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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.

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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:

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  • $\begingroup$ 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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