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.

1 Answer 1


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).

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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.] $\endgroup$ Commented Jun 26, 2017 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.
    Commented Jun 26, 2017 at 18:53
  • 1
    $\begingroup$ Yeah, I think that was my guess, but I just wasn't super sure about it. $\endgroup$ Commented Jun 26, 2017 at 19:21

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