In an Evans aldol reaction, the enolate is typically generated by addition of a boron Lewis acid (e.g. $\ce{(n-Bu)2BOTf}$) and an amine, which acts as a weak base. I'd like to know whether there's any difference between

  1. Triisobutylamine, $\ce{(i-Bu)3N}$
  2. N,N-Diisopropylethylamine (Hünig's base), $\ce{(i-Pr)2NEt}$
  3. Triethylamine, $\ce{Et3N}$

in the selectivity of the enolate formation and subsequent aldol reaction. I'm thinking about the different $\mathrm pK_\mathrm a$ values (1>2>3) and the different steric demand (1>2>3). Probably the base with the highest $\mathrm pK_\mathrm a$ will be kinetically fastest. Also, the cis-enolate should be favoured as it avoids a destabilising syn-pentane interaction between the terminal methyl group and the auxiliary. However, do all of the bases lead to the same ratio of cis/trans-enolate formation (and hence diastereoselectivity in the subsequent aldol reaction)?

Generation of enolate for Evans aldol

  • 2
    $\begingroup$ I'll have to dig up my notes on this, but I believe amides are always stereoselective for (Z)-enolate formation. It's likely due to an A1,3 strain. If you have access, check JACS 1982, 104, 1737. $\endgroup$ – jerepierre Nov 17 '14 at 17:05
  • 1
    $\begingroup$ Not 100% sure, but first thing that occurs to me is that the less bulkier the amine the more proportion of cis-enolate formed? If we are talking about a thermodynamic product, the one that should yield more Z proportion, is likely to be the one that gives rise to an enolate of the less bulky ammonium salt. For the same reason I also think that lithium amides will yield high Z isomer proportion, since OLi group is small and will cause lower repulsion to the methyl group than any amine. $\endgroup$ – Altered State Nov 18 '14 at 7:12

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