See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: Grignard reagents are strong nucleophiles that add to carbonyl compounds. Dimethyl carbonate (CH3OCOOCH3) is a diester of carbonic acid. When a Grignard reagent reacts with an ester, the first addition gives a ketone intermediate (after loss of the alkoxide leaving group). The ketone then reacts with a second equivalent of the Grignard reagent to give a tertiary alcohol. Step 1 - First addition of PhMgBr to dimethyl carbonate: Dimethyl carbonate CH3OCOOCH3 has two methoxy (OCH3) groups. PhMgBr attacks the carbonyl carbon, and one OCH3 leaves as methoxide. This gives methyl phenyl carbonate (CH3O-CO-Ph), which is effectively an ester. Step 2 - Second addition of PhMgBr to the ester intermediate: PhMgBr attacks the carbonyl of CH3O-CO-Ph again, the second OCH3 leaves, giving a ketone: Ph-CO-Ph (benzophenone) after workup? Wait - let me re-examine. Actually, dimethyl carbonate is (CH3O)2C=O. First PhMgBr addition: Ph attacks C=O, one OCH3 leaves → methyl benzoate (PhCOOCH3) intermediate equivalent. Second PhMgBr addition to methyl benzoate: Ph attacks C=O of PhCOOCH3, OCH3 leaves → gives benzophenone (Ph2C=O) after loss of OCH3. Third equivalent of PhMgBr adds to benzophenone: Ph attacks C=O → gives Ph3C-OMgBr → after H2O workup → triphenylmethanol Ph3COH, i.e., Ph-C(OH)(Ph)-Ph. Since excess PhMgBr is used, all three additions occur: dimethyl carbonate → (via two additions and loss of two OCH3 groups) benzophenone → (third addition) triphenylmethanol. The product is Ph-C(OH)(Ph2), i.e., the central carbon bears OH and three phenyl groups = triphenylmethanol, which matches option (c): Ph-C(OH)(Ph)-Ph. Why other options fail: (a) CH3-C(OH)(Ph)2 would require a methyl group on the central carbon, which would come from CH3MgBr, not PhMgBr. (b) CH3-CH(OH)-Ph is a secondary alcohol inconsistent with excess Grignard on a carbonate. (d) CH3-CO-Ph (acetophenone) is a ketone; Grignard reactions with excess reagent go beyond the ketone stage. Therefore, the correct answer is C.