5
$\begingroup$

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

enter image description here

$\endgroup$
2
  • 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

9
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ +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
  • 1
    $\begingroup$ @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

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.