$word_{count}$: 200

The key question: does the sugar have a FREE anomeric carbon $\frac{hemiacetal}{hemiketal OH}$? YES = reducing sugar (can open to expose C=O for oxidation). NO = non-reducing. All monosaccharides are reducing (glucose, fructose, galactose). Fructose (a ketose) reduces Tollens'/Fehling's via enediol tautomerism in basic medium — this is a classic JEE trap. For disaccharides, check the glycosidic bond: Sucrose = alpha-1,2 bond locks BOTH anomeric carbons → non-reducing (the only common non-reducing disaccharide). Maltose = alpha-1,4 → one free anomeric C on second glucose → reducing. Lactose = beta-1,4 → one free anomeric C → reducing. Cellobiose = beta-1,4 → reducing. For polysaccharides: technically reducing (one free end per chain) but practically non-reducing (one reducing end among thousands of units). JEE strategy: if asked "which is non-reducing?" — answer is sucrose. If asked about Tollens'/Fehling's: all common sugars are positive EXCEPT sucrose. If asked "does fructose reduce Fehling's?" — YES (enediol mechanism).