See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 – Determine x (moles of CH3MgBr consumed in Reaction-1). The starting material has two acid chloride (–COCl) groups on the ring plus a –CH(SH)–CH2OH side chain. The thiol (–SH) and alcohol (–OH) groups on the side chain are acidic protons; each will protonate (and therefore destroy) one equivalent of CH3MgBr before the Grignard addition can occur. • Side-chain acidic protons: –SH (1 equiv) + –CH2OH (1 equiv) = 2 equiv CH3MgBr consumed as base. • Each –COCl reacts with CH3MgBr to form a ketone intermediate (–CO–CH3), which then reacts with a second equivalent of CH3MgBr to give a tertiary alkoxide; after workup this gives –C(CH3)2OH. So each –COCl requires 2 equivalents of CH3MgBr. • Two –COCl groups: 2 × 2 = 4 equiv CH3MgBr for addition. Total x = 2 (for acidic protons) + 4 (for two acid-chloride groups) = 6. Step 2 – Determine y (moles of CH3MgBr consumed in Reaction-2). The starting material is ethyl 4-(chloromethyl)benzoate. • The –CH2Cl (benzyl chloride) reacts with Mg to form a Grignard reagent (ArCH2MgCl); this Grignard then reacts with the ester group in the same molecule – but in practice the reagent added is CH3MgBr. • The –CH2Cl group: benzyl chloride reacts with 1 equiv CH3MgBr in a substitution (SN2-type with organocuprate-like behaviour) or, more directly, the –CH2Cl acts as an electrophile toward CH3MgBr, consuming 1 equiv to give –CH2CH3. • The ester (–COOEt) requires 2 equiv CH3MgBr: first addition gives a ketone intermediate (–CO–CH3, releasing EtO–), second addition gives –C(CH3)2O– (tertiary alkoxide), giving –C(CH3)2OH after workup. Total y = 1 (for –CH2Cl) + 2 (for ester) = 3. Step 3 – Calculate x/y. x/y = 6/3 = 2. Why other options fail: • 1.5 would require x=3, y=2 – inconsistent with the functional group count. • 2.5 would require a non-integer total – not possible with whole equivalents. • 3 would require x=9 or y=2 – overcounts or undercounts the reactive sites. Therefore, the correct answer is B.