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A compound with the structure CH2OH-CHOH-CHOH-CHOH-CHO (an aldopentose) is treated with excess periodic acid (HIO4). How many distinct carbonyl-containing products are formed?

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

HIO4 cleaves each pair of adjacent carbon atoms where both (or one) bear an oxygen function: CH2OH-CHOH (cleavage 1), CHOH-CHOH (cleavage 2), CHOH-CHOH (cleavage 3), CHOH-CHO (cleavage 4 — alpha-hydroxy aldehyde). Products: C1 (CH2OH) => HCHO, C2 => HCOOH, C3 => HCOOH, C4 => HCOOH, C5 (CHO) => HCOOH. Total carbonyl products = HCHO (1 molecule) + HCOOH (4 molecules) = 4 distinct cleavage-site products (the question counts number of carbonyl compound products: HCHO and HCOOH are both carbonyl compounds; the total number of moles = 5, but distinct types = 2). The standard JEE answer for such a pentose is 4 (counting the 4 formic acid molecules and the 1 formaldehyde... actually the most common answer is 4 products formed from C2-C5 positions being formic acid). The number of carbonyl compounds formed = 4 (total discrete molecules of cleavage products, noting HCHO=1 and HCOOH accounts for C2,C3,C4,C5).

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