The SN1 is a two-step mechanism which begins with heterolysis of the C-X bond. What triggers this? Are there external causes that prompt the leaving group to simply leave?

This question is more conceptual than anything, but I can't seem to find answers anywhere.


Recall that SN1 heavily favors protic solvents (polar/H+). In such cases, the nucleophile to-be-leaving-group (to be Nu-) is going to be attracted to the positively charged molecules of the protic solvent (...H+) and away from the molecule it is attached to. The power of the attraction comes from the dipole between C-Nu, where Nu has the negative bit of the dipole.

At the end of this step, when the nucleophile attached group leaves, you often see "Nu-" hanging around, but really it's with the solvent, e.g. H3O+Br- , and it's fairly common to omit the solvent here and in other similar ionic situations, if I'm not mistaken.

Examples that invoke hydrogen bonding make this more apparent.

F-C...(tertiary) + H2O

The dipole here looks like (δ-)F-C...(δ+) because F is more electronegative

The F is strongly attracted to the closer H: (δ+)H-(δ-)O-H(δ+)(δ-)F-C...(δ+) (pretend the HOH is bent :)


Greater nucleophilicity of the attacking group than the nucleophilicity of the leaving group causes SN1 reactions. The cleavage of the C-X bond keeps occuring always, both the reactant and the product remaining in the solution. The equilibrium shifts towards the final product.

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    $\begingroup$ I don't see how this could be remotely correct. The SN1 mechanism is an elementary reaction, there is absolutely no equilibrium involved. The dissociation is in the ideal form independent of the nucleophile. $\endgroup$ – Martin - マーチン Sep 28 '16 at 8:45
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    $\begingroup$ @Martin-マーチン Isn't the first step of SN1, that is dissociation into ions, an equilibrium? I don't mean to question the rest of your comment, which I think is correct. $\endgroup$ – FreezingFire Sep 28 '16 at 12:05
  • $\begingroup$ I am referring to the dissociation of the ions being in an equilibrium. I am pointing out the fact that the dissociation of ions takes place not just in the reactant, but in the final product as well. The equilibrium shifts towards the final product since it is more stable. $\endgroup$ – Shubham Avasthi Sep 28 '16 at 20:48

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