I was reading my textbook and this website. I just cannot find the mechanism for what happens when we go from CH3CH2OSO3H to ethanol by the addition of water and heat.

I understand the electrophillic addition part.

But, what happens next? Is there some kind of Sn2 reaction to make HSO4- the leaving group? Then, why does the EtOH2+ molecule give out one of its acidic protons to HSO4- and not to H2O which is a more powerful base?

I hate when my textbook leaves out the details and I obsess on them :(


Yes, I believe you are correct. Water attacks at the sulfate containing carbon by an SN2 mechanism. You are also right that sulfuric acid is a stronger acid than hydronium (or protonated ethanol), which means that in solution the bisulfate anion will predominate. Often people (including textbook authors) will draw a mechanism that goes to neutral products. Most likely water will carry out the final deprotonation (especially given that water is the solvent).

Incidentally, most likely this reaction is self catalyzed, as one sulfate protonates another, activating the sulfate for SN2 reaction by a weak nucleophile.

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  • $\begingroup$ Great answer. Small questions: we said HSO4- won't be taking the final proton from protic ethanol. Did you draw it like that just to point out the fact that most textbook authors try to put neutral products? Then, in that catalyzed mechanism, why does the sulphate part NEED to get a proton before acting as a leaving group? $\endgroup$ – yolo123 Aug 14 '14 at 15:17
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    $\begingroup$ Yes, I drew the protonation of bisulfate just to show how you would get to the indicated products. In the catalyzed version, the extra proton acts to make the sulfate into a better leaving group. Water displacing the protonated sulfate will be faster than the neutral sulfate. You will see this regularly in organic chemistry: when there is a weak nucleophile (like water and many neutrals) the elecrophile will be activated by protonation. $\endgroup$ – jerepierre Aug 14 '14 at 19:52

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