From Clayden et al.

However, iodine is expensive and a way round that problem is to use a catalytic amount of iodide. The next phosphonium salt is formed slowly from benzyl bromide but the addition of a small amount of LiI speeds up the reaction considerably.

The iodide reacts as a better nucleophile than Ph3P and then as a better leaving group than Br–. Each iodide ion goes round and round many times as a nucleophilic catalyst.

In a private conversation with ron I asked him how this catalyzing effect worked, he said that question would be more suited here and I agreed with him. It seems a bit strange at first that a strong nucleophile can also be a good leaving group (he really wants that nucleus, right, why would he leave?).

This is the explanation I came up with (probably wrong, but it is always better to first think self): Iodine is a stronger nucleophile here than bromine because of its higher energy HOMO, which lies closer to the energy of the $\ce{C-X}$ $\sigma^{*}$ (here $\ce{X}$ = $\ce{Br}$) LUMO. So it will react more rapidly than bromine, resulting in an equilibrium.

Now iodine is also a good leaving group, firstly because the $\ce{C-I}$ bond is higher is energy than the $\ce{C-Br}$ bond (less stable bond). And secondly (quoting Ron from another answer. Well explained, no need to rewrite it again): the larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. So $\ce{I^{-}}$ will be more stable, making it easier for iodine to leave.

My thought was that iodine can easily knock of bromide but not remain in that position for very long (therefore the equilibrium). Now, if $\ce{PPh_3}$ happens to meet the benzyl iodide it is more eager to react with it than with the benzyl bromide (lower energy $\ce{C-X}$ $\sigma^{*}$). The resulting phosphonium salt is either to stable to be attacked by iodine again or precipitates out (or both?), so that benzyl bromide continues to be transformed into benzyl iodide.

Is this correct?


1 Answer 1


Iodine is a stronger nucleophile here than bromine

I agree with your HOMO-LUMO reasoning, but this doesn't really matter too much. As long as iodide is a strong enough nucleophile to displace bromide at least some of the time, then the reaction we want to happen, will happen. Of course the more nucleophilic iodide is (here's where your HOMO-LUMO argument comes into play), the faster the desired reaction will happen.

I pretty much agree with the rest of what you say Jori, bit I might say it just a bit differently.

I'm not sure how triphenyl phosphine (TPP) is hybridized, I'd guess that it is unhybridized just like phosphine ($\ce{PH3}$). If it is unhybridized than our lone pair on phosphorous is in an s-orbital. Since the s-orbital has little "directionality", this lone pair on phosphorous will be a weak nucleophile. On the other hand, if TPP is hybridized (like triphenyl amine), then although the lone pair is now in an orbital with high p-character (and therefore very directional) it will be extensively delocalized into the 3 aromatic rings (just like triphenyl amine which is a very weak base for just that reason), which again means that the lone pair isn't very available and consequently TPP will be a weak nucleophile. So now matter how TPP is hybridized, we expect it to be a weak nucleophile.

Iodide is, as you note, a much better leaving group then bromide. In fact, iodide is such a good leaving group that it can be readily displaced by even a poor nucleophile like TPP. Iodide is also a good nucleophile and can displace bromide.

The reaction is run with just a small (catalytic) amount of iodide. As soon as iodide replaces a bromide (nucleophilic displacement), the iodide, being such a good leaving group, will now be quickly displaced by the weakly nucleophilic TPP. During this last step, iodide is regenerated and the cycle starts again.

  • $\begingroup$ I just did the calculation, $\ce{PPh3}$ gives a nicely directionalized $\ce{sp}$-lone pair on DF-BP86/def2-SVP. $\angle(\ce{Ph-P-Ph}$ is about $102^\circ$. The $\ce{C-P}$ bonds are $38\%~\ce{sp^{$4.5$}~P}$ and $62\%~\ce{sp^{$2.6$}~C}$. I would argue, that this is due to steric effects. But as you pointed out, this has little to no effect on your explanation. $\endgroup$ Aug 19, 2014 at 2:32
  • $\begingroup$ Is there much delocalization of the lone pair into the benzene rings? $\endgroup$
    – ron
    Aug 19, 2014 at 2:38

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