Textbook Question
For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.
(a) cyclohexane
(b) methylcyclopentane
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Ch.4 - The Study of Chemical Reactions
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 4, Problem 47e
For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.
(e) 
Verified step by step guidance1
Step 1: Recognize that the problem involves free-radical bromination, which is highly selective and favors the formation of the most stable radical intermediate. Bromine reacts preferentially at the position that forms the most stable radical.
Step 2: Analyze the structure of ethylbenzene. It consists of a benzene ring attached to an ethyl group (CH2CH3). The ethyl group has two types of hydrogens: the benzylic hydrogens (on the CH2 group) and the hydrogens on the CH3 group.
Step 3: Determine the stability of the radicals formed by removing a hydrogen atom from each position. Removing a benzylic hydrogen (from CH2) forms a benzylic radical, which is highly stabilized due to resonance with the benzene ring. Removing a hydrogen from the CH3 group forms a primary radical, which is less stable.
Step 4: Conclude that bromination will occur at the benzylic position (CH2 group) because the benzylic radical is the most stable intermediate. Bromine will replace one of the hydrogens on the CH2 group.
Step 5: The major product of free-radical bromination of ethylbenzene will be bromine attached to the benzylic position, resulting in 1-bromo-2-phenylethane.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Free-Radical Bromination
Free-radical bromination is a reaction where bromine (Br2) reacts with alkanes or alkenes in the presence of heat or light to form brominated products. This process involves the formation of bromine radicals, which abstract hydrogen atoms from the substrate, leading to the generation of carbon-centered radicals. The stability of these radicals influences the selectivity of the reaction, favoring the formation of the most stable radical intermediate.
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05:20
Using the Hammond Postulate to describe radical bromination.
Radical Stability
The stability of radicals is a key factor in determining the major product of free-radical bromination. Radicals are more stable when they are tertiary (attached to three carbon atoms), followed by secondary (two carbon atoms), and then primary (one carbon atom). This stability hierarchy arises from hyperconjugation and the inductive effect, which help to delocalize the unpaired electron, making the radical less reactive and more stable.
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03:43The radical stability trend.
Selectivity in Bromination
Bromination is known for its selectivity, meaning it preferentially reacts with the most stable radical formed during the reaction. This selectivity is due to the energy barrier associated with forming different radicals; more stable radicals require less energy to form and are thus favored. In the case of ethylbenzene, the presence of the ethyl group can influence the stability of the radical formed, leading to specific products based on the structure of the substrate.
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00:54Mechanism of Allylic Bromination.
Related Practice
Textbook Question
When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as unreacted methane.
a. Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the formation of these compounds from chloromethane.
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Textbook Question
In the presence of a small amount of bromine, the following light-promoted reaction has been observed.
a. Write a mechanism for this reaction. Your mechanism should explain how both products are formed. (Hint: Notice which H atom has been lost in both products.)
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Textbook Question
In the presence of a small amount of bromine, the following light-promoted reaction has been observed.
b. Explain why only this one type of hydrogen atom has been replaced, in preference to any of the other hydrogen atoms in the starting material.
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Textbook Question
For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.
(f)
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Textbook Question
When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as unreacted methane.
b. How would you run this reaction to get a good conversion of methane to CH3Cl? Of methane to CCl4?
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