Sodium periodate (NaIO₄) is astrong oxidizing agent mainly used for theoxidative cleavage of 1,2-diols (vicinal diols),forming aldehydesand ketones depending on the structure of the alcohol.

So, there are two things happening here: 1) the OH group is oxidized to a carbonyl, and 2) the C-C bond with the oxygens is cleaved. And the pattern is that a primary OH group gives formaldehyde, secondary OH groups produce a ketone, while a tertiary alcohol results in a ketone.

The Mechanism of NaIO₄ Oxidative Cleavage

The reaction starts with a nucleophilic addition of the OH groups to the iodine, forming a cyclic iodate ether, which eventually, after proton transfer steps, is cleaved into two carbonyl groups:

An indirect confirmation of the mechanism going via a five-membered intermediate is the selective oxidation of thecis-decalindiol. Thetrans isomer is not oxidized by HIO₄ because the opposite orientation of the OH groups does not allow for the formation of a five-membered cyclic periodate:

Although the mechanisms are not entirely identical, this is somewhat similar to the oxidative cleavage of alkenes withpotassium permanganate (KMnO₄) orozonolysis followed by treatment with hydrogen peroxide.

However, the difference is that KMnO₄ oxidizes the resulting aldehydes and/or ketones further to carboxylic acids:

There is a name reaction called The Lemieux–Johnson or Malaprade–Lemieux–Johnsonoxidation,which consists of two steps of syn dihydroxylation of alkenes, followed by oxidative cleavage of the diol with NaIO₄:

Remember, in thesyn dihydroxylation of alkenes with OsO₄,NMO (N-methylmorpholine N-oxide) is often added to the reaction, and it deoxidizes the Os⁶⁺ species back into OsO_4, which can perform another oxidation of the alkene:

Check this66-question,Multiple-ChoiceQuiz with a 2-hour Video Solutionon the naming, preparation, and reactions of Alcohols.

Free

Alcohols Quiz – Naming, Preparation, and Reactions

Take Now

Check Also

• Ozonolysis of Alkenes with Practice Problems
• Syn Dihydroxylation of Alkenes with KMnO4 and OsO4
• Anti-Dihydroxylation of Alkenes with MCPBA and Other Peroxides with Practice Problems

Reactions of Alcohols

• Nomenclature of Alcohols: Naming Alcohols based on IUPAC Rules with Practice Problems
• Preparation of Alcohols via Substitution or Addition Reactions
• Reaction of Alcohols with HCl, HBr, and HI Acids
• Mesylates and Tosylates as Good Leaving Groups
• SOCl₂ and PBr₃ for Conversion of Alcohols to Alkyl Halides
• Alcohols in Substitution Reactions  Practice Problems
• POCl₃ for Dehydration of Alcohols
• Dehydration of Alcohols by E1 and E2 Elimination
• The Oxidation States of Organic Compounds
• LiAlH4 and NaBH4 Carbonyl Reduction Mechanism
• Alcohols from Carbonyl Reductions – Practice Problems
• Grignard Reaction in Preparing Alcohols with Practice Problems
• Grignard Reaction in Organic Synthesis with Practice Problems
• Protecting Groups For Alcohols in Organic Synthesis
• Oxidation of Alcohols: PCC, PDC, CrO₃, DMP, Swern, and All of That
• Diols: Nomenclature, Preparation, and Reactions
• The Pinacol Rearrangement
• The Williamson Ether Synthesis
• Alcohol Reactions Practice Problems
• Naming Thiols and Sulfides
• Reactions of Thiols
• Alcohols Quiz – Naming, Preparation, and Reactions
• Reactions Map of Alcohols

4 thoughts on “NaIO4 Oxidative Cleavage of Diols”

1. Is thenPotassium permanganate equal to NAIO4
   In cleavage of Diols ?

   Reply

   • Potassium permanganate can oxidize a double bond to a diol in basic media and at lower temperatures. In acidic conditions and elevated temperatures, it cleaves the C-C bond and further oxidizes the C-O bonds to a ketone and carboxylic acid.
     NaIO4 doesn’t usually oxidize a double bond, at least in undergraduate textbooks. It is gentler with diols too as it stops their oxidation at an aldehyde and a ketone.

     Reply

2. Hi, nice demo drawings!
   Stil I have to point out that the writing doesn’t show clearly that IO4(-) turns into IO3(-).
   Cleaning C-C an oxydise 2 OH into C=O (ketons and/or aldehydes means oxydation and concomitant reduction hidden by (NaO-)I(-OH)2=O because discretly = NaOI(=O)2 +H2O (NaIO3)

   Reply

   • Not “cleaning” but instead “cutting” damn IA corrector.

     Reply