See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

To solve this matching problem, we need to determine how many moles of PhMgBr (phenylmagnesium bromide) are required to convert each substrate into a tertiary (3°) alcohol upon aqueous workup. **Concept:** Grignard reagents add to carbonyl compounds. Esters react with 2 equivalents of Grignard reagent (first addition gives a ketone intermediate, second addition gives a tertiary alcohol). Ketones react with 1 equivalent. Acid chlorides react with 2 equivalents (first addition gives a ketone, second gives a tertiary alcohol). Hydroxyl groups (–OH) are acidic and will protonate (destroy) the Grignard reagent before it can react with the carbonyl — so one extra mole of PhMgBr is needed per –OH group just to deprotonate it. **Case (c): PhMgBr + acetone (CH3-C(=O)-CH3) → 3° alcohol** Acetone is a ketone. One mole of PhMgBr adds to give Ph-C(CH3)2-OH (triphenylcarbinol precursor; here it's Ph-C(OH)(CH3)2). Only 1 mole of PhMgBr is needed. Answer: (p) 1 **Case (b): PhMgBr + HO-CH2-C(=O)-CH3 → 3° alcohol** The substrate is hydroxyacetone (1-hydroxy-2-propanone). It has both a –OH group and a ketone (C=O). The –OH is acidic: 1 mole of PhMgBr is consumed by protonation (destroyed). Then the ketone requires 1 more mole of PhMgBr to add and give a tertiary alcohol (since Ph adds to the ketone carbon which already has CH3 and CH2O- groups, giving a tertiary alcohol). Total = 1 (for OH) + 1 (for ketone addition) = 2 moles. Answer: (q) 2 **Case (a): PhMgBr + diethyl carbonate (Et-O-C(=O)-O-Et) → 3° alcohol** Diethyl carbonate is a carbonate ester. Esters react with 2 equivalents of Grignard: the first equivalent adds to give a ketone (in this case, after loss of EtO-, a phenyl ester or a ketone intermediate: first addition gives Ph-C(=O)-OEt after loss of one OEt, which is an ester; this ester reacts with a second PhMgBr to give Ph2C(OH)OEt... Actually, diethyl carbonate: first PhMgBr adds and loses EtO- to give Ph-C(=O)-OEt (ethyl benzoate). Ethyl benzoate then reacts with a second PhMgBr to give Ph2C(=O) (benzophenone) after loss of OEt-, then a third PhMgBr adds to give Ph3COH (triphenylmethanol) — a tertiary alcohol. Total = 3 moles of PhMgBr. Answer: (r) 3 **Case (d): PhMgBr + 3,5-dihydroxybenzoyl chloride → 3° alcohol** The substrate is a benzoyl chloride with two –OH groups on the ring (at positions 3 and 5). The acid chloride reacts like an ester needing 2 equivalents of PhMgBr (first gives ketone, second gives tertiary alcohol). However, two –OH groups are present, each consuming 1 mole of PhMgBr by protonation. Total = 2 (for two –OH groups) + 2 (for acid chloride to tertiary alcohol) = 4 moles. Answer: (s) 4 **Summary of matches:** - (a) → (r) 3 - (b) → (q) 2 - (c) → (p) 1 - (d) → (s) 4 Therefore, the correct answer is {"a": ["R"], "b": ["Q"], "c": ["P"], "d": ["S"]}.