See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: KH is a strong, non-nucleophilic base that fully deprotonates ketones to form enolates. With excess KH and excess MeI, exhaustive alpha-methylation occurs. Step 1 - Identify the alpha positions of cyclohexanone: Cyclohexanone has two alpha carbons (C2 and C6), each bearing two alpha hydrogens. Step 2 - First methylation: KH deprotonates cyclohexanone at an alpha carbon (C2) to form the enolate, which reacts with MeI to give 2-methylcyclohexanone. Step 3 - Continued methylation with excess reagents: Because KH and MeI are both in excess, the reaction does not stop at mono- or dimethylation. Each alpha carbon can be methylated twice (since each alpha carbon originally has 2 H's). With excess base and electrophile, all four alpha hydrogens are replaced by methyl groups. Step 4 - Final product: Both alpha carbons (C2 and C6) become fully methylated (gem-dimethyl at each), giving 2,2,6,6-tetramethylcyclohexanone, which corresponds to structure (c). Step 5 - Why other options fail: - (a) 2,6-dimethylcyclohexanone would result from only one methylation at each alpha carbon; this would require controlled (1 equiv) conditions, not excess. - (b) 2,2,6-trimethylcyclohexanone is an intermediate that would be further methylated under excess conditions. - (d) The structure shown does not correspond to exhaustive alpha-methylation of cyclohexanone. The 81% yield is consistent with the known literature preparation of 2,2,6,6-tetramethylcyclohexanone from cyclohexanone using excess KH and excess MeI. Therefore, the correct answer is C.