See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The Robinson annulation involves a Michael addition of an enolate to methyl vinyl ketone (MVK), followed by an intramolecular aldol condensation, to form a cyclohexenone ring fused to the existing ring. Step 1 – Identify the target product: The product shown is a bicyclic diketone with two fused six-membered rings (a Hajos-Parrish / Wieland-Miescher type ketone precursor). It has a methyl group at the ring junction (C8a), a ketone in each ring, and a double bond in the newly formed ring. This is consistent with Robinson annulation of a 2-substituted 1,3-diketone enolate with MVK. Step 2 – Determine the required starting enolate: To produce a bicyclic product with a methyl group at the ring junction and two ketone groups, the starting material must be a cyclic 1,3-diketone bearing a methyl group at the active methine carbon (C2). The enolate forms at C2 (between the two carbonyls), which is the most acidic position. This enolate performs Michael addition to MVK (1,4-addition) at C2, appending a –CH2CH2COCH3 chain. The subsequent intramolecular aldol condensation forms the new six-membered ring with the enone, giving the observed bicyclic diketone with the methyl at the ring junction. Step 3 – Match to the options: Option (c) shows the enolate of 2-methylcyclohexane-1,3-dione with the carbanion (negative charge, Theta) at C2 (the carbon bearing the methyl group and flanked by both ketone carbonyls). Both ketone oxygens are explicitly drawn, and the methyl group is at the anionic carbon. This is exactly the enolate needed. Step 4 – Eliminate other options: - Option (a): Enolate of 2-methylcyclohexan-1-one at C2 — only one ketone, would not give a diketone bicycle with the required methyl at the ring junction via Robinson annulation in one step. - Option (b): Appears similar to (c) but the charge placement or structural depiction is at a different carbon (alpha carbon without the methyl), which would not place the methyl at the ring junction in the product. - Option (d): Enolate of 2-methylcyclohexan-1-one at C6 (no methyl at the carbanion carbon) — monoketone enolate; Michael addition would not give the observed diketone bicyclic product with methyl at the junction. Therefore, the correct answer is C.