Diol cleavage with periodate: practice and mechanism
NaIO₄ cleaves the C–C bond between the two OH groups. Each carbon becomes a carbonyl group: CH(OH) gives an aldehyde, C(OH) with two groups a ketone.
Example
Periodate forms a cyclic ester with the 1,2-diol and cleaves the C–C bond between the two OH groups. Each of the two carbons becomes a carbonyl group (aldehyde or ketone), the same result as ozonolysis of the matching alkene.
Mechanism step by step
- 1
Periodate (IO₄⁻) binds to both OH groups of the 1,2-diol and forms a cyclic periodate ester.
- 2
The five-membered ring falls apart: the C–C bond breaks and both C–O bonds become C=O.
- 3
Iodine is reduced from +7 to +5 (iodate) in the process; two carbonyl compounds remain.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Periodate does not oxidise further to the carboxylic acid. Cleavage gives aldehydes or ketones, like ozonolysis with reductive workup.
- No epoxide forms. Periodate forms a cyclic ester with the diol, and that ester falls apart by cleaving the C–C bond.
- Oxidising just one OH group would be e.g. PCC. Periodate engages both OH groups at once and breaks the C–C bond between them.
- Periodate does not remove the oxygens. It is an oxidant: the two carbons become C=O, not a C=C bond.
Often confused with: Ozonolysis (reductive work-up) · Pinacol rearrangement
Frequently asked questions
Why only 1,2-diols?
The cyclic ester needs two adjacent OH groups. 1,3-diols cannot form a suitable ring and are not cleaved.
How does this relate to ozonolysis?
OsO₄ followed by NaIO₄ gives the same as ozonolysis with reductive workup: first the cis diol, then cleavage.