See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: LDA is a strong, bulky, non-nucleophilic base that deprotonates ketones kinetically (at the less substituted alpha carbon) to form the kinetic enolate. It is used in excess to ensure complete and irreversible enolate formation before the electrophile is added. Step 1 - Identify the two alpha positions of 2-methylcyclohexan-1-one: - C2 (more substituted alpha carbon, already bearing a methyl group, so it has one alpha hydrogen) - C6 (less substituted alpha carbon, bearing two alpha hydrogens) Step 2 - Determine which enolate forms with LDA: LDA is a kinetic base. It preferentially removes the more accessible (less hindered) proton. In 2-methylcyclohexan-1-one, C6 has two protons and is less sterically hindered compared to C2 which already has a methyl substituent. LDA therefore deprotonates at C6 to give the kinetic (less substituted) enolate. Step 3 - Alkylation with ethyl iodide (CH3CH2I): The enolate at C6 attacks ethyl iodide via SN2, placing an ethyl group at C6. Step 4 - Product formed: The product retains the methyl group at C2 and gains an ethyl group at C6, with the ketone at C1. This is 2-methyl-6-ethylcyclohexan-1-one, which corresponds to option (b). Why other options fail: (a) Would require double alkylation or methylation at C2, not consistent with LDA/EtI conditions. (c) Shows two methyl groups - would require methylation, not ethylation. (d) Shows substitution at C3 and C4, inconsistent with alpha-alkylation adjacent to the carbonyl. Therefore, the correct answer is B.