Can someone explain to me the precise mechanism of 1,2 cleavage by periodic acid? I understand that a cyclic periodite ester is formed, but when I count the oxygens, one seems to be missing.

 

 

HO OH

\ / HIO4 \

--C---C-- --------> 2x C=0

/ \ H2O, THF /

 

The periodite ester has four oxygens (two double bonded, two single) and one OH group - water must be involved, but I'm not sure how. I am assuming that in an h20 solution the periodite acid goes to IO4-, and that the H+ isn't playing a huge role. I have looked around on the net, but I haven't found anything that answers my curiosity very well - all the sites just gloss over the actual mechanism and say it's complicated.

oh wow, posting screwed up the orientation of my atoms and bonds. I suppose anyone who can answer this already knows what I'm talking about though, and doesn't need my silly diagrams.

The water plays an essential role.

 

HIO4 is metaperiodic acid. With water, orthoperiodic acid is formed

 

H2O + HIO4 <---> H3IO5

 

I'm not sure, it may even go further to H5IO6.

 

Now, you have multiple HO-.. groups per molecule of acid, which can form a bridging (cyclic) periodate ester with more than one ester-group.