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Draw the product of this reaction. Ignore inorganic byproducts. Br Br 1. NaNH2, A 2. Нзо* Please

Draw the product of this reaction. Ignore inorganic byproducts. Br Br 1. NaNH2, A 2. Нзо* Please

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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**Title: Understanding Organic Reactions: Formation of Alkyne** **Introduction:** In this exercise, we will explore the transformation of a vicinal dibromide into an alkyne using sodium amide (NaNH₂) as a base followed by treatment with an acid (H₃O⁺). This type of reaction is commonly encountered in organic synthesis to form carbon-carbon triple bonds. **Reaction Details:** 1. **Starting Material:** - The initial compound is a six-carbon chain with two bromine atoms attached to adjacent (vicinal) carbon atoms (vicinal dibromide). 2. **Reagents:** - NaNH₂ (sodium amide), Δ (heat) - H₃O⁺ (acidic workup) 3. **Mechanism:** - The reaction proceeds through an elimination mechanism. Sodium amide (NaNH₂) acts as a strong base, abstracting hydrogen atoms and facilitating the removal of bromine atoms as bromide ions. - The elimination of two equivalents of HBr leads to the formation of an alkyne. 4. **Product:** - The final product is a linear alkyne, characterized by a carbon-carbon triple bond. **Diagram Explanation:** - **Reaction Scheme:** - The starting dibromide on the left is transformed into an alkyne on the right. - The arrow indicates the reaction progression, driven by the applied reagents. **Conclusion:** This transformation from a vicinal dibromide to an alkyne highlights the utility of strong bases in organic synthesis, allowing chemists to build complex molecular architectures with precise control over functional group placement.
Transcribed Image Text:**Title: Understanding Organic Reactions: Formation of Alkyne** **Introduction:** In this exercise, we will explore the transformation of a vicinal dibromide into an alkyne using sodium amide (NaNH₂) as a base followed by treatment with an acid (H₃O⁺). This type of reaction is commonly encountered in organic synthesis to form carbon-carbon triple bonds. **Reaction Details:** 1. **Starting Material:** - The initial compound is a six-carbon chain with two bromine atoms attached to adjacent (vicinal) carbon atoms (vicinal dibromide). 2. **Reagents:** - NaNH₂ (sodium amide), Δ (heat) - H₃O⁺ (acidic workup) 3. **Mechanism:** - The reaction proceeds through an elimination mechanism. Sodium amide (NaNH₂) acts as a strong base, abstracting hydrogen atoms and facilitating the removal of bromine atoms as bromide ions. - The elimination of two equivalents of HBr leads to the formation of an alkyne. 4. **Product:** - The final product is a linear alkyne, characterized by a carbon-carbon triple bond. **Diagram Explanation:** - **Reaction Scheme:** - The starting dibromide on the left is transformed into an alkyne on the right. - The arrow indicates the reaction progression, driven by the applied reagents. **Conclusion:** This transformation from a vicinal dibromide to an alkyne highlights the utility of strong bases in organic synthesis, allowing chemists to build complex molecular architectures with precise control over functional group placement.
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