Textbook Question
Calculate the percentage of isopropylcyclohexane molecules that have the isopropyl substituent in an equatorial position at equilibrium. (Its ∆G° value at 25 °C is -2.1 kcal/mol.)
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Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics
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
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Bruice 8th EditionCh. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and KineticsProblem 19
Bruice 8th EditionCh. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and KineticsProblem 19Chapter 6, Problem 19
The ∆G° for conversion of “axial” fluorocyclohexane to “equatorial” fluorocyclohexane at 25 °C is -0.25kcal/mol. Calculate the percentage of fluorocyclohexane molecules that have the fluoro substituent in an equatorial position at equilibrium.
Verified step by step guidance1
Step 1: Understand the relationship between Gibbs free energy (∆G°) and the equilibrium constant (K). The equation to use is ∆G° = -RT ln(K), where R is the gas constant (1.987 cal/(mol·K)) and T is the temperature in Kelvin (25 °C = 298 K).
Step 2: Rearrange the equation to solve for K, the equilibrium constant. This gives K = e^(-∆G° / RT). Substitute the given values for ∆G° (-0.25 kcal/mol, converted to -250 cal/mol), R (1.987 cal/(mol·K)), and T (298 K) into the equation.
Step 3: Once K is calculated, recognize that K represents the ratio of the concentrations of the equatorial form to the axial form at equilibrium. Specifically, K = [equatorial]/[axial].
Step 4: Use the relationship between K and the percentage of molecules in the equatorial position. The percentage of equatorial molecules can be calculated as: % equatorial = (K / (1 + K)) × 100.
Step 5: Substitute the value of K obtained in Step 2 into the percentage formula to determine the percentage of fluorocyclohexane molecules with the fluoro substituent in the equatorial position.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gibbs Free Energy (∆G)
Gibbs Free Energy (∆G) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. A negative ∆G indicates that a reaction is spontaneous, while a positive ∆G suggests non-spontaneity. In this context, the negative value of -0.25 kcal/mol implies that the conversion from axial to equatorial fluorocyclohexane is favored, leading to a higher proportion of equatorial molecules at equilibrium.
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05:02
Breaking down the different terms of the Gibbs Free Energy equation.
Equilibrium Constant (K)
The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It is related to the Gibbs Free Energy change by the equation ∆G° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. By calculating K from the given ∆G°, we can determine the ratio of equatorial to axial fluorocyclohexane, which helps in finding the percentage of equatorial molecules.
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02:19The relationship between equilibrium constant and pKa.
Percentage Calculation at Equilibrium
To find the percentage of molecules in a specific conformation at equilibrium, we use the equilibrium constant (K) derived from the Gibbs Free Energy. The formula K = [equatorial]/[axial] allows us to express the fraction of equatorial molecules as [equatorial] / ([equatorial] + [axial]). By converting this fraction into a percentage, we can determine how many of the fluorocyclohexane molecules are in the equatorial position at equilibrium.
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03:23How to solve for the percentage of each enantiomer.
Related Practice
Textbook Question
For each of the reactions in Problem 15, indicate which reactant is the nucleophile and which is the electrophile.
a.
b.
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Textbook Question
Calculate ∆G° for the conversion of “axial” methylcyclohexane to “equatorial” methylcyclohexane at 25 °C.
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Textbook Question
Why is the percentage of molecules with the substituent in an equatorial position greater for isopropylcyclohexane than for fluorocyclohexane?
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Textbook Question
a. For which reaction in each set will ∆S° be more significant?
b. For which reaction will ∆S° be positive?
1. A ⇌ B or A + B ⇌ C
2. A ⇌ B + C or A + B ⇌ C + D
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Textbook Question
Use curved arrows to show the movement of electrons in the following reaction steps
a.
b.
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