6) 25pis Draw the structure of the major alkene product (or products) formed by treatment of each of the following haloalkanes with sodium ethoxide in ethanol. Assume the mechanism is E2 elimination. t-BuO K t-BUOH Br CH3 Eto Na F ELOH CH2CH3 CI- Eto Na -CH2CH3 ELOH H- ČH3 Br EtO Na ETOH CH3 CI H CH2CH3 Eto Na H3C ELOH H D

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**Exercise: E2 Elimination and Alkene Formation**

**Instructions:**
Draw the structure of the major alkene product (or products) formed by the treatment of each of the following haloalkanes with sodium ethoxide in ethanol. Assume the mechanism is E2 elimination.

1. **First Reaction:**
   - **Haloalkane:** (Image shows a haloalkane with a Br (bromine) group attached.)
   - **Reagents:** t-BuO⁻ K⁺, t-BuOH
   - **Box for Drawing:** (Students should draw the alkene product.)

2. **Second Reaction:**
   - **Haloalkane:** (Image shows a cyclopentane with a CH₃ (methyl) and F (fluorine) group.)
   - **Reagents:** EtO⁻ Na⁺, EtOH
   - **Box for Drawing:** (Students should draw the alkene product.)

3. **Third Reaction:**
   - **Haloalkane:** (Image shows a haloalkane with a Cl (chlorine) atom and ethyl groups.)
   - **Reagents:** EtO⁻ Na⁺, EtOH
   - **Box for Drawing:** (Students should draw the alkene product.)

4. **Fourth Reaction:**
   - **Haloalkane:** (Image shows a cyclohexane with a Br (bromine) and CH₃ (methyl) group.)
   - **Reagents:** EtO⁻ Na⁺, EtOH
   - **Box for Drawing:** (Students should draw the alkene product.)

5. **Fifth Reaction:**
   - **Haloalkane:** (Image shows a molecule with Cl (chlorine), H (hydrogen), CH₂CH₃ (ethyl), H, D (deuterium), and CH₃ (methyl) groups.)
   - **Reagents:** EtO⁻ Na⁺, EtOH
   - **Box for Drawing:** (Students should draw the alkene product.)

**Note:** Make sure to apply the Zaitsev's Rule where applicable, which suggests that the more substituted alkene (more stable) is generally the major product in E2 eliminations.
Transcribed Image Text:**Exercise: E2 Elimination and Alkene Formation** **Instructions:** Draw the structure of the major alkene product (or products) formed by the treatment of each of the following haloalkanes with sodium ethoxide in ethanol. Assume the mechanism is E2 elimination. 1. **First Reaction:** - **Haloalkane:** (Image shows a haloalkane with a Br (bromine) group attached.) - **Reagents:** t-BuO⁻ K⁺, t-BuOH - **Box for Drawing:** (Students should draw the alkene product.) 2. **Second Reaction:** - **Haloalkane:** (Image shows a cyclopentane with a CH₃ (methyl) and F (fluorine) group.) - **Reagents:** EtO⁻ Na⁺, EtOH - **Box for Drawing:** (Students should draw the alkene product.) 3. **Third Reaction:** - **Haloalkane:** (Image shows a haloalkane with a Cl (chlorine) atom and ethyl groups.) - **Reagents:** EtO⁻ Na⁺, EtOH - **Box for Drawing:** (Students should draw the alkene product.) 4. **Fourth Reaction:** - **Haloalkane:** (Image shows a cyclohexane with a Br (bromine) and CH₃ (methyl) group.) - **Reagents:** EtO⁻ Na⁺, EtOH - **Box for Drawing:** (Students should draw the alkene product.) 5. **Fifth Reaction:** - **Haloalkane:** (Image shows a molecule with Cl (chlorine), H (hydrogen), CH₂CH₃ (ethyl), H, D (deuterium), and CH₃ (methyl) groups.) - **Reagents:** EtO⁻ Na⁺, EtOH - **Box for Drawing:** (Students should draw the alkene product.) **Note:** Make sure to apply the Zaitsev's Rule where applicable, which suggests that the more substituted alkene (more stable) is generally the major product in E2 eliminations.
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