4
$\begingroup$

I'm having problems when trying to know the outcome of this reaction.

The first step must be a Michael addition of the 1,3-dicarbonyl compound to the α,β-unsaturated carbonyl compound, followed by a acid hydrolisis and a decarboxylation of the ester.

LDA must deprotone the least hindered hydrogen, that must be the $\alpha$-$\ce{CH3}$ to the non cyclic ketone. But at this point, I don't know how this reaction can continue to give one of the above products.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Intramolecular aldol reaction? $\endgroup$
    – orthocresol
    Commented Jun 27, 2019 at 16:58
  • $\begingroup$ @user55119 So it was as easy as a Robinson annulation, but the LDA step was driving me crazy, it makes sense now, thanks. $\endgroup$
    – user78790
    Commented Jun 27, 2019 at 18:49
  • $\begingroup$ I don't think you need LDA to finish an aldol condensation. The first step probably effect the Robinson annulation with the ester group intact. Acid hydrolysis would lead to CO2, methanol and compound C because the precursor is a vinylogous beta-ketoester. Probably a made-up problem. – $\endgroup$
    – user55119
    Commented Jun 27, 2019 at 20:05

2 Answers 2

Reset to default
3
$\begingroup$

Although the work of Wilds and Werth1 does not align perfectly with the conditions of the OP's question, it does illustrate that Robinson annulation and decarbomethoxylation can occur under alkaline conditions. Catalytic base was found to be optimal to effect the Michael addition with methyl vinyl ketone (MVK). The aldol condensation was conducted in aqueous KOH at reflux. Whether decarbomethoxylation occurs prior to aldolization (2-->3-->5) or subsequently (2-->4-->5) was not addressed. α'α-Dialkylated β-ketoesters generally undergo attack at the keto group rather the carboxylate by base. This generality may be tempered by the aromatic nature of the keto group in 2. Initial aldolization of 2 to 4 affords a structure that is a vinylogous β-ketoester; it is susceptible to base initiated decarboxylation.

enter image description here

1) A. L. Wilds and R. G. Werth, J. Org. Chem. 1952, 17, 1154.

Improve this answer
$\endgroup$
2
$\begingroup$

The OP's original question has some flaws. Since he is asking the most possible product among the given four, I'd give reasonable mechanism to achieve that product, which would be C. Following is the suggested mechanism:

Aldol-Michael addition

Late addition: Evidently, lithium diisopropylamide (LDA) has been used to selectively abstract proton from the least steric position among three available acidic hydrogens (irreversible generation of the kinetic enolate). Such a non-nucleophilic, sterically-demanding, strong base will always abstract a proton from the least hindered side. The resultant lithium enolates will avoide Proton transfer at low temperatures in ethereal solvents, so that addition of a second carbonyl group will produce the desired aldol product (Organic Chemistry Portal). For an example for use of lithium dialkyl amide in aldol condensation, see the given reference:

Atsushi Seki, Fusae Ishiwata, Youichi Takizawa, Masatoshi Asami, "Crossed aldol reaction using cross-linked polymer-bound lithium dialkylamide," Tetrahedron 2004, 60(23), 5001-5011 (https://doi.org/10.1016/j.tet.2004.04.026).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Mathew Mahindaratne: Do you have a reference for the use of LDA in the context of your example? If so, could you post it in your answer? Conceptually, it works! Practically, does it? I'm curious. PS: In your fourth structure, the keto form is undoubtedly better than the enol form. $\endgroup$
    – user55119
    Commented Jun 28, 2019 at 11:47
  • $\begingroup$ @user55119: I have added a paragraph and a reference for your request. $\endgroup$
    – Mathew Mahindaratne
    Commented Jun 28, 2019 at 21:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.