# What is the result when the polarity of the solvent is increased in nucleophilic substitution?

I read in the Guidebook to Mechanism in Organic Chemistry that increasing the polarity of the solvent slightly reduces the rate of an $S_N2$ reaction. Why does this happen?

Another question: We know that in $S_N2$, using a polar protic solvent would reduce the rate as the nucleophile gets solvated by hydrogen bonding, but I found in Solomon's Organic Chemistry that the $S_N1$ reaction rate increases in a polar protic solvent. Why does this occur? The nucleophile still gets solvated right?

The standard textbook explanation is based on a comparison of charge separation in the ground state and the transition state (TS). If there is more charge separation in the TS than the ground state, then polar solvents stabilize the TS more than the ground state and the reaction will be accelerated. On the other hand, if there is more charge in the ground state than in the TS, then polar solvents will stabilize the ground state more than the TS and consequently the reaction will slow down.

Let's look further at the effect of charge separation and solvent on these two specific reactions and see how they are affected. Let's start with the $\ce{S_{N}1}$ reaction.

I found in another book that the SN1 reaction rate increases in polar protic solvent. Why does this occur?

In the $\ce{S_{N}1}$ reaction, the rate determining step (rds) is formation of the carbocation intermediate. Let's say we start with a charged nucleophile and a neutral compound. Moving along the reaction coordinate, by the time we reach the carbocation intermediate, the starting materials (charged nucleophile plus neutral compound) have transformed into a charged carbocation plus a charged leaving group plus the charged nucleophile. Moving from starting materials to the intermediate has created additional charged species (carbocation plus leaving group).

$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [R^+ +X^{-} + Nu^- ]~~ (the~ TS)}$$

Since the transition state leading to these intermediate species will strongly resemble the intermediate, it will also have more charge separation then the starting materials and therefor, will be more stabilized by polar solvents then the starting materials. This stabilization of the transition state will lower its energy and speed up the rds. In other words a polar solvent has sped up the rate of the $\ce{S_{N}1}$ because more charge exists in the transition state than in the ground state.

I read in a (book) that increasing the polarity of the solvent slightly reduces the rate of SN2 reaction. Why does this happen?

Now, lets perform the same analysis on the $\ce{S_{N}2}$ reaction. Again we start out with a charged nucleophile and a neutral compound. The rds in this case is the formation of a trigonal bipyramid species.

$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [\delta^{-} ~~ X...R...Nu ~~\delta^{-} ]~~ (the~ TS)}$$

In the ground state there was one unit of negative charge residing fully on the nucleophile. In the TS that unit of charge is now spread out over both the nuceophile and the leaving group; the charge has become much more diffuse, the charge density per unit volume has decreased in the TS compared to the ground state. In this case, we would expect a polar solvent to stabilize the ground state (starting materials) slightly more than the still charged - but more diffusely charged - TS. Stabilizing the starting materials slightly more than the TS would slightly increase the energy required to pass over the TS. Hence we would expect polar solvents to slightly slow down the rate of an $\ce{S_{N}2}$ reaction.

I read in a (book) that increasing the polarity of the solvent slightly reduces the rate of SN2 reaction. Why does this happen?

Increasing the polarity of the solvent increases the solvation of the nucleophile. There will be increased dipole-dipole interactions between a solvent and a nucleophile as polarity of the solvent increase.

This will reduce the mobility of the nucleophile and in turn reduce the rate of reaction in an SN2 reaction. The SN2 mechanism's rate is determined by the rate at which the nucleophile and the substrate can randomly collide into each other (in the right way - i.e. you want the nucleophile to attack the backside of the leaving group).

But I found in another book that the SN1 reaction rate increases in polar protic solvent. Why does this occur? The nucleophile still gets solvated right?

Yes, the nucleophile is still solvated, but in an SN1 reaction, the rate-determining step is the rate at which the leaving group leaves. That is by far the slowest step in an SN1 reaction because of the high activation energy involved in creating a carbocation, which is highly unstable. So it doesn't really matter if the nucleophile is heavily solvated or not so heavily solvated - the slowest step determines the rate.

Also, regarding why polar protic solvent increase the rate of SN1 reactions - think about the carbocation. There is a high activation energy barrier in forming this carbocation - it's extremely unstable. Protic solvents can stabilize this carbocation through hydrogen bond formation. This lowers the activation energy barrier.