[quantumnumber]

Hi guys,
the other day I found in a book an exercise about Claisen condensations... And I saw a reaction that I did not understand well. The reaction is showed in the picture I attach.

It is supposed that the ketone enolizes, and then it attacks the ethyl formate, leading to the condensation. My question is... Due to the fact that the ketone is more reactive towards nucleophiles, shouldn't the enolate attack the ketone instead of the ester? As far as I know, the carbonyl group in a ketone is more electrophilic than in an ester...

Thanks!

[discodermolide]

The formyl group is more reactive and the OEt is a good leaving group.

[quantumnumber]

But the charge in the ester is more spread because the carbonyl group is stabilized by conjugation. This doesn't happen in the ketone, the charge is more localized, so the carbonilic carbon in the ketone should be more electrophilic...

[discodermolide]

Check out the reactivity of ketones and aldehydes towards nucleophiles. Look for experiments where a mixture of ketone and aldehyde is reacted with a nucleophile. Which group is the most reactive?

[quantumnumber]

Aldehydes are more reactive, because of two factors:
1. Steric reasons (aldehyde is less hindered)
2. The electronic effect of the substituent; alkyl groups can weakly donate electronic density to the C=O carbon, making it less electrophilic.

But in the case of the ester, the electron density is spread between the O=C-OEt atoms, making the carbon also less electrophilic than in the ketone... Or that I thought...

[orgopete]

There is more to this than indicated simply by the arrow, namely order of addition. There are three components. What is added to what and what effect will that have? By the way, that isn't the product. No acid has been added and I think the enol is preferred.

[quantumnumber]

By the way, that isn't the product. No acid has been added and I think the enol is preferred.

Yeah, I forgot to mention that the last step in which the acid is added is omitted, but you have to take it into account.

[orgopete]

Yeah, I forgot to mention that the last step in which the acid is added is omitted, but you have to take it into account.

Well, in that case was anything else left out? Perhaps your text mentions something about crossed aldols and Claisen condensations. This would explain why the self condensation is not the major product. That was the question.

[quantumnumber]

No, that's all. I think it's just a simple exercise in which they want you to write the Claisen condensation between an ester and a ketone. But in my view, the self condensation should be also a product... I know that this ester in particular is quite reactive towards nucleophile attack, but I'm not so sure that it is more reactive than the ketone.

In Clayden's "Organic chemistry" I've read that certain esters (such as carbonates, or the formate ester showed in the exercise) could be more reactive than ketones towards nucleophilic attack, but usually ketones are more reactive than esters. But how to tell these small differences?

Sometimes looks like it is mandatory to carry out quantum mechanical calculations to be completely sure of this stuff...

[clarkstill]

The product of the reaction is entirely determined by thermodynamics: you don't need to worry about which is more reactive, just which product is lower in energy. The base isn't strong enough to completely deprotonate the ketone, so there will be a mixture of ketone and enolate forms in solution. Undoubtedly some of this undergoes a self-aldol condensation to generate a beta-hydroxy ketone, but this reaction will be reversible (via regeneration of the ketone and kicking out of the enolate) so doesn't really need consideration.  The Claisen condensation, however, generates a beta keto ester with a low enough pKa to be formally and fully deprotonated by ethoxide. This deprotonation of the 1,3-dicarbonyl makes it unreactive to nucleophilic hydroxide or ethoxide, preventing reversibility in the reaction and ultimately (after acidic workup) making the beta ketoester the final product.

[orgopete]

If ethyl formate were mixed with ethoxide, no apparent reaction takes place (ethoxide adds, but elimination returns ethyl formate). Adding the ketone slowly to the mixture will result in some enolization. However, because the molar concentration of formate is much greater than the ketone, it reacts faster.

If the reverse were attempted, that is add the formate to ketone and ethoxide, the aldol reaction takes place. Granted there is a stoichiometric effect also taking place. If an aldol reaction were catalyzed with hydroxide, elimination returns hydroxide so the reaction could be catalytic in base. The Claisen or the crossed Claisen in this case yields a more acidic product than the reactants and thus consumes the base. A Claisen requires at least a stoichiometric amount of base.

[quantumnumber]

Thanks for all your answers