Question

What is the major product formed when 2-methyl-3-chloropentane undergoes alcoholic KOH elimination (E2 reaction)?

• A. 2-methyl-1-pentene
• B. 2-methyl-2-pentene
• C. 3-methyl-1-pentene
• D. 3-methyl-2-pentene

Hint:

In an E2 elimination reaction, the most stable alkene (the one with the most substituted double bond) is usually the major product.

Answer 1

The Correct Option is B

In an E2 reaction, the elimination of a leaving group and a proton occurs simultaneously. When 2-methyl-3-chloropentane undergoes alcoholic KOH elimination, the base (OH\(^-\)) abstracts a hydrogen atom from the carbon adjacent to the carbon bearing the chlorine (which is in the beta position). This leads to the formation of the double bond. The major product will be 2-methyl-2-pentene, as it is the more stable alkene due to the formation of a more substituted, stable double bond. - Option (A) is incorrect as the elimination does not result in 1-pentene.
- Option (C) is incorrect because the product will have a double bond between carbon 2 and 3, not between carbon 1 and 2.
- Option (D) is incorrect because the double bond will not form at position 3.
Therefore, the correct answer is option (B).