Reducing aldose with chiral centers

Steps:

• Fehling's reagent gives precipitate with reducing sugars, indicating an aldehyde or alpha-hydroxy ketone group.
• Presence of stereoisomers requires at least two chiral centers, typical in aldoses with 4+ carbons.
• Compound X must be an aldose (aldehyde sugar) to satisfy both reducing property and chirality.
• Eliminate non-aldoses or those lacking chirality/reducing ability.

Why D is correct:

• D is an aldopentose (e.g., ribose) with an aldehyde group for Fehling's reaction and multiple chiral carbons for stereoisomers.

Why the others are wrong:

• A: Ketose (e.g., fructose) lacks free aldehyde, so weak or no Fehling's precipitate.
• B: Achiral compound (e.g., glyceraldehyde) has only one chiral center, limiting stereoisomers.
• C: Non-reducing disaccharide (e.g., sucrose) fails Fehling's test due to no free aldehyde.

Final answer: D