See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: Phthalic anhydride is a cyclic anhydride with two carbonyl groups (C1 and C3) bridged by an oxygen. Step 2 - Reactivity of anhydrides with Grignard reagents: When a Grignard reagent (RMgX) attacks an anhydride, the first equivalent adds to one of the carbonyl carbons to give, after tetrahedral collapse, a ketone intermediate (one carboxylate leaves as the other acyl unit). For phthalic anhydride, the first mole of MeMgBr attacks one carbonyl, the ring opens, and after collapse of the tetrahedral intermediate, a keto-acid salt (or keto-ester equivalent) is formed. Specifically, the first equivalent of MeMgBr adds to one carbonyl of phthalic anhydride; the anhydride ring opens to give an intermediate that is essentially an ortho-(methylcarbonyl)benzoate (a ketone with an adjacent carboxylate). However, with cyclic anhydrides and excess Grignard, the commonly accepted mechanism is: the first MeMgBr adds to the anhydride to open it giving a magnesium carboxylate salt of an acyl product. The product of ring-opening of phthalic anhydride with first equivalent of MeMgBr gives 2-(acetyl)benzoate salt (ortho-acetylbenzoic acid salt), because one carbonyl becomes a ketone (C-Me bond formed) and the other becomes a carboxylate. Step 3 - Second equivalent of MeMgBr: The second mole of MeMgBr then adds to the ketone carbonyl of the 2-acetylbenzoate intermediate, forming a tertiary alkoxide on that carbon (now bearing two methyl groups and an OH after workup). The carboxylate does not react with the second equivalent under these conditions because carboxylate salts are resistant to Grignard addition (they already bear a negative charge). Step 4 - Workup with H3O+: Acidic workup protonates the alkoxide to give a tertiary alcohol C(OH)(CH3)2, and protonates the carboxylate to give -COOH. Step 5 - Final product: The product is an ortho-disubstituted benzene bearing a -C(OH)(CH3)2 group (from double methyl addition via ketone intermediate) and a -COOH group. This matches option (d): 2-(2-hydroxypropan-2-yl)benzoic acid, i.e., ortho-substituted benzene with C(OH)(CH3)2 and COOH groups. Step 6 - Why other options fail: - Option (a) would require both carbonyls to react with one equivalent each, giving a diol-lactol; this is not what 2 moles of MeMgBr on phthalic anhydride gives. - Option (b) shows only one addition and an intact lactone, but with 2 moles of Grignard this would not be the final product. - Option (c) shows gem-dimethyl without OH groups, which would require elimination/different conditions. Therefore, the correct answer is D.