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#1 2009-01-25 22:55:12


Why is iodide a better nucleophile than fluoride?

(twister) wrote:

I'm reading the nucleophilic substitution chapter in an organic chemistry book , and it states that:

"For atoms in a family (a column in the periodic table), the larger atoms (those with larger diameters) are better nucleophiles in polar protic solvents: I- > Br- > Cl- > F-."

Yet, any chemistry textbook will tell you that HI is the strongest halogen acid and HF is the weakest... which means HI gives up an H+ (a nucleus) much more easily.

How can it be, then, that I- is more attracted to a nucleus (i.e., is a stronger nucleophile) than F-? What's the point of giving up an H+ if you want it back?

The question can also be stated as: why is HI a stronger acid whereas I- is a stronger base? I thought strong acids had weak conjugate bases!

Excellent question and excellent analysis. How indeed can a pair of electrons that cannot bind to a proton, bind to a carbon?

I cannot answer that question nor am I certain anyone actually can. I can provide you with the rationalizations I have seen, but I cannot verify their truth.

Looking at the data, in dipolar aprotic solvents, fluoride is the better nucleophile. In polar protic solvents, the nucleophilicity reverses. Arguably, fluoride forms hydrogen bonds to the solvent and that slows its rate of attack. Iodide, with its greater numbers of protons in its nucleus pulls its electrons in so they are less affected by hydrogen bonding. That is the hydrogen bonding argument. However, iodide is actually little affected by the solvent while fluoride is. Therefore, I remain puzzled as to why iodide should have been fast in the first place.

Argument number two, polarizability. While it is true that the electrons of iodide are not available for protonation, the outer electrons face a sea of inner electrons that resist their being pulled into the nucleus. Consequently, they are able to extend further and therefore are more nucleophilic. This argument doesn't make sense. It contradicts the original premise of iodide making its electrons less available for bonding, but I have to chose one, so I chose polarizability.

At the heart of this question is a certain realization that we don't know as much about the behavior of electrons as we would profess. If two electrons can form a lower energy pair if they are of opposite spin, one could argue the property of spin must have an attractive force equal to or greater than the coulombic repulsion. Nope, don't want to do that as any physicist would show you I am dead wrong. Then why are they at a lower energy? Quantum theory!

If we were to apply Pearson's Hard and Soft theory, he argues that reactions of equal kind are best, hard w hard and soft w soft. I am not too fond of H and S theory, but I do favor some parts of this. Fluorine is hard. It's valence electrons are only repulsed by the 1s electrons. Iodine is soft. It's valence electrons are repulsed by several shells. Comparatively, because the volume consumed by electrons is large and variable, the further out they are, the greater the variability might exist. This is in essence a polarizability argument.

I want to add one more element to this discussion, kinetic versus thermodynamic. I haven't thought this through enough to search for data on this, but I have been thinking that the pKa measurements are essentially thermodynamic in nature. However, the SN2 substitution reactions are essentially kinetic in nature. That is, protonation is an "on" and "off" measurement while the alkylation is an "on" measurement. It isn't that these collisions don't occur, it is just that in measuring the pKa we are measuring the ratio of those that are attached versus those not. Even though the ratio may greatly favor no attachment, it could still mean that microscopically, the protons are attaching themselves to the iodine electrons very rapidly, but just as rapidly becoming unattached. You may recognize this as a variant of the polarizability argument.


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