[Ch3m1st]

I am having trouble figuring out why this statement is true, especially the bolded sentence:

"When the nucleophile and/or electrophile have a formal charge, S_N2 reaction rate is fastest when the solvent is of modest polarity. When the nucelophile and electrophile are both neutral, S_N2 reaction rate is fastest when the solvent is highly polar."

[orgopete]

I think the answer is the same for both. The first would actually be the fastest in a non-polar solvent (except the concentrations are so low the reaction fails, but the principle should still hold). Then the second question should follow as the opposite. So, the question is, "How could an intermediate be stabilized in an SN2 reaction?" ("And what is the rate limiting step?")

[Ch3m1st]

I think the answer is the same for both. The first would actually be the fastest in a non-polar solvent (except the concentrations are so low the reaction fails, but the principle should still hold). Then the second question should follow as the opposite. So, the question is, "How could an intermediate be stabilized in an SN2 reaction?" ("And what is the rate limiting step?")

The answer above comes from my professor's own solutions, so they must stand as is. Also, I believe the mechanism is elementary and that there are no intermediates (only the transition state). That being said, what charge(s) would the transition state of rxn consisting of neutral electrophile and nucleophile have to be stabilized by the "highly polar solvent?"

[orgopete]

You are correct. I should have said transition states.

If a neutral compound donated electrons, would the transition state be charged? What solvent would enable that process, polar or non-polar?

[Ch3m1st]

You are correct. I should have said transition states.

If a neutral compound donated electrons, would the transition state be charged? What solvent would enable that process, polar or non-polar?

I would assume that the whole transition state would carry a entire negative formal charge? I'm having a hard to visualizing why though. I know that the transition state of a negative nucleophile with a neutral electrophile has two partial negative charges...

[Dan]

Consider the changes in polarisation of charge in the starting materials vs the transition state.

If you consider the transition state for e.g. displacement of iodide from methyl iodide with cyanide ion:

Starting materials (SM):

^-CN + CH₃-I

Transition state (TS):

[NC^d-...CH₃^...d-]^-

While both SM an TS have an overall charge of -1, that charge is more dispersed in the TS where it is shared between the CN and I ligands. In the SM the charge is localised on cyanide.

One can argue that: Since stabilisation of the TS relative to the SM will result in a lower activation energy, a solvent that stabilises the TS more effectively than the SM will result in a higher reaction rate. As the charge is more dispersed in the TS than the SM, we can say the TS is less polar than the SM, and that a solvent of lower polarity will result in an enhanced reaction rate.

Note: As has been mentioned, in a practical setting the solvent must still be polar enough to dissolve the starting materials.

If you follow the same steps for the reaction:

Ph₃P + CH₃I --> Ph₃P⁺CH₃ I^-

You should be able to argue that a more polar solvent will give higher reaction rates. Give it a try.

[Ch3m1st]

Consider the changes in polarisation of charge in the starting materials vs the transition state.

If you consider the transition state for e.g. displacement of iodide from methyl iodide with cyanide ion:

Starting materials (SM):

^-CN + CH₃-I

Transition state (TS):

[NC^d-...CH₃^...d-]^-

While both SM an TS have an overall charge of -1, that charge is more dispersed in the TS where it is shared between the CN and I ligands. In the SM the charge is localised on cyanide.

One can argue that: Since stabilisation of the TS relative to the SM will result in a lower activation energy, a solvent that stabilises the TS more effectively than the SM will result in a higher reaction rate. As the charge is more dispersed in the TS than the SM, we can say the TS is less polar than the SM, and that a solvent of lower polarity will result in an enhanced reaction rate.

Note: As has been mentioned, in a practical setting the solvent must still be polar enough to dissolve the starting materials.

If you follow the same steps for the reaction:

Ph₃P + CH₃I --> Ph₃P⁺CH₃ I^-

You should be able to argue that a more polar solvent will give higher reaction rates. Give it a try.

Thanks Dan! I got it now.