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MattC

1,2 diol cleavage mechanism?

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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.

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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.

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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.

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