Textbook Question
Using the given starting material and any necessary organic or inorganic reagents, indicate how the desired compounds could be synthesized:
a.
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Ch. 12 - Radicals
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
Chapter 13, Problem 44
A chemist wanted to determine experimentally the relative ease of removing a hydrogen atom from a tertiary, a secondary, and a primary carbon by a chlorine radical. He allowed 2-methylbutane to undergo chlorination at 300 °C and obtained as products 36% 1-chloro-2-methylbutane, 18% 2-chloro-2-methylbutane, 28% 2-chloro-3-methylbutane, and 18% 1-chloro-3-methylbutane. What values did he obtain for the relative ease of removing a hydrogen atom from tertiary, secondary, and primary hydrogen carbons by a chlorine radical under the conditions of his experiment?
Verified step by step guidance1
Step 1: Identify the types of hydrogens in 2-methylbutane. The molecule contains primary (1°), secondary (2°), and tertiary (3°) hydrogens. Primary hydrogens are attached to carbons bonded to only one other carbon, secondary hydrogens are attached to carbons bonded to two other carbons, and tertiary hydrogens are attached to carbons bonded to three other carbons.
Step 2: Determine the number of each type of hydrogen in 2-methylbutane. Count the primary, secondary, and tertiary hydrogens in the molecule. For example, the primary hydrogens are on the terminal carbons, the secondary hydrogens are on carbons bonded to two other carbons, and the tertiary hydrogen is on the central carbon.
Step 3: Calculate the relative reactivity of each type of hydrogen. Use the product distribution percentages provided (36%, 18%, 28%, and 18%) and the number of each type of hydrogen to determine the relative ease of abstraction. The formula to use is: \( \text{Relative Reactivity} = \frac{\text{Product Percentage}}{\text{Number of Hydrogens of that Type}} \).
Step 4: Normalize the relative reactivity values. Divide each calculated relative reactivity by the smallest value to express the relative ease of abstraction as a ratio. This will give you the relative reactivity of tertiary, secondary, and primary hydrogens.
Step 5: Interpret the results. The normalized values will indicate how much more reactive tertiary hydrogens are compared to secondary and primary hydrogens under the experimental conditions. This reflects the stability of the resulting radicals formed after hydrogen abstraction.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Radical Halogenation
Radical halogenation is a reaction where halogens, such as chlorine, react with alkanes in the presence of heat or light to form alkyl halides. This process involves the generation of radicals, which are highly reactive species with unpaired electrons. The reaction proceeds through three main steps: initiation, propagation, and termination, with the propagation step being crucial for determining the relative reactivity of different hydrogen atoms.
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Radical Chain Reaction Mechanism.
Stability of Carbocations
The stability of carbocations is a key factor in understanding the reactivity of hydrogen atoms in alkane chlorination. Tertiary carbocations are more stable than secondary, which are more stable than primary due to hyperconjugation and inductive effects. This stability influences the likelihood of hydrogen abstraction by chlorine radicals, with tertiary hydrogens being removed more easily than secondary or primary hydrogens.
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Product Distribution and Reactivity
The distribution of products in a radical halogenation reaction reflects the relative ease of hydrogen abstraction from different types of carbon. The percentages of the various chlorinated products indicate the preference of chlorine radicals for tertiary, secondary, and primary hydrogens. By analyzing the product distribution, one can deduce the relative reactivity of these hydrogen types, allowing for the determination of the ease of removal under the experimental conditions.
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Related Practice
Textbook Question
At 600 °C, the ratio of the relative rates of formation of a tertiary, a secondary, and a primary radical by a chlorine radical is 2.6 : 2.1 : 1. Explain the change in the degree of regioselectivity compared to what was found in Problem 44.
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Textbook Question
Given that ∆H° for the reaction is -42 kcal/mol and the bond dissociation enthalpies for the C−H, C−Cl, and O−H bonds are 101, 85, and 105 kcal/mol respectively, calculate the bond dissociation enthalpy of the O−Cl bond.
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Textbook Question
Using the given starting material and any necessary organic or inorganic reagents, indicate how the desired compounds could be synthesized:
c.
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Textbook Question
Draw the products of the following reactions, including all stereoisomers:
c.
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Textbook Question
Draw the products of the following reactions, including all stereoisomers:
d.
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