In this Mukaiyama aldol reaction, taken from a 1999 paper by D.A. Evans,1 when $\ce{TiCl4}$ was used as the Lewis acid catalyst, the aldehyde 4a formed a chelated intermediate whereas 4b did not (due to the bulk of the TBS group). This was demonstrated by the stereochemistry of the product. Aldehyde 4a produced a syn:anti ratio of 3:97, whereas aldehyde 4b led to a ratio of 93:7:

Reaction scheme

The chelated intermediate which aldehyde 4a goes through is a standard six-membered ring with a half-chair conformation. Perplexingly, the α-methyl group has to be placed pseudo-axial instead of pseudo-equatorial (or else the syn aldol product will be obtained):

Conformational analysis

Ordinarily, substituents would adopt a pseudo-equatorial position in the dominant conformer. Why is this an exception?


  1. Evans, D. A.; Allison, B. D.; Yang, M. G. Chelate-controlled carbonyl addition reactions. The exceptional chelating ability of dimethylaluminum chloride and methylaluminum dichloride. Tetrahedron Lett. 1999, 40 (24), 4457–4460. DOI: 10.1016/S0040-4039(99)00739-X.

Having the methyl group psuedo-axial means that when the nucleophile approaches, the transitiation state is chair-like. The energy penalty for having the methyl group initially in the less favourable position is compensated for the low energy TS.

The ring-flipped version, in which the methyl group is psuedo-equatorial, would open via a high energy twisted TS (the opposite of the argument presented above, i.e. the low energy conformation of the starting material doesn't compensate for the high energy of the twisted TS).

enter image description here

Looking at the Evans' paper specifically, the TS he draws has an addition isopropyl group on the beta-carbon, which is equatorial (this would further stabilise the starting material, since the isopropyl would be far higher in energy if axial compared to the methyl).

  • $\begingroup$ In your second TS, you can have the nucleophile attack from below, no? And that would give the syn aldol product, but still via a chair TS. [I saw the TS with the iPr group, but he repeated it with a different substrate lacking the iPr group (the one I drew above), and the result was similar.] $\endgroup$
    – orthocresol
    Jun 26 '17 at 18:10
  • $\begingroup$ I suppose theres a steric argument, that the nucleophile (in this case a bulky silyl enol ether) doesn't want to approach across the methyl group $\endgroup$
    – NotEvans.
    Jun 26 '17 at 18:53
  • 1
    $\begingroup$ Yeah, I think that was my guess, but I just wasn't super sure about it. $\endgroup$
    – orthocresol
    Jun 26 '17 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.