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 is 5-methyl-3-oxohexanoic acid ethyl ester.

Step-by-step solution

Identify the nucleophile and electrophile

The Claisen condensation reaction is an organic reaction in which an ester molecule reacts with another ester molecule to form a β-keto ester. In this case, ethyl 3-methylbutanoate is the ester molecule, and sodium ethoxide is the base used to generate the nucleophile, which will attack the carbonyl group of the ester, acting as the electrophile.

Determine the reaction mechanism

The mechanism for this reaction consists of three major steps: 1. Deprotonation of the ester (ethyl 3-methylbutanoate) by the base (sodium ethoxide) to form an enolate anion. 2. Nucleophilic attack of the enolate anion on the carbonyl group of another ester molecule. 3. Acidification with the addition of aqueous HCl.

Show the product formation

Let's go through the reaction step by step: 1. Deprotonation: Sodium ethoxide deprotonates the α-hydrogen of ethyl 3-methylbutanoate, forming the enolate anion and ethoxide ion. \(\mathrm{CH_{3}CH_{2}O^{-}Na^{+}} + \mathrm{CH_{3}CH_{2}COCH(CH_{3})C(O)OCH_{2}CH_{3}} \rightarrow \mathrm{CH_{3}CH_{2}OH} + \mathrm{CH_{3}CH=C(O^{-})CH(CH_{3})C(O)OCH_{2}CH_{3}}\) 2. Nucleophilic attack: The enolate anion acts as a nucleophile and attacks the carbonyl group (C=O) of another ethyl 3-methylbutanoate molecule. The carbonyl group reforms, and the electrons from the original π-bond (C=O) are pushed onto the oxygen atom, which then takes a proton from an ethoxide ion to form the intermediate product. \(\mathrm{CH_{3}CH=C(O^{-})CH(CH_{3})C(O)OCH_{2}CH_{3}} + \mathrm{CH_{3}CH_{2}COCH(CH_{3})C(O)OCH_{2}CH_{3}} \rightarrow \mathrm{CH_{3}CH=C(O)CH(CH_{3})C(O)OCH_{2}COCH(CH_{3})C(O)OCH_{2}CH_{3}}\) 3. Acidification: Finally, the addition of aqueous HCl protonates the oxyanion (O-) and forms the final product, which is a β-keto ester. \(\mathrm{CH_{3}CH=C(O)CH(CH_{3})C(O)OCH_{2}COCH(CH_{3})C(O)OCH_{2}CH_{3}} + \mathrm{H^{+}} \rightarrow \mathrm{CH_{3}CH=C(O)CH(CH_{3})C(O)OCH_{2}COCH(CH_{3})C(O)OCH_{3}}\) The product of the Claisen condensation of ethyl 3-methylbutanoate in the presence of sodium ethoxide followed by acidification with aqueous HCl is 5-methyl-3-oxohexanoic acid ethyl ester.