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What is the mechanism of the hydrolysis reaction of an ester using water and HCl as the acid catalyst, including the intermediate steps involved?

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ago by (139k points)
The hydrolysis of an ester using water and HCl as the acid catalyst involves the following steps:1. Protonation of the ester carbonyl oxygen:The reaction begins with the protonation of the ester carbonyl oxygen by the HCl, which acts as an acid catalyst. This step increases the electrophilicity of the carbonyl carbon, making it more susceptible to nucleophilic attack.Ester + HCl  Protonated Ester2. Nucleophilic attack by water:Water, acting as a nucleophile, attacks the electrophilic carbonyl carbon of the protonated ester, forming a tetrahedral intermediate.Protonated Ester + H2O  Tetrahedral Intermediate3. Deprotonation of the tetrahedral intermediate:The tetrahedral intermediate undergoes deprotonation, losing a proton to form a neutral tetrahedral intermediate. This step can be facilitated by the conjugate base of the acid catalyst  Cl- .Tetrahedral Intermediate  Neutral Tetrahedral Intermediate + HCl4. Protonation and leaving group departure:The neutral tetrahedral intermediate undergoes protonation at the oxygen atom of the former ester group, which increases its leaving group ability. The protonated oxygen then leaves as a neutral alcohol molecule, resulting in the formation of a carbocation.Neutral Tetrahedral Intermediate + HCl  Carbocation + Alcohol5. Deprotonation of the carbocation:Finally, the carbocation undergoes deprotonation, facilitated by the conjugate base of the acid catalyst  Cl- , to form the carboxylic acid product.Carbocation + Cl-  Carboxylic Acid + HClOverall, the hydrolysis of the ester using water and HCl as the acid catalyst results in the formation of a carboxylic acid and an alcohol.
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