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rite the mechanism for the reaction below. -OH + CH3OH

rite the mechanism for the reaction below. -OH + CH3OH

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Reaction Mechanism: Ester Hydrolysis in Basic Conditions** In the given reaction, an ester is being hydrolyzed in the presence of a hydroxide ion (−OH). This is a nucleophilic substitution reaction that results in the formation of a carboxylate ion and methanol. **Reaction Steps:** 1. **Nucleophilic Attack**: The hydroxide ion (−OH) acts as a nucleophile and attacks the carbonyl carbon of the ester. This results in the formation of a tetrahedral intermediate. 2. **Intermediate Formation**: The negatively charged oxygen of the tetrahedral intermediate re-forms the carbonyl double bond, causing the methoxy group (−OCH3) to leave. 3. **Product Formation**: The carboxylate ion (with a negative charge on the oxygen) and methanol (CH3OH) are formed. **Diagram Explanation:** - On the left side of the arrow, the ester structure is depicted, with a carbonyl group (C=O) and an attached methoxy group (−OCH3). - The hydroxide ion (−OH) is shown as the reactant leading to the right side of the equation. - On the right side of the arrow, the products include a carboxylate ion (indicated by the negatively charged oxygen) and methanol (CH3OH). This mechanism illustrates how esters are converted into their corresponding carboxylate ions and alcohols under basic conditions.
Transcribed Image Text:**Reaction Mechanism: Ester Hydrolysis in Basic Conditions** In the given reaction, an ester is being hydrolyzed in the presence of a hydroxide ion (−OH). This is a nucleophilic substitution reaction that results in the formation of a carboxylate ion and methanol. **Reaction Steps:** 1. **Nucleophilic Attack**: The hydroxide ion (−OH) acts as a nucleophile and attacks the carbonyl carbon of the ester. This results in the formation of a tetrahedral intermediate. 2. **Intermediate Formation**: The negatively charged oxygen of the tetrahedral intermediate re-forms the carbonyl double bond, causing the methoxy group (−OCH3) to leave. 3. **Product Formation**: The carboxylate ion (with a negative charge on the oxygen) and methanol (CH3OH) are formed. **Diagram Explanation:** - On the left side of the arrow, the ester structure is depicted, with a carbonyl group (C=O) and an attached methoxy group (−OCH3). - The hydroxide ion (−OH) is shown as the reactant leading to the right side of the equation. - On the right side of the arrow, the products include a carboxylate ion (indicated by the negatively charged oxygen) and methanol (CH3OH). This mechanism illustrates how esters are converted into their corresponding carboxylate ions and alcohols under basic conditions.
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