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 final products for the Claisen condensation in each case are (a) C_6H_5C(O)CH_2C(O)OCH_2C_6H_5, and (b) CH_3(CH_2)_4C(O)CH_2C(O)O(CH_2)_4CH_3.

Step-by-step solution

(a) Ethyl phenylacetate in the presence of sodium ethoxide

(1) Write down the structure and reactant: Ethyl phenylacetate is an ester with the following structure: C_6H_5C(O)CH_2C(O)OCH_2CH_3. Sodium ethoxide (NaOEt) is the strong base used to aid in the Claisen condensation. (2) Formation of the enolate ion: The sodium ethoxide (NaOEt) donates an electron pair to the α-carbon between the carbonyl group and the ester in ethyl phenylacetate, forming an enolate ion and releasing a sodium ion (Na^+). (3) Nucleophilic addition: The enolate ion attacks the carbonyl carbon of another ethyl phenylacetate molecule, leading to the formation of an alkoxide intermediate. This step is a nucleophilic acyl substitution reaction. (4) Proton transfer: The alkoxide intermediate abstracts a proton from a molecule of ethanol, which is generated in the reaction, forming a β-keto ester as the final product. The final product for this reaction is C_6H_5C(O)CH_2C(O)OCH_2C_6H_5.

(b) Methyl hexanoate in the presence of sodium methoxide

(1) Write down the structure and reactant: Methyl hexanoate is an ester with the following structure: CH_3(CH_2)_4C(O)OCH_3. Sodium methoxide (NaOMe) is the strong base used to aid in the Claisen condensation. (2) Formation of the enolate ion: The sodium methoxide (NaOMe) donates an electron pair to the α-carbon between the carbonyl group and the ester in methyl hexanoate, forming an enolate ion and releasing a sodium ion (Na^+). (3) Nucleophilic addition: The enolate ion attacks the carbonyl carbon of another methyl hexanoate molecule, leading to the formation of an alkoxide intermediate. This step is a nucleophilic acyl substitution reaction. (4) Proton transfer: The alkoxide intermediate abstracts a proton from a molecule of methanol, which is generated in the reaction, forming a β-keto ester as the final product. The final product for this reaction is CH_3(CH_2)_4C(O)CH_2C(O)O(CH_2)_4CH_3.