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I'm having a hard time determining regio-selectivity in those two reactions.

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At the image above I've drawn one example for each on of them. My question is how do we choose where the deprotonation will occur. My approach was that the base will deprotonate the substrate in order to produce the most stable enolate. In the Claisen condensation, A should be the main product since the -CH2- group is present thus corresponding the mechanism. For B there will be no acidic hydrogens so it is not favoured, or at least that's what I thought.

Moving to the aldol condensation I can't figure out which product will be the main. Theoretically in C, the enolate is more stable than that of D, but the final unsaturated ketone isn't more stable in D?

Either way, I can't comprehend how can we determine what enolate will be produced each time.

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Claisen and aldol condensations are thermodynamically controlled,* so it is not a question of which enolate forms, it is a question of which product forms.

In your first example, product A can be deprotonated by ethoxide as there is still an acidic hydrogen between the two carbonyls. Product B, on the other hand, has a quaternary carbon between the two carbonyls and cannot be deprotonated.

Because the pKa of a 1,3-dicarbonyl (~10) is significantly smaller than the pKa of ethanol (~15),** this deprotonation of A is almost complete and it therefore pulls the equilibrium position over to one which favours product A, simply by le Chatelier's principle. In fact if you somehow make product B via a different route and then treat it with ethoxide, you should exclusively get product A out of your flask, via retro-Claisen and Claisen condensation.

For the second example I would expect product D (tetrasubstituted alkene) to be preferentially formed over C (trisubstituted).


* Unless you use kinetic deprotonation conditions such as LDA. In this case you will then have to look at the stability of the transition states leading to enolate formation, but that's a story for another day.

** Very rough values.

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  • $\begingroup$ Thank you for your answer. Another way I think to justify the Claisen product A is the stability of the anion. The one produced in A has one alkyl group while the B has two. Given that alkyl are electron donors, A should be favoured. Regarding aldol condensation, some of my colleagues support that while D forms a more stable C=C bond it is also the most sterically hindered so C is favoured, basically two things competing each other. $\endgroup$ – Αντώνιος Κελεσίδης Jan 31 '18 at 10:28
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    $\begingroup$ No, enolate stability really has nothing to do with it (and even if it did, the more substituted enolate is more stable: note that the predominant resonance form of an enolate is $\ce{C=C-O-}$, not $\ce{^-C-C=O}$). $\endgroup$ – orthocresol Jan 31 '18 at 11:21
  • $\begingroup$ Why do you say that these reactions are thermodynamically controlled? Do you mind elaborating on that? $\endgroup$ – Tan Yong Boon Aug 7 '18 at 12:16
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    $\begingroup$ @TanYongBoon all steps are generally considered to be reversible under the typical conditions - NaOEt and heat for the Claisen, for example. Except for the deprotonation of the a,b-unsaturated ketone product - that is irreversible. Likewise for the aldol the simplest procedures just add Ba(OH)2 or NaOH to the substrate and heat. Everything there is reversible. $\endgroup$ – orthocresol Aug 7 '18 at 12:37
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    $\begingroup$ @TanYongBoon, that's what I was very briefly alluding to with LDA. Generally, you add carbonyl A to LDA to generate enolate A, and then you add carbonyl B to enolate A. Funnily enough, the rate of the aldol addition is faster than the rate of proton exchange between A and B, which allows you to get good yields of the desired aldol product; C–H deprotonations are generally slower than one might think. I've never actually had to do an aldol myself, so I don't know how much enolisation of B you see (if any), but there's an example procedure here: orgsyn.org/demo.aspx?prep=CV7P0185 $\endgroup$ – orthocresol Aug 8 '18 at 2:04

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