Determine the product in the following conjugated addition reaction.
A
B
C
D
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1
Identify the reactants: The reaction involves a conjugated addition (Michael addition) between a cyclic diketone and an α,β-unsaturated ketone.
Recognize the role of the base: Sodium ethoxide (NaOEt) in ethanol (EtOH) is used to deprotonate the diketone, generating an enolate ion.
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.
Consider the role of the acid: The subsequent acid workup with H3O+ protonates the alkoxide intermediate, resulting in the final product.
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.