## Step 1: Identify the E2 Reaction and Antiperiplanar Requirement In an E2 reaction, a strong base (here, CH₃O⁻) abstracts a proton from a β-carbon that is **antiperiplanar** to the leaving group (Br in this case). The leaving group and the hydrogen to be abstracted **must be on opposite sides** (180° apart) for the reaction to occur efficiently. - The Br is on a wedge (coming out of the page), so the abstracted β-hydrogen must be on a dash (going into the page) on an adjacent carbon. - Examine all β-hydrogens. The methyl group on the adjacent carbon is on a dash (into the page), so that hydrogen is antiperiplanar to Br. **Explanation:** E2 eliminations require antiperiplanar geometry for the base to abstract a hydrogen and the leaving group to leave simultaneously, resulting in a double bond. --- ## Step 2: Draw the Alkene Product with Correct Stereochemistry - The elimination of H and Br forms a double bond between the α-carbon (where Br was) and the β-carbon (where H was abstracted). - The resulting alkene will be between the carbon attached to Br and the carbon with the methyl group. - The most substituted alkene (Zaitsev product) is favored, and the stereochemistry of the product should be depicted. The phenyl group and methyl group will be on opposite sides of the double bond (E-configuration) due to the antiperiplanar elimination. **Explanation:** The major product is the more stable, more substituted (Zaitsev) alkene, and the E2 reaction proceeds with stereospecificity, leading to the (E)-alkene. --- ## **Final Answer:** The major product is: ``` Ph \ C==C / \ H3C CH3 ``` **Stereochemistry:** The phenyl group and one methyl group are on opposite sides of the double bond (E-alkene). The product is **(E)-1-phenyl-1-butene**. --- ### **Summary Diagram (with stereochemistry):** - Draw the double bond between the benzylic carbon and the β-carbon. - Place phenyl (Ph) and methyl (CH₃) on opposite sides of the double bond. **This is the E2 elimination product with correct stereochemistry.**

# E2 Reaction Mechanism and Product Structure ## Step 1: Identify the E2 Reaction and Antiperiplanar Requirement In an E2 reaction, a strong base (here, CH₃O⁻) abstracts a proton from a β-carbon that is **antiperiplanar** to the leaving group (Br in this case). The leaving group and the hydrogen to be abstracted must be on opposite sides (180° apart) for the reaction to occur efficiently. - The Br is on a wedge (coming out of the page), which means the abstracted β-hydrogen must be on a dash (going into the page) on an adjacent carbon. - Identify the β-hydrogens on the adjacent carbon. The methyl group on the adjacent carbon has a hydrogen on a dash (into the page), which is antiperiplanar to Br. **Explanation:** E2 eliminations require antiperiplanar geometry for the base to abstract a hydrogen while the leaving group departs, resulting in a double bond formation. --- ## Step 2: Draw the Alkene Product with Correct Stereochemistry - The elimination of H and Br creates a double bond between the α-carbon (where Br was) and the β-carbon (where H was abstracted). - The double bond forms between the carbon attached to Br and the carbon with the methyl group. - The more substituted alkene (Zaitsev product) is favored. The stereochemistry of the product should be indicated; the phenyl group and the methyl group will be on opposite sides of the double bond, resulting in the (E)-configuration. **Explanation:** The major product is the more stable, more substituted (Zaitsev) alkene, and the E2 reaction is stereospecific, leading to the formation of an (E)-alkene. --- ## Final Answer The major product of the reaction is: ``` Ph \ C==C / \ H3C CH3 ``` **Stereochemistry:** The phenyl group (Ph) and one methyl group (CH₃) are positioned on opposite sides of the double bond, indicating the (E)-configuration. The product formed is **(E)-1-phenyl-1-butene**. --- ### Summary Diagram - Illustrate the double bond between the benzylic carbon and the β-carbon. - Ensure the phenyl (Ph) and methyl (CH₃) groups are on opposite sides of the double bond. This structure represents the E2 elimination product with appropriate stereochemistry.