Question

The following compound results from base-catalyzed aldol cyclization of a 2-substituted cyclohexanone.

1. Show the diketone that would cyclize to give this product.
2. Propose a mechanism for the cyclization

Short answer

When there are two carbonyl group present in a molecule there is a probability of forming intramolecular aldol cyclization product. Atleast one of the carbonyl group must have enolizable alpha protons for the reaction to proceed.

Aldol cyclization reaction mainly favour five membered and six memebered rings. Rings with higher number of carbons are not common as their energy and entropy do not favour their formation

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Step-by-step solution

Aldol Cyclizations

When there are two carbonyl group present in a molecule there is a probability of forming intramolecular aldol cyclization product. Atleast one of the carbonyl group must have enolizable alpha protons for the reaction to proceed.

Aldol cyclization reaction mainly favour five membered and six memebered rings. Rings with higher number of carbons are not common as their energy and entropy do not favour their formation

Diketone inovled in cycliziation

The diketone involved in the production of the given compound is 2-(3_oxobutyl) cyclohexan-1-one. Its structure is shown below.

Structure of 2-(3-oxobutyl) cyclohexan-1-one

Mechanism of base-catalyzed intramolecular aldol cyclization

In the first step of the reaction the base extracts an alpha proton to give the enolate ion. The enolate ion is resonance stabilized as shown in the resonance structures below. The enolate acts as a nucleophile and attacks the electrophilic carbon resulting in cyclization. Then oxygen abstracts a proton from the environment to form β-hydroxyketone. Then elimination of results in the formation of the desired product which is an α,β-unsaturated ketone.

Formation of 4,4a,5,6,7,8-hexahydronapththalen-2(3H)-one