Fundamentholfundamenthol
JEE Advanced2022Paper 2CHEM-III
Q.

Treatment of D-glucose with aqueous NaOH results in a mixture of monosaccharides, which are

  1. A
  2. B
  3. C
  4. D
Solution

This is the Lobry de Bruyn–van Ekenstein transformation. In dilute aqueous base, D-glucose enolises through its C1–C2 enediol intermediate. The enediol then re-tautomerises three different ways:

1. Back to D-glucose itself.

2. To D-mannose (epimerisation at C2; the C2 stereocenter inverts because the enediol erases C2 stereochemistry).

3. To D-fructose (ketose isomer; the enediol can collapse onto either C1 or C2 carbonyl).

So the equilibrium mixture contains D-glucose, D-mannose, and D-fructose. Option (C) matches this mixture.

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