The major product formed in the following reaction
is:
Show Hint
When sodium ethoxide is used in elimination reactions, the major product typically involves the formation of a triple bond, with a corresponding shift in the position of the functional group.
In this reaction, the NaOEt (sodium ethoxide) promotes the elimination of the leaving group from the substrate, leading to the formation of a product with a triple bond. The major product here is non-3-yn-8-one, which involves the formation of an alkyne at position 3 and a ketone at position 8.
Step 1: Understand the reaction mechanism.
Sodium ethoxide induces elimination, which leads to the formation of an alkyne. The structure formed will have the triple bond at position 3 and the ketone at position 8.
Step 2: Analyze the options.
- (A) non-6-yn-2-one: This structure does not match the expected product.
- (B) non-3-yn-8-one: This is the correct product, as the triple bond forms at position 3 and the ketone at position 8.
- (C) non-2-yn-6-one: This structure does not match the product.
- (D) non-3-en-8-one: This structure contains a double bond instead of a triple bond, so it is incorrect.
