# Why glucose and other similar carbohydrates are oxidised fully by HIO4?

$$\ce{HIO4_{(aq)}}$$ requires the vicinal diols to have syn configuration, as it forms a cyclic periodate ester.

But glucose does not have syn configuration at C3, and if you consider cyclic forms of glucose it shouldn't oxidise fully at all because alternate hydroxy groups with anti configuration will be unreactive. But it still gets oxidised by $$\ce{HIO4}$$ to give 5 molecules of methanoic acid, $$\ce{HCOOH}$$, and one molecule of formaldehyde, $$\ce{HCHO}$$.

• I googled "syn anti configuration" and it said "did you mean syn anti conformation". So the search engine has a good point: In the linear form, you can get a conformation where the hydroxyls point the same way. – Karsten Theis Dec 20 '19 at 16:50
• "Anti" looks impossible in a Haworth projection. Use a chair conformation of glucose. Anti is slower than syn both with an ~60 deg. dihedral angle. HIO4 can catalyze beta to alpha. Once syn C1-C2 is cleaved in alpha-D-glucose, the ring is gone and there is no anti/syn issue. Or it may all happen in the open form. Look at my answer here: chemistry.stackexchange.com/questions/270/… – user55119 Dec 20 '19 at 21:31

Glucose can react with $$\ce{HIO4}$$, which can be understood easily by looking at chair conformation of glucose.