# Dehydration of secondary alcohol via E1 mechanism

Write the mechanism for this reaction:

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:

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.

• 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 – K_P Nov 7 '14 at 23:57

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.
$$\ce{CH3CO2H + H2SO4 <=> CH3C(OH)2^{+} + HSO4-}$$
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 $\ce{H2O}$.