Disaccharides and glycosidic linkage

The general reaction. Every glycosidic linkage forms the same way: the anomeric –OH of one sugar (the hemiacetal –OH — C1 in an aldose like glucose, C2 in a ketose like fructose) condenses with an –OH group on a second sugar, releasing water.

where: Sugar₁–OH is the monosaccharide contributing its anomeric carbon, HO–Sugar₂ is the partner monosaccharide (which may contribute an ordinary –OH, or its own anomeric –OH), and the C–O–C bridge left behind is the glycosidic linkage. The two labels a disaccharide's name needs are (i) which carbons are joined, e.g. C1 to C4, and (ii) the anomeric configuration at the linking carbon, α or β.

Maltose = glucose + glucose, α-1,4-linked. The α-anomeric C1 of one glucose bonds to the (non-anomeric) C4–OH of a second glucose. That second glucose still has its own C1 free — a free anomeric carbon that can open to a reactive –CHO — so maltose is a reducing sugar and shows mutarotation. It is obtained by partial hydrolysis of starch.

Lactose = β-D-galactose + D-glucose, β-1,4-linked. Galactose's anomeric C1 (β-form) bonds to glucose's C4. The glucose unit's own C1 is untouched, so lactose too is a reducing sugar. It is the natural sugar of milk, present at about 4–5%.

Sucrose = α-D-glucose + β-D-fructose, 1,2-linked. Here the linkage runs between glucose's C1 and fructose's C2 — the anomeric carbon of both partners is consumed in forming the single bond. With no free anomeric carbon left on either ring, sucrose is a non-reducing sugar and shows no mutarotation. Acid or the enzyme invertase hydrolyses it to an equimolar glucose–fructose mixture called invert sugar.

Reducing power traces back to a free anomeric carbon that can open its ring and expose a reactive carbonyl (–CHO or C=O); that open form is what donates electrons to Fehling's or Tollens' reagent, giving the colour changes used as the test.

Step 1. In maltose the glycosidic bond consumes only one anomeric carbon — C1(α) of the first glucose, which links to C4 (an ordinary, non-anomeric position) of the second glucose. The second glucose's own C1 is never touched by the linkage, so it can still open to a free aldehyde. That free end gives maltose a positive Fehling's/Tollens' test and lets it mutarotate.

Step 2. Lactose works identically in structural logic, even though the two rings are different sugars: galactose donates its anomeric C1 to the bond, but glucose's C4 (not its C1) receives it. Glucose's own anomeric carbon is left free, so lactose is reducing for exactly the same reason as maltose — one ring keeps an open door.

Step 3. Sucrose is built differently: the bond forms between glucose's C1 and fructose's C2, so both partners spend their only anomeric carbon on the same linkage. There is no third carbon left over to open — neither ring has an unused anomeric position, so neither can generate a free carbonyl.

Step 4. With no anomeric carbon free on either half of sucrose, it cannot reduce Fehling's or Tollens' reagent, and it cannot mutarotate (mutarotation itself requires a ring that can open and re-close at a free anomeric centre). This single structural fact — one anomeric carbon consumed versus two — is the entire reason maltose and lactose are reducing sugars while sucrose is not.

Question. Sucrose is dextrorotatory, $[\alpha]_D = +66.5°$. On hydrolysis it gives an equimolar mixture of D-glucose ($[\alpha]_D = +52.5°$) and D-fructose ($[\alpha]_D = -92°$). (a) Explain why the mixture's rotation is called "inverted," and (b) estimate the net specific rotation of the 1:1 mixture.

Step 1. Write the hydrolysis: $\text{Sucrose} + \text{H}_2\text{O} \xrightarrow{\text{H}^+ / \text{invertase}} \text{D-glucose} + \text{D-fructose}$. One mole of sucrose gives one mole each of glucose and fructose, so the product is a 1:1 (equimolar) mixture.

Step 2. For an equimolar mixture, the net optical rotation is approximately the average of the two components' specific rotations (equal moles contribute equally): $[\alpha]_{\text{mix}} \approx \dfrac{(+52.5°) + (-92°)}{2} = \dfrac{-39.5°}{2} = -19.75°$.

Step 3. Compare signs: sucrose itself is dextrorotatory ($+66.5°$), but the hydrolysis product is laevorotatory ($-19.75°$, negative) because fructose's laevorotation ($-92°$) outweighs glucose's dextrorotation ($+52.5°$) in magnitude. The sign of rotation has flipped from $+$ to $-$.

Step 4. This sign flip on hydrolysis is called inversion, which is why the acid- or invertase-hydrolysed 1:1 glucose–fructose mixture is named invert sugar — sweeter than sucrose and a natural component of honey. Answer: the mixture is laevorotatory with net specific rotation ≈ −19.75°, opposite in sign to sucrose's +66.5°, which is why it is called invert sugar.

• Assuming every disaccharide is non-reducing because sucrose (the everyday example) is — maltose and lactose are both reducing; sucrose is the exception among these three, not the rule.
• Mixing up the linkage numbering: maltose and lactose are both 1,4-linked (C1 of one ring to C4 of the other), while sucrose is 1,2-linked, precisely because both its anomeric carbons (C1 of glucose, C2 of fructose) must be used.
• Writing lactose as glucose + glucose — it is glucose + galactose; glucose + glucose is maltose.
• Treating invert sugar as some new compound rather than simply the equimolar glucose + fructose mixture, named for the sign reversal of optical rotation on hydrolysis.

Ask the tutor: want to see how the same anomeric-carbon logic extends to polysaccharides like starch and cellulose?