Question

The major product formed in the reaction of (2R,3R)-2-bromo-3-methylpentane with NaOMe is:

• A. (Z)-3-methylpent-2-ene
• B. (E)-3-methylpent-2-ene
• C. (2R,3R)-2-methoxy-3-methylpentane
• D. (2S,3R)-2-methoxy-3-methylpentane

Hint:

In elimination reactions (E2), the product is determined by the anti-periplanar geometry, and the more stable (E)-alkene is typically the major product.

Answer 1

The Correct Option is B

In this reaction, NaOMe (sodium methoxide) is used, which is a strong nucleophile and can promote elimination reactions. The reaction proceeds via an E2 mechanism, where the leaving group (Br\(^-\)) is eliminated in a concerted process along with a proton (H\(^+\)) from the adjacent carbon. Since the configuration at the 2 and 3 positions is (2R,3R), the elimination will follow anti-periplanar geometry to form the alkene. The major product will be the (E)-isomer, as it is the most stable configuration. Therefore, the correct answer is (B).