Problem 139 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\) on reaction with sodium ethoxide in ethanol gives A, which on heating in the presence of acid gives B. Compound \(\mathrm{B}\) is (a) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\)

Step-by-Step Solution

Answer

Compound B would be acetone \( \mathrm{CH}_3 \mathrm{COCH}_3 \), not listed in the options.

Step 1: Identify the reaction components and conditions

The exercise states that ethyl acetate \( \mathrm{CH}_3 \mathrm{CO}_2 \mathrm{C}_2 \mathrm{H}_5 \) reacts with sodium ethoxide. Sodium ethoxide is a strong base, and the solvent is ethanol.

Step 2: Understand the reaction mechanism with sodium ethoxide

When ethyl acetate reacts with sodium ethoxide, it undergoes a Claisen condensation. This involves the deprotonation of an alpha hydrogen in the ester by the strong base, forming an enolate ion. This enolate then attacks the carbonyl carbon of another ester molecule, resulting in a β-keto ester.

Step 3: Determine the product A from the reaction

The product \( A \) formed is the β-keto ester \( \mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{COOC}_2 \mathrm{H}_5 \), specifically ethyl acetoacetate. This is because two molecules of ethyl acetate react to form this compound.

Step 4: Heating with acid to identify product B

Upon heating with acid, ethyl acetoacetate undergoes hydrolysis followed by decarboxylation. The hydrolysis removes the ester group, and the decarboxylation removes carbon dioxide, leaving an acetyl group. This results in the formation of \( \mathrm{CH}_3 \mathrm{COCH}_3 \), which is acetone.

Step 5: Match compound B with the given options

Option (a) is \( \mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{COOH} \), which is not acetone. Option (b) \( \mathrm{CH}_3 \mathrm{OCH}_3 \) is dimethyl ether. Neither matches our identified product \( \mathrm{CH}_3 \mathrm{COCH}_3 \). It appears there is a mismatch in the options given or an error in the question or it might actually involve a multi-step reaction leading to different compounds not listed here, considering heating and acid conditions could lead to different derivatives not matching standard outcomes.

Key Concepts

Sodium Ethoxide Reaction

Sodium ethoxide is a popular reagent in organic chemistry known for its ability to deprotonate compounds, instigating significant transformations. In the case of the Claisen condensation, sodium ethoxide plays a crucial role. When sodium ethoxide reacts with an ester like ethyl acetate, it acts as a strong base. It deprotonates the hydrogen atom adjacent to the carbonyl group, forming an enolate ion.
This enolate is a powerful nucleophile that subsequently attacks the carbonyl carbon of another ester molecule. The result is the formation of a new carbon-carbon bond—the foundation of a β-keto ester. This initial stage sets the stage for the subsequent formation of larger, more complex molecules.

• The base, sodium ethoxide, is essentially driving the initial reaction step by forming an enolate ion.
• This enolate is poised for further reaction, leading to the eventual formation of β-keto esters.
• This feature of sodium ethoxide is exploited in many synthetic pathways to create intricate molecular frameworks.

This reaction step is known for its role in building larger molecular structures from simpler starting materials, and sodium ethoxide is central to facilitating this transformation.

β-Keto Ester Formation

During a Claisen condensation, the reaction between two ester molecules leads to the formation of an important class of compounds known as β-keto esters. This reaction, specifically, begins with the generation of an enolate ion, which occurs through the deprotonation by a strong base such as sodium ethoxide.
The enolate ion then attacks the carbonyl carbon of another ester molecule, leading to the formation of a tetrahedral intermediate. This intermediate subsequently collapses, releasing an alcohol molecule and forming the β-keto ester bond.

• β-Keto esters are characterized by having a keto (carbonyl) group along with an ester group, separated by one carbon.
• The bond formation in β-keto esters is driven by the nucleophilic nature of the enolate ion which targets the electrophilic carbonyl groups.
• These compounds are often used as precursors in many types of synthetic pathways, including those that have applications in pharmaceuticals and agrochemicals.

Understanding this transformative stage offers insights into how carbon-carbon bonds can be strategically formed, aiding the construction of more complex molecules through accessible means.

Decarboxylation Process

After the formation of the β-keto ester, introducing heat and an acidic environment prompts a series of reactions known as hydrolysis and decarboxylation. The β-keto ester, in this scenario ethyl acetoacetate, undergoes hydrolysis which results in the breakdown of the ester group back to a carboxylic acid, releasing ethanol.
The subsequent decarboxylation part describes the removal of a carboxyl group as carbon dioxide, converting the molecule to a simpler ketone derivative. This process often leads to the formation of compounds like acetone from β-keto esters.

• Decarboxylation is driven by the instability of carboxylic acids when heat is applied, particularly when adjacent to a carbonyl group.
• This transformation is significant in synthetic organic chemistry for reducing molecules to smaller, economically valuable fragments.
• The final compound often has a more stable configuration post-decarboxylation, showcasing why this reaction is optimally used for simplifying molecular structures.

Recognizing the importance of decarboxylation processes lies in their frequent utility to yield commercially valuable or synthetically useful smaller ketone compounds efficiently.