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: Evidence for a cyclic form.
Glucose sometimes behaves as if it has no free -CHO group. It gives two anomers with different melting points and specific rotations, and a fresh solution changes its optical rotation until it settles at a fixed value. Such behaviour is only possible if the aldehyde is locked inside a ring for most of the time.
Step 2: The pyranose ring (Haworth picture).
The ring is drawn as a flat six-membered Haworth ring. The -OH on the fifth carbon adds across the C-1 carbonyl to make a hemiacetal, so C-1 and C-5 become joined through an oxygen bridge. Because the ring resembles pyran, the form is named glucopyranose.
Step 3: Alpha and beta anomers.
The carbon that was the aldehyde (C-1) is now bonded to -H, -OH, the ring oxygen and the rest of the chain, making it a fresh stereocentre. When its -OH lies below the ring plane we get alpha-D-glucose; when it lies above we get beta-D-glucose. Switching between them through the tiny amount of open chain form is mutarotation.
Step 4: Bromine water test.
Bromine water performs a gentle oxidation. Only the terminal aldehyde is affected and becomes a carboxylic acid, so the product is the six-carbon gluconic acid, CH₂OH(CHOH)₄COOH. Since ketones are untouched by bromine water, this proves the group present is an aldehyde.
Step 5: Hydroxylamine test.
Hydroxylamine supplies an -NH₂ that condenses with the carbonyl carbon, expelling water and building a C=N-OH linkage. The product is glucose oxime. Formation of an oxime is a standard proof that a free carbonyl group is available in the open chain form.