Let us say that we have water reacting with t-butyl chloride. When the chlorine atom departs we have a t-butyl cation which becomes an attractive substrate to the nucleophilic oxygen in water. My question is in the reaction that follows, does the hydrogen break off from water first allowing oxygen to attack the carbocation or does water immediately form a bond with the carbocation and then deprotinates the extra hydrogen?

  • $\begingroup$ I think that OH which is already dissociated( because water inherently dissociates to give H and OH so that pH is 7) will replace Cl , since OH concentration is now decreased then according to le chatliers principle more water will dissociative to give more H and OH and this goes on eventually the mixture becomes acidic. $\endgroup$ – JM97 Sep 5 '16 at 5:10
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    $\begingroup$ First one molecule of water attacks the carbocation, then it loses a proton. Check SN1 mechanisms for more information. $\endgroup$ – Variax Sep 5 '16 at 7:20
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    $\begingroup$ @JM97 no, the concentration of free OH- in water is so small, it is much more likely that H2O attacks. This can be confirmed by practically any textbook that describes SN1/SN2 mechanisms. $\endgroup$ – orthocresol Sep 5 '16 at 8:25

Surely there are both $\ce{H2O}$ and $\ce{OH-}$ which attacks the carbocation, as well as, heaven forbid, $\ce{H3O+}$ (and countless other species formed by the auto-ioniziation of water).

Usually textbooks record $\ce{H2O}$ as the attacker because it is much more abundant in the mixture.

After that, usually another $\ce{H2O}$ comes and take away the extra $\ce{H}$ with itself converting to $\ce{H3O+}$, but as I mentioned above, this step could also be done by $\ce{OH-}$ among many more species, but the concentration of those other species are too small that they are not worth mentioning.

PS: the last step can also be done by the product, i.e. t-butyl alcohol.


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