Curved arrows are used to illustrate the flow of electrons. Follow the curved arrows and draw the product of the E2 reaction shown below. Include all lone pairs. Ignore byproducts. :Br: H CH3OH heat -CH3 Q

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**Title: Understanding the E2 Reaction Mechanism**

In this exercise, we explore the elimination reaction mechanism (E2) depicted in the diagram above. **Curved arrows** are used to illustrate the movement of electrons during the reaction. Follow these arrows to understand the changes taking place in the molecule. The goal is to identify the product formed by the E2 reaction and note all lone pairs. Byproducts should be ignored.

### Diagram Explanation:

1. **Chemical Structure:**
   - The starting molecule is a bromoalkane. The carbon chain is bonded to a bromine atom (\(\text{Br}\)) with two lone pairs, and adjacent to it is a hydrogen atom (\(\text{H}\)).

2. **Curved Arrows:**
   - **First Arrow:** Indicates the removal of the bromine atom as a leaving group, taking its electron pair with it.
   - **Second Arrow:** Shows the abstraction of a hydrogen (\(\text{H}\)) atom by the negatively charged methoxide ion (\(\text{O}^\ominus\)—\(\text{CH}_3\)).
   - **Third Arrow:** Demonstrates the formation of a double bond between the two carbon atoms as the electrons from the C-H bond shift into the space between them.

3. **Reactants and Conditions:**
   - The reaction occurs in the presence of methanol (\(\text{CH}_3\text{OH}\)) and heat, facilitating the elimination process.

### Objective:

By carefully analyzing and executing the E2 mechanism, draw the resulting alkene by forming a double bond and including all relevant lone pairs on atoms. This exercise will enhance your understanding of electron movement and bond formation in chemical reactions.

---

**Note:** The E2 mechanism usually proceeds in a single, concerted step, where the base abstracts a proton, simultaneous to the leaving group departure, resulting in the formation of a double bond.
Transcribed Image Text:**Title: Understanding the E2 Reaction Mechanism** In this exercise, we explore the elimination reaction mechanism (E2) depicted in the diagram above. **Curved arrows** are used to illustrate the movement of electrons during the reaction. Follow these arrows to understand the changes taking place in the molecule. The goal is to identify the product formed by the E2 reaction and note all lone pairs. Byproducts should be ignored. ### Diagram Explanation: 1. **Chemical Structure:** - The starting molecule is a bromoalkane. The carbon chain is bonded to a bromine atom (\(\text{Br}\)) with two lone pairs, and adjacent to it is a hydrogen atom (\(\text{H}\)). 2. **Curved Arrows:** - **First Arrow:** Indicates the removal of the bromine atom as a leaving group, taking its electron pair with it. - **Second Arrow:** Shows the abstraction of a hydrogen (\(\text{H}\)) atom by the negatively charged methoxide ion (\(\text{O}^\ominus\)—\(\text{CH}_3\)). - **Third Arrow:** Demonstrates the formation of a double bond between the two carbon atoms as the electrons from the C-H bond shift into the space between them. 3. **Reactants and Conditions:** - The reaction occurs in the presence of methanol (\(\text{CH}_3\text{OH}\)) and heat, facilitating the elimination process. ### Objective: By carefully analyzing and executing the E2 mechanism, draw the resulting alkene by forming a double bond and including all relevant lone pairs on atoms. This exercise will enhance your understanding of electron movement and bond formation in chemical reactions. --- **Note:** The E2 mechanism usually proceeds in a single, concerted step, where the base abstracts a proton, simultaneous to the leaving group departure, resulting in the formation of a double bond.
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