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K* -OCH,CH3 Br-

K* -OCH,CH3 Br-

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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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For the following reaction, choose the most likely reaction pathway and draw the major product. Only draw the resulting organic species, omit counterions.

 

 

The image depicts a chemical reaction involving a haloalkane and an ethoxide ion. **Reactants:** 1. On the left side, there is a molecule of 2-bromobutane. This is a secondary haloalkane with the bromine (Br) atom attached to the second carbon of a four-carbon chain. 2. The arrow indicates that the reaction is proceeding from left to right. 3. The reagent above the arrow is potassium ethoxide (K⁺ -OCH₂CH₃). It is a strong base and a nucleophile, often used in nucleophilic substitution or elimination reactions. **Reaction Type:** This setup typically suggests a potential nucleophilic substitution (Sₙ2) or elimination (E2) reaction, depending on additional conditions such as solvent and temperature. Given the setup, an elimination reaction (E2) might occur, leading to the formation of an alkene. **Expected Products:** In an E2 reaction, the potassium ethoxide would abstract a proton from the β-carbon (the carbon adjacent to the one bonded to bromine), leading to the removal of the bromine atom and the formation of a double bond: - The product is likely to be an alkene, specifically but-2-ene, due to the formation of a double bond between the second and third carbon atoms. This reaction showcases the utility of strong bases in facilitating elimination reactions, illustrating concepts in organic chemistry such as reactivity, mechanism, and structural transformation.
Transcribed Image Text:The image depicts a chemical reaction involving a haloalkane and an ethoxide ion. **Reactants:** 1. On the left side, there is a molecule of 2-bromobutane. This is a secondary haloalkane with the bromine (Br) atom attached to the second carbon of a four-carbon chain. 2. The arrow indicates that the reaction is proceeding from left to right. 3. The reagent above the arrow is potassium ethoxide (K⁺ -OCH₂CH₃). It is a strong base and a nucleophile, often used in nucleophilic substitution or elimination reactions. **Reaction Type:** This setup typically suggests a potential nucleophilic substitution (Sₙ2) or elimination (E2) reaction, depending on additional conditions such as solvent and temperature. Given the setup, an elimination reaction (E2) might occur, leading to the formation of an alkene. **Expected Products:** In an E2 reaction, the potassium ethoxide would abstract a proton from the β-carbon (the carbon adjacent to the one bonded to bromine), leading to the removal of the bromine atom and the formation of a double bond: - The product is likely to be an alkene, specifically but-2-ene, due to the formation of a double bond between the second and third carbon atoms. This reaction showcases the utility of strong bases in facilitating elimination reactions, illustrating concepts in organic chemistry such as reactivity, mechanism, and structural transformation.
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