Textbook Question
Draw the products of each of the following SN2/E2 reactions. If the products can exist as stereoisomers, show which stereoisomers are formed.
c. (3S,4R)-3-bromo-4-methylhexane + CH3O−
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Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
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Bruice 8th EditionCh. 9 - Substitution and Elimination Reactions of Alkyl HalidesProblem 116
Bruice 8th EditionCh. 9 - Substitution and Elimination Reactions of Alkyl HalidesProblem 116Chapter 10, Problem 116
Two elimination products are obtained from the following E2 reaction:
a. What are the elimination products?
b. Which is formed in greater yield?
Verified step by step guidance1
Identify the type of reaction: The problem specifies an E2 elimination reaction. In an E2 reaction, a strong base removes a β-hydrogen (a hydrogen atom on a carbon adjacent to the carbon bearing the leaving group) while the leaving group departs, forming a double bond (alkene).
Analyze the structure of the substrate: The substrate is CH3CH2CHDCH2Br. The β-hydrogens are located on the carbon adjacent to the carbon bonded to the bromine (leaving group). In this case, there are two β-hydrogens: one is a regular hydrogen (H), and the other is a deuterium (D, a heavier isotope of hydrogen).
Determine the possible elimination products: The elimination can occur in two ways, depending on which β-hydrogen is removed. If the regular hydrogen (H) is removed, the product will be CH3CH=CHD. If the deuterium (D) is removed, the product will be CH3CH=CH2. These are the two possible alkenes formed.
Consider the regioselectivity and isotope effect: The E2 reaction typically follows Zaitsev's rule, which states that the more substituted alkene is favored. However, the isotope effect must also be considered. The C-D bond is stronger than the C-H bond, so the removal of the regular hydrogen (H) is kinetically favored over the removal of deuterium (D).
Predict the major product: Based on the isotope effect, the product CH3CH=CHD (formed by removing the regular hydrogen) will be formed in greater yield compared to CH3CH=CH2 (formed by removing deuterium).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
E2 Reaction Mechanism
The E2 (bimolecular elimination) reaction is a concerted mechanism 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 reaction typically requires strong bases and is characterized by a single transition state, leading to the simultaneous removal of the leaving group and proton.
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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 is generally the major product. This is due to the stability of alkenes; more substituted alkenes have greater hyperconjugation and lower energy, making them more favorable in terms of yield compared to less substituted alkenes.
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01:18Defining Zaitsev’s Rule
Regioselectivity in Elimination Reactions
Regioselectivity refers to the preference of a chemical reaction to yield one structural isomer over others. In E2 reactions, the orientation of the leaving group and the hydrogen being abstracted can lead to different alkene products, and understanding the sterics and electronics of the substrate can help predict which product will be favored.
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00:40Recognizing Elimination Reactions.
Related Practice
Textbook Question
Draw the structures of the products obtained from the following reaction:
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Textbook Question
How could you prepare the following compounds from the given starting materials?
b.
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Textbook Question
cis-4-Bromocyclohexanol and trans-4-bromocyclohexanol form the same elimination product but a different substitution product when they react with HO−.
a. Why do they form the same elimination product?
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Textbook Question
Draw the products of each of the following SN2/E2 reactions. If the products can exist as stereoisomers, show which stereoisomers are formed.
a. (3S,4S)-3-bromo-4-methylhexane + CH3O−
1524
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Textbook Question
Draw the products of each of the following SN2/E2 reactions. If the products can exist as stereoisomers, show which stereoisomers are formed.
b. (3R,4R)-3-bromo-4-methylhexane + CH3O−
927
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