Option 1: Establishing the glucose structure from its reactions.
Step 1: Glucose is an aldohexose with formula C6H12O6 and molar mass 180 g/mol.
Step 2: Long reduction with hydroiodic acid strips every functional group and leaves n-hexane, so the six carbons form one continuous, unbranched chain.
Step 3: Oxime formation with NH2OH and cyanohydrin formation with HCN reveal a C=O group; because bromine water (too weak to touch a ketone) still oxidises glucose to the C6 acid gluconic acid, that C=O must be a terminal aldehyde.
Step 4: Acetylation to a penta-ester (glucose pentaacetate) fixes the count of hydroxyls at five, each on a separate carbon, since a gem-diol would be unstable.
Step 5: Vigorous oxidation with HNO3 yields the diacid saccharic acid, proving the far end of the chain carries a primary –CH2OH that becomes the second –COOH.
Step 6: These facts give the linear formula OHC–(CHOH)4–CH2OH, whose natural configuration is D-(+)-glucose.
Step 7: The open chain cannot explain the failed Schiff and bisulphite tests or the two anomers that inter-convert (mutarotation). So C-1 and C-5 close into a six-membered glucopyranose ring; the fresh C-1 hydroxyl is the anomeric OH, giving the alpha and beta Haworth forms.
Option 2:
i) Carbohydrates are graded by how many sugar units they release on hydrolysis. A monosaccharide releases none because it is already the smallest unit (glucose, fructose, galactose). A disaccharide releases two units (sucrose, maltose, lactose). A polysaccharide releases many hundreds to thousands of units (starch, glycogen, cellulose).
ii) When the carboxyl group of one amino acid condenses with the amino group of the next, water is eliminated and a –CO–NH– amide link forms: this peptide bond strings amino acids into proteins. When the hemiacetal OH of one sugar condenses with an OH of another sugar, water is eliminated and an oxygen bridge C–O–C forms: this glycosidic bond joins monosaccharides into di- and polysaccharides.