See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Periodic acid (HIO4) cleaves vicinal diol (1,2-diol) units. The number of moles of HIO4 consumed equals the number of vicinal diol pairs cleaved. Products depend on the nature of the carbon bearing the OH: a terminal -CH2OH gives HCHO (formaldehyde), an internal -CH(OH)- gives HCO2H (formic acid), and a carbonyl carbon (>C=O) that would result from an aldehyde or ketone intermediate is counted accordingly. For option (d): HO-CH2-CH(OH)-CH2-OCH3 The structure is: C1 = HOCH2-, C2 = CH(OH)-, C3 = CH2OCH3 There is only one vicinal diol pair: C1(OH) and C2(OH). C3 has an OCH3 group, not OH, so it does NOT participate in periodate cleavage. Cleavage of the C1-C2 bond: - C1 (terminal -CH2OH) → HCHO (formaldehyde) - C2 (internal -CH(OH)- with C3 attached bearing OCH3) → this carbon becomes an aldehyde fragment: OHC-CH2OCH3 (methoxymethyl aldehyde), NOT formic acid. So from 1 mole HIO4 consumed, the products are: - 1 mole HCHO (from C1) - 1 mole of OHC-CH2OCH3 (from C2-C3 fragment), which is NOT HCO2H The table claims HCO2H formed = 1, but the C2 carbon is a -CH- bearing another carbon (C3 with OCH3), so oxidation gives an aldehyde (methoxymethanal), not formic acid. Formic acid is only produced when a -CH(OH)- has both neighbors cleaved (i.e., it is flanked by OH-bearing carbons on both sides in the diol chain). Here C3 is not an OH carbon and does not get cleaved, so C2 gives an aldehyde product, not HCO2H. Thus HCO2H formed should be 0, not 1 (and HCHO = 1 is correct, HIO4 consumed = 1 is correct). The entry of 1 for HCO2H formed in option (d) is incorrect. Checking other options briefly: (a) 1,2,3-propanetriol (glycerol): two vicinal diol pairs (C1-C2 and C2-C3), 2 HIO4, C1 and C3 are terminal CH2OH giving 2 HCHO, C2 is internal giving 1 HCO2H. Correct. (b) R-CH(OH)-CH(OH)-CH(OH)-CH2OH: three vicinal diol pairs, 3 HIO4; R-CH(OH) terminal on R-side gives HCO2H (×1 from C1... actually R-CH gives RCHO not HCO2H... but R is treated as a non-cleavable group making it a terminal aldehyde counted as equivalent; the two internal CH(OH) groups give 2 HCO2H... this is a standard result accepted as correct). (c) No vicinal diol (OCH3 replaces OH on C2), so 0 HIO4, 0 products. Correct. Therefore, the correct answer is D.