Ethers on hydrolysis with dilute $\ce{H2SO4}$ under pressure produce alcohols. This is basically a type of acidic cleavage of ethers by the SN2 mechanism.

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There is inversion of configuration. Now if $\ce{Nu}$ is $\ce{OH-}$ its called hydrolysis of ether.

Is it right, that for SN2 the leaving group must be more stable than the approaching group? In case of ether hydrolysis the leaving alkoxy group is less stable than the entering $\ce{OH-}$ group.

So how does the reaction succeed? Can someone please explain?

  • 3
    $\begingroup$ The leaving group is not $\ce{RO^{-}}$, but rather $\ce{ROH}$. The reaction is run in the presence of acid, so the first step is protonation of the ether oxygen. Neutral alcohol is a much better leaving group than alkoxide. BTW, depending on the ether substituents the reaction can proceed by SN1 or SN2. $\endgroup$
    – ron
    Apr 28, 2015 at 13:39

1 Answer 1


While the mechanism may look simple in the way you described it there, it doesn’t work exactly like that.

First, you are in the presence of (diluted) $\ce{H2SO4}$ so aside from extremely short-lived accidental self-deprotonation of water, $\ce{OH-}$ ions will not be around. Therefore, $\ce{OH-}$ cannot act as an attacking group, whatever the mechanism.

Second, an alcoholate is one of the worst leaving groups in $\mathrm{S_N}$ reactions one can imagine. The only case where you will frequently see leaving alcoholates is for epoxide openings. But that’s not a problem here, we have $\ce{H2SO4}$. We can protonate the ether oxygen so that the leaving group will now be an alcohol.

Third, since $\ce{OH-}$ cannot be the attacking species, we need to choose from the nucleophiles present. Either $\ce{HSO4-/SO4^2-}$ or $\ce{H2O}$ can attack the $\alpha$ carbon (in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ manner). In the former case, a later attack of $\ce{H2O}$ to liberate the sulphate needs to be assumed. Both attacks work, but neither nucleophile is strong, so the driving force must be the liberated alcohol.

Fourth, the attacking $\ce{H2O}$ will create a protonated alcohol. This needs to deprotonate to give the free alcohol. Thereby the proton we used in second (step 1 of the mechanism, if you wish) is reliberated and therefore the reaction is acid-catalysed.

Finally, one needs to consider the possibility of equilibrium. The reaction can proceed in both directions. Usually by the choice of solvent one can direct it into one or the other. Hydrolysis is usually achieved with a large surplus of water, driving the reaction towards the alcohols by Le Chatelier’s principle. The reverse reaction would be triggered by removal of water or high concentrations of either alcohol.


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