Reducing aldose with chiral center

Steps:

• Fehling's reagent detects reducing sugars with free aldehyde or alpha-hydroxy ketone groups, forming a red Cu2O precipitate.
• Positive test indicates compound X is a reducing sugar, likely an aldose.
• Stereoisomers require at least one chiral carbon, common in monosaccharides like aldoses.
• Match to options: D has both aldehyde for reduction and chiral center for isomers.

Why D is correct:

• D is an aldose (e.g., glyceraldehyde, R-CHO with -CHOH-), reducing via aldehyde oxidation (Fehling's reaction) and chiral at C2 for D/L stereoisomers.

Why the others are wrong:

• A: Ketone without alpha-OH, non-reducing, no Fehling's precipitate.
• B: Achiral symmetric structure, no stereoisomers.
• C: Non-reducing disaccharide, fails Fehling's test despite possible chirality.

Final answer: D