See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Periodic acid (HIO4) cleaves vicinal diol (and polyol) systems. Each C-C bond between two carbons each bearing an -OH (or =O equivalent) is cleaved. The products depend on the carbon's substitution: a terminal -CH2OH gives HCHO (formaldehyde), an internal -CH(OH)- gives HCOOH (formic acid), and a -CHO (aldehyde end) gives HCOOH (formic acid). Reaction (i): The compound is arabinose-type structure: CHO | (CHOH)3 | CH2OH — this is a 5-carbon open-chain aldopentose (e.g., arabinose). It has 5 carbons total. - Carbon 1: CHO (aldehyde) → cleaved to give HCOOH - Carbons 2, 3, 4: each is -CHOH- (internal) → each gives HCOOH - Carbon 5: CH2OH (terminal) → gives HCHO So formic acid from reaction (i) = 4 moles (from C1 as CHO and C2, C3, C4 as internal CHOH). Moles of HIO4 consumed = 4 (consistent with 4HIO4 given). Reaction (ii): The compound is CH2OH | (CHOH)4 | CH2OH — this is a 6-carbon open-chain alditol (sugar alcohol, e.g., sorbitol/mannitol). It has 6 carbons total. - Carbon 1: CH2OH (terminal) → gives HCHO - Carbons 2, 3, 4, 5: each is -CHOH- (internal) → each gives HCOOH - Carbon 6: CH2OH (terminal) → gives HCHO So formic acid from reaction (ii) = 4 moles (from C2, C3, C4, C5). Moles of HIO4 consumed = 5 (consistent with 5HIO4 given). Ratio = moles of formic acid in (i) / moles of formic acid in (ii) = 4/4 = 1. Why other options fail: - (a) 3/4: would imply 3 moles in (i), incorrect since the CHO carbon also yields formic acid. - (b) 4/5: would imply 5 moles in (ii), which is not the case since the two terminal CH2OH groups give formaldehyde, not formic acid. - (d) 5/4: would imply more formic acid in (i) than (ii), which is not the case. Therefore, the correct answer is C.