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

When an aldose containing an $$\alpha$$-hydrogen (such as D-glucose) is treated with a dilute aqueous base, it undergoes the Lobry de Bruyn-van Ekenstein transformation.

Step 1 - Formation of an enediol
Under basic conditions the C-2 hydrogen is abstracted, giving an enolate that rapidly protonates at oxygen to form a common enediol (or enediolate) intermediate.

Step 2 - Re-protonation at C-1 or C-2
The enediol can tautomerise in three different directions:
  • Protonation at C-1 with the original stereochemistry regenerates the starting aldose, D-glucose.
  • Protonation at C-1 with opposite stereochemistry gives the C-2 epimer of glucose, namely D-mannose.
  • Protonation at C-2 followed by keto-enol rearrangement converts the aldose into the ketose D-fructose.

Step 3 - Equilibrium mixture
All three monosaccharides interconvert through the same enediol, so the reaction mixture at equilibrium contains

$$\text{D-glucose} \;\;+\;\; \text{D-mannose} \;\;+\;\; \text{D-fructose}$$

Hence, aqueous NaOH converts D-glucose into a mixture of its C-2 epimer (D-mannose) and its corresponding keto-isomer (D-fructose), besides the unreacted D-glucose.

Option C which is: D-glucose, D-fructose and D-mannose