Textbook Question
a. Propose a mechanism for the free-radical chlorination of ethane,
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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 8a
For each reaction, estimate whether ΔS° for the reaction is positive, negative, or impossible to predict.
(a) 
Verified step by step guidance1
Step 1: Analyze the reaction. The reaction involves the catalytic cracking of n-decane (C₁₀H₂₂) into propene (C₃H₆) and heptane (C₇H₁₆). This process breaks a single large molecule into two smaller molecules.
Step 2: Consider the change in the number of molecules. The reactant is one molecule of n-decane, while the products are two molecules (propene and heptane). An increase in the number of molecules generally leads to an increase in entropy (ΔS° > 0).
Step 3: Evaluate the physical states of the reactants and products. n-Decane is a liquid at standard conditions, while propene is a gas and heptane is a liquid. The formation of a gaseous product (propene) from a liquid reactant contributes to an increase in entropy.
Step 4: Consider molecular complexity. Breaking a larger molecule into smaller fragments typically increases the disorder of the system, further supporting a positive ΔS°.
Step 5: Conclude the estimation. Based on the increase in the number of molecules, the formation of a gaseous product, and the increase in disorder, ΔS° for this reaction is likely positive.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Entropy (ΔS°)
Entropy, represented as ΔS°, is a measure of the disorder or randomness in a system. In chemical reactions, an increase in the number of gas molecules or a transition from a more ordered to a less ordered state typically results in a positive ΔS°, indicating greater disorder. Conversely, a decrease in the number of gas molecules or a transition to a more ordered state results in a negative ΔS°.
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Explaining what entropy is.
Catalytic Cracking
Catalytic cracking is a process used in the petroleum industry to break down large hydrocarbon molecules into smaller, more valuable products like gasoline and olefins. This reaction typically involves the use of heat and a catalyst, which facilitates the breaking of carbon-carbon bonds, leading to the formation of smaller alkanes and alkenes, such as propene and heptane in the given reaction.
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Reaction Products and Their States
The products of a reaction can significantly influence the entropy change (ΔS°). In the provided reaction, n-decane (C10H22) is converted into propene (C3H6) and heptane (C7H16). Since the reaction produces a gaseous product (propene) alongside a liquid product (heptane), the overall disorder of the system increases, suggesting a positive ΔS° due to the generation of more gaseous molecules.
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Related Practice
Textbook Question
When ethene is mixed with hydrogen in the presence of a platinum catalyst, hydrogen adds across the double bond to form ethane. At room temperature, the reaction goes to completion. Predict the signs of ΔH° and ΔS° for this reaction. Explain these signs in terms of bonding and freedom of motion.
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Textbook Question
Draw a reaction-energy diagram for the propagation steps of the free-radical addition of HBr to isobutylene. Draw curves representing the reactions leading to both the Markovnikov and the anti-Markovnikov products. Compare the values of ∆Gº and Ea and for the rate-limiting steps, and explain why only one of these products is observed.
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Textbook Question
For each reaction, estimate whether ΔS° for the reaction is positive, negative, or impossible to predict.
(c)
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Textbook Question
For each reaction, estimate whether ΔS° for the reaction is positive, negative, or impossible to predict.
(b)
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Textbook Question
The dehydrogenation of butane to trans-but-2-ene has ΔH° = +116 kJ/mol (+27.6 kcal/mol) and ΔS° = +117J/kelvin-mol (+28.0 cal/kelvin-mol). a. Compute the value of ΔG° for dehydrogenation at room temperature (25 °C or 298 °K). Is dehydrogenation favored or disfavored?HINT: When you are doing synthesis problems, avoid using these high-temperature industrial methods. They require specialized equipment, and they produce variable mixtures of products.