Question

The major product formed in the following reaction is 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

For elimination reactions (E2 mechanism), the base removes a hydrogen atom from the carbon adjacent to the one bearing the leaving group.

Answer 1

The Correct Option is A

Mechanism:

E2 elimination at α-carbon: NaOMe (strong base) abstracts the α-proton next to the carbonyl, eliminating Br to form an enolate that converts to an α,β-unsaturated ketone

Nucleophilic substitution: MeOH/MeOH⁻ replaces the Br on the CH₂Br group with OMe, forming a methyl ester

Starting material analysis:

• α-carbon: CHBr with Ph and Me substituents
• β-carbon: CH₂Br

Product formation:

• Elimination creates C=C between α and β carbons
• The geometry places Ph and Me on opposite ends of the double bond
• CH₂Br → CH₂OMe → converts ketone to ester

Result: Option (A)

• Ph-C(Me)=CH-CO-OMe
• An α,β-unsaturated methyl ester with Ph and Me groups on the double bond

The answer is (A).