Question

Show the product of Claisen condensation of ethyl 3 -methylbutanoate in the presence of sodium ethoxide followed by acidification with aqueous \(\mathrm{HCl}\).

Short answer

Answer: The product of Claisen condensation of ethyl 3-methylbutanoate in the presence of sodium ethoxide followed by acidification with aqueous HCl is CH3CHCHCOOCH2C(CH3)COOH.

Step-by-step solution

Identify the starting material and reagent

In this problem, the starting material is ethyl 3-methylbutanoate and the reagent is sodium ethoxide.

Identify the nucleophile and electrophile

In Claisen condensation, an ester molecule acts as both a nucleophile and an electrophile. The ethoxide ion from the sodium ethoxide serves as the base, which deprotonates the ester's alpha-hydrogen, forming a resonance-stabilized enolate ion. The carbonyl oxygen of another ester molecule acts as the electrophile.

Forming the enolate ion

The ethoxide ion attacks the alpha-hydrogen of ethyl 3-methylbutanoate, which results in the formation of the resonance-stabilized enolate ion. The intermediate enolate ion looks like this: \(\mathrm{CH_3CHCHCOOCH_2CH_3}\)

Nucleophilic attack on the ester carbonyl carbon

The generated enolate ion attacks the carbonyl carbon of a second ethyl 3-methylbutanoate molecule, and the carbonyl double bond shifts to the oxygen, forming a new carbon-carbon bond. The product after nucleophilic attack will be \(\mathrm{CH_3CHCHCOO(CH_2)_3CH_3}\)

Regeneration of carbonyl group and leaving group formation

The oxygen reforms the double bond with the carbonyl carbon, forcing the ethoxide group to leave as the leaving group. The product will now be \(\mathrm{CH_3CHCHCOOCH_2C(CH_3)COOCH_2CH_3}\)

Acidification with aqueous HCl

Finally, the reaction mixture is treated with aqueous HCl, which protonates the ester oxygen and forms the final product. The final product after acidification is \(\mathrm{CH_3CHCHCOOCH_2C(CH_3)COOH}\)