Carbohydrates

Classify by carbonyl type. A carbohydrate is an aldose if its carbonyl is an aldehyde (glucose) or a ketose if it is a ketone (fructose). Both glucose and fructose share the molecular formula $C_6H_{12}O_6$ — they are structural isomers, not the same molecule.

Classify by hydrolysis. Monosaccharides cannot be split further (glucose, fructose). Oligosaccharides yield 2 to 10 monosaccharide units on hydrolysis — the examinable disaccharides are sucrose, maltose, and lactose. Polysaccharides yield many units and include starch, cellulose, and glycogen; these are not sweet and are mostly non-reducing.

Glucose closes into a ring. The open-chain form, $\text{CH}_2OH-(\text{CHOH})_4-\text{CHO}$, is not the whole story: the $-OH$ on C-5 attacks the C-1 aldehyde carbon intramolecularly, closing a six-membered pyranose ring. This creates a brand-new stereocentre at C-1, called the anomeric carbon, and gives two ring forms — $\alpha$-D-glucose and $\beta$-D-glucose — that differ only in the spatial orientation of the C-1 $-OH$. Such a pair is called a set of anomers. The D/L label is unrelated to this: it is fixed once, at the highest-numbered chiral carbon, by comparison with D/L-glyceraldehyde, and almost every natural sugar is a D-sugar.

The glycosidic linkage is the join. Two sugar units combine when the anomeric $-OH$ of one attacks an $-OH$ of another, releasing water — a condensation exactly analogous to ester or amide formation:

where: $\text{Sugar}_1\text{-OH}$ is the anomeric hydroxyl of the first ring, $\text{HO-Sugar}_2$ is any hydroxyl (anomeric or otherwise) on the second ring, and the new $\text{-O-}$ bridge is the glycosidic linkage. Whether that linkage uses up one anomeric carbon or both decides everything about the product's reducing behaviour — the reasoning in the next section.

The syllabus asks for reasoning, not a formula, so the key logic is worth stating plainly. A reducing sugar is one whose ring can pop open back into the free aldehyde (or ketone) form, because it still has an anomeric $-OH$ not tied up in a glycosidic bond; that open carbonyl is what reduces Tollens' reagent (silver mirror) or Fehling's solution (brick-red $\text{Cu}_2O$).

Why sucrose is non-reducing. Sucrose forms when $\alpha$-D-glucose (C-1) links to $\beta$-D-fructose (C-2) — an $\alpha,\beta$-1,2-glycosidic bond. Crucially, both partners contribute their anomeric carbon to this single bond, so neither ring has a free carbonyl left to open. No free carbonyl means no reduction, so sucrose is non-reducing and, unlike glucose or maltose, gives a negative Fehling's test.

Why maltose and lactose are reducing. Maltose joins two $\alpha$-D-glucose units through a C-1(first)–C-4(second) linkage: only the first unit's anomeric carbon is consumed, leaving the second unit's C-1 free to open. The same logic applies to lactose (glucose + galactose). One free anomeric carbon is enough for a positive test.

Why starch and cellulose behave so differently despite the same monomer. Both are polymers of D-glucose, but starch's amylose chain uses $\alpha$-1,4-linkages that coil into a helix digestible by human enzymes, while cellulose uses $\beta$-1,4-linkages that pack into rigid, hydrogen-bonded fibres. Human digestive enzymes are shaped to fit only the $\alpha$-linkage geometry, so we can metabolise starch but not cellulose — one stereochemical flip in the linkage separates "food" from "fibre."

Question: A student is given three white powders — sucrose, maltose, and starch — all giving a positive Molisch test (confirming they are carbohydrates). Using only the Fehling's test and the linkage logic above, work out how to identify each, and justify starch's result.

Step 1 — Predict sucrose. Sucrose has both anomeric carbons locked in its single $\alpha,\beta$-1,2-glycosidic bond (glucose C-1 to fructose C-2). No free carbonyl exists, so sucrose must give a negative Fehling's test.

Step 2 — Predict maltose. Maltose links two glucose units through only one anomeric carbon (C-1 of unit 1 to C-4 of unit 2); the second unit's C-1 stays free. So maltose must give a positive Fehling's test, just like glucose itself.

Step 3 — Predict starch. Starch (amylose plus amylopectin) is a long chain of many glucose units joined by $\alpha$-1,4- (and $\alpha$-1,6- at amylopectin branch points) linkages. Although one terminal glucose unit technically retains a free anomeric carbon, it is a vanishingly small fraction of the huge polymer's mass, so the reducing test is effectively negative in practice — polysaccharides are treated as non-reducing on this syllabus.

Step 4 — Assign the results. Running Fehling's test on all three: the sample giving a positive brick-red precipitate is maltose; the two giving a negative result are sucrose and starch, which can then be told apart because starch, unlike sucrose, gives a deep blue colour with iodine solution (a structural test, not a reducing-sugar test).

Answer: maltose gives a positive Fehling's test (free anomeric carbon); sucrose and starch both give a negative Fehling's test, and starch is distinguished from sucrose by the iodine colour test.

• Writing glucose only as an open chain and forgetting the cyclic pyranose form with its anomeric carbon — most board questions expect both representations.
• Confusing "reducing sugar" with "sweet sugar": reducing character depends purely on a free anomeric $-OH$, not on taste.
• Swapping $\alpha$ and $\beta$ glycosidic linkages between starch and cellulose — this single difference is why starch is digestible and cellulose is not, so it is a favourite exam trap.
• Assuming every monosaccharide is automatically a D-sugar "because it's a sugar" — D/L is a configurational label fixed by comparison with glyceraldehyde, not an inherent property of sugars.

Ask the tutor: Want to see how glucose's open chain actually folds into the pyranose ring, or trace exactly which bond breaks when maltose is hydrolysed back to two glucose units? Ask the AI tutor to walk through either mechanism step by step.