[orgo814]

I am given the question: Which wquld you expect to be the stronger nucleophile in a polar APROTIC solvent?
a) H2O or H2S
b) (CH3)3P or (CH3)3N

Since we are not dealing with protic solvent, we don't need to worry about polarizability. The trend is "upward" the periodic table (at least for the halogens) as nucleophilicity in aprotic solvent (opposite of protic solvent since the more polarizable the less it will be encumbered the hydrogen bonding in protic solvent). I thought the answer would be H2O and (CH3)3N. The answer in the solutions manual to the book is the opposite for both. Either the solutions manual is wrong or I'm clearly missing something. My solutions manual does have a tendency to be wrong a lot but still... I'm not sure. We're not dealing with halogens here so maybe the trend is different?

Please help.

[Schrödinger]

That reasoning is true for halides, probably because halides are single atom-ions and their sizes and hence solvation change evidently as you move down the group. On the contrary, when we start comparing larger groups like (CH₃)₃P or (CH₃)₃N, the solvation shouldn't matter much since both are approximately solvated equally. So, the nucleophilicity being greater as you move down the group probably applies to groups other than the halogens. Probably. Please correct me if I'm wrong.

[orgopete]

There had been an earlier question about why iodide should both be a good leaving group and a good nucleophile. I find this bahavior difficult to understand. Why should the electrons of iodide be pulled toward its larger nucleus and away from a proton to explain its acidity and the electrons also to be able to attack another nucleophile?

When I have compared reaction rates of the halides, the rate changes the most for fluoride if the solvent is changed from protic to aprotic. Iodide is relatively less sensitive to similar solvent effects. This seems consistent with the solvation suggested by Schrödinger. Solvation (protonation) does explain why fluoride should change by a large amount. That leaves why iodide should be a good leaving group and a good nucleophile.

If we assume that iodide is a good nucleophile with or without solvent effects, then it follows that S>O and P>N. I think that is correct and your answer book is correct.

The explanation that I find most suitable is one based upon polarizability. This is how I rationalize it. The acidity is a thermodynamic effect. The larger charge of the iodide nucleus should pull its electrons in more and increases the acidity. Although this does occur, electrons are not stationary, their position is not fixed. The inner electrons also create a repulsive force to the outer electrons. If we rationalize this could result in electrons temporarily taking excursions further from the nucleus, then the atoms may have both a thermodynamic and kinetic effect. As a thermodynamic effect, it can be more acidic and as a kinetic effect, be more nucleophilic.