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Use the E2 mechanism to explain why when I is mixed with sodium ethoxide (NaOEt) in ethanol, the major product is III, but when II is mixed with sodium ethoxide in ethanol, the major product is IV. H3 C H3 C CH3 CI CH3 CH3 1 CI CH3 || H3C. CH3 H3C CH3 CH3 = ||| CH3 IV

Use the E2 mechanism to explain why when I is mixed with sodium ethoxide (NaOEt) in ethanol, the major product is III, but when II is mixed with sodium ethoxide in ethanol, the major product is IV. H3 C H3 C CH3 CI CH3 CH3 1 CI CH3 || H3C. CH3 H3C CH3 CH3 = ||| CH3 IV

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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### Transcription for Educational Use: **E2 Mechanism and Product Formation** Use the E2 mechanism to explain why when compound I is mixed with sodium ethoxide (NaOEt) in ethanol, the major product is compound III, but when compound II is mixed with sodium ethoxide in ethanol, the major product is compound IV. **Compound Structures:** 1. **Compound I:** - Structure: Cyclohexane ring with a chlorine (Cl) substituent and two methyl (CH₃) groups, one oriented above the plane and one below. 2. **Compound II:** - Structure: Similar to I, but with the chlorine (Cl) substituent in a different position on the cyclohexane ring, with two methyl groups. 3. **Compound III:** - Structure: Cyclohexene ring without chlorine, features a double bond and two methyl groups. 4. **Compound IV:** - Structure: Cyclohexene ring without chlorine, also features a double bond and two methyl groups, but differs in the position of the double bond compared to III. **Explanation:** The diagram indicates a focus on the stereochemistry and regiochemistry in E2 reactions influenced by the positioning of substituents and the corresponding major products. The products (III and IV) are determined by the more stable alkene formed through the elimination reaction using sodium ethoxide in ethanol. This explanation is to help students understand how different arrangements of the same substituents in cyclohexanes can influence the outcome of elimination reactions, showcasing the concept of Zaitsev’s rule where the more substituted alkene is typically the major product.
Transcribed Image Text:### Transcription for Educational Use: **E2 Mechanism and Product Formation** Use the E2 mechanism to explain why when compound I is mixed with sodium ethoxide (NaOEt) in ethanol, the major product is compound III, but when compound II is mixed with sodium ethoxide in ethanol, the major product is compound IV. **Compound Structures:** 1. **Compound I:** - Structure: Cyclohexane ring with a chlorine (Cl) substituent and two methyl (CH₃) groups, one oriented above the plane and one below. 2. **Compound II:** - Structure: Similar to I, but with the chlorine (Cl) substituent in a different position on the cyclohexane ring, with two methyl groups. 3. **Compound III:** - Structure: Cyclohexene ring without chlorine, features a double bond and two methyl groups. 4. **Compound IV:** - Structure: Cyclohexene ring without chlorine, also features a double bond and two methyl groups, but differs in the position of the double bond compared to III. **Explanation:** The diagram indicates a focus on the stereochemistry and regiochemistry in E2 reactions influenced by the positioning of substituents and the corresponding major products. The products (III and IV) are determined by the more stable alkene formed through the elimination reaction using sodium ethoxide in ethanol. This explanation is to help students understand how different arrangements of the same substituents in cyclohexanes can influence the outcome of elimination reactions, showcasing the concept of Zaitsev’s rule where the more substituted alkene is typically the major product.
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