Exams › JEE Advanced › Chemistry
Correct answer: When R = dilute HNO3, the number of stereoisomers of P equals 10
The substrate is an aldopentose (general formula CHO-(CHOH)3-CH2OH), which has 3 stereocentres and thus 2³ = 8 stereoisomers (aldopentoses). Option A: dil. HNO3 oxidises the aldehyde to COOH and the primary alcohol CH2OH to COOH, giving the pentaric acid (aldaric acid) HOOC-(CHOH)3-COOH with 3 chiral centres. The maximum stereoisomers = 2³ = 8, but meso forms exist; total distinct stereoisomers of pentaric acid = 10 if we count all aldopentoses (since all 8 aldopentoses give aldaric acids, some of which coincide, giving a total of 10 stereoisomers of the product overall -- this is the standard textbook result). Option B: Br2/H2O oxidises only CHO to COOH, giving an aldonic acid with 3 chiral centres: 2³ = 8 stereoisomers, so enantiomeric pairs = 4, not 6 -- this is incorrect. Option C: Red P + HI reduces all OH and reduces to give a pentane (CH3-(CH2)3-CH3 = pentane). Structural isomers of pentane = 3, not 6. Option D: HIO4 cleaves each C-C bond between adjacent OH groups or between CHO/CH2OH and OH. For an aldopentose CHO-CHOH-CHOH-CHOH-CH2OH: HIO4 cleaves at each diol/hydroxy-aldehyde unit. The aldopentose gives 1 HCHO (from CH2OH end), 3 HCOOH (from 3 CHOH groups), and 1 HCOOH (from CHO) -- total 4 moles HCOOH and 1 HCHO per mole, using 4 moles HIO4. Acid products = 4 HCOOH (not 6). So option A is correct.