[Charkol]

Na^{+ -}:I   +   CH₃CH₂--Br   --->   CH₃CH₂--I   +   Na^{+ -}:Br

This is the equation that has given me the question. 

I am wondering what is the idea that allows the iodide ion to *Ignore me, I am impatient* out the bromide ion. 

This is my thinking: I understand Iodine is less electronegative than Bromine.  So the extra electron in I:^- should be "shared" more so than the extra electron in Br:^-.  As the closer that electron is to the bonding carbon, the stronger the bond will be, and thus the Iodide ion will push over the Bromide ion.
Is this right?

[AndersHoveland]

This is known as the Finkelstein reaction. The NaBr only solidifies out because it is less soluble in the (typically acetone) solvent. The reaction tends to be very slow, taking days or months without an appropriate catalyst.
Otherwise, formation of the iodide ion is somewhat more favorable than that of the bromine ion.

One catalyst idea is a beta-diketiminate complex.

condense acetylacetone with methylamine, using a strong base, to form a double imine.
 
This has a structure of  CH3C(=NCH3)CH2C(=NCH3)CH3
 
Now add Cu+2 ions. This could potentially form an unusual aromatic complex.
 
 
 
             (+)
CH3--N--Cu--N--CH3
         ll         l
CH3--C--CH=C--CH3    H(+)
 
 
As you can hopefully see, the acetylacetone imine becomes deprotonated (acetylacetone is known to sometimes behave as an acid)
 
The aromatic copper complex would have a net charge of +1
The ring has several resonance states.
 
Copper acetylacetonate, for example, is an interesting complex with a surprisingly high vapor pressure. However, the two methyl groups from the CH3NH2 would be expected to sterically hinder a second ring from forming, which is the case with the plain copper acetylacetonate complex salt.

[Doc Oc]

This is my thinking: I understand Iodine is less electronegative than Bromine.  So the extra electron in I:^- should be "shared" more so than the extra electron in Br:^-.

Yes.  To put it more simply, Iodide is more nucleophilic.

[orgopete]

This is one of those really good questions and I don't know that I can give a really good answer. When you compare the acidity of the haloacids, you are comparing a thermodynamic property. HI is a stronger acid, and iodide is a weaker base. That suggests its electrons are not available for reaction.

The formation of a C-I bond is essentially irreversible. That it, its formation is a kinetic process. In protic solvents, iodide reacts faster. A common explanation is hydrogen bonding to the other possible halide anions slows their reaction. That would be consistent with the acidity data as well. However, iodide is also relatively immune to a solvent effect. In an aprotic solvent, fluoride must just over iodide in rates. While iodide does slow somewhat, it is still relatively reactive in aprotic solvents.

On that account, I have come to think that iodide, with its 54 valence electrons, has two properties. One is that the electrons are relatively unavailable, thermodynamically. The effect of the protons pulling the electron toward it. The other is that this shell of electrons may also move or jump out to other positive centers, a kinetic process. This would be the so-called polarizability feature. I think of this as a feature that all heavier atoms have. As more electrons are added to the shell, the easier it is for them to react kinetically. That is consistent with sulfur, phosphorous, etc.

[AndersHoveland]

"orgopete": while I admire your method of thinking, which can often be useful, I am unsure that you are correct in this case. Larger atoms certainly do seem to be more reactive, but this may have more to do with other factors.
Larger atoms can more easily transiently accept or donate electrons into or from their respective molecules. Often the reactivity of a molecule can be increased if the molecule is substituted with a group that can serve as an electron transport route. Carbon tetrachloride is fairly inert, but phosgene can easily hydrolyze as an electron can enter the molecule through the double bond. SF6 is extremely inert, but SF5Cl is highly reactive. Alkanes do not readily react with nitrogen dioxide, but if substituted with a phenyl group, the adjoining carbon can ionize off one of its hydrogens atoms, eventually being replaced with a nitro group. This also explains why N-phenyl substitution of 1,2,3-triazole renders the triazole ring vulnerable to nitration.

This probably has little to do with the Finkelstein reaction, which is dependant on equilibrium and solubilities.
Sodium bromide is virtually insoluble in acetone (the solvent) so it eventually solidifies out. The bromine and iodine do not need to have any unusual chemical properties to explain the reaction. However, iodide ions can form covalent bonds, and this may help facilitate the exchange. It should be noted that bromomethane can ionize, but the equilibrium is extremely low. This is, however the reason for bromomethane's reactivity in substitution reactions.

[spirochete]

This is one of those really good questions and I don't know that I can give a really good answer. When you compare the acidity of the haloacids, you are comparing a thermodynamic property. HI is a stronger acid, and iodide is a weaker base. That suggests its electrons are not available for reaction.

The formation of a C-I bond is essentially irreversible. That it, its formation is a kinetic process. In protic solvents, iodide reacts faster. A common explanation is hydrogen bonding to the other possible halide anions slows their reaction. That would be consistent with the acidity data as well. However, iodide is also relatively immune to a solvent effect. In an aprotic solvent, fluoride must just over iodide in rates. While iodide does slow somewhat, it is still relatively reactive in aprotic solvents.

On that account, I have come to think that iodide, with its 54 valence electrons, has two properties. One is that the electrons are relatively unavailable, thermodynamically. The effect of the protons pulling the electron toward it. The other is that this shell of electrons may also move or jump out to other positive centers, a kinetic process. This would be the so-called polarizability feature. I think of this as a feature that all heavier atoms have. As more electrons are added to the shell, the easier it is for them to react kinetically. That is consistent with sulfur, phosphorous, etc.

 I also disagree with orgopete in this situation.  Iodide is a good nucleophile but also a good leaving group - It's the classic example of a reversable nucleophilic addition process.  And because C-Br bonds are stronger than C-I bonds (due to bromine being closer in size to carbon, and also it's greater electronegativity relative to carbon), the reaction is not thermodynamically favorable without some additional driving force.  In the Finklestein reaction, that driving force is the precipitation of NaBr.  That's why solvent choice is so critical in this reaction - we need something that will crash out NaBr and drive the rxn forward by Le Chatlier's principle.

[orgopete]

I concede my answer was off target. This is a simple Le Chatelier's principle. It wouldn't matter what the bond strengths were if the reaction is being driven by the equilibrium.

I was trying to answer a question not being asked, namely, if iodide is a good leaving group, how can it be a good nucleophile? That didn't matter to explain the product.