Question

Discuss the structure of glucose in detail. (5)
OR
i) What do you understand by mono, di and polysaccharides? Explain with examples. (3)
ii) What are peptide bonds and glycosidic bonds? (2)

Hint:

For the structure, recall the six pieces of evidence (n-hexane on HI, oxime and cyanohydrin, gluconic acid with bromine water, pentaacetate, saccharic acid with HNO₃) and then the cyclic pyranose form (anomers, mutarotation). For the OR part, classify sugars by the number of units released on hydrolysis.

Answer 1

Option 1: Structure of glucose

Step 1: Molecular formula. Elemental analysis and molar mass determination show glucose is C₆H₁₂O₆ (molar mass 180 g/mol).

Step 2: A straight chain of six carbons. On prolonged heating with HI, glucose gives n-hexane. This proves all six carbon atoms are joined in a straight (unbranched) chain.

Step 3: Presence of a carbonyl group. Glucose reacts with hydroxylamine (NH₂OH) to form an oxime and adds one molecule of HCN to form a cyanohydrin. Both reactions confirm a carbonyl (>C=O) group.

Step 4: The carbonyl is an aldehyde. On mild oxidation with bromine water, glucose gives gluconic acid, a six-carbon monocarboxylic acid. Since a weak oxidant converts the carbonyl to a –COOH, the group must be a terminal aldehyde (–CHO).

Step 5: Five hydroxyl groups. Glucose reacts with acetic anhydride to form glucose pentaacetate. Formation of five acetate groups shows glucose contains five –OH groups, one each on five different carbon atoms (two OH on the same carbon would be unstable).

Step 6: A primary alcohol group. On oxidation with nitric acid, glucose gives saccharic acid, a dicarboxylic acid. This shows a primary alcohol group (–CH₂OH) at the end of the chain, which is oxidised to a second –COOH.

Step 7: Open-chain structure. Combining all the evidence, the open-chain structure is CHO–CHOH–CHOH–CHOH–CHOH–CH₂OH, i.e. an aldohexose. Its natural spatial arrangement is that of D-(+)-glucose.

Step 8: Cyclic (pyranose) structure. The open chain could not explain that glucose does not give the Schiff test, does not react with NaHSO₃, and exists as two forms (alpha and beta) that show mutarotation. These facts are explained by a cyclic hemiacetal formed when the –OH on C-5 adds to the –CHO on C-1. This gives a six-membered pyranose ring and a new –OH at C-1 (the anomeric carbon). The two positions of this OH give alpha-D-glucose and beta-D-glucose, drawn as Haworth projections.

Option 2:

i) Mono, di and polysaccharides.
Monosaccharides: the simplest carbohydrates that cannot be hydrolysed into smaller carbohydrate units. Examples: glucose, fructose, ribose.
Disaccharides: carbohydrates that on hydrolysis give two monosaccharide units. Examples: sucrose (gives glucose + fructose), maltose (gives two glucose), lactose (gives glucose + galactose).
Polysaccharides: carbohydrates that on hydrolysis give a large number of monosaccharide units. Examples: starch, cellulose, glycogen.

ii) Peptide bond and glycosidic bond.
A peptide bond is the amide linkage (–CO–NH–) formed between the –COOH group of one amino acid and the –NH₂ group of another amino acid, with the loss of a water molecule.
A glycosidic bond is the C–O–C linkage formed between two monosaccharide units when they join together with the loss of a water molecule (for example, the linkage between glucose and fructose in sucrose).