Devise a detailed mechanism for formation of the major product of the elimination reaction below. CH3 H2SO4 -CHH Draw curved arrows to show electron reorganization for the mechanism step below. Make the ends of your arrows specify the origin and destination of reorganizing electrons. Arrow-pushing Instructions H. H. :0-H CH3 H

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Chapter1: Chemical Foundations
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**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.
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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