Question

Claisen condensation between diethyl phthalate and ethyl acetate followed by saponification, acidification, and decarboxylation forms a diketone, \(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{O}_{2}\). Diethyl phthalate Ethyl acetate Propose structural formulas for compounds \(A\) and \(B\) and the diketone.

Short answer

Question: Determine the structural formulas for the intermediate compounds A and B, as well as the diketone, in a reaction sequence involving Claisen condensation, saponification, acidification, and decarboxylation. Answer: The structural formulas for the compounds are: - Compound A: \(\mathrm{C}_{6} \mathrm{H}_{4}(COOCH_{2}CH_{3})_{2}(COCH_{2}COCH_{3})\) - Compound B: \(\mathrm{C}_{6} \mathrm{H}_{4}(COO^{-})_{3}\) - Diketone: \(\mathrm{C}_{6} \mathrm{H}_{4}(COCH_{3})_{2}\)

Step-by-step solution

Claisen Condensation

The Claisen condensation is a reaction between an ester (ethyl acetate in this case) and a suitable enolizable compound (diethyl phthalate) resulting in the formation of a β-keto ester (compound \(A\)) and a small molecule byproduct (ethanol). In this case, the ester compound is ethyl acetate (\(\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\)) and diethyl phthalate (\(\mathrm{C}_{6} \mathrm{H}_{4}(COOCH_{2}CH_{3})_{2}\)). The Claisen condensation between these two compounds will lead to the formation of compound \(A\) with the formula \(\mathrm{C}_{6} \mathrm{H}_{4}(COOCH_{2}CH_{3})_{2}(COCH_{2}COCH_{3})\).

Saponification

Saponification is the process of hydrolyzing an ester under basic conditions to form an alcohol and a carboxylate salt. After saponification, both the ester groups of compound \(A\) are converted to carboxylate anions, giving compound \(B\) with the formula \(\mathrm{C}_{6} \mathrm{H}_{4}(COO^{-})_{3}\).

Acidification

Acidification is the process of adding an acid to convert a carboxylate anion into a carboxylic acid group. Acidification of compound \(B\) converts all three carboxylate groups to carboxylic acid groups, giving the tricarboxylic acid compound with the formula \(\mathrm{C}_{6} \mathrm{H}_{4}(COOH)_{3}\).

Decarboxylation

Decarboxylation is the process of removing a carboxyl group from a molecule as carbon dioxide (\(\mathrm{CO}_{2}\)). As there are three carboxyl groups in the tricarboxylic acid compound, we need to decarboxylate two of them to get our final diketone product with the formula \(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{O}_{2}\). The two carboxyl groups that are adjacent to each other will have a higher tendency to decarboxylate since the resulting product will be more stable. The resulting diketone will have the structure \(\mathrm{C}_{6} \mathrm{H}_{4}(COCH_{3})_{2}\). In conclusion, the structural formulas for the compounds are: - Compound \(A\): \(\mathrm{C}_{6} \mathrm{H}_{4}(COOCH_{2}CH_{3})_{2}(COCH_{2}COCH_{3})\) - Compound \(B\): \(\mathrm{C}_{6} \mathrm{H}_{4}(COO^{-})_{3}\) - Diketone: \(\mathrm{C}_{6} \mathrm{H}_{4}(COCH_{3})_{2}\)

Key concepts

Saponification

Saponification is a fundamental organic chemistry process often encountered in both laboratory and industrial settings. It involves the hydrolysis of ester compounds in the presence of a base, typically sodium hydroxide (NaOH) or potassium hydroxide (KOH), to produce an alcohol and a carboxylate salt. The name 'saponification' literally means 'soap making' as it is the traditional method used to make soap from animal fats (which contain esters).

During the saponification step of the Claisen condensation exercise, the ester groups in compound A react with a strong base. This causes the ester linkage to break, forming an alcohol (ethanol in this case) and carboxylate anions (a salt form of carboxylic acids). Carboxylate anions are more nucleophilic than esters, which allows subsequent chemical reactions to occur more readily. Mastery of the saponification concept is crucial for students as it sets the stage for further chemical transformations, such as acidification and decarboxylation, in multi-step synthesis problems.

Acidification

Acidification is the next step after saponification in many chemical syntheses, where it allows us to convert carboxylate salts back to their corresponding carboxylic acids. The process is straightforward: add an acid, often hydrochloric acid (HCl) or sulfuric acid (H2SO4), to the carboxylate salt solution. The acid donates a proton (H+) to the negatively charged carboxylate ion (RCOO-), neutralizing the charge and forming a carboxylic acid (RCOOH).

In the context of the Claisen condensation exercise, acidification is employed to transform compound B, which contains carboxylate groups, into a tricarboxylic acid. This compound is particularly important because it is the precursor to the desired diketone product. By understanding the acidification process, students can appreciate its utility in preparing compounds for subsequent steps such as decarboxylation. It showcases the interplay between acids and bases in chemistry, as well as the importance of pH control in synthesis.

Decarboxylation

Decarboxylation, as the term implies, involves the removal of a carboxyl group (–COOH) from a molecule, releasing carbon dioxide (CO2). This chemical reaction is significant in organic synthesis because it can simplify complex molecules and is often used in the preparation of aromatic ketones, which are valuable in various industries.

In our textbook problem, decarboxylation serves as the final step to produce the diketone with the formula \(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{O}_{2}\). The tricarboxylic acid formed during acidification undergoes heat-induced decarboxylation, where two of the three carboxyl groups are eliminated. This process is driven by the stability of the resulting diketone structure, which is less energy-intensive for the molecule to maintain. Students should note how decarboxylation is influenced by not only temperature but also by the molecular stability of the products formed. Grasping decarboxylation helps students predict the outcomes of similar reactions and understand their relevance in creating more complex chemical compounds.