Which of the following pairs of starting materials would result in the formation of 2-methyl-1,2,3,4-tetralone via a Robinson annulation?

• A. Cyclohexanone and ${\mathrm{CH}}_2\mathrm{CHCHO}$ (acrolein)
• B. Cyclopentanone and ${\mathrm{CH}}_3\mathrm{COCH}{2}_{}\mathrm{CH}{3}_{}$ (methyl vinyl ketone)
• C. Cyclohexanone and ${\mathrm{CH}}_3\mathrm{COCH}{3}_{}$ (acetone)
• D. Cyclohexanone and ${\mathrm{CH}}_3\mathrm{COCH}{2}_{}\mathrm{CH}{3}_{}$ (methyl vinyl ketone)

Answer: (D)

Verified step by step guidance

1. Understand the Robinson annulation reaction: it is a tandem reaction involving a Michael addition followed by an intramolecular aldol condensation, typically between a cyclic ketone and an α,β-unsaturated carbonyl compound to form a fused ring system.
2. Identify the target molecule: 2-methyl-1,2,3,4-tetralone is a bicyclic ketone with a six-membered ring fused to a cyclohexanone ring and a methyl substituent at the 2-position, indicating the formation of a new six-membered ring fused to the original cyclohexanone.
3. Analyze the starting materials: cyclohexanone is a six-membered cyclic ketone suitable for Robinson annulation. The Michael acceptor must be an α,β-unsaturated carbonyl compound capable of undergoing conjugate addition and subsequent cyclization.
4. Evaluate each pair: (1) cyclohexanone and acrolein (an α,β-unsaturated aldehyde) can undergo Michael addition but typically forms a different product; (2) cyclohexanone and methyl vinyl ketone (an α,β-unsaturated ketone) are classic substrates for Robinson annulation leading to a fused bicyclic ketone with a methyl substituent; (3) cyclohexanone and acetone (a saturated ketone) cannot undergo Michael addition as acetone lacks the α,β-unsaturation.
5. Conclude that the pair of cyclohexanone and methyl vinyl ketone is the correct choice because it provides the necessary α,β-unsaturated ketone for Michael addition and subsequent aldol condensation to form 2-methyl-1,2,3,4-tetralone.