Textbook Question
Keto–enol tautomerism is a reaction we discuss in detail in Chapter 19. Estimate the equilibrium constant of this reaction (BDE for C―C π bond = 65 kcal/mol ; for C―O π bond = 85 kcal/mol).
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Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics
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
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Mullins 1st EditionCh. 5 - Chemical Reaction Analysis: Thermodynamics and KineticsProblem 63b
Mullins 1st EditionCh. 5 - Chemical Reaction Analysis: Thermodynamics and KineticsProblem 63bChapter 4, Problem 63b
Calculate Keq for the following acid–base reaction.
Verified step by step guidance1
Step 1: Identify the acid and base on both sides of the reaction. In this reaction, the methyl anion (CH₃⁻) acts as the base, and isopropanol (CH₃CHOHCH₃) acts as the acid on the left side. On the right side, methane (CH₄) is the conjugate acid, and the isopropoxide ion (CH₃CH⁻OCH₃) is the conjugate base.
Step 2: Recall the relationship between pKa values and equilibrium constants (Keq). The equilibrium constant for an acid-base reaction can be calculated using the formula: Keq = 10^(pKa(acid on left) - pKa(acid on right)).
Step 3: Look up the pKa values for the acids involved. The pKa of isopropanol (CH₃CHOHCH₃) is approximately 16. The pKa of methane (CH₄) is approximately 50.
Step 4: Substitute the pKa values into the formula for Keq. Using the formula Keq = 10^(pKa(acid on left) - pKa(acid on right)), substitute pKa(acid on left) = 16 and pKa(acid on right) = 50.
Step 5: Simplify the expression to determine the magnitude of Keq. The calculation will show that Keq is a very small number, indicating that the equilibrium strongly favors the reactants (left side of the reaction).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acid-Base Reactions
Acid-base reactions involve the transfer of protons (H+) between reactants. In organic chemistry, these reactions are fundamental for understanding the behavior of various compounds, particularly in determining their reactivity and stability. The strength of acids and bases is often quantified using the pKa scale, which helps predict the direction of the reaction.
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The Lewis definition of acids and bases.
Equilibrium Constant (K_eq)
The equilibrium constant (K_eq) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction. It provides insight into the favorability of a reaction; a K_eq greater than 1 indicates that products are favored, while a K_eq less than 1 suggests that reactants are favored. Calculating K_eq is essential for understanding the extent of acid-base reactions.
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02:19The relationship between equilibrium constant and pKa.
Relationship Between pKa and K_eq
The relationship between pKa and K_eq is crucial in acid-base chemistry. The pKa of an acid is the negative logarithm of its acid dissociation constant (Ka), and it can be used to derive K_eq for acid-base reactions. Specifically, K_eq can be calculated using the difference in pKa values of the acids and bases involved, allowing for a quantitative assessment of the reaction's equilibrium position.
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02:19The relationship between equilibrium constant and pKa.
Related Practice
Textbook Question
Reaction (c), on the other hand, is favored (∆G° < 0). Identify the bonds formed and broken and explain this result in light of (a) and (b).
(c)
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Textbook Question
In light of your answers to parts (b) and (c), where both were shown to be quite favorable, imagine a scenario where either reaction is possible. Of the two, which would you expect to be faster? Which would you expect to be more favored? Explain each in the context of the important thermodynamic and/or kinetic parameters.
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Textbook Question
The hydrogenation of alkenes is a reaction we study in Chapter 9.
(b) Is this reaction favored or disfavored in terms of entropy?
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Textbook Question
Parts (a)–(d) of this assessment assist in the development of what will become a common theme in organic reactions and should be worked in order. [Think carefully about how each question relates to the others.]
(c) Without worrying about the mechanism of the reaction, estimate an equilibrium constant for the following carbonyl addition reaction based on the relative stability of the Lewis bases.
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Textbook Question
In Chapter 13, we discuss the ring-opening reactions of epoxides, such as the one shown here.
(a) Based on the bonds formed and the bonds broken, calculate ∆H°.
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