See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The structure shown is a six-membered ring containing one ring oxygen, with two carbonyl groups — one is a lactone (ester C=O, since the oxygen is in the ring) and one is a ketone C=O. This is tetrahydro-2H-pyran-2,5(6H)-dione (glutaric anhydride or a related cyclic structure). More carefully: the ring has O in the ring, a C=O adjacent to O (lactone), and another C=O on the opposite side (ketone). This is dihydro-2H-pyran-2,5(3H)-dione, i.e., glutaric anhydride is a five-membered ring; this six-membered version is glutaric anhydride analog — actually this is 1,3-dioxan-2-one? Re-examining: the structure is a six-membered ring with one O in the ring, carbonyl at C2 (lactone) and carbonyl at C5 (ketone). This is tetrahydropyran-2,5-dione, a cyclic structure with one lactone and one ketone. Step 2: Reaction with LiAlH4. LiAlH4 reduces both the lactone and the ketone. The lactone is reduced to give a diol (the ring opens: the C=O of the lactone becomes CH2OH and the ring oxygen gives an OH after workup — actually lactone reduction gives a primary alcohol and opens the ring, and the ketone gives a secondary alcohol). Specifically, reduction of the lactone gives ring opening: the ester oxygen becomes an OH (on the carbon that was the acyl carbon, giving primary -CH2OH) and the alkoxy oxygen becomes a free OH. The ketone C=O is reduced to CHOH (secondary alcohol). Product (A): HOCH2-CH2-CH(OH)-CH2-CH2OH? Let me think more carefully. The six-membered ring: O-C(=O)-CH2-C(=O)-CH2-CH2- (ring closed back to O). So numbering: O(1)-C2(=O)-C3H2-C4(=O)-C5H2-C6H2-back to O1. LiAlH4 reduces: C2 lactone carbonyl → the C-O bond breaks, C2 becomes CH2OH (primary alcohol), and O1 becomes a free hydroxyl on C6. C4 ketone → C4 becomes CHOH (secondary alcohol). Product (A) = HO-CH2-CH2-CH(OH)-CH2-CH2OH: 1,3,5-pentanetriol (specifically pentane-1,3,5-triol). Step 3: Count OH groups in (A). Pentane-1,3,5-triol has 3 hydroxyl groups: one primary at C1, one secondary at C3, one primary at C5. Step 4: Reaction with Ac2O. Acetic anhydride (Ac2O) acetylates hydroxyl groups. Each mole of Ac2O reacts with one OH group (since Ac2O reacts with ROH to give ROAc + AcOH, consuming 1 mole Ac2O per OH). With 3 OH groups, x = 3 moles of Ac2O are consumed. Step 5: Why other options fail. (a) 1 and (b) 2 undercount the OH groups. (d) 4 would require 4 OH groups, but there are only 3. Only 3 OH groups are present in (A), so x = 3. Therefore, the correct answer is C.