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 46a
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.)
Verified step by step guidance1
Step 1: Recognize that the reaction is a free radical halogenation reaction promoted by light (hv). Bromine (Br2) undergoes homolytic cleavage under light to form two bromine radicals (Br•). This is the initiation step.
Step 2: Identify the hydrogen atoms that are abstracted in the products. In the first product, the hydrogen atom from the methyl group is abstracted, while in the second product, the hydrogen atom from the allylic position (next to the double bond) is abstracted. These positions are key to understanding the mechanism.
Step 3: Explain the propagation steps. The bromine radical (Br•) abstracts a hydrogen atom from the methyl group or the allylic position of the starting compound, forming a carbon radical. The carbon radical then reacts with another Br2 molecule to form the brominated product and regenerate a bromine radical.
Step 4: Discuss the stability of the intermediates. The allylic radical is stabilized by resonance, making it more likely to form the second product. The methyl radical is less stable but still forms the first product due to the availability of the methyl hydrogen.
Step 5: Conclude with the termination step. Two bromine radicals can combine to form Br2, or other radical combinations can occur, but these are less relevant to the formation of the observed products. The reaction overall produces two brominated products and HBr as a byproduct.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Allylic Bromination
Allylic bromination is a reaction where bromine (Br2) adds to the allylic position of an alkene or alkane, typically in the presence of light or heat. This process involves the abstraction of a hydrogen atom from the allylic position, leading to the formation of a radical intermediate. The radical can then react with bromine to form brominated products, resulting in a mixture of products due to the possibility of bromination at different allylic positions.
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Mechanism of Allylic Bromination.
Radical Mechanism
The radical mechanism involves three main steps: initiation, propagation, and termination. In the initiation step, light or heat generates bromine radicals from Br2. During propagation, these radicals abstract hydrogen atoms from the substrate, forming new radicals that can react with bromine to create brominated products. This mechanism is crucial for understanding how both products in the reaction are formed, as it explains the formation of different brominated species.
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Selectivity in Radical Reactions
Selectivity in radical reactions refers to the preference for a radical to abstract a hydrogen atom from a specific position, leading to different products. In allylic bromination, the stability of the resulting radical influences which hydrogen is abstracted. More stable radicals (like tertiary over secondary or primary) are favored, which can lead to a mixture of products, as seen in the reaction where both products are formed from different hydrogen atoms being lost.
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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
Draw the important resonance forms of the following free radicals.
a.
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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
Draw the important resonance forms of the following free radicals.
(c)
(d)
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Textbook Question
Draw the important resonance forms of the following free radicals.
b.
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