On the topic of the ionisation mechanism (i.e. $\mathrm{S_N1}$), Carey & Sundberg (2007) discuss the effect of polarity of solvent on the rate of $\mathrm{S_N1}$ reactions of different charge types (p. 392). It is well-known that the rate of $\mathrm{S_N1}$ reactions are increased by the use of polar, protic solvents. Hence, the energy profile diagram shown in a is no surprise. This is type of $\mathrm{S_N1}$ involves the initial ionisation step $\ce{RX -> R^+ + X^-}$ and it is typically seen.

However, the energy profile diagram shown in b is less straightforward. The reaction that it describes has the first ionisation step as $\ce{RX^+ -> R^+ + X}$. In the caption, it is written:

Polar solvents decrease the rate because stabilisation of the [transition state] is less than for the more polar reactant.

Firstly, I would like to ask if "decrease the rate" here means that this type of $\mathrm{S_N1}$ reactions occur more slowly in polar solvents compared to in nonpolar solvents. If we look at the activation energy $E_\mathrm a$ and assuming they are relatively drawn to scale, then we can conclude that for such reaction type, the rate would be faster in nonpolar solvents since it has a comparatively smaller $E_\mathrm a$.

Actually, we do notice that the TS is still lower in energy for the polar solvent, which is logical since although it may be less charged overall, the TS is nevertheless charged and would obviously be stabilised by a polar solvent.

Finally, I would like to ask if the reason why this type of $\mathrm{S_N1}$ reactions proceed faster in nonpolar solvents because, the charged reactants $\ce{RX^+}$ are in fact stabilised in polar solvents to a much greater extent compared to in nonpolar solvents. Thus, the $E_\mathrm a$ in polar solvents is thus much larger. This point was not discussed in the text and it is purely my deduction. Could I confirm that it is a correct interpretation?

Effect of solvent polarity


Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.

  • $\begingroup$ Pretty much yes to everything you've written. $\endgroup$ – orthocresol Dec 25 '18 at 2:02

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