# Rate-determining step of Claisen condensation

If we use a strong base such as LDA or NaH, will the rate-determining step change from enolate ion formation to nucleophillic acyl substitution step?

• If you use a strong base, you're likely generating the enolate in full before adding the electrophile, so I believe this question is moot. – Zhe Jul 11 '17 at 16:21
• @Zhe If the reaction were to be carried out in protic solvent, won't it also aid carbonyl bond polarisation. – Ayushmaan Jul 11 '17 at 16:55
• But you choose when to add the electrophile. The rate effects are totally irrelevant. – Zhe Jul 11 '17 at 17:23

Let me get one thing out of the way: the hydride ion is not a strong base. However, once a hydride is protonated to form hydrogen the acid-base equilibrium breaks down which makes it seem like a much stronger base than it is.

Your actual question — whether the rate determining step shifts from enolisation to nucleophilic attack — is harder to answer as it depends primarily on the reaction procedure. It depends on how you perform the Claisen condensation and what you want to condense.

If, for example, you want to perform a hetero Claisen condensation (i.e. couple two different esters), then you would typically make sure to deprotonate quickly and kinetically first before adding the second ester as Zhe implied in a comment. For example, you might use a very strong base such as $\ce{BuLi}$ or $\ce{LDA}$ for complete deprotonation followed by a waiting period before adding another ester. In this case, obviously you are already dealing with a fully deprotonated product and the question becomes moot.

However, you might be interested in a homocoupling product. For that, you might decide to add $\pu{0.5eq}$ of a strong base to a solution of the ester. Here, it depends on whether you add the strong base rapidly, (meaning the acid-base equilibrium is practically immediately established, in turn meaning that now the attack becomes rate determining) or whether you add it slowly and dropwise, in which case you, the experimentor, are rate-limiting deprotonation and causing it to remain the rate-determining step.

Finally, in response to a comment you gave under your question: You cannot use a protic solvent if you intend to use a strong base. Protic solvents are characterised by a relatively high acidity, meaning they will be deprotonated long before the ester ($\mathrm{p}K_\mathrm{a} \approx 25$). This will completely inhibit the reaction.

If a strong base is used, it will result in complete formation of the enolate ion, while in weaker bases, the RDS is enolate ion formation as it is a very weak acid.