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Step 1: Deprotonation and Enolate Formation
- The hydroxide ion (OH⁻) acts as a base and deprotonates the alpha carbon of one of the ketone groups, forming an enolate anion. This is the most acidic proton due to resonance stabilization of the resulting enolate.
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Step 2: Nucleophilic Attack (Intramolecular Aldol)
- The enolate anion acts as a nucleophile and attacks the electrophilic carbonyl carbon of the *other* ketone group within the same molecule.
- This forms a new carbon-carbon bond and creates a six-membered ring containing an alkoxide intermediate.
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Step 3: Protonation (Workup)
- The alkoxide intermediate is protonated by water (H₂O), yielding a beta-hydroxy ketone (an aldol addition product).
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Step 4: Dehydration (Under Acidic Conditions and Heat)
- The beta-hydroxy ketone is protonated on the hydroxyl group under acidic conditions (H⁺).
- Water leaves as a leaving group, forming a carbocation at the beta position.
- A neighboring alpha proton is removed by a weak base (e.g., H₂O), forming a conjugated enone system (alpha,beta-unsaturated ketone). Heat drives this dehydration step to completion.
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Final Product:
- A fused bicyclic system: a cyclohexane ring fused to a cyclopentene ring, with a ketone group on the cyclohexane ring and a double bond between the two rings (specifically, between the former alpha and beta carbons of the aldol product).
Parent Tip: Review the logic above to help your child master the concept of worksheet missing items.
