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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Biomolecules

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Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are
  1. (A)(see structure)
    Structure A
    A
    Structure B
    B
    Structure C
    C
  2. (B)(see structure)
    Structure A
    A
    Structure B
    B
    Structure C
    C
  3. (C)(see structure)
    Structure A
    A
    Structure B
    B
    Structure C
    C
    ✓ Correct
  4. (D)(see structure)
    Structure A
    A
    Structure B
    B
    Structure C
    C

Answer: (C)

Solution

Treatment of D-glucose with aqueous NaOH causes base-catalyzed epimerization through an enediol intermediate. This process, known as Lobry de Bruyn-van Ekenstein transformation, converts D-glucose into an equilibrium mixture of D-glucose, D-mannose (C-2 epimer), and D-fructose (ketose isomer). Option (C) correctly shows these three monosaccharides.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Single correct option

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