In polar aprotic solvent, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?
What I think: If I talk about their basicity, $\ce{F-}$ would be more basic due to instability from its high electron density. Is it applied same when talking about nucleophiliicty?
Why is an atom with high polarizability a strong nucleophile?
What I heard: an increased atom size means a high polarizability, and a high polarizability means a high nucleophilicty because the atom can give its weakly-held $e^-$ easier.
What I think: above sentence makes sense to me. But in another point of view, isn't it also right that the atom with properties like that would be stable, making it poor nucleophile?
When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?
What I heard: when deciding the nucleophilicty of an atom in a polar protic solvent, there are two factors that come into play: solvation and polarizability.
What I think: in a polar protic solvent, since the effect of solvation is significant, the polarizability shouldn't mean so much. Rather, its effect should be significant in polar aprotic solvent since there's no hydrogen bonding. Is this wrong?
1 Answer
Why is an atom with high polarizability a strong nucleophile?
An atom with a higher polarizability will share his electrons more easier since they are less attracted by the nucleus. The more the electrons are near the nucleus, the more it's hard the share them with an electrophile. $\mathrm{I^-}$ is thus more nucleophile than $\mathrm{F^-}$.
As you'll se below, the solvent can affect the nucleophilicity and reverse what I just said.
In polar aprotic solvent, why is $\mathrm{F^-}$ a stronger nucleophile than $\mathrm{I^-}$?
When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?
First, basicity and nucleophilicity are not the same at all. For instance, a strong base can be non-nucleophile, such as LDA for instance.
Then, let's talk about the polarizability of the $\mathrm{F^-}$ and $\mathrm{I^-}$ anions. The fluoride has a small polarizability: its charge is dense, very localized. It's called an hard anion. Iodide has a great polarizability with its dispersed charge. It's called a soft anion.
Using a protic polar solvent, the anions will be well solvated, even better if they are hard anions because the hydrogen bondings will be more efficient. The anion $\mathrm{F^-}$ is thus more solvated than $\mathrm{I^-}$ and less reactive.
In the case of an aprotic polar solvent, the anion is not so well solvated and the fluoride will be more reactive.
