Question:medium

When 2-Chlorobutane is boiled with concentrated alcoholic solution of $\mathrm{KOH}$, the major product formed is

Show Hint

Remember the crucial reagent difference for $\mathrm{KOH}$: 1. $\text{Alcoholic } \mathrm{KOH} \rightarrow \text{Elimination (forms alkenes via Saytzeff's rule)}.$ 2. $\text{Aqueous } \mathrm{KOH} \rightarrow \text{Substitution (forms alcohols)}.$
Updated On: Jun 11, 2026
  • But-1-ene
  • But-2-ene
  • Butan-2-ol
  • Butan-1-ol
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Recognise the reaction type.
An alkyl halide heated with concentrated alcoholic $KOH$ undergoes dehydrohalogenation, a $\beta$ elimination that makes an alkene.
Step 2: Map the substrate.
2-chlorobutane is $CH_3-CHCl-CH_2-CH_3$. The chlorine carbon is the $\alpha$ carbon, flanked by two different $\beta$ carbons.
Step 3: List the two possible alkenes.
Removing an $H$ from the terminal methyl gives but-1-ene $CH_2=CH-CH_2-CH_3$. Removing an $H$ from the inner $CH_2$ gives but-2-ene $CH_3-CH=CH-CH_3$.
Step 4: Count substituents on each double bond.
But-1-ene is monosubstituted, but-2-ene is disubstituted, so but-2-ene is the more substituted alkene.
Step 5: Apply Saytzeff's rule.
The more substituted, more stable alkene is the major product because of greater hyperconjugation.
Step 6: Choose the major product.
But-2-ene wins as the major product, with but-1-ene only minor.
\[ \boxed{\text{But-2-ene (option B)}} \]
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