Ring forms, anomers and the reactions of glucose

A puzzle in the evidence

The last lesson ended with a puzzle. Glucose gives no Schiff's test and does not add sodium hydrogensulphite. Another puzzle: when glucose is crystallised from different solvents, we get two different crystals. One melts at 146 °C and the other at 150 °C. They also rotate polarised light by different amounts.

An open chain has no way to make two forms like this. A ring does.

The ring closes

The $\text{-}\text{OH}$ on C-5 lies close to the aldehyde carbon, C-1. The oxygen of that $\text{-}\text{OH}$ attacks the carbonyl carbon. The chain curls, and a six-membered ring forms. It holds five carbon atoms and one oxygen atom.

The new group on C-1 is both an alcohol and an ether. It is a hemiacetal. The new $\text{-}\text{OH}$ is called the anomeric hydroxyl group, and C-1 is the anomeric carbon.

A ring with five carbons and one oxygen looks like pyran, so the sugar form is a pyranose. A five-membered ring with four carbons and one oxygen looks like furan and is a furanose.

Two rings from one chain

When C-1 becomes a stereocentre, its $\text{-}\text{OH}$ can point either way. This gives two forms that differ only at C-1. They are anomers.

We draw the ring as a Haworth projection. The ring lies flat, seen from the side, with the ring oxygen at the back on the right. The thick edge is nearest to you. We read the groups from a Fischer projection like this:

• A group on the right in the Fischer projection points down in the Haworth ring.
• A group on the left points up.
• The $\text{-}\text{CH}_{2}\text{OH}$ group on C-5 points up in every D sugar.

Take D-glucose, with the pattern R L R R. C-2 is on the right, so its $\text{-}\text{OH}$ points down. C-3 is on the left, so its $\text{-}\text{OH}$ points up. C-4 is on the right, so its $\text{-}\text{OH}$ points down. $\text{-}\text{CH}_{2}\text{OH}$ is up.

The two anomers differ at C-1:

• In α-D-glucopyranose the $\text{-}\text{OH}$ on C-1 points down, on the opposite side from the $\text{-}\text{CH}_{2}\text{OH}$ group.
• In β-D-glucopyranose the $\text{-}\text{OH}$ on C-1 points up, on the same side as $\text{-}\text{CH}_{2}\text{OH}$.

Mutarotation

Pure α-D-glucose has a specific rotation of $+112.2$ degrees. Pure β-D-glucose has $+18.7$ degrees. Dissolve either one in water, and the rotation slowly changes until it reaches $+52.7$ degrees. This change is mutarotation.

The explanation is simple. In water each ring can open to the aldehyde form, and close again to either ring. The mixture settles to a fixed composition. We find the shares from the rotation, because the final value is a weighted mean:

This gives $x = 0.364$. So the equilibrium mixture is about 36% α and 64% β. The open-chain form is a very small trace, far below 1%. That explains the puzzle. Almost no free aldehyde is there, which is why the usual tests for an aldehyde fail.

Standard values and NCERT. NCERT prints $+111$, $+19.2$ and $+52.5$ degrees for these three rotations, with melting points of 419 K and 423 K. The values used here are the ones listed in the standard reference tables. The difference is about one degree and does not change any conclusion.

Why is β the main form? In the chair shape of the ring, every group of β-D-glucopyranose can sit in an equatorial position. The bulky groups are far from one another, so β is the more stable anomer.

Start with 100% α and read the first reading. Now start with 0% α. The curve climbs instead of falling. Both curves meet the same dashed line.

Practise with new numbers

Fructose, a ketose

Fructose is the ketohexose. It has the carbonyl group on C-2. It closes into a ring in the same way, but the $\text{-}\text{OH}$ on C-5 or on C-6 attacks C-2. Free fructose is mostly the six-membered pyranose ring in solution, with a good share of the five-membered furanose ring. In sucrose it is held in the furanose form.

Fructose is levorotatory. Its specific rotation at equilibrium is $-92.4$ degrees. It is the sweetest of the common sugars.

Reactions of glucose

The reactions follow from the structure. The table keeps them together.

• $\text{HI}$, heated for a long time: n-hexane. This shows six carbons in a straight chain.
• $(\text{CH}_{3}\text{CO})_{2}\text{O}$: glucose pentaacetate. This shows five $\text{-}\text{OH}$ groups.
• $\text{NH}_{2}\text{OH}$: glucose oxime. This shows a carbonyl group.
• $\text{HCN}$: a cyanohydrin. This shows a carbonyl group.
• Bromine water: gluconic acid. The aldehyde group is oxidised to $\text{-}\text{COOH}$.
• Dilute $\text{HNO}_{3}$: saccharic acid. Both ends are oxidised.
• Benedict's, Fehling's or Tollens' reagent: a red solid or a silver mirror. Glucose is a reducing sugar.
• Excess phenylhydrazine: glucosazone. Only C-1 and C-2 react.

Fructose also reduces Tollens' and Fehling's reagents, even though it is a ketose. In the alkaline medium of these tests, fructose rearranges through an enediol to glucose and mannose, which have aldehyde groups. So all common monosaccharides are reducing sugars.

Osazones. Three moles of phenylhydrazine react with one mole of an aldose or a ketose. The balanced equation is $\text{C}_{6}\text{H}_{12}\text{O}_{6} + 3\text{C}_{6}\text{H}_{8}\text{N}_{2} \to \text{C}_{18}\text{H}_{22}\text{N}_{4}\text{O}_{4} + \text{C}_{6}\text{H}_{7}\text{N} + \text{NH}_{3} + 2\text{H}_{2}\text{O}$. Only C-1 and C-2 change. So glucose, fructose and mannose, which have the same groups from C-3 to C-6, give the same osazone.

Fermentation. Yeast enzymes turn glucose into ethanol: $\text{C}_{6}\text{H}_{12}\text{O}_{6} \to 2\text{C}_{2}\text{H}_{5}\text{OH} + 2\text{CO}_{2}$.

Try it in Python

The program starts from pure α-glucose and follows the rotation hour by hour. The rate constant is made up. The final value is the real one.

Common mistakes

• Saying α and β glucose are different sugars. They are two forms of one sugar, joined by an open chain.
• Taking α to mean that the $\text{-}\text{OH}$ points up. In D sugars, α has the anomeric $\text{-}\text{OH}$ down in the Haworth ring.
• Thinking mutarotation changes D to L. Only C-1 changes. C-5 keeps its arrangement.
• Thinking a ketose cannot reduce Tollens' reagent. Fructose can, through the enediol.

Remember these

• The C-5 $\text{-}\text{OH}$ attacks C-1 and closes a six-membered pyranose ring. C-1 becomes the anomeric carbon, a hemiacetal.
• Right in the Fischer projection is down in the Haworth ring. Left is up. α has the anomeric $\text{-}\text{OH}$ down in a D sugar, and β has it up.
• Mutarotation: $+112.2$ (α) and $+18.7$ (β) both go to $+52.7$ degrees, about 36% α and 64% β.
• Glucose and fructose both reduce Tollens' and Fehling's reagents. Glucose, fructose and mannose give one osazone.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In D-glucose, the $\text{-}\text{OH}$ on C-4 is on the right in the Fischer projection. Does it point up or down in the Haworth ring?

2. A solution of pure β-D-glucose slowly changes its rotation. Does it go up or down, and to what value?

3. Why does glucose give a red precipitate with Fehling's solution, even though the open-chain form is such a small share?

4. Name the product when glucose reacts with bromine water, and say what that shows.

References

• Nelson and Cox, Lehninger Principles of Biochemistry, 8th edition, Chapter 7 (monosaccharides, anomers and mutarotation).
• NCERT Chemistry for Class 12, Part II, the chapter Biomolecules (structure of glucose, anomers, fructose).
• The rotations $+112.2$, $+18.7$ and $+52.7$ degrees are the standard tabulated values (for example in the CRC Handbook of Chemistry and Physics).