Henry Rzepa

This is another of those textbook reactions, involving reaction of a carbonyl compound with a phosphonium ylid to form an alkene and a phosphine oxide. The reaction continues to be frequently used, in part because it can be highly stereospecific. 

Thus the standard version tends to give Z-alkenes with good specificity, and is thought to proceed via an oxaphosphatane 4-ring intermediate. The reaction and its stereochemistry is sensitive to the reagent (including the nature of the R group), and so one model cannot capture all the aspects of this transform. Here I am starting with the very simple model shown above, where R=H (ωB97XD/6-311G(d,p)/SCRF=tetrahydrofuran). There are four transition states to consider; whether the  rate-determining (stereochemical determining) step is TS1 or TS2, and whether the relative orientation of the two (in this example methyl) groups are syn or anti, resulting in E- or Z- alkenes. The most interesting issue would be whether the mechanism can account for why the apparently more sterically hindered route leading to the Z-alkene is often the actual outcome. 

Leading to E-alkene
TS1 0.0 kcal/mol	TS2 -3.9
Leading to Z-alkene
TS1 0.0 kcal/mol	TS2 -2.6

Key comments about these results:

1. TS1 is higher than TS2 in both cases, and so (for these substituents) is rate determining.
2. At this transition state, the two methyl groups are moving apart for the E-isomer but together for the Z-isomer. But at the transition states themselves, the steric interaction of these two groups is fairly similar, and the Z-transition state has much better antiperiplanar bond alignments compensating for the methyl clash. To put it in a nutshell, the increased steric clash for formation of the Z-isomer comes only AFTER the transition state is passed.
   

   E-alkene forming	Z-alkene forming

3. The gradients of the IRC profile for this step of the Wittig reveal that much of the action occurs after the transition state is passed, at IRC=3 for the E and IRC=4 for Z, this comprising rotation around the first formed C-C bond in order to create the P-O bond. This is where the steric clash of methyls for the Z-isomer really kicks in, but it has no impact upon the energy of the transition state, coming too late for that.
   

   E-alkene forming	Z-alkene forming

4. The model we have built is sterically incomplete; we have used PH₃ rather than eg PPh₃ (done so as to allow an IRC to be computed in a reasonable time). If we look at the models above (click on the images to get a 3D model), then it is clear that the E-transition state will suffer the greater steric clash of a methyl with one of the phenyl groups on the phosphorus than the Z-isomer will. This probably accounts for why this latter isomer is the normal stereochemical outcome.

Much more could be done here, but even a fairly simple model of the Wittig reaction can bring a lot of insight into its unique characteristics.