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Do Claisen Condensation products prefer the di-keto form, or a keto/enol form due to increased conjugation and hydrogen bonding?

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The Claisen reaction yields a product with an acidified proton in $\alpha$-position of carbonyl and ester, right at the same carbon atom of R depicted in blue.enter image description here

In presence of a base, yes, $\beta$-dicarbonyl compounds undergo deprotonation at their methylene unit to yield a keto-enolate, stabilized by mesomerism with the secound carbonyl group in conjugation. Compare the $pK_a$-values of $\beta{}$-dicarbonyl compounds with ketones, for example (one reference), to estimate if the equilibrium is on the left, or on the right. enter image description here.

I think cations of the right size and charge equally favour the formation of the keto-enolate. If you synthesize $\beta$-dicarbonyl compounds, the risk to loose product due to chelation is lowered if initial work-up is concluded by drying the washed organic phase with sodium sulphate ($\ce{Na2SO4}$), rather than by magnesium sulphate ($\ce{MgSO4}$).

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Generally, diketo compounds can react to form an enol-keto form and back again. This tautomery can not be considered a [1,3] proton shift due to orbital chemistry and has to be thought of as a combination of deprotonation/reprotonation or protonation/deprotonation steps — ask another question if you’re interested in details (assuming they are not already posted somewhere on the site).

Notice, that in its enol-keto form, the compound can form an intramolecular hydrogen bond. If desired, the $\mathrm{p}K_{\mathrm{a}}$ of the molecule itself is enough to protonoate another one, so the process can be thought of as self-catalysed. Thus, there is a tendency of these compounds to equilibrate to a desired ratio. Only minor differences need to be considered, if $\mathrm{R} = \ce{OR''}$

But what is that ratio going to be? It depends on the surroundings. If you dissolve it in solvents that are capable of donating hydrogen bonds, then both keto oxygens will accept H-bonds, and the diketo tautomer will be preferred. On the other hand, if the solvent cannot form any hydrogen bonds, the enol tautomer, which is capable of forming an intramolecular H-bond, is less disfavoured for that exact reason. Further parameters like other molecules dissolved, their $\mathrm{p}K_{\mathrm{a}}$’s, the dryness of the solvent, the pH-value etc. also factor in.

Which tautomer dominates in which solvent and what the ratio between the two is (it is extremely unlikely for one form to be the only one present) cannot be predicted exactly and needs to be measured experimentally.

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  • $\begingroup$ The question is about $\beta$ keto-esters not 1,3 diketones though. $\endgroup$ – bon May 14 '15 at 17:32
  • $\begingroup$ @bon See last line of second paragraph. Does it change much? $\endgroup$ – Jan May 14 '15 at 17:43
  • $\begingroup$ Missed that. I'm not sure how much it changes though so have to let someone else answer that. $\endgroup$ – bon May 14 '15 at 18:39

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