Glycoside hydrolysis: practice and mechanism
Acid cleaves the glycoside (an acetal) at the anomeric position: methanol is released and the sugar remains as the cyclic hemiacetal.
Example
Glycosides are acetals: stable to base, but cleaved in aqueous acid. The hemiacetal (the free sugar) forms again and methanol is released. In water the α/β equilibrium is then established.
Mechanism step by step
- 1
The acid protonates the OCH₃ oxygen on the anomeric carbon.
- 2
Methanol leaves, assisted by the ring O: the oxocarbenium ion forms again.
- 3
Water attacks; after deprotonation the free sugar is present as the hemiacetal (α/β mixture).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The open-chain aldehyde form exists in equilibrium, but only to a tiny extent. In water sugars are almost entirely present as the ring (hemiacetal).
- Dilute acid is not an oxidant. The carboxylic acid (aldonic acid) would need e.g. bromine water.
- Without a reducing agent (e.g. NaBH₄) no sugar alcohol forms. Hydrolysis only cleaves the acetal.
Often confused with: Acetal hydrolysis · Ether cleavage with HI
Frequently asked questions
Why are glycosides stable to base?
Like all acetals they need protonation to create a leaving group. Hydroxide cannot displace CH₃O⁻.
How does the body cleave glycosides?
With enzymes (glycosidases), e.g. amylase for starch. They are specific for α or β, which is why humans can digest starch but not cellulose.