Question

Show the product formed from each Michael product in the solution to Example \(19.13\) after (1) hydrolysis in aqueous \(\mathrm{NaOH}\), (2) acidification, and (3) thermal decarboxylation of each \(\beta\)-ketoacid or \(\beta\)-dicarboxylic acid. These reactions illustrate the usefulness of the Michael reaction for the synthesis of 1,5-dicarbonyl compounds.

Short answer

Answer: The final product is a 1,5-dicarbonyl compound with the structure CH₃C(=O)CH₂CH₂C(=O)H.

Step-by-step solution

Hydrolysis in aqueous NaOH

In the first step, we will hydrolyze the ester and the enol ether present in the molecule using aqueous NaOH. This reaction will convert the ester (\(\text{OEt}\)) to a carboxylate ion, and convert the enol ether CH(OMe) to a ketone. The product after hydrolysis will be: \( \text{CH}_3\text{C}(= {\text{O}})\text{CH}_2\text{C}(= {\text{O}}) \text{CH}_2\text{CH}_{2}\text{C}(= {\text{O}}) \text{ONa}\)

Acidification

In the second step, we will acidify the product from step 1 using a strong acid (e.g., HCl). This action will protonate the carboxylate ion, converting it back into a carboxylic acid. The product after acidification will be: \( \text{CH}_3\text{C}(= {\text{O}})\text{CH}_2\text{C}(= {\text{O}}) \text{CH}_2\text{CH}_{2}\text{C}(= {\text{O}}) \text{OH}\)

Thermal Decarboxylation

In the third step, we will perform thermal decarboxylation on the product from step 2. During this reaction, the β-ketoacid will lose a molecule of carbon dioxide (\(\text{CO}_2\)) to form a 1,5-dicarbonyl compound. The final product will be: \( \text{CH}_3\text{C}(= {\text{O}})\text{CH}_2\text{CH}_{2}\text{C}(= {\text{O}}) \text{H}\) Thus, the final product after hydrolysis, acidification, and thermal decarboxylation is a 1,5-dicarbonyl compound, which demonstrates the utility of the Michael reaction in synthesizing such compounds.