JEE Advanced 2022 Paper 2 · Q18 · Carbohydrates
Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are:
Reveal answer + step-by-step solution
Correct answer:C
Solution
Treatment of D-glucose with dilute aqueous NaOH effects the Lobry de Bruyn–Van Ekenstein rearrangement: an enediol intermediate is formed at C1–C2, which can re-tautomerise to give:
(i) D-glucose itself (aldose, CHO at C1, OH on C2-right Fischer) (ii) D-fructose (ketose, CH$_2$OH-CO- at C1–C2) (iii) D-mannose (aldose, C-2 epimer of D-glucose; the H/OH at C2 are swapped relative to glucose).
The equilibrium mixture is therefore glucose + fructose + mannose. The set must contain one ketohexose (fructose) and two aldohexoses that are C-2 epimers (glucose and mannose).
Option (A): three aldoses, no ketose — WRONG. Option (B): one ketose + two aldoses, but the aldoses shown are not the correct glucose/mannose pair — WRONG. Option (C): glucose + fructose + mannose — CORRECT. Option (D): three aldoses, no ketose — WRONG.
Answer: (C).
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