Textbook Question
Use the information in Table 4-2 (p. 167) to rank the following radicals in decreasing order of stability.
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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 43(3)a,b
For each alkane, which monobrominated derivatives could you form in good yield by free-radical bromination?
a. Cyclopentane
b. Methylcyclopentnae
Verified step by step guidance1
Identify the structure of the given alkanes: Cyclopentane is a cyclic alkane with five carbon atoms in a ring, and Methylcyclopentane is a cyclopentane ring with a single methyl group attached to one of the carbons.
Understand the concept of free-radical bromination: Bromination occurs at the most stable radical intermediate, which is determined by the stability of the carbon radical formed during the reaction. Radical stability follows the order: tertiary > secondary > primary > methyl.
For Cyclopentane: All hydrogens in cyclopentane are equivalent because of the symmetry of the molecule. Bromination at any carbon will yield the same product, 1-bromocyclopentane.
For Methylcyclopentane: The hydrogens are not equivalent due to the presence of the methyl group. Identify the different types of hydrogens: (1) Hydrogens on the methyl group (primary hydrogens), (2) Hydrogens on the carbon adjacent to the methyl group (secondary hydrogens), and (3) Hydrogens on the other carbons in the ring (secondary hydrogens). Bromination will preferentially occur at the secondary carbon adjacent to the methyl group because it forms the most stable radical.
Conclude the products: For Cyclopentane, the product is 1-bromocyclopentane. For Methylcyclopentane, the major product is bromination at the secondary carbon adjacent to the methyl group, forming 1-bromo-2-methylcyclopentane. Minor products may form from bromination at other secondary or primary positions, but these are less favored.
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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 mechanism where bromine (Br2) reacts with alkanes in the presence of heat or light to form brominated products. This process involves the generation of bromine radicals, which abstract hydrogen atoms from the alkane, leading to the formation of alkyl radicals. These alkyl radicals can then react with bromine molecules to yield monobrominated derivatives. The selectivity of this reaction is influenced by the stability of the resulting radicals.
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Using the Hammond Postulate to describe radical bromination.
Alkane Structure and Reactivity
Alkanes are saturated hydrocarbons characterized by single carbon-carbon bonds. Their reactivity in free-radical bromination depends on the structure of the alkane, including the presence of branching and cyclic structures. For example, cyclopentane and methylcyclopentane have different reactivity patterns due to their unique structures, which affect the stability of the radicals formed during bromination. Understanding these structural features is crucial for predicting the products of the reaction.
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Monobrominated Derivatives
Monobrominated derivatives are organic compounds formed when one hydrogen atom in an alkane is replaced by a bromine atom. The formation of these derivatives during free-radical bromination can lead to multiple products, especially in branched or cyclic alkanes, due to the presence of different hydrogen environments. Identifying the possible monobrominated products requires analyzing the alkane's structure and the positions where bromination can occur, which is essential for understanding the yields of the reaction.
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Related Practice
Textbook Question
Use bond-dissociation enthalpies (Table 4-2, p. 167) to calculate values of ΔH° for the following reactions.
c. (CH3)3C—OH + HCl → (CH3)3C—Cl + H2O
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Textbook Question
For each alkane,
3. which monobrominated derivatives could you form in good yield by free-radical bromination?
c. 2-methylpentane
d. 2,2,3,3-tetramethylbutane
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Textbook Question
Use bond-dissociation enthalpies (Table 4-2, p. 167) to calculate values of ΔH° for the following reactions.
d. CH3CH2CH3 + H2 → CH3CH3 + CH4
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Textbook Question
For each alkane,
1. draw all the possible monochlorinated derivatives.
a. Cyclopentane
b. Methylcyclopentane
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Textbook Question
For each alkane,
2. determine whether free-radical chlorination would be a good way to make any of these monochlorinated derivatives. Will the reaction give mostly one major product?
a. Cyclopentane
b. Methylcyclopentane
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