Question

What is the major product formed when 2-bromopentane reacts with alcoholic KOH?

• A. Pent-1-ene
• B. Pent-2-ene
• C. Pentane
• D. Pentan-2-ol

Hint:

Alcoholic KOH causes dehydrohalogenation and the more substituted alkene forms (Saytzeff rule).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Alcoholic KOH removes HBr from an alkyl halide (dehydrohalogenation, E2 elimination) and gives an alkene. Aqueous KOH would give substitution instead.

Step 2: Structure
2-Bromopentane is \(\text{CH}_3\text{-CHBr-CH}_2\text{-CH}_2\text{-CH}_3\). The hydrogen can be lost from C1 or from C3.

Step 3: Two possible alkenes
Loss of H from C1 gives pent-1-ene, a terminal alkene with one alkyl group on the double bond. Loss of H from C3 gives pent-2-ene, an internal alkene with an alkyl group on each double-bond carbon.

Step 4: Saytzeff rule
The more substituted, more stable alkene is the major product. So pent-2-ene (mostly trans) is major.

Step 5: Other options
Pentane would need reduction, and pentan-2-ol needs aqueous KOH.

Final Answer:
By the Saytzeff rule the more substituted alkene, pent-2-ene, is the major product, option (B). \[ \boxed{\text{Pent-2-ene}} \]