Textbook Question
Explain the dramatic difference in rotational energy barriers of the following three alkenes. (Hint: Consider what the transition states must look like.)
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Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 7, Problem 69a
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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09:36
Drawing the E2 Mechanism.
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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01:18Defining Zaitsev’s Rule
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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Related Practice
Textbook Question
One of the following dichloronorbornanes undergoes elimination much faster than the other. Determine which one reacts faster, and explain the large difference in rates.
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Textbook Question
When (±)−2,3−dibromobutane reacts with potassium hydroxide, some of the products are (2S,3R)-3-bromobutan-2-ol and its enantiomer and trans-2-bromobut-2-ene. Why is no cis-2-bromobut-2-ene formed?
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Textbook Question
When (±)−2,3−dibromobutane reacts with potassium hydroxide, some of the products are (2S,3R)-3-bromobutan-2-ol and its enantiomer and trans-2-bromobut-2-ene. Give mechanisms to account for these products.
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Textbook Question
When 2-bromo-3-phenylbutane is treated with sodium methoxide, two alkenes result (by E2 elimination). The Zaitsev product predominates.
b. When one pure stereoisomer of 2-bromo-3-phenylbutane reacts, one pure stereoisomer of the major product results. For example, when (2R,3R)-2-bromo-3-phenylbutane reacts, the product is the stereoisomer with the methyl groups cis. Use your models to draw a Newman projection of the transition state to show why this stereospecificity is observed.
c. Use a Newman projection of the transition state to predict the major product of elimination of (2S,3R)-2-bromo-3-phenylbutane.
d. Predict the major product from elimination of (2S,3S)-2-bromo-3-phenylbutane. This prediction can be made without drawing any structures, by considering the results in part (b).
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Textbook Question
Pure (S)-2-bromo-2-fluorobutane reacts with methoxide ion in methanol to give a mixture of (S)-2-fluoro-2-methoxybutane and three fluoroalkenes.
b. Propose a mechanism to show how (S)-2-bromo-2-fluorobutane reacts to give (S)-2-fluoro-2-methoxybutane. Has this reaction gone with retention or inversion of configuration?
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