Textbook Question
Each of the following proposed mechanisms for the free-radical chlorination of methane is wrong. Explain how the experimental evidence disproves each mechanism.
b.
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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 4
The bromination of methane proceeds through the following steps:1. Br2 + 2 Br• ΔH° (per mole)/+190 kJ (45 kcal)Ea (per mole)/ 190 kJ (45 kcal)2. CH4 + Br• —> CH3+ HBr +73 kJ (17 kcal) 79 kJ (19 kcal) 3. • CH3 + Br2 —> CH3Br + Br -112 kJ (-27 kcal) 4 kJ (1 kcal) d. Compute the overall value of ΔH° for the bromination
Verified step by step guidance1
Identify the individual steps involved in the bromination of methane and their respective enthalpy changes (\( \Delta H^\circ \)).
Write down the enthalpy change for each step: Step 1: \( \Delta H^\circ = +190 \text{ kJ/mol} \), Step 2: \( \Delta H^\circ = +73 \text{ kJ/mol} \), Step 3: \( \Delta H^\circ = -112 \text{ kJ/mol} \).
To find the overall \( \Delta H^\circ \) for the reaction, sum the enthalpy changes of all the steps: \( \Delta H^\circ_{\text{overall}} = \Delta H^\circ_1 + \Delta H^\circ_2 + \Delta H^\circ_3 \).
Substitute the values from each step into the equation: \( \Delta H^\circ_{\text{overall}} = 190 + 73 - 112 \).
Calculate the sum to determine the overall \( \Delta H^\circ \) for the bromination of methane.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Enthalpy Change (ΔH°)
Enthalpy change (ΔH°) is a measure of the heat content of a system at constant pressure. It indicates whether a reaction is exothermic (releases heat, ΔH° < 0) or endothermic (absorbs heat, ΔH° > 0). In the context of the bromination of methane, calculating the overall ΔH° involves summing the enthalpy changes of each step in the reaction mechanism.
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Activation Energy (Ea)
Activation energy (Ea) is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to form products. In the bromination of methane, understanding Ea helps in analyzing the rate of the reaction and the feasibility of each step, as higher Ea values typically correlate with slower reaction rates.
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Reaction Mechanism
A reaction mechanism is a step-by-step description of the pathway taken by reactants to convert into products. It includes the individual elementary steps, intermediates, and transition states involved in the overall reaction. For the bromination of methane, recognizing the mechanism is crucial for understanding how the enthalpy changes and activation energies of each step contribute to the overall reaction energetics.
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Related Practice
Textbook Question
Deuterium (D) is the hydrogen isotope of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly stronger than the C―H bond by 5.0 kJ/mol (1.2 kcal/mol). Reaction rates tend to be slower when a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step.This effect, called a kinetic isotope effect, is clearly seen in the chlorination of methane. Methane undergoes free-radical chlorination 12 times as fast as tetradeuteriomethane (CD4)Faster: CH4 + Cl⋅ —> CH3Cl + HCl relative rate= 12Slower: CD4 + Cl⋅ —> CD3Cl + DClrelative rate= 1 c. Consider the thermodynamics of the chlorination of methane and the chlorination of ethane, and use the Hammond postulate to explain why one of these reactions has a much larger isotope effect than the other.
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Textbook Question
The reaction of tert-butyl chloride with methanol(CH3)3C—Cl Tert-butylchloride + CH3—OH methanol —> (CH3)C—OCH3 methyltert-butylether + HCl is found to follow the rate equation rate= Kt[(CH3)3C—Cl] a. What is the kinetic order with respect to tert-butyl chloride?
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Textbook Question
Use bond-dissociation enthalpies [TABLE 4-2], p. 167) to calculate values of ΔH° for the following reactions.a. CH3—CH3 + I2 —> CH3CH2I + HI
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Textbook Question
The bromination of methane proceeds through the following steps:1. Br2 + 2 Br• ΔH° (per mole)/+190 kJ (45 kcal)Ea (per mole)/ 190 kJ (45 kcal)2. CH4 + Br• —> CH3+ HBr +73 kJ (17 kcal) 79 kJ (19 kcal) 3. • CH3 + Br2 —> CH3Br + Br -112 kJ (-27 kcal) 4 kJ (1 kcal) a. Draw a complete reaction-energy diagram for this reaction. b. Label the rate-limiting step.
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Textbook Question
Each of the following proposed mechanisms for the free-radical chlorination of methane is wrong. Explain how the experimental evidence disproves each mechanism.
a.
1
views