I have a problem removing the benzoyl groups of my glucose moiety using sodium methylate (standard procedure). Indeed, the benzoyl groups are removed. However, a significant amount of glucose gets hydrolysed leaving an alcohol group. I have tried different equivalents of sodium methylate and also other bases like potassium carbonate and ammonia. However, in all cases hydrolysis takes place.


Does anyone have an idea why hydrolysis takes place? Could it have something to do with the TMSOTf (strong acid) in the previous step? The only workup I did for compound 1 was quenching the reaction using the same molar amount of TEA as TMSOTf, then evaporating $\ce{CHCl3}$, then silica column chromatography ($\ce{CHCl3 : MeOH}$ in $10 : 1$, see picture). Is it possible that TMSOTf is still present in my compound?

Column of compound 1: Column after TMSOTf reaction:

  • $\begingroup$ The mildest conditions I know of for ester hydrolysis is LiOH in 19/1 THF/water at room temp. Could be worth trying. $\endgroup$ – Waylander Dec 11 '17 at 13:57
  • $\begingroup$ I very much doubt TMSOTf would survive the presence of the MeOH in your eluent $\endgroup$ – Waylander Dec 11 '17 at 16:35
  • $\begingroup$ What's the water content of the methanol? $\endgroup$ – Beerhunter Dec 11 '17 at 21:37
  • $\begingroup$ How do you know that the first step was complete. Was the benzoylglucoside pure? $\endgroup$ – user55119 Dec 11 '17 at 23:57
  • $\begingroup$ MeOH is of AR quality. $\endgroup$ – John Dec 12 '17 at 9:51

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