Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Question
**Detailed Mechanism for Elimination Reaction** In this exercise, we will devise a mechanism for the formation of the major product in the elimination reaction depicted below. **Reaction Overview:** The reaction involves the starting material, a cyclic alcohol with a methyl group, reacting with sulfuric acid (\( \text{H}_2\text{SO}_4 \)) to undergo dehydration, forming an alkene. **Mechanism Step:** 1. **Protonation Step:** - The hydroxyl group (\(-OH\)) on the cyclic structure is protonated by sulfuric acid, resulting in water (\( \text{H}_2\text{O} \)) being a good leaving group. **Arrow-Pushing Instructions:** To understand the electron reorganization, curved arrows should be used. The arrows indicate the movement of electrons. **Diagram Explanation:** - The first carbon structure represents the cyclic alcohol with a methyl group and an adjacent hydrogen. - Upon interaction with the acid, a positive charge is indicated on the carbon atom where the water will leave. - The lone pairs of electrons on the oxygen of the incoming water molecule are shown. The ultimate goal is to demonstrate the formation of a carbocation intermediate, where an elimination of the water molecule occurs, leading to the formation of a double bond and resulting in an alkene as the product. Remember to use the correct direction for arrow pushing; typically, arrows start from electron-rich areas (like lone pairs or negative charges) and move towards electron-deficient areas (like positive charges). **Instruction Note:** Ensure the placement of arrows correctly follows electron flow conventions for a clear understanding of the mechanistic steps.
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Transcribed Image Text:**Detailed Mechanism for Elimination Reaction** In this exercise, we will devise a mechanism for the formation of the major product in the elimination reaction depicted below. **Reaction Overview:** The reaction involves the starting material, a cyclic alcohol with a methyl group, reacting with sulfuric acid (\( \text{H}_2\text{SO}_4 \)) to undergo dehydration, forming an alkene. **Mechanism Step:** 1. **Protonation Step:** - The hydroxyl group (\(-OH\)) on the cyclic structure is protonated by sulfuric acid, resulting in water (\( \text{H}_2\text{O} \)) being a good leaving group. **Arrow-Pushing Instructions:** To understand the electron reorganization, curved arrows should be used. The arrows indicate the movement of electrons. **Diagram Explanation:** - The first carbon structure represents the cyclic alcohol with a methyl group and an adjacent hydrogen. - Upon interaction with the acid, a positive charge is indicated on the carbon atom where the water will leave. - The lone pairs of electrons on the oxygen of the incoming water molecule are shown. The ultimate goal is to demonstrate the formation of a carbocation intermediate, where an elimination of the water molecule occurs, leading to the formation of a double bond and resulting in an alkene as the product. Remember to use the correct direction for arrow pushing; typically, arrows start from electron-rich areas (like lone pairs or negative charges) and move towards electron-deficient areas (like positive charges). **Instruction Note:** Ensure the placement of arrows correctly follows electron flow conventions for a clear understanding of the mechanistic steps.
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