Question

Explain the ring structure of glucose. What happens when D-glucose reacts with the following reagents? (i) Bromine water (ii) Hydroxylamine.

Hint:

Picture the C-5 -OH closing onto the C-1 -CHO to give a pyranose hemiacetal with alpha and beta anomers; bromine water gives gluconic acid and hydroxylamine gives the oxime.

Answer 1

Step 1: Why glucose needs a ring structure.
The open chain formula of glucose is an aldohexose, CH₂OH(CHOH)₄CHO. A few facts are not explained by this open chain: glucose does not give the Schiff test, it does not form the hydrogensulphite addition product with NaHSO₃, and it exists in two crystalline forms whose rotation slowly changes in water (mutarotation).
Step 2: Formation of the ring.
The oxygen of the C-5 hydroxyl group attacks the C-1 aldehyde carbon within the same molecule. This intramolecular addition forms a cyclic hemiacetal and a stable six-membered ring that contains one oxygen atom. This six-membered ring is called the pyranose form (glucopyranose).
Step 3: The anomeric carbon.
During ring closure C-1 becomes a new chiral centre called the anomeric carbon. The new -OH at C-1 can point below the ring (alpha form) or above it (beta form), giving alpha-D-glucose and beta-D-glucose, which are anomers. Their interconversion in solution explains mutarotation.
Step 4: (i) Reaction with bromine water.
Bromine water is a mild oxidising agent. It oxidises only the -CHO group to -COOH, giving gluconic acid, which confirms an aldehyde group.
CH₂OH(CHOH)₄CHO + Br₂ + H₂O → CH₂OH(CHOH)₄COOH + 2HBr
Step 5: (ii) Reaction with hydroxylamine.
The -CHO group reacts with hydroxylamine (NH₂OH) to form glucose oxime with loss of water, confirming a carbonyl (aldehyde) group.
CH₂OH(CHOH)₄CHO + NH₂OH → CH₂OH(CHOH)₄CH=NOH + H₂O