Question

{"given":"The reaction sequence is given as: D-Mannose $\\xrightarrow{\\text{HO}^-}$ D-Glucose $\\xrightarrow{\\text{HO}^-}$ (A);","key_observation":"This reaction represents the Lobry de Bruyn–van Ekenstein transformation, an alkali-catalyzed interconversion of aldoses and ketoses via an enediol intermediate.","option_analysis":[{"label":"(A)","text":"D-glucose","verdict":"incorrect","explanation":"D-glucose is the intermediate in the second step, not the final product (A) of the transformation from D-glucose."},{"label":"(B)","text":"D-fructose","verdict":"correct","explanation":"Under basic conditions, D-glucose undergoes tautomerization to form D-fructose (a ketose) and D-mannose (an epimer) via an enediol intermediate. Thus, (A) is D-fructose."},{"label":"(C)","text":"D-talose","verdict":"incorrect","explanation":"D-talose is an aldohexose but is not directly formed from D-glucose in this specific interconversion."},{"label":"(D)","text":"D-idose","verdict":"incorrect","explanation":"D-idose is an aldohexose but is not directly formed from D-glucose in this specific interconversion."}],"answer":"B","formula_steps":[]}