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.
(e)
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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 47f
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) 
Verified step by step guidance1
Step 1: Analyze the structure of the compound provided. The compound is a bicyclic structure with two fused cyclohexane rings. Identify all the possible hydrogen atoms that can be abstracted to form radicals.
Step 2: Recall that bromination is highly selective and prefers the formation of the most stable radical. Stability of radicals follows the order: tertiary > secondary > primary > methyl. Look for tertiary carbons in the structure where a radical can form.
Step 3: Identify the positions where bromination can occur. In this case, the tertiary carbons in the bicyclic structure are the most likely sites for radical formation. These positions will lead to the major products.
Step 4: Consider the stereochemistry of the compound. Since the structure is bicyclic, bromination at different tertiary positions may lead to two distinct products due to the spatial arrangement of the molecule.
Step 5: Predict the major products by substituting bromine at the identified tertiary positions. Ensure that you account for the selectivity of bromination and the stability of the resulting radicals.
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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 to form alkyl bromides through a radical mechanism. This process involves three main steps: initiation, propagation, and termination. The reaction begins with the homolytic cleavage of the Br-Br bond, generating bromine radicals that can abstract hydrogen atoms from alkanes, leading to the formation of more stable radicals.
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05:20
Using the Hammond Postulate to describe radical bromination.
Radical Stability
The stability of radicals is crucial in predicting the products of free-radical bromination. Radicals are more stable when they are tertiary (attached to three carbon atoms) compared to secondary or primary radicals. This stability influences which hydrogen atoms are abstracted during the reaction, as the formation of the most stable radical intermediate is favored, leading to the major product.
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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. This selectivity arises from the energy differences between the various radical intermediates formed during the reaction. As a result, the major product of bromination will be the one derived from the most stable radical, which is typically the one that forms from the most substituted carbon in the alkane.
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00:54Mechanism of Allylic Bromination.
Related Practice
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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Textbook Question
The chlorination of pentane gives a mixture of three monochlorinated products.
a. Draw their structures.
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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.
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.
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
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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