When an aqueous solution of D-fructose is treated with a dilute alkali and allowed to stand for some time, it undergoes the Lobry de Bruyn-van Ekenstein rearrangement. Which of the following represents the major monosaccharide components present in the resulting equilibrium mixture?

Step-by-step solution

In the presence of a dilute alkali, D-fructose undergoes deprotonation at the -carbon to form a planar enediol intermediate. When this enediol intermediate undergoes reprotonation to form an aldose, the proton can add from either face of the double bond. This results in the formation of two C-2 epimers: D-glucose and D-mannose. Since the reaction is reversible, the final equilibrium mixture contains the unreacted D-fructose along with D-glucose and D-mannose. Answer: D-fructose, D-glucose and D-mannose