Concept 01 / 05Biomolecules
Carbohydrates: Classification and Structure of Glucose
Classification of carbohydrates (mono-, oligo- and polysaccharides; reducing and non-reducing sugars; aldoses and ketoses), preparation and open-chain structure of glucose with the chemical evidence, D and L configuration, the cyclic hemiacetal structure, anomers and mutarotation.
Carbohydrates are optically active polyhydroxy aldehydes or ketones, or compounds that give them on hydrolysis. Many have the formula , but not all compounds of this form are carbohydrates (for example, ethanoic acid ) and some carbohydrates do not fit it (2-deoxyribose, ).
Classification by hydrolysis.
Monosaccharides cannot be hydrolysed further: glucose, fructose, ribose.
Oligosaccharides give 2 to 10 monosaccharide units: disaccharides such as sucrose, maltose and lactose.
Polysaccharides give a large number of units: starch, cellulose, glycogen.
Sugars are sweet, crystalline and water-soluble (monosaccharides and oligosaccharides); polysaccharides are not sweet.
Reducing sugars reduce Tollens' reagent and Fehling's solution because they have a free aldehyde or ketone group (or a free hemiacetal carbon). All monosaccharides are reducing; sucrose is non-reducing.
Monosaccharides are named by carbonyl type and carbon count: glucose is an aldohexose, fructose a ketohexose, ribose an aldopentose.
Preparation of glucose. From sucrose by boiling with dilute HCl or in alcoholic solution (glucose and fructose in equal amounts); commercially by hydrolysis of starch with dilute at 393 K under pressure: .
Evidence for the open-chain structure ().
Prolonged heating with HI gives n-hexane: the six carbons are in a straight chain.
Reaction with hydroxylamine (oxime) and HCN (cyanohydrin): a carbonyl group is present.
Mild oxidation with bromine water gives gluconic acid, , with the same number of carbons: the carbonyl is an aldehyde.
Acetylation with acetic anhydride gives a pentaacetate: five OH groups, on different carbons since the molecule is stable.
Oxidation with nitric acid gives saccharic (glucaric) acid, a dicarboxylic acid: glucose also has a primary alcohol group, which is oxidised along with the aldehyde.
Configuration. D and L refer to the configuration at the lowest chiral carbon relative to glyceraldehyde: in D-glucose the OH on C5 is on the right in the Fischer projection. D/L is unrelated to the sign of rotation; natural glucose is D-(+)-glucose.
Cyclic structure. The open-chain formula fails to explain several facts: glucose gives no 2,4-DNP derivative and no Schiff's test, and does not form a hydrogensulfite adduct with ; the pentaacetate does not react with hydroxylamine (no free CHO); and glucose exists in two crystalline forms, α (m.p. 419 K) and β (m.p. 423 K). These are explained by the C5-OH adding to the C1 aldehyde, forming a six-membered hemiacetal ring (pyranose, by analogy with pyran). C1 becomes a new chiral centre, giving two anomers: α (C1-OH on the same side as the C5 ring oxygen in the Fischer form; down in the Haworth form) and β (up in the Haworth form).
Mutarotation. A freshly prepared solution of either anomer changes its specific rotation until it reaches . The ring opens and closes through the small amount of open-chain form, and an equilibrium mixture is formed.
Epimers differ in configuration at only one chiral carbon (other than the anomeric one): glucose and mannose (C2), glucose and galactose (C4). Anomers are a special pair of epimers that differ at the anomeric carbon.
Key Takeaways:
- Glucose is an aldohexose with a straight six-carbon chain, 5 OH groups, a CHO and a primary alcohol group.
- The open-chain formula cannot explain the missing NaHSO₃ and Schiff reactions, the unreactive pentaacetate or the two crystalline forms.
- α and β anomers differ only at C1; mutarotation leads to .
- Pitfall: D/L describes configuration, not the direction of rotation.
- Pitfall: glucose's pentaacetate shows all 5 OH groups are free, but its failure to react with shows that the aldehyde is tied up in the ring.
Brain-Teaser 1: The specific rotations of α- and β-D-glucose are and . The equilibrium value is . Find the percentage of each anomer at equilibrium (neglect the open-chain form).
Solution: Let the fraction of α be . Then , so and . The mixture is about α and β. The β anomer dominates because all its bulky groups, including the C1-OH, are equatorial in the chair.
Brain-Teaser 2: How many moles of acetic anhydride react with one mole of glucose, and how many moles of are consumed by one mole of open-chain glucose?
Solution: Five OH groups: 5 mol of acetic anhydride give the pentaacetate. Periodic acid cleaves every C-C bond between adjacent carbons carrying OH or CHO: glucose has 5 such bonds (C1-C2 to C5-C6), so 5 mol of are consumed, giving 5 HCOOH and 1 HCHO.
Brain-Teaser 3: Which pair is anomers and which epimers: (a) α-D-glucose and β-D-glucose; (b) D-glucose and D-galactose; (c) D-glucose and D-fructose?
Solution: (a) Anomers (they differ only at the anomeric C1). (b) C4 epimers. (c) Neither: they are functional (aldose-ketose) isomers with the same formula .
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