What happens when : n-butyl chloride is treated with alcoholic KOH ?
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Solution and Explanation
Dehydrohalogenation ($\beta$-elimination) reaction of alkyl halides.
Step 2: Meaning
Alcoholic KOH acts as a strong base, favoring elimination over substitution.
Step 3: Analysis
When n-butyl chloride ($CH_{3}-CH_{2}-CH_{2}-CH_{2}-Cl$) is heated with alcoholic KOH, it loses a molecule of hydrogen chloride ($HCl$). The chlorine atom is removed from the $\alpha$-carbon, and a hydrogen atom is removed from the adjacent $\beta$-carbon, resulting in the formation of a double bond at the terminal position.
Final Answer: Dehydrohalogenation takes place, and But-1-ene ($CH_{3}-CH_{2}-CH=CH_{2}$) is formed as the major product along with potassium chloride and water.
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