Textbook Question
What is the major elimination product obtained from an E2 reaction of each of the following alkyl halides with hydroxide ion?
f.
9. Alkenes and Alkynes
Zaitsev Rule
6:31 minutesProblem 69a
Textbook Question
When 2-bromo-3-phenylbutane is treated with sodium methoxide, two alkenes result (by E2 elimination). The Zaitsev product predominates.
a. Draw the reaction, showing the major and minor products.
Verified step by step guidance1
Step 1: Identify the substrate and the base. The substrate is 2-bromo-3-phenylbutane, which is a secondary alkyl halide. The base is sodium methoxide (NaOCH₃), a strong base that promotes E2 elimination.
Step 2: Recall the mechanism of E2 elimination. In this mechanism, the base abstracts a proton from a β-carbon (a carbon adjacent to the carbon bonded to the leaving group), while the leaving group (Br⁻) departs simultaneously, forming a double bond.
Step 3: Analyze the β-hydrogens available for elimination. In 2-bromo-3-phenylbutane, there are two β-carbons: one on the methyl group and one on the phenyl-substituted carbon. Each β-carbon has hydrogens that can be abstracted by the base.
Step 4: Apply Zaitsev's rule to predict the major product. Zaitsev's rule states that the more substituted alkene (the one with more alkyl groups attached to the double bond) will be the major product. Elimination of a β-hydrogen from the phenyl-substituted carbon leads to the more substituted alkene, which is the major product.
Step 5: Draw the reaction products. The major product is the Zaitsev alkene, formed by elimination at the phenyl-substituted β-carbon. The minor product is the less substituted alkene, formed by elimination at the methyl β-carbon. Ensure the stereochemistry of the double bonds is considered, as E2 elimination can lead to both E and Z isomers.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
E2 Elimination Mechanism
The E2 elimination mechanism is a bimolecular reaction where a base abstracts a proton from a β-carbon while a leaving group departs from the α-carbon, resulting in the formation of a double bond. This concerted process requires strong bases and typically leads to the formation of alkenes. The stereochemistry of the substrate and the orientation of the leaving group influence the product distribution.
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Zaitsev's Rule
Zaitsev's Rule states that in elimination reactions, the more substituted alkene (the Zaitsev product) is generally favored over the less substituted one. This is because more substituted alkenes are typically more stable due to hyperconjugation and the inductive effect. Understanding this rule helps predict the major product in elimination reactions involving multiple possible alkenes.
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Sodium Methoxide as a Base
Sodium methoxide (NaOCH3) is a strong, non-bulky base commonly used in organic reactions, particularly in E2 eliminations. Its ability to effectively abstract protons from β-carbons facilitates the formation of alkenes. The choice of base can influence the reaction pathway and the selectivity of the products, making it crucial to understand its role in the reaction mechanism.
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