Textbook Question
Classify each reaction as a substitution, an elimination, or neither. Identify the leaving group in each reaction, and the nucleophile in substitutions.
b.
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Ch.6 - Alkyl Halides; Nucleophilic Substitution
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 6, Problem 10c
Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why we expect to get a single major product.
(c) 
Verified step by step guidance1
Step 1: Identify the target compound, 1-bromo-1-phenylbutane, and recognize that it is formed via free-radical halogenation. The bromine atom is attached to the carbon adjacent to the phenyl group, which is the benzylic position.
Step 2: Understand the mechanism of free-radical halogenation. This reaction involves three steps: initiation, propagation, and termination. The initiation step generates bromine radicals (Br•) by homolytic cleavage of Br₂ under UV light or heat.
Step 3: Recognize that the benzylic position is highly reactive in free-radical halogenation due to the stability of the benzylic radical. The phenyl group stabilizes the radical through resonance, making this position the preferred site for bromination.
Step 4: During the propagation step, the bromine radical abstracts a hydrogen atom from the benzylic position of 1-phenylbutane, forming a benzylic radical. This radical then reacts with another Br₂ molecule to form the desired product, 1-bromo-1-phenylbutane, and regenerates the bromine radical.
Step 5: Explain why a single major product is expected. The benzylic position is the most reactive site due to radical stability, and there are no other competing positions with similar stability. This ensures selective bromination at the benzylic position, yielding 1-bromo-1-phenylbutane as the major product.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Free-Radical Halogenation
Free-radical halogenation is a reaction where alkanes react with halogens (like bromine) in the presence of heat or light to form alkyl halides. This process involves the generation of free radicals, which are highly reactive species that can initiate a chain reaction. The reaction typically proceeds through three steps: initiation, propagation, and termination, leading to the substitution of hydrogen atoms with halogen atoms.
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Radical Chain Reaction Mechanism.
Selectivity in Free-Radical Reactions
Selectivity in free-radical reactions refers to the preference for the formation of certain products over others. In the case of halogenation, the stability of the resulting radical intermediates plays a crucial role. More stable radicals (like tertiary radicals) are formed preferentially, leading to a single major product when the structure allows for minimal competition among possible sites for substitution.
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Stability of Radicals
The stability of radicals is a key factor in determining the outcome of free-radical reactions. Radicals can be classified as primary, secondary, or tertiary based on the number of carbon atoms attached to the carbon bearing the unpaired electron. Tertiary radicals are the most stable due to hyperconjugation and inductive effects from adjacent carbon atoms, which influences the selectivity of the halogenation process and explains why a single major product is expected.
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Related Practice
Textbook Question
Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why we expect to get a single major product.
(d)
Textbook Question
Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why we expect to get a single major product.
(a) 1-chloro-2,2-dimethylpropane (neopentyl chloride)
(b) 2-bromo-2-methylbutane
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Textbook Question
The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products:
a. Give a mechanism for this reaction, showing how the two products arise as a consequence of the resonance-stabilized intermediate.
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Textbook Question
Classify each reaction as a substitution, an elimination, or neither. Identify the leaving group in each reaction, and the nucleophile in substitutions.
a.
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Textbook Question
The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products:
b. The bromination of cyclohexene using NBS gives only one major product, as shown on the previous page. Explain why there is no second product from an allylic shift.
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