See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify vicinal diol units in each structure. HIO4 cleaves each pair of adjacent -OH groups (vicinal diols), consuming 1 mol HIO4 per cleavage. Step 2: Structure X (methyl alpha-D-glucopyranoside): Free OH groups at C2, C3, C4, and CH2OH at C6. The ring oxygen prevents C1 from being a free diol. Vicinal diol pairs: C2-C3 (1 mol), C3-C4 (1 mol), C4-C6 via C5 (the C5-C6 unit: CH-CH2OH = 1 mol). Total = 3 mol HIO4 consumed for X. Step 3: Structure Y (methyl glucoside with oxygen on the other face): The ring oxygen is on C2 side, making C2 an acetal (no free OH at C2). Free OH at C3, C4, C6 (CH2OH). Vicinal diol pairs: C3-C4 (1 mol), C4-C5-C6 unit (1 mol). Total = 2 mol HIO4 consumed for Y. So answer A: 3, 2 → option (d). Step 4 (Part B - Formic acid): Each internal CHOH that is cleaved on both sides gives HCOOH. For X: C2, C3, C4 each give HCOOH → 2 mol HCOOH (internal carbons C3, C4 give formic acid; C2 gives formic acid too, but check: 3 cleavages on X yield fragments from C2, C3, C4 as HCOOH = 2 mol formic acid). For Y: only C3, C4 are internal → 1 mol HCOOH. Answer B: 2, 1 → option (b). Step 5 (Part C - Formaldehyde): CH2OH (primary alcohol) cleaved gives HCHO. Both X and Y have one CH2OH (C6). Each gives 1 mol HCHO. Answer C: 1, 1 → option (a). Therefore, the correct answer is {"A": ["D"], "B": ["B"], "C": ["A"]}.