Acetaldehyde reacts with malonic acid in presence of sodium ethoxide at high temperature to give (a) acetic acid (b) propanoic acid (c) tartaric acid (d) crotonic acid

Short Answer

Expert verified
Answer: Propanoic acid.

Step by step solution

01

Write down the reagents and their structures

First, let's write down the structures of the reactants: acetaldehyde, malonic acid, and sodium ethoxide. - Acetaldehyde: CH3CHO - Malonic acid: HOOC-CH2-COOH - Sodium ethoxide: C2H5O-Na+
02

Identify the nucleophile and electrophile

In the Claisen condensation, a nucleophile (in this case, malonic acid) attacks an electrophile (in this case, acetaldehyde). The sodium ethoxide serves as a base to deprotonate the nucleophile, making it more nucleophilic.
03

Deprotonation of malonic acid

Malonic acid has acidic protons at its carbonyl alpha position, which can be deprotonated by sodium ethoxide: \[C_2H_5O^{-}Na^+ + HOOC-CH_2-COOH \rightarrow C_2H_5OH + NaOOC-CH^-_2-COOH\]
04

Nucleophilic attack on acetaldehyde

Now that the malonic acid is deprotonated, it can act as a nucleophile and attack the electrophilic carbonyl carbon of acetaldehyde: \[CH_3CHO + NaOOC-CH^-_2-COOH \rightarrow CH_3CH(O^-)CH_2-COONa\]
05

Protonation of the oxygen

The oxygen anion generated in the previous step then takes a proton from either water or another molecule of malonic acid. The resulting product is a beta-keto ester: \[CH_3CH(O^-)CH_2-COONa + HOOC-CH_2-COOH \rightarrow CH_3CH(OH)CH_2-COONa + HOOC-CH_2-COO^-\]
06

Identify the final product

Upon heating, the beta-keto ester undergoes a decarboxylation reaction, in which the carboxylic acid group linked with an alpha-keto ester is removed as carbon dioxide (CO2). The final product is propanoic acid: \[CH_3CH(OH)CH_2-COONa \rightarrow CH_3CH_2COOH + NaOH + CO_2\] Based on the reaction mechanism and the resulting product, the correct option is: (b) propanoic acid

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