Predict the major product obtained when ethyl acetate (${\mathrm{CH}}_3\mathrm{COOEt}$) undergoes a Claisen condensation.

• A. ${\mathrm{CH}}_3\mathrm{COOCH}3$ (methyl acetate)
• B. ${\mathrm{CH}}_3\mathrm{COCH}3$ (acetone)
• C. ${\mathrm{CH}}_3\mathrm{COCH}2\mathrm{COOEt}$ (ethyl acetoacetate)
• D. ${\mathrm{CH}}_3\mathrm{COOH}$ (acetic acid)

Answer: (C)

Verified step by step guidance

1. Identify the structure of ethyl acetate, which is an ester with the formula \(\mathrm{CH_3COOEt}\), where the alpha hydrogens (hydrogens on the carbon adjacent to the carbonyl) are acidic and can be deprotonated under basic conditions.
2. Recall that the Claisen condensation involves the formation of an enolate ion from one ester molecule by deprotonation at the alpha carbon using a strong base, typically an alkoxide ion matching the ester alkoxy group to avoid transesterification.
3. Write the enolate formation step: the base abstracts an alpha hydrogen from ethyl acetate, generating an enolate ion \(\mathrm{CH_2^-COOEt}\), which is resonance stabilized between the alpha carbon and the carbonyl oxygen.
4. Next, the enolate ion attacks the carbonyl carbon of a second molecule of ethyl acetate in a nucleophilic acyl substitution, forming a tetrahedral intermediate that collapses to expel an ethoxide ion, resulting in a β-ketoester intermediate.
5. Finally, the product formed is ethyl acetoacetate (\(\mathrm{CH_3COCH_2COOEt}\)), which is a β-ketoester, and the leaving group is ethoxide ion that can pick up a proton to form ethanol or react further; the reaction also produces acetic acid as a byproduct after workup.