Determine the product in the following conjugated addition reaction.

• A.
• B.
• C.
• D.

Answer: (D)

Verified step by step guidance

1. Identify the reactants: The reaction involves a conjugated addition (Michael addition) between a cyclic diketone and an α,β-unsaturated ketone.
2. Recognize the role of the base: Sodium ethoxide (NaOEt) in ethanol (EtOH) is used to deprotonate the diketone, generating an enolate ion.
3. Understand the conjugate addition mechanism: The enolate ion acts as a nucleophile and attacks the β-carbon of the α,β-unsaturated ketone, forming a new carbon-carbon bond.
4. Consider the role of the acid: The subsequent acid workup with H3O+ protonates the alkoxide intermediate, resulting in the final product.
5. Analyze the product structure: The product will have a new bond between the β-carbon of the α,β-unsaturated ketone and the α-carbon of the diketone, forming a 1,5-diketone structure.