Textbook Question
From which of the following compounds can HO− remove a proton in a reaction that favors product formation?
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Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry
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. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 68a
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 68aChapter 3, Problem 68a
For each of the following pairs of reactions, indicate which one has the more favorable equilibrium constant (that is, which one most favors products):
1. 
2. 
Verified step by step guidance1
Step 1: Understand the problem. The question asks us to compare the equilibrium constants for two pairs of reactions. The equilibrium constant depends on the relative acidity of the acids involved, which can be determined using their pKa values. Lower pKa values indicate stronger acids.
Step 2: Analyze the first pair of reactions: CH3CH2OH + NH3 ⇌ CH3CH2O− + NH4 and CH3OH + NH3 ⇌ CH3O− + NH4. Compare the acidity of CH3CH2OH and CH3OH using their pKa values. CH3CH2OH has a slightly higher pKa than CH3OH, meaning CH3OH is the stronger acid. Therefore, the reaction involving CH3OH will have a more favorable equilibrium constant.
Step 3: Analyze the second pair of reactions: CH3CH2OH + NH3 ⇌ CH3CH2O− + NH4 and CH3CH2OH + CH3NH2 ⇌ CH3CH2O− + CH3NH3+. Compare the acidity of NH4+ and CH3NH3+ using their pKa values. NH4+ has a lower pKa (9.40) compared to CH3NH3+ (10.70), meaning NH4+ is the stronger acid. Therefore, the reaction involving NH4+ will have a more favorable equilibrium constant.
Step 4: Summarize the reasoning. For each pair of reactions, the equilibrium constant favors the reaction where the acid involved is stronger (lower pKa). This is because stronger acids donate protons more readily, driving the equilibrium toward the products.
Step 5: Conclude the comparison. For the first pair, the reaction with CH3OH is more favorable. For the second pair, the reaction with NH4+ is more favorable. Use the provided pKa table to justify these conclusions.
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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) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. A larger K value indicates a greater favorability for the formation of products, while a smaller K suggests that reactants are favored. Understanding K is crucial for predicting the direction of a reaction and its extent.
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02:19
The relationship between equilibrium constant and pKa.
pKa and Acid Strength
pKa is a measure of the strength of an acid in solution; it is the negative logarithm of the acid dissociation constant (Ka). Lower pKa values correspond to stronger acids, which dissociate more completely in solution. In acid-base reactions, the relative pKa values of the acids involved can help predict which direction the equilibrium will favor.
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07:45Identifying pKa values
Acid-Base Reactions
Acid-base reactions involve the transfer of protons (H+) from an acid to a base. The strength of the acids and bases involved determines the position of equilibrium. In the context of the given reactions, comparing the pKa values of the acids (like CH3OH and CH3CH2OH) helps to ascertain which reaction will favor product formation more strongly.
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02:49The Lewis definition of acids and bases.
Related Practice
Textbook Question
Tenormin, a member of the group of drugs known as beta-blockers, is used to treat high blood pressure and improve survival after a heart attack. It works by slowing down the heart to reduce its workload. Which atom in Tenormin is the most basic?
Textbook Question
a. Without using a calculator, estimate the pH of each of the following solutions:
1. [HO−] = 3.2 × 10−5
2. [H3O+] = 8.3 × 10−1
3. [H3O+] = 1.7 × 10−3
b. Determine the exact pH, using a calculator.
Textbook Question
b. Determine the exact pKa values, using a calculator.
c. Which is the strongest acid?
1. nitrous acid (HNO2), Ka = 4.0 × 10−4
2. nitric acid (HNO3), Ka = 22
3. bicarbonate (HCO3−), Ka = 6.3 × 10−11
4. hydrogen cyanide (HCN), Ka = 7.9 × 10−10
5. formic acid (HCOOH), Ka = 2.0 × 10−4
6. phosphoric acid (H3PO4), Ka = 2.1
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Textbook Question
You are planning to carry out a reaction that produces protons. The reaction will be buffered at pH = 10.5. Would it be better to use a protonated methylamine/methylamine buffer or a protonated ethylamine/ethylamine buffer? (pKa of protonated methylamine = 10.7; pKa of protonated ethylamine = 11.0)
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Textbook Question
Which of the four reactions has the most favorable equilibrium constant?
1.
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