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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    $\begingroup$ 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 '13 at 17:32
  • $\begingroup$ @Philipp thank you . u actually answerd my question :-) $\endgroup$ – nilanjana Aug 24 '13 at 19:29
  • $\begingroup$ I guess I'll make an answer out of my comment then :) $\endgroup$ – 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.

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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 '13 at 20:10
  • $\begingroup$ @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 '13 at 20:27

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