I came across the reaction of dimethyl sulfate with water to yield methanol. I was curious how that works as $\ce{H2O}$ isn't a strong nucleophile.

From my understanding the $\ce{H2O}$ should pop off a $\ce{-CH3}$ and you have $\ce{[CH3OH2]+}$ which has a $\mathrm{p}K_\mathrm{a}$ of -2.5 so a relatively strong base. Does the $\ce{[CH3OH2]+}$ protonate the $\ce{H2O}$ to form $\ce{H3O+}$ and $\ce{CH3OH}$?

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    $\begingroup$ @Joel There's no redox reaction, also I wouldn't be so sure with this mechanism, please be more careful with your posts. $\endgroup$
    – Mithoron
    Commented Nov 30, 2015 at 1:20
  • $\begingroup$ @Mithoron , I'll be more careful thank you I should have checked. Apparently the DMS is hydrolyzed first to methyl sulfuric acid then to methanol and sulfuric acid in a two step process occurring slowly in cold water and rapidly in warm water or acidic solution. (Ref:Chemours/dimethyl sulfate/chemical properties) $\endgroup$
    – Technetium
    Commented Nov 30, 2015 at 1:30

1 Answer 1


The mechanism of methyl transfer of $\ce{Me2SO4}$ is generally thought to be an $\mathrm{S_N2}$ reaction. Thus, water would attack the methyl group, releasing $\ce{MeO-SO3-}$, a good leaving group generating methanolium $\ce{H3C-OH2+}$. The proton is later transferred back to a methyl sulphate to give methyl hydrogen sulphate $\ce{HSO4Me}$ and methanol.

One could imagine the proton from methanolium to initially be transferred to water to generate hydronium. That’s actually pretty probable and it will be hard to tell where it will end up. Water, methanol and hydrogen sulphate (and therefore likely methyl sulphate, too) all have similar $\mathrm{p}K_\mathrm{b}$ values and are equally eager to displace it/non-eager to accept it. adding some base, e.g. sodium carbonate should provide a good ‘resting point’ for such an unhappy proton.

The question is why water can do this even though it is a lousy nucleophile. Well, it turns out that for one, $\ce{MeSO4-}$ is a great leaving group enhancing the reaction rate and for two, there is nothing else around if you mix just $\ce{Me2SO4}$ and water that would be a better nucleophile. ($\ce{OH-}$ is present in too low concentrations to actually count.) Thus, the two reagents have the choice of doing nothing or slowly doing something thermodynamically favourable (= reacting). Kinetics dictate that it must happen slowly, but it can happen.


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