One of the factors that favor kinetic products is bulky bases. So, I'd expect that, for a certain substrate, the use of a more bulky base to favor even more the kinetic product.

However, in the table below we see that as the base becomes even more sterically hindered, the thermodynamic product is formed. It's never the main product but it does get produced a little bit more (1% to 10%).

Why is that? Why in the case of Ph3CLi we get more B than in the case of LDA?

enter image description here

  • $\begingroup$ Consider an even larger base where the steric interactions in the transition state to form A or B would be close to identical. I think you're getting closer to that limit. $\endgroup$
    – Zhe
    Commented Nov 7, 2018 at 18:31
  • $\begingroup$ I thought that since either can be formed, the more the reaction is delayed the more the thermodynamic (stabler) product would form. So since bigger bases means more difficult and slower displacement, then more time is given to the reaction to form B. $\endgroup$ Commented Nov 8, 2018 at 12:57
  • $\begingroup$ The temperature is low enough that you're likely not getting any equilibration. What you really want to see is the relative rates between the first 3 rows. $\endgroup$
    – Zhe
    Commented Nov 8, 2018 at 14:02
  • $\begingroup$ As I've mentioned, I don't expect the rate to become vice versa for the two products, since there is not much energy for it. Regarding my comment it's the only reasonable approach I thought. $\endgroup$ Commented Nov 8, 2018 at 19:52
  • $\begingroup$ It's frustratingly difficult to find a source for the information (it's in Carey/Sundberg Vol A, which in turn cites Augustine's Carbon-Carbon Bond Formation, which I can't access)... In the absence of further information, I think both arguments make sense: either the reaction is purely under kinetic control and increasing the steric bulk decreases $\Delta \Delta G^\ddagger$, or there is a thermodynamic component to it arising from equilibration between kinetic enolate + unreacted ketone. $\endgroup$ Commented Nov 17, 2018 at 22:04

1 Answer 1


One problem with comparing product ratios using $\ce{LiN($i$Pr)2}$ vs $\ce{Ph3CLi}$ is that it's not really an apples-to-apples comparison - one is a lithium amide base, whereas the other is a alkyllithium that is highly stabilised by conjugation.

I think this illustrates two important points:

  1. Things are more complicated than they first appear, ie. are the species we draw really representative of the species undergoing reaction?
  2. Analysis of substrate and reagent structures will only get you so far, eg. for kinetically controlled processes, analysis of competing transition states is necessary.

Regarding the first point, there have been dozens of papers looking into the structure and reactivity of lithium amide and alkyllithium bases, particularly by the research group of David Collum. LDA is probably the most heavily studied. It turns out that these species are much more complex than what is implied by the structural formula. For instance, LDA in a coordinating solvent, such as THF, exists as a disolvated dimer:

enter image description here

How does the steric bulk of this species compare to the LDA structure we normally draw? Likewise, what is the actual structure of $\ce{Ph3CLi}$ in solution?

Even with a particular solvent and conditions, transition states have been put forward for the deprotonation event that vary markedly in structural features between substrates. For example, in a review of LDA (Angew. Chem. Int. Ed. 2007, 46, 3002-3017), Collum and coworkers show the following two different transition state structures for deprotonation of an ester using LDA in $\ce{tBuOMe}$ vs THF:

enter image description here

So, if we can get a good idea of what the transition state for LDA vs $\ce{Ph3CLi}$ deprotonation looks like, the complexity is such that high level calculations, eg. DFT, would be needed to explore the difference in selectivity. Even if we could qualitatively say which transition state involves the more bulky species, there may be features of the transition state, eg. conformational, stereoelectronic, that only become apparent through the modelling. Sometimes, it's not just which reagent is bigger, but details of how the two reactants come together at the transition state that determines the favoured outcome.

  • $\begingroup$ I wouldn’t consider (monomeric) Ph3CLi particularly resonance-stabilised, because there’s no way that the lone pair can be conjugated with all three phenyl groups at once. But the inductive withdrawal of the phenyl groups definitely pulls the pKa down quite a bit. Unrelated: Collum’s papers have an amazing sense of humour sometimes. $\endgroup$ Commented Nov 22, 2018 at 13:02
  • 1
    $\begingroup$ Yes, he can be quite conversational in some of them. He's also quite forthright on certain issues on social media. Point taken re: degree of resonance stabilisation of trityllithium, although there will be partial overlap of each phenyl ring pi system, so I wouldn't say that the lone pair can't be conjugated with all three at once. Stability of the trityl cation is evidence that the conjugation afforded by partial overlap due to lack of co-planarity of the phenyl rings is still significant. $\endgroup$ Commented Nov 22, 2018 at 14:08
  • $\begingroup$ Fair point re trityl cation! $\endgroup$ Commented Nov 22, 2018 at 14:46
  • $\begingroup$ Goodness me. 2018 me did NOT know about Collum's tweets at all. This is certainly offtopic for SE but I will leave a warning for people to not go near it, and I 100% don't endorse any of it... $\endgroup$ Commented Jun 5, 2020 at 13:46

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