Question

Show the product of Claisen condensation of each ester. (a) Ethyl phenylacetate in the presence of sodium ethoxide (b) Methyl hexanoate in the presence of sodium methoxide

Short answer

Answer: The products of Claisen condensation for the given esters are: (a) CH_3COOCH(C_6H_5)CH(O)COCH_3C_6H_5 (a β-keto ester), and (b) CH_3(CH_2)_4COOCH_2(CH_2)_4COOCH_3 (another β-keto ester).

Step-by-step solution

(a) Ethyl phenylacetate and sodium ethoxide

Ethyl phenylacetate is an ester with the structure CH_3COOCH_2C_6H_5. Sodium ethoxide is a strong base with the formula C_2H_5ONa.

(b) Methyl hexanoate and sodium methoxide

Methyl hexanoate is an ester with the structure CH_3(CH_2)_4COOCH_3. Sodium methoxide is a strong base with the formula CH_3ONa. #Phase 2: Following the mechanism of Claisen condensation and obtaining the products# Now, we will follow the Claisen condensation mechanism for each case to obtain the products.

(a) Claisen condensation of ethyl phenylacetate in the presence of sodium ethoxide

The mechanism involves the deprotonation of the α-carbon of an ester molecule by the strong base, followed by the nucleophilic attack on the carbonyl carbon of another ester molecule, and finally, protonation of the alkoxide formed. 1. Deprotonation: C_2H_5O^-(Na^+) + CH_3COOCH_2C_6H_5 -> C_2H_5OH + CH_3COOCH(-)C_6H_5 2. Nucleophilic attack: CH_3COOCH(-)C_6H_5 + CH_3COOCH_2C_6H_5 -> CH_3COOCH(C_6H_5)CH(-O^-)COCH_3C_6H_5 3. Protonation: CH_3COOCH(C_6H_5)CH(-O^-)COCH_3C_6H_5 + C_2H_5OH -> CH_3COOCH(C_6H_5)CH(O)COCH_3C_6H_5 The final product is the β-keto ester: CH_3COOCH(C_6H_5)CH(O)COCH_3C_6H_5.

(b) Claisen condensation of methyl hexanoate in the presence of sodium methoxide

Similarly, we follow the Claisen condensation mechanism for methyl hexanoate and sodium methoxide. 1. Deprotonation: CH_3O^-(Na^+) + CH_3(CH_2)_4COOCH_3 -> CH_3OH + CH_3(CH_2)_4COOCH_2^- 2. Nucleophilic attack: CH_3(CH_2)_4COOCH_2^- + CH_3(CH_2)_4COOCH_3 -> CH_3(CH_2)_4COOCH_2(CH_2)_4COOCH_2^- 3. Protonation: CH_3(CH_2)_4COOCH_2(CH_2)_4COOCH_2^- + CH_3OH -> CH_3(CH_2)_4COOCH_2(CH_2)_4COOCH_3 The final product is the β-keto ester: CH_3(CH_2)_4COOCH_2(CH_2)_4COOCH_3.

Key concepts

β-keto ester

In the Claisen condensation reaction, a significant product is the β-keto ester. This molecule is formed when two esters combine, with one of the esters contributing its β-position (or β-carbon) to the keto group of the other ester. The β-keto ester has a ketone group, which is characterized by a carbon atom doubly bonded to an oxygen (C=O), positioned at the β-carbon relative to the ester group.
This structure is crucial as it provides the stability required for further transformation in different chemical reactions. The β-keto ester's stability is partly due to resonance, which allows electron delocalization across the molecule, thereby lowering its energy.

Nucleophilic attack

A nucleophilic attack is a key step in the Claisen condensation reaction mechanism. Here, a nucleophile, which is an electron-rich species, attacks an electron-deficient area, often a carbon atom bonded to an electronegative atom like oxygen.
During Claisen condensation, after deprotonation of an ester, the resulting carbanion acts as the nucleophilic species. This carbanion attacks the carbonyl carbon of another ester molecule, forming a new carbon-carbon bond and leading towards the product formation. This step is crucial as it sets the foundation for the molecular framework of the final β-keto ester product.

Deprotonation

In chemical terms, deprotonation refers to the removal of a proton (H⁺ ion) from a molecule, facilitated by a base. In the Claisen condensation, deprotonation occurs at the α-carbon of the ester which is slightly acidic due to its proximity to the carbonyl group. The ester is deprotonated by a strong base, forming a carbanion.
This carbanion is reactive and plays a vital role in attacking the carbonyl of another ester in the subsequent nucleophilic attack. This step is vital because it activates the ester making it more reactive and ready for the formation of the new carbon-carbon bond in the reaction.

Sodium ethoxide

Sodium ethoxide is a strong base represented by the formula C₂H₅ONa and is commonly used in organic reactions such as Claisen condensation. It acts by deprotonating the α-hydrogen of an ester, generating a carbanion.
Its role doesn't only stop at deprotonation, but it also enables the reaction medium by being soluble in organic solvents. With its alkoxide ion, sodium ethoxide ensures effective deprotonation leading to the formation of a stable intermediate conducive for further reaction steps.

Sodium methoxide

Similar to sodium ethoxide, sodium methoxide (CH₃ONa) is a robust base used in various organic transformations, particularly in the Claisen condensation of methyl esters. This compound effectively deprotonates α-hydrocarbons in an ester, aiding in the generation of a carbanion that partakes in the nucleophilic attack on another ester.
It is important for reactions involving methoxide to handle it carefully as it is moisture-sensitive. Its effectiveness comes from its strong base properties and solubility in organic solvents, which facilitate the formation of essential intermediates in the Claisen condensation pathway.