On: Nov 12, 2026
- Butan-1-ol
- Butan-2-ol
- But-2-ene
- But-1-ene
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Alcoholic \(\text{KOH}\) brings about \(\beta\)-elimination (dehydrohalogenation) of haloalkanes.
Step 2: Locate the groups
In 1-chlorobutane, \(\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{2}\text{Cl}\), chlorine is on \(\text{C-1}\) and the only \(\beta\)-hydrogens are on \(\text{C-2}\).
Step 3: Elimination
Removing \(\text{HCl}\) forms a double bond between \(\text{C-1}\) and \(\text{C-2}\).
\(\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{2}\text{Cl}\rightarrow \text{CH}_{3}\text{CH}_{2}\text{CH}=\text{CH}_{2}\)
Final Answer:
The product is but-1-ene. \[ \boxed{\text{(D) }\text{But-1-ene}} \]
Alcoholic \(\text{KOH}\) brings about \(\beta\)-elimination (dehydrohalogenation) of haloalkanes.
Step 2: Locate the groups
In 1-chlorobutane, \(\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{2}\text{Cl}\), chlorine is on \(\text{C-1}\) and the only \(\beta\)-hydrogens are on \(\text{C-2}\).
Step 3: Elimination
Removing \(\text{HCl}\) forms a double bond between \(\text{C-1}\) and \(\text{C-2}\).
\(\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{2}\text{Cl}\rightarrow \text{CH}_{3}\text{CH}_{2}\text{CH}=\text{CH}_{2}\)
Final Answer:
The product is but-1-ene. \[ \boxed{\text{(D) }\text{But-1-ene}} \]
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