See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: Grignard reagents (RMgBr) react with acid chlorides and ketones to give tertiary alcohols after aqueous workup. With excess MeMgBr, all electrophilic carbonyl groups react. Step 1 - Identify the electrophilic sites in the starting material: The starting material is a cyclobutane ring bearing two carbonyl groups: - An acid chloride (-C(=O)Cl) exocyclic on one ring carbon - A ketone (C=O) where the ring carbon itself is a ketone on the opposite carbon Step 2 - Reaction of MeMgBr with the acid chloride: MeMgBr adds to the acid chloride first (acid chlorides are more reactive than ketones). The first equivalent of MeMgBr displaces Cl- and forms a ketone intermediate (methyl ketone), then a second equivalent of MeMgBr adds to this ketone intermediate to give, after workup, a tertiary alcohol: the exocyclic carbon becomes -C(CH3)2OH (dimethyl carbinol, i.e., a carbon bearing two methyl groups and one OH). Step 3 - Reaction of MeMgBr with the ring ketone: With excess MeMgBr, the ring ketone also reacts. MeMgBr adds to the ring ketone carbon, and after aqueous workup gives a tertiary alcohol: the ring carbon that was C=O becomes C(OH)(CH3) - bearing OH and one methyl group (and two ring carbons). Step 4 - Determine the final product: - The acid chloride end: -C(=O)Cl → after 2 equivalents MeMgBr and H2O → -C(CH3)2OH (tertiary alcohol with two methyl groups) - The ring ketone end: ring C=O → after 1 equivalent MeMgBr and H2O → ring C(OH)(CH3) (tertiary alcohol with one methyl group) This matches option (d): one end of the cyclobutane has H3C-C(CH3)(OH)- (from the acid chloride reacting with 2 MeMgBr) and the other end has -C(OH)(CH3) (from the ketone reacting with 1 MeMgBr), with OH groups on both carbons. Why other options fail: - (a) shows a methyl ester, but MeMgBr does not form esters; it adds nucleophilically - (b) shows a carboxylic acid, which would require oxidation, not Grignard addition - (c) shows the ring ketone unreacted (still C=O remaining), but excess MeMgBr would react with all carbonyls Therefore, the correct answer is D.