Glycoside formation (Fischer glycosylation): practice and mechanism
Only the anomeric OH group (on the carbon bearing two oxygens) is swapped for OCH₃. It is special because the ring O stabilises the cation at this carbon. The other OH groups remain.
Example
Only the OH group on the anomeric carbon (the hemiacetal carbon bearing two oxygens) is exchanged with methanol and acid. Via the oxocarbenium ion a full acetal forms, the glycoside. All other OH groups stay unchanged.
Mechanism step by step
- 1
The acid protonates the anomeric OH group and water leaves.
- 2
An oxocarbenium ion forms: the ring O stabilises the positive charge on the anomeric carbon with its lone pair.
- 3
Methanol attacks the cation from above or below; after deprotonation the methyl glycoside forms, mostly as the α anomer (anomeric effect).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Primary OH groups are not etherified by methanol and acid. Only at the anomeric carbon does a cation form (oxocarbenium ion) that methanol can attack.
- Ordinary secondary OH groups stay unchanged; a cation forms only at the anomeric carbon. Methylating every OH group would need CH₃I and Ag₂O.
- The ring does not open to the open-chain dimethyl acetal. The ring O stays bonded and only the anomeric OH is replaced: the cyclic glycoside forms.
Often confused with: Acetal formation (protecting group) · Permethylation
Frequently asked questions
Why is the glycoside an acetal?
The anomeric carbon now carries two O-alkyl groups (ring O and OCH₃). It is therefore no longer reducing and does not mutarotate.
Where do glycosidic bonds occur?
In all di- and polysaccharides: sucrose, lactose, starch (α-1,4) and cellulose (β-1,4). There the second "alcohol" molecule is another sugar.