Textbook Question
A misguided chemist attempted to synthesize trans-1,2-dimethylcyclohexane via the hydrogenation of 1,2-dimethylcyclohexene. Explain why this is not possible.
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Ch. 9 - Alkenes II: Oxidation and Reduction
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471
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Table of contents
- Ch. 2 - General Chemistry Translated: Finding the Electrons114
- Ch. 3 - Alkanes and Cycloalkanes: Properties and Conformational Analysis109
- Ch. 4 - Acids and Bases: Electron Flow144
- Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics118
- Ch. 6 - Stereoisomerism: Arrangement of Atoms in Space112
- Ch. 7 - Principles of Green (Organic) Chemistry3
- Ch. 8 - Alkenes I: Properties and Electrophilic Additions158
- Ch. 9 - Alkenes II: Oxidation and Reduction150
- Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions101
- Ch. 11 - Properties and Synthesis of Alkyl Halides: Radical Reactions108
- Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes172
- Ch. 13 - Alcohols, Ethers and Related Compounds: Substitution and Elimination239
- Ch. 14 - Structural Identification I: Infrared Spectroscopy and Mass Spectrometry54
- Ch. 15 - Structural Identification II: Nuclear Magnetic Resonance Spectroscopy93
- Ch. 16 - Metals in Organic Chemistry48
- Ch. 17 - Carbonyl Addition Reactions: Aldehydes and Ketones45
- Ch. 18 - Nucleophilic Acyl Substitution I: Carboxylic Acids18
- Ch. 19 - Nucleophilic Acyl Substitution II: Carboxylic Acid Derivatives39
- Ch. 20 - Enolates: Carbonyl Addition and Substitution13
- Ch. 21 - Conjugated Systems I: Stability and Addition Reactions56
- Ch. 22 - Conjugated Systems II: Pericyclic Reactions54
- Ch. 23 - Benzene I: Aromatic Stability and Substitution Reactions64
- Ch. 24 - Benzene II: Reactions Influenced by the Aromatic Ring62
- Ch. 25 - Amines: Structure, Reactions, and Synthesis27
- Ch. 26 - Amino Acids, Proteins, and Peptide Synthesis9
- Ch. 27 - Carbohydrates, Nucleic Acids, and Lipids54
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Not the one you use?Change textbook
Chapter 8, Problem 37a
Without concerning yourself with the mechanism of the reaction, calculate the equilibrium constant for the following equilibrium processes. (Assume T = 298 K.)
(a) 
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Step 1: Identify the relationship between the equilibrium constant (K) and the Gibbs free energy change (ΔG°) using the equation ΔG° = -RT ln(K), where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K).
Step 2: Determine the value of ΔG° for the given equilibrium process. If ΔG° is not provided, it may need to be calculated or derived from other thermodynamic data such as enthalpy (ΔH°) and entropy (ΔS°) using the equation ΔG° = ΔH° - TΔS°.
Step 3: Rearrange the equation ΔG° = -RT ln(K) to solve for K. This gives ln(K) = -ΔG° / (RT).
Step 4: Substitute the known values of ΔG°, R, and T into the equation ln(K) = -ΔG° / (RT) to calculate ln(K).
Step 5: Exponentiate both sides of the equation to solve for K, using K = e^(ln(K)). This will yield the equilibrium constant for the reaction.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Constant (K)
The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It is temperature-dependent and provides insight into the extent of a reaction; a large K indicates a reaction that favors products, while a small K suggests a preference for reactants.
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The relationship between equilibrium constant and pKa.
Le Chatelier's Principle
Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change and restore a new equilibrium. This principle helps predict how changes in concentration, pressure, or temperature will affect the position of equilibrium in a reaction.
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Standard Conditions
Standard conditions refer to a set of specific conditions (usually 1 atm pressure and 298 K temperature) under which thermodynamic measurements are made. These conditions provide a consistent basis for comparing the equilibrium constants of different reactions, ensuring that variations in temperature or pressure do not affect the calculated values.
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Related Practice
Textbook Question
Conjugated dienes, molecules containing two alkenes separated by one single bond, are discussed in detail in Chapter 21.
(b) How might you account for this difference in stability?
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Textbook Question
Questions (a)–(d) all refer to the following reaction, which has been engineered to produce one enantiomer to the exclusion of the other.
(c) Suppose the difference in activation energy is 1.6 kcal/mol. At what temperature would you produce C in 99% ee?
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Textbook Question
Because deuterium behaves like hydrogen in chemical reactions yet is detected differently, chemists use the incorporation of deuterium to better understand the subtleties of reaction mechanisms. Deuterium is incorporated by replacing H₂ with D₂ in the hydrogenation reaction. Identify the product expected when the alkenes in Assessment 9.34 react with D₂ and Pd/C. [Don't worry about showing all diastereomers.]
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Textbook Question
What product results when the following molecules are treated with H₂ Pd/C? Be sure to indicate the relative stereochemical outcome. Draw both enantiomers of any racemic mixtures. [It is difficult to control the stoichiometry of gases so there is enough H₂ to reduce all alkenes.]
(b)
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Textbook Question
Conjugated dienes, molecules containing two alkenes separated by one single bond, are discussed in detail in Chapter 21.
(a) Considering the observed ∆H° of hydrogenation, is hexa-1,3-diene (conjugated) or hexa-1,4-diene (unconjugated) more stable?
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