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I'm wondering if the addition of a base of type $\ce{NR_3}$ (R: alkyl) to a ketone (e.g. 3-pentanone or 4-heptanone) results in the cis or trans-enolate.

I'd like to ask here about four cases:

  1. $\ce{NMe_3}$ (small)
  2. $\ce{NMeEt_2}$ (medium)
  3. $\ce{NEt_3}$ (big)
  4. $\ce{N(iPr)_2Et}$ (very bulky)

Maybe smaller bases as 1 or 2 usually give the trans-enolate, and bulkier bases such as 3 or even 4 affords the cis-enolate. (That's what I read, but I'm not able to understand why.) Is there a good explanation that rationalizes which enolate is formed, the cis or the trans?

Sterically hindered bases forms the the kinetic enolate. For deprotonation of 3-Methyl-4-heptanone I assume the enolate at the linear unbranched chain site, that's clear for me, but kinetic/thermodynamic enolate has nothing to do with cis/trans enolate or does it?


Related: Does the addition of base to 3-pentanone give the cis or trans enolate?

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  • $\begingroup$ With weak bases like the amines shown, the conditions are necessarily thermodynamic, so I would expect the thermodynamic Z-enolate to be favored over the E-enolate. Usually there is a trapping agent (e.g., TMSCl) available, which could change the outcome. $\endgroup$ – jerepierre Jul 9 '15 at 19:40
  • $\begingroup$ Why is the Z-enolate thermodynamically favoured? (Can we say the same without the carbonyl, that cis-propene is thermodynamically favoured over trans-propene?) $\endgroup$ – laminin Jul 9 '15 at 20:29
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    $\begingroup$ Tertiary amines are hardly basic enough to extract the alpha-proton of ketone (pKa ~20) to form enolate. Do you mean organolithium bases like (1) LiMe2 ...(4) Li(iPr)2 etc for a meaningful comparison with LDA or NaH? NEt3 was used to form enolate in the ACID-catalyzed enol formation. HNEt3(+) has a pKa of ~10. $\endgroup$ – SYK Jul 17 '15 at 23:15
  • $\begingroup$ @SYK Your comment is very worthful for me! Let's take the smallest Lewis Acid possible, for carbonyl activation, I would try LiI or NaI. Then I suggest a), b), c) and d) all give the cis-enolate. $\endgroup$ – laminin Jul 17 '15 at 23:41
  • $\begingroup$ I doubt LiI or NaI with a tertiary amine will deprotonate a carbonyl, so this question doesn't really make sense unless you are using some kind of boron activation (as your comment on the answer suggests), or silicon (as jerepierre suggested). $\endgroup$ – orthocresol Mar 3 '18 at 15:32
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The keto-enol tautomerization cannot take place without at least a trace of acid or base, meaning that there is no direct shift of proton from alpha-carbon to oxygen or vice versa.

The two most common enolation mechanisms are:

  1. Base-catalyzed (with organolithium reagent such as LDA (secondary amine))

In the base-catalyzed enol formation, the first step, proton extraction from the alpha-carbon, is slow. (The subsequent enolate formation is fast (due to resonance)). The base is proposed to form a chair-like transition state (TS) with the ketone, extracting the alpha-carbon proton and at the same time donating the Li to the oxygen. The resulting product is the classical lithium enolate.

Based on the structure of TS, you can then predict whether the “side chain” of the base would form a unfavorable syn-pentane interaction.

https://en.wikipedia.org/wiki/File:Ireland_model_enolate1.svg

  1. Acid-catalyzed

In the acid-catalyzed enol formnation, the first step, the protonation of oxygen and subsequent “charge re-localization” to the hydroxyl carbon, is fast. (The subsequent enol formation is slow). Since the protonation can occur from any free proton in the environment, the size of the acid plays little direct role in the stereochemistry of the resulting enolate. (However, the size of the “side chain” in the ketone reactant does).

However, there is a 3rd mechanism when the “base” is a tertiary amine (as you appear to be very interested in) but the reaction is acid-catalyzed. In this mechanism, the amine behaves like a nucleophile and a carbinolamine intermediate is detected.

See Bruice, J. Am. Chem. Soc. 1983, 105, 4982

http://pubs.acs.org/doi/abs/10.1021/ja00353a023

(Also see Bruice, J. Am. Chem. Soc. 1989, 111, 962 and 1990, 112, 7361)

In drawing this carbinolamine intermediate, you can predict a bulky tertiary amine to favor a cis-enolate due to steric hindrance. However, as suggested in the paper, "severely sterically hindered" tertiary amine can be too bulky to follow this 3rd mechanism and just catalyzes enolation via the general base-catalyzed mechanism.

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  • $\begingroup$ Thank you very much for your answer. Probably there is a fourth mechanism which I'm more interested in than the other three: Addition of a lewis acid. a1) Bu2BOTf / (iPr)2NEt gives the cis enolate, b1) Cy2BCl / NEt3 gives the trans-enolate because of the bulky Cy. What does change when you instead use a2) Bu2BOTf / NEt3 and b2) Cy2BCl / (iPr)2NEt ? And last but not least I'd like to ask as a second question, if the addition of c) LiI and d) NaI also give the cis-enolate? $\endgroup$ – laminin Jul 18 '15 at 13:28
  • $\begingroup$ chemistry.stackexchange.com/questions/34153/… $\endgroup$ – laminin Jul 18 '15 at 20:41
  • $\begingroup$ (−1) OP is asking about the generation of stoichiometric enolate, but you seem to be talking about keto-enol tautomerism. Plus, LDA isn't going to "catalyse" keto-enol tautomerism. Your answer seems to be confused about this: "In the base-catalyzed enol formation [...] The resulting product is the classical lithium enolate." $\endgroup$ – orthocresol Mar 3 '18 at 15:36

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