# Halogenation of alcohols in the presence of ZnCl2

I know that $\ce{OH-}$ is usually a bad leaving group, but why does the following reaction occur via SN1 in presence of anhydrous $\ce{ZnCl2}$?

• Is $\ce{HCl}$ really part of your reaction equation? Because if it is you have no problem with $\ce{HO-}$ as leaving group as $\ce{HCl}$ would protonate the hydroxyl group converting it into a good leaving group. – Philipp Aug 24 '13 at 17:32
• I guess I'll make an answer out of my comment then :) – Philipp Aug 24 '13 at 19:31

You are right, $\ce{OH-}$ is a bad leaving group. But the $\ce{HCl}$ in the reaction mixture protonates it. Thus, the bad leaving group $\ce{OH-}$ is converted to the good leaving group $\ce{H2O}$. Anhydrous $\ce{ZnCl2}$ might additionally facilitate the reaction by acting as a Lewis acid and it binds the expelled water thus preventing the back reaction.
• @GregE. Oops, sorry. You are completely right. I answered a question about $\text{S}_{\text{N}}2$ four hours ago and seem to have mixed that in. I've corrected my answer. – Philipp Aug 24 '13 at 20:27