Draw the product of the E2 reaction shown below. Include all lone pairs. Ignore byproducts. -:O-CH3 CH3OH heat Select to Draw the E2 Product

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Chapter1: Chemical Foundations
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**E2 Reaction Product Determination**

**Instructions:**
Draw the product of the E2 reaction shown below. Include all lone pairs. Ignore byproducts.

**Reaction Diagram:**
- The diagram depicts a substrate featuring a benzene ring attached to a carbon chain. The carbon chain has a bromine (Br) atom attached to one carbon, and a hydrogen (H) atom that will be abstracted in the E2 elimination process.
- The E2 reaction mechanism is indicated with curved arrows:
  - One arrow starts from a base (methoxide ion, CH₃O⁻), pointing towards the hydrogen atom (H) that it will abstract.
  - Another arrow indicates the formation of a double bond between the carbon atoms as the hydrogen is abstracted.
  - The last arrow shows the departure of the bromine atom as a leaving group.

**Reagents and Conditions:**
- Reagent: CH₃OH (methanol)
- Condition: Heat

**Interactive Section:**
- A dashed box is provided for the user to draw the E2 product. It is labeled: "Select to Draw the E2 Product."

**Explanation of the Graphical Mechanism:**
1. The base (CH₃O⁻) abstracts a proton (H) from the β-carbon (the carbon adjacent to the one bonded to bromine).
2. Simultaneously, the electrons from the C-H bond form a double bond (π-bond) between the α and β carbons.
3. As the double bond forms, the bromine (Br) atom leaves, taking its bonding electrons with it.

Students are expected to use these details to draw the final product of the E2 reaction, ensuring they showcase all lone pairs on atoms where applicable.
Transcribed Image Text:**E2 Reaction Product Determination** **Instructions:** Draw the product of the E2 reaction shown below. Include all lone pairs. Ignore byproducts. **Reaction Diagram:** - The diagram depicts a substrate featuring a benzene ring attached to a carbon chain. The carbon chain has a bromine (Br) atom attached to one carbon, and a hydrogen (H) atom that will be abstracted in the E2 elimination process. - The E2 reaction mechanism is indicated with curved arrows: - One arrow starts from a base (methoxide ion, CH₃O⁻), pointing towards the hydrogen atom (H) that it will abstract. - Another arrow indicates the formation of a double bond between the carbon atoms as the hydrogen is abstracted. - The last arrow shows the departure of the bromine atom as a leaving group. **Reagents and Conditions:** - Reagent: CH₃OH (methanol) - Condition: Heat **Interactive Section:** - A dashed box is provided for the user to draw the E2 product. It is labeled: "Select to Draw the E2 Product." **Explanation of the Graphical Mechanism:** 1. The base (CH₃O⁻) abstracts a proton (H) from the β-carbon (the carbon adjacent to the one bonded to bromine). 2. Simultaneously, the electrons from the C-H bond form a double bond (π-bond) between the α and β carbons. 3. As the double bond forms, the bromine (Br) atom leaves, taking its bonding electrons with it. Students are expected to use these details to draw the final product of the E2 reaction, ensuring they showcase all lone pairs on atoms where applicable.
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