[jenny]

I'm a little confused about how Iodine, being a weak base is a good leaving group for sn reactions and a good nucleophille for sn2 reactions. I understand that a good nucleophile carries a negative charge and has electrons farther from nucleus (lower on per. table). I also understand that a good leaving group is a weak base so it is stable. But it makes no sense to me. that the same molecule can attack the carbon that can be good for leaving the carbon. Is there ever a time in an sn2 reaction where Iodine is leaving? Sorry if this is a stupid question, but it's confusing. Thanks.

[movies]

Yes, there are many examples of S_N2 reactions with I- as a leaving group.  The thing that you should think about is whether or not the leaving group is stable as an ion.  You know that I- is stable because it has a noble gas electron configuration.  In terms of being a good nucleophile, I- meets all the criteria you listed.

I know it seems strange, but it is the case!

[maxyoung]

That Iodine is a good leaving group and nucleophile just means it reacts more rapidly. This is from kinetic consideration (iodione is more polarizable).
the transition state is stabilized for both forward and backward reaction.

but thermodynamically, iodine may not be a good nucleophile. Consider the case of iodine displacing OH group from an alcohol

[movies]

I don't think that I- would ever displace OH- (since hydroxide is such a poor leaving group).  I think a better comparison might be I- displacing Br-.  I think that if you were actually going to run that reaction you would use an excess of I- to drive the reaction to completion.  Does that assuage some of your concerns?

[maxyoung]

I used the example of I- displacing OH- just to demonstrate that I- is not a good nucleophile in thermodynamic sense. The reaction is reversible, so there is indeed some I- displacing OH-, but the reverse reaction(OH- displace I-) will be favored. Maybe this is not a good example.

[jenny]

Thanks to both of you, I get it.  

[cbee]

Remeber as well that most of the time you use I- as a Nu- it is using something like NaI to displace Alk-Br in an organic solvent.  While I- is a better leaving group than Br-, the resulting NaBr is much less soluble in organic solvents than NaI, crashes out, and therefore drives the equilibrium to the products.


Cisco

[dexangeles]

hmmm I thought you answered your own question

best leaving group because of the length of its bonds with others
best nucleophile because of it's size, minimizing it's nucleus' repulsion on the other, and making it a really good nucleophile