Question

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

• A. But-1-ene
• B. But-2-ene
• C. Butan-2-ol
• D. Butan-1-ol

Hint:

Remember the crucial reagent difference for $\mathrm{KOH}$: 1. $\text{Alcoholic } \mathrm{KOH} \rightarrow \text{Elimination (forms alkenes via Saytzeff's rule)}.$ 2. $\text{Aqueous } \mathrm{KOH} \rightarrow \text{Substitution (forms alcohols)}.$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The reaction involves boiling an alkyl halide, 2-chlorobutane, with a concentrated alcoholic solution of potassium hydroxide ($\mathrm{KOH}$). We need to predict the major organic product formed.

Step 2: Key Formula or Approach:
Heating an alkyl halide with alcoholic $\mathrm{KOH}$ causes a dehydrohalogenation reaction ($\beta$-elimination). In this pathway, a hydrogen atom from a $\beta$-carbon and the halogen atom from the $\alpha$-carbon are eliminated to form a carbon-carbon double bond.
When alternative $\beta$-carbons are available, the regioselectivity is governed by

Saytzeff's Rule (Zaitsev's rule), which states that the highly substituted, more stable alkene will be the major product.

Step 3: Detailed Explanation:
Let's analyze the structure of 2-chlorobutane: $$\mathrm{\overset{\beta_1}{C}H_3 - \overset{\alpha}{C}H(Cl) - \overset{\beta_2}{C}H_2 - \mathrm{CH_3}}$$ The carbon holding the chlorine atom is the $\alpha$-carbon. It is flanked by two distinct sets of $\beta$-carbons:

• Elimination involving the $\beta_1$ methyl carbon yields But-1-ene: $$\mathrm{CH_2 = CH - CH_2 - CH_3} \quad (\text{monosubstituted alkene})$$

• Elimination involving the $\beta_2$ methylene carbon yields But-2-ene: $$\mathrm{CH_3 - CH = CH - CH_3} \quad (\text{disubstituted alkene})$$
According to Saytzeff's rule, the more highly alkylated alkene is thermodynamically more stable due to hyperconjugation. Therefore, the disubstituted alkene, But-2-ene, forms as the dominant major product ($\approx 80\%$), while But-1-ene forms only as a minor product.

Step 4: Final Answer:
The major product formed is But-2-ene, which corresponds to option (B).