Write the mechanism for this reaction:

enter image description here

I am trying to do a dehydration of an alcohol using the reagents sulfuric acid and acetic acid. I don't know how I am supposed to use the acetic acid, instead I've been using water. This is my proposed mechanism:

enter image description here

But it doesn't include the acetic acid, and it doesn't say in the question whether it is done in solution with water or not but I only know how to do it with $\ce{H2O}$. My main issue is that I need to put the acetic acid into the mechanism.

  • 5
    $\begingroup$ You can use $\ce{HSO_4^-}$ to pick up the proton in the bottom drawing. Going back to the original question I can't see what the role of AcOH is. I would have thought that $\ce{AcOH/H_2SO_4}$ is a very nice way to make an acetate (esterification) but if you plan to dehydrate then I am not sure what AcOH helps with. Not much help sorry, hopefully someone else could answer this $\endgroup$
    – K_P
    Nov 7, 2014 at 23:57

1 Answer 1


Your proposed mechanism looks reasonable. I think that acetic acid primarily serves as an acidic solvent for the reaction. However, in the presence of concentrated sulfuric acid, $\ce{CH3CO2H}$ can be protonated by $\ce{H2SO4}$, like in the first step of this esterification. The resulting resonance-stabilized cation is strongly acidic ($\mathrm{p}K_\mathrm{a} = -6.1$ according to this source), and should readily protonate the alcohol. This would be a possibility to include acetic acid into the mechanism.

$$\ce{CH3CO2H + H2SO4 <=> CH3C(OH)2^{+} + HSO4-}$$

enter image description here

(Image source: ChemGuide)

So in the first step, the alcohol can either be protonated by $\ce{H2SO4}$ or $\ce{CH3C(OH)2^{+}}$, and in the end, the eliminated proton can either be picked up by $\ce{HSO4-}$ or $\ce{H2O}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.