**Title: Understanding Mixed Claisen Ester Condensation****Objective:** Learn how to draw the most stable form of the major mixed Claisen product formed in the given reaction.**Reaction Overview:**- **Reactants:** - Two esters, a propionate ester and an ethyl acetate. - **Reagents:** 1. Sodium ethoxide (NaOC₂H₅) 2. Acidic workup (H₃O⁺)- **Solvent:** Ethanol (C₂H₅OH)**Mechanism Explanation:**1. **Deprotonation:** Sodium ethoxide (NaOC₂H₅) deprotonates the alpha-hydrogen of one of the esters, forming an enolate ion.2. **Nucleophilic Attack:** The enolate ion attacks the carbonyl carbon of the other ester, leading to the formation of a β-keto ester intermediate.3. **Protonation:** Acidic workup (H₃O⁺) protonates the alkoxide intermediate, resulting in the stable Claisen condensation product.**Structural Insight:**- Both ester groups participate in the reaction, but the more substituted enolate is typically favored in mixed Claisen reactions.- The major product is a β-keto ester.**Key Learning Points:**- Recognize how mixed Claisen condensations form key carbon-carbon bonds.- Understand the role of each reagent and solvent in the mechanism.- Predict the structure of the Claisen product by considering enolate formation and nucleophilic acyl substitution.This reaction is integral in organic synthesis, providing a pathway to create complex molecules from simple ester precursors. Understanding this process aids in mastering broader synthetic strategies in chemistry.

Claisen condensation is the reaction in which enolates are added to an ester molecule. Enolate is formed from the carbonyl compound or ester molecule with alpha hydrogens. The main three steps involved in the Claisen condensation reaction are: 1) Reaction of ester having alpha hydrogen with a base. This forms an enolate ion. 2)Enolate ion reacts with the carbonyl carbon of the another ester molecule. 3)In the third step elimination of alkoxide group from the above addition product takes place. Example:The given reactants are: In Claisen condensation, the most acidic hydrogen is eliminated from the alpha position of the ester by the base. Between the given esters, the enolate ion is formed from ethyl acetate and not from the ethyl isobutyrate. The reason is the alpha hydrogens in ethyl acetate are more acidic than ethyl isobutyrate. Since in ethyl isobutyrate the two methyl groups reduce the acidity of the alpha hydrogens.The product formed with mechanism is as shown below:Major product formed in the reaction is: