Let's solve the problem **step-by-step**: --- ## 1. **Understanding the Question** You are asked to find **the major E2 product** (with correct E/Z stereochemistry) formed from the halide shown in the Newman projection. - **E2 elimination** requires the leaving group (Cl here) and a β-hydrogen to be **anti-periplanar** (180° apart). - You must then draw the *alkene* product, indicating whether it is E (trans) or Z (cis). --- ## 2. **Interpreting the Newman Projection** The Newman projection is typically viewed **down the C–C bond**: - **Front carbon (left):** - Up: H - Down-left: Cl - Down-right: C₆H₅ (phenyl) - **Back carbon (right):** - Up: CH₃ - Down-left: C₂H₅ (ethyl) - Down-right: C₆H₅ (phenyl) --- ## 3. **Identifying the Antiperiplanar Pair** For E2, we need: - **Cl** (leaving group, front carbon) - **H** (β-hydrogen, back carbon) **anti to the Cl** **Looking at the Newman:** - Cl (front, down-left) - H (front, up) - The group anti to Cl (back, up): **CH₃** But we need a β-H **anti to Cl**. Since the only hydrogens on the back carbon are on the **CH₃** and **C₂H₅** groups, but the only group anti to Cl is **CH₃** (up, back). So, the β-hydrogen that gets eliminated is from the CH₃ group. --- ## 4. **Draw the Resulting Alkene** When the base abstracts the **anti-periplanar β-H** from the **CH₃ group**, and Cl leaves, a double bond forms between the two carbons. ### The two carbons (alkene): - **Left carbon:** Attached to H, Cl, and C₆H₅ - **Right carbon:** Attached to CH₃, C₂H₅, and C₆H₅ After elimination: - **Left carbon:** Attached to H, C₆H₅, and the new double bond - **Right carbon:** Attached to C₂H₅, C₆H₅, and the new double bond ### Substituents on double bond: - **Left alkene C:** H and C₆H₅ - **Right alkene C:** CH₂ (from CH₃ after loss of H) and C₆H₅ --- ## 5. **Assign E/Z Stereochemistry** Assign priorities at each alkene carbon: - **Left:** C₆H₅ > H - **Right:** C₆H₅ > CH₂ (ethyl group) **From the Newman projection**, C₆H₅ groups are on **opposite sides** → **E (trans) isomer**. --- ## 6. **Draw the Product** The major E2 product is: ``` C₆H₅ C₂H₅ \ / C == C / \ H C₆H₅ ``` This is (E)-1,2-diphenyl-1-butene. --- ## 7. **Final Answer** ### **The major E2 product is:** #### **(E)-1,2-diphenyl-1-butene** **Structure:** ``` C₆H₅ C₂H₅ \ / C==C / \ H C₆H₅ ``` - The phenyl (C₆H₅) groups are on opposite sides (E configuration). --- ### **Summary Table** | Carbon | Substituents after elimination | |--------|-------------------------------| | Left | H, C₆H₅, alkene | | Right | C₂H₅, C₆H₅, alkene | --- ### **Explanation** - E2 elimination requires anti-periplanar geometry (Cl anti to CH₃). - Remove Cl and a β-H from CH₃ group. - Form double bond between these carbons. - Assign E/Z: highest priority (C₆H₅) opposite → **E isomer**. --- If you need the skeletal structure drawn, let me know!