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

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  • 3
    $\begingroup$ 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. $\endgroup$
    – Philipp
    Aug 24, 2013 at 17:32
  • $\begingroup$ I guess I'll make an answer out of my comment then :) $\endgroup$
    – Philipp
    Aug 24, 2013 at 19:31

1 Answer 1


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.

enter image description here

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    $\begingroup$ +1, good answer. I just want to make one minor point: since the mechanism is SN1 and the carbocation intermediate is planar, one should get a racemic mixture of products. The representation of stereochemistry is therefore somewhat inaccurate and misleading, I think. $\endgroup$
    – Greg E.
    Aug 24, 2013 at 20:10
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    $\begingroup$ @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. $\endgroup$
    – Philipp
    Aug 24, 2013 at 20:27
  • $\begingroup$ @Philipp How does ZnCl2 acts as lewis acid ? Does it uses it's empty 4s orbital? $\endgroup$
    – Lalit
    Jan 26, 2022 at 17:48
  • $\begingroup$ @Lalit chemistry.stackexchange.com/questions/14028/… $\endgroup$
    – Shub
    Jan 7, 2023 at 15:42

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