Baeyer–Villiger oxidation: practice and mechanism
The peracid inserts an O atom between the carbonyl carbon and one group: the ketone becomes an ester. The group that better stabilises positive charge migrates (tertiary > secondary ≈ phenyl > primary > methyl).
Example
The peracid adds to the ketone, then one group migrates from the carbonyl carbon to the oxygen: an ester forms (a lactone from cyclic ketones). The group that best carries a positive charge migrates: tertiary > secondary ≈ phenyl > primary > methyl.
Mechanism step by step
- 1
The peracid attacks the carbonyl carbon (often after protonation): a tetrahedral Criegee intermediate forms.
- 2
One group moves from the carbon to the neighbouring O while the carboxylate leaves.
- 3
Deprotonation gives the ester (a lactone from cyclic ketones).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The methyl group is the worst migrator. The oxygen is inserted on the side of the more substituted group.
Often confused with: Epoxidation
Frequently asked questions
Which group migrates?
The one that best stabilises positive charge: tertiary > secondary ≈ phenyl > primary > methyl. With aldehydes H migrates and the carboxylic acid forms.
Why does the double bond not react?
With isolated alkenes it often does (epoxidation). For ketones without C=C only the Baeyer–Villiger reaction remains.