See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: Reaction of a carbonate ester with excess Grignard reagent. Step 1: Identify the starting material. Diethyl carbonate (Et-O-C(=O)-O-Et) is a carbonate ester with a carbonyl flanked by two ethoxy groups. Step 2: First equivalent of CH3MgBr attacks the carbonyl carbon of diethyl carbonate. The tetrahedral intermediate collapses, expelling one ethoxide (OEt) as a leaving group, generating an ester: CH3-C(=O)-O-Et (ethyl acetate). This is the intermediate after the first addition. Step 3: Since CH3MgBr is in excess, a second equivalent of CH3MgBr attacks the carbonyl of the newly formed ethyl acetate. The tetrahedral intermediate collapses, expelling OEt-, generating a ketone intermediate (acetone, CH3-C(=O)-CH3) — but this is not isolated because excess Grignard is still present. Step 4: A third equivalent of CH3MgBr adds to acetone (CH3-C(=O)-CH3) at the carbonyl, giving a magnesium alkoxide: CH3-C(OMgBr)(CH3)-CH3. Step 5: Workup with H3O+ (aqueous acid) protonates the alkoxide to give the tertiary alcohol: CH3-C(OH)(CH3)-CH3, which is 2-methyl-2-propanol (tert-butanol). The central carbon bears one OH and three CH3 groups. Why other options fail: - (a) CH3-C(=O)-O-Et is only the first intermediate, not the final product with excess Grignard. - (b) CH3-C(=O)-CH3 (acetone) is a second intermediate; excess Grignard reacts further with it. - (d) CH3-CH2-CH3 (propane) is not a product of this reaction sequence. Therefore, the correct answer is C.