Why is the Still-Gennari reaction Z-selective? enter image description here

I guess dipole interactions have the biggest influence here. They somehow destabilize the reversion of the oxaposhpetane to the betaine and prevents formation of the more stable E isomer but I am not sure.


As you hopefully realise, the stereochemical outcome of the HWE reaction (and variations) is due to the relative orientation of the R-groups in the phosphaoxetane intermediate.

If the R groups are syn as in 5A then the (Z)-olefin is formed, with an anti arrangement 5B favouring the (E)-olefin as the product.

enter image description here

Source: Modern Carbonyl Olefination, Wiley, 2004

The pertinent question when looking at the selectivity is why do we get dominant elimination from the syn phosphaoxetane (leading to the correct olefin), rather than dominant elimination from what is arguably the sterically most favourable anti phosphaoxetane (in which the R groups are pointing away from one another).

As it turns out, the initial 'aldol' addition into the aldehyde is rate determining, with the anti addition being favoured over the syn addition on steric grounds.

enter image description here

Source: Stereoselective alkene synthesis, Springer-Verlag, 2012 (a really useful book, FYI)

Once the aldol addition has taken place, the subsequent steps are rapid due to the high electrophilicity of the Still-Gennari phosphonate (with the additional electron withdrawing groups).

This means that the entire process is essentially controlled by the initial addition to give a syn or an anti adduct. Once formed, the adducts are sufficiently short lived that despite the cis phosphaoxetane being less favoured due to the clash between R-groups, the (Z) olefin is formed via quick and irreversible elimination.

The HWE therefore is (E)-selective, in part, due to the fact that elimination is less rapid and as such there is equilibration to the more favourable phosphaoxetane from which elimination may occur.

  • $\begingroup$ Why is the (Z) olefin faster obtained? $\endgroup$ – RBW Jul 28 '16 at 17:14
  • $\begingroup$ The other thing I learnt is that, Still-Gennari conditions usually have something to sequester the metal ion - typically 18-crown-6 for K+. In a normal HWE the metal ion would stabilise the starting materials via chelation (as your diagram shows) and this would make the oxaphosphetane formation "more reversible" i.e. lower activation energy for reverse reaction. $\endgroup$ – orthocresol Jul 31 '16 at 23:16
  • $\begingroup$ The sequestering thing is pretty interesting. There are some really nice papers where they probe the effects of using various crown ethers etc. $\endgroup$ – NotEvans. Jul 31 '16 at 23:19
  • $\begingroup$ In the first image you defined syn and anti as the relationship between the two R groups, in the second image it is between R and phosphonate group. Could you define in both cases syn and anti in the same way for clarity? $\endgroup$ – RBW Jul 31 '16 at 23:25
  • $\begingroup$ No. This would be incorrect notation. Though admittedly you would be understood $\endgroup$ – NotEvans. Aug 1 '16 at 6:49

I am not convinced by the transitions states given in "Stereoselective Alkene Synthesis" because they rely on chelation to restrict the conformation of the phosphonate anion AND the orientation of the aldehyde.

And yet, the reaction conditions are KHMDS/18-crown-6 which are clearly designed to eliminate the possibility of coordination to the potassium cation and maximise the rate of addition of the "naked" alkoxide anion to the electron deficient P.

My own suggestion for the intrinsic "anti" (in this example) selectivity of the fluorophosphonate anion is that it simply mirrors the intrinsic selectivity of the unstabilised Wittig ylid and Horner-Wittig (Ph2PO) stabilised anions both of which produce Z-alkenes stereoselectively.


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