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Chapter 19: Problem 29

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

Expert verified
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

01

(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.
02

(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.
03

(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.
04

(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.

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Most popular questions from this chapter

Show how to convert benzoic acid to 3-methyl-1-phenyl-1-butanone (isobutyl phenyl ketone) by the following synthetic strategies, each of which uses a different type of reaction to form the new carbon-carbon bond to the carbonyl group of benzoic acid. (a) A lithium diorganocopper (Gilman) reagent (b) A Claisen condensation

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2-Ethyl-1-hexanol was needed for the synthesis of the sunscreen octyl \(p\)-methylcinnamate (See Chapter 23, Chemical Connections: "Sunscreens and Sunblocks"). Show how this alcohol could be synthesized (a) by an aldol condensation of butanal and (b) by a malonic ester synthesis starting with diethyl malonate.

In 1887 , the Russian chemist Sergei Reformatsky at the University of Kiev discovered that treatment of an \(\alpha\)-haloester with zinc metal in the presence of an aldehyde or ketone followed by hydrolysis in aqueous acid results in formation of a \(\beta\)-hydroxyester. This reaction is similar to a Grignard reaction in that a key intermediate is an organometallic compound, in this case a zinc salt of an ester enolate anion. Grignard reagents, however, are so reactive that they undergo selfcondensation with the ester. Show how a Reformatsky reaction can be used to synthesize these compounds from an aldehyde or ketone and an \(\alpha\)-haloester.
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