Structures of glucose and fructose

Glucose is an aldohexose. Its open-chain (Fischer) structure is a straight six-carbon chain with an aldehyde at one end:

Here C1 is the –CHO carbon, C2 through C5 each carry an –OH fixed in the D-glucose stereochemistry, and C6 is the terminal –CH₂OH.

Fructose is a ketohexose. Its open-chain structure places the carbonyl one carbon in from the end:

Here C2 carries the ketone, C1 and C6 are both –CH₂OH, and C3, C4, C5 are the three chiral –CHOH centres. Neither formula was written down by guesswork — each detail traces to a specific reaction:

• HI on prolonged heating reduces every –OH and the carbonyl away, and both sugars collapse to the same product, n-hexane. This proves both have an unbranched six-carbon backbone.
• Hydroxylamine ($\text{NH}_2\text{OH}$) reacts with a carbonyl to form an oxime. Both sugars form a mono-oxime, so each has exactly one carbonyl group — no more, no less.
• Acetic anhydride acetylates every free –OH. Both give a penta-acetate, so each open-chain form carries exactly five free hydroxyls.
• Bromine water, a mild oxidant, oxidises only an aldehyde, not a ketone. It attacks glucose but leaves fructose untouched — the reaction that tells an aldose from a ketose (Worked example below).
• Dilute nitric acid, a stronger oxidant, attacks glucose at both ends, turning –CHO and the far –CH₂OH into –COOH together, giving the dicarboxylic saccharic acid. This locates the terminal –CH₂OH directly opposite the –CHO.
• Excess phenylhydrazine consumes the carbonyl carbon and its neighbouring –CHOH to build an osazone. Glucose and fructose give the same osazone, proving they share identical configuration from C3 to C6 and differ only at C1–C2.

The open-chain picture above explains five of the six reactions cleanly, but it cannot explain two stubborn observations: glucose gives only a weak, inconsistent Schiff's test despite owning a free aldehyde, and pure glucose can be crystallised in two distinct forms — one melting at 146 degrees Celsius with specific rotation $+112°$, the other at 150 degrees Celsius with specific rotation $+19°$ — that slowly drift to the same equilibrium value of $+52.7°$ when dissolved in water. That drift is called mutarotation, and no straight-chain structure with a fixed, permanent aldehyde group can produce two stable, interconvertible forms with different optical rotations. The resolution is that the open-chain form is not the resting structure at all. Glucose's own C5 hydroxyl is positioned to attack its own C1 carbonyl intramolecularly, and an –OH attacking a carbonyl carbon is exactly the reaction that builds a hemiacetal. That internal attack closes a six-membered ring (the pyranose form) and, critically, creates a brand-new stereocentre at C1 — the anomeric carbon — because the incoming –OH can end up on either face. The two outcomes are the α- and β-anomers, and they are precisely the two crystalline forms observed: different compounds by rotation and melting point, yet freely interconverting in solution by briefly reopening to the open-chain form and reclosing. That equilibrium reopening also explains the weak Schiff's test — at any instant only a small fraction of glucose molecules sit in the reactive open-chain form, so the aldehyde signal is diluted rather than absent. Fructose undergoes the same logic with its own geometry: the C5 hydroxyl attacks the C2 ketone, and because that spans one fewer atom, the ring that closes is five-membered (the furanose form), again with a new anomeric centre at C2. So the cyclic hemiacetal is not a separate, competing idea from the open-chain structure — it is the same molecule folding on itself, and it is the only structure consistent with everything the six reagents and the mutarotation data show together.

Question. Bromine water is added separately to solutions of glucose and fructose. Explain what is observed in each case and state precisely what this establishes.

Step 1. Identify what bromine water can and cannot do. It is a mild oxidising agent, strong enough to convert an aldehyde (–CHO) into a carboxylic acid (–COOH), but too weak to touch a ketone (>C=O), since oxidising a ketone would require breaking a carbon–carbon bond rather than just adding oxygen at an existing terminal carbon.

Step 2. Apply this to glucose. Glucose's C1 is an aldehyde, so bromine water oxidises it smoothly to gluconic acid, $\text{COOH–(CHOH)}_4\text{–CH}_2\text{OH}$. The bromine water is consumed and its orange colour decolourises — a clear positive result.

Step 3. Apply this to fructose. Fructose's carbonyl sits at C2 as a ketone, which bromine water cannot oxidise. The colour of the bromine water persists — no reaction.

Step 4. State the conclusion. The contrasting results establish unambiguously that glucose's carbonyl is an aldehyde and fructose's is a ketone. Answer: glucose decolourises bromine water, forming gluconic acid, while fructose gives no reaction — the standard test distinguishing an aldose from a ketose.

Competitive stretch

If instead excess phenylhydrazine is used on both sugars, they give the identical osazone, which looks like it contradicts the bromine-water result. It does not: osazone formation only consumes C1 and C2 together (the carbonyl carbon plus the adjacent –CHOH), so it is blind to whether the carbonyl was originally an aldehyde or a ketone — it merely shows C3 to C6 are configured identically in both sugars. Bromine water, by contrast, probes only the carbonyl itself. The two reactions are testing different parts of the molecule, so both can be true at once.

• Writing fructose's ketone at C1 — fructose is specifically a 2-ketohexose; the carbonyl sits at C2, one carbon in from the terminal –CH₂OH.
• Concluding from the shared osazone that glucose and fructose are the "same" molecule — the osazone reaction only proves C3–C6 match; C1–C2 differ, which is exactly why bromine water and hydroxylamine (mono-oxime, but at different carbons) still tell the sugars apart.
• Treating the cyclic structure as optional extra detail rather than a requirement — it exists specifically because the open-chain form cannot explain mutarotation or the weak Schiff's test; always cite one of those two observations as the reason a ring is needed.
• Confusing the two oxidants: bromine water (mild) oxidises only –CHO and distinguishes aldose from ketose; dilute nitric acid (strong) oxidises both the –CHO and the terminal –CH₂OH of glucose to the dicarboxylic saccharic acid.

Ask the tutor: want a step-by-step walkthrough of how the α- and β-anomers actually differ in 3D, or why mutarotation stops changing once it reaches equilibrium?