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1. Draw the product of the following reaction with an arrow-pushing mechanism. CH3ONA CH;OH

1. Draw the product of the following reaction with an arrow-pushing mechanism. CH3ONA CH;OH

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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Draw the product of the following reaction with an arrow-pushing mechanism.

**Question 1: Reaction Mechanism Description** **Instruction:** Draw the product of the following reaction with an arrow-pushing mechanism. **Reactants:** - **Starting Material:** An epoxide attached to a phenyl group (benzene ring). - **Reagents:** Sodium methoxide (\( \text{CH}_3\text{ONa} \)) in methanol (\( \text{CH}_3\text{OH} \)). **Reaction Explanation:** In this reaction, the sodium methoxide acts as a nucleophile, attacking the less substituted carbon of the epoxide. This leads to the opening of the epoxide ring, resulting in the formation of an alkoxide intermediate. Subsequently, protonation occurs from the methanol, yielding the final product, which is a methoxy alcohol attached to a phenyl group. **Note on Mechanism:** Arrow-pushing mechanisms involve the following steps: 1. **Nucleophilic Attack:** The lone pair of electrons from the methoxide ion (\( \text{CH}_3\text{O}^- \)) attacks the less substituted carbon atom of the epoxide ring. 2. **Ring Opening:** The carbon-oxygen bond breaks, opening the epoxide ring and forming an alkoxide intermediate. 3. **Protonation:** The alkoxide ion abstracts a proton from methanol, resulting in the formation of a methoxy alcohol. This detailed mechanism illustrates how functional groups are transformed during the chemical synthesis.
Transcribed Image Text:**Question 1: Reaction Mechanism Description** **Instruction:** Draw the product of the following reaction with an arrow-pushing mechanism. **Reactants:** - **Starting Material:** An epoxide attached to a phenyl group (benzene ring). - **Reagents:** Sodium methoxide (\( \text{CH}_3\text{ONa} \)) in methanol (\( \text{CH}_3\text{OH} \)). **Reaction Explanation:** In this reaction, the sodium methoxide acts as a nucleophile, attacking the less substituted carbon of the epoxide. This leads to the opening of the epoxide ring, resulting in the formation of an alkoxide intermediate. Subsequently, protonation occurs from the methanol, yielding the final product, which is a methoxy alcohol attached to a phenyl group. **Note on Mechanism:** Arrow-pushing mechanisms involve the following steps: 1. **Nucleophilic Attack:** The lone pair of electrons from the methoxide ion (\( \text{CH}_3\text{O}^- \)) attacks the less substituted carbon atom of the epoxide ring. 2. **Ring Opening:** The carbon-oxygen bond breaks, opening the epoxide ring and forming an alkoxide intermediate. 3. **Protonation:** The alkoxide ion abstracts a proton from methanol, resulting in the formation of a methoxy alcohol. This detailed mechanism illustrates how functional groups are transformed during the chemical synthesis.
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