Permethylation: practice and mechanism
CH₃I and Ag₂O methylate every OH group to the methyl ether. All OH become OCH₃, the anomeric one becoming a methyl glycoside.
Example
Ag₂O makes the OH groups more nucleophilic and captures iodide; CH₃I then methylates every OH group to the methyl ether (Williamson-like). At the anomeric carbon a methyl glycoside forms. Historically important for sugar structure determination.
Mechanism step by step
- 1
Ag₂O acts as a mild base and halide scavenger; the OH group attacks CH₃I.
- 2
Iodide leaves and precipitates as AgI, which drives the reaction.
- 3
With excess CH₃I every OH group becomes a methyl ether, the anomeric one a methyl glycoside.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which reagent achieves this transformation?
Typical mistakes
- With excess CH₃I not just the primary but every OH group is methylated.
Often confused with: Glycoside formation (Fischer glycosylation) · Peracetylation
Frequently asked questions
How was this used to determine ring size?
Permethylate, then cleave the glycoside with acid (only the anomeric OCH₃ comes off). The one free OH left besides C1 shows which O was in the ring (C5 → pyranose, C4 → furanose).
Why not NaOH and CH₃I?
Strong base would degrade the sugar via the open form (enolisation, rearrangements). Ag₂O is mild enough. Today NaH/CH₃I in DMSO is also used.