Textbook Question
Does methanol behave as an acid or a base when it reacts with methylamine?
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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 16a(1,2)
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 16a(1,2)Chapter 3, Problem 16a(1,2)
For each of the acid–base reactions in [Section 2.3], compare the pKa values of the acids on either side of the equilibrium arrows to prove that the equilibrium lies in the direction indicated.
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Verified step by step guidance1
Step 1: Identify the acids and bases in the given reactions. In the first image, CH3O−H acts as the acid, donating a proton to H−O−. In the second image, CH3O−H acts as the base, accepting a proton from H3O+.
Step 2: Compare the pKa values of the acids on both sides of the equilibrium. For the first reaction, the pKa of CH3O−H2 (protonated methanol) is −2.5, and the pKa of H2O is 15.7. For the second reaction, the pKa of H3O+ is −1.7, and the pKa of CH3O−H2 is −2.5.
Step 3: Recall that equilibrium favors the side with the weaker acid (higher pKa value). In the first reaction, the equilibrium lies toward the formation of H2O (pKa = 15.7) because it is a weaker acid compared to CH3O−H2 (pKa = −2.5). In the second reaction, the equilibrium lies toward the formation of CH3O−H2 (pKa = −2.5) because it is a weaker acid compared to H3O+ (pKa = −1.7).
Step 4: Analyze the bond-breaking and bond-forming events. In the first image, the bond between CH3O−H breaks, and the lone pair on H−O− forms a new bond with the proton. In the second image, the bond between H3O+ and one of its hydrogens breaks, and the lone pair on CH3O−H forms a new bond with the proton.
Step 5: Summarize the direction of equilibrium based on pKa values and the chemical interactions. The equilibrium in the first reaction favors the formation of H2O and CH3O−, while the equilibrium in the second reaction favors the formation of CH3O−H2 and H2O.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
pK<sub>a</sub> Values
The pK<sub>a</sub> value is a quantitative measure of the strength of an acid in solution. It is the negative logarithm of the acid dissociation constant (K<sub>a</sub>), indicating how readily an acid donates protons (H<sup>+</sup>). Lower pK<sub>a</sub> values correspond to stronger acids, as they dissociate more completely in water.
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06:27
Calculating Chair Equilibrium
Acid-Base Equilibrium
In acid-base reactions, equilibrium refers to the state where the rates of the forward and reverse reactions are equal. The position of equilibrium can be predicted by comparing the pK<sub>a</sub> values of the acids involved; the equilibrium will favor the formation of the weaker acid (higher pK<sub>a</sub>), as it is less likely to donate protons.
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04:00Determining Acid/Base Equilibrium
Bond Formation and Breaking
During acid-base reactions, bonds between atoms are broken and formed. When an acid donates a proton, a bond between the hydrogen and the acid molecule breaks, while a new bond is formed between the proton and the base. Understanding this process is crucial for visualizing how equilibrium shifts based on the strength of the acids and bases involved.
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Related Practice
Textbook Question
Ethyne (HC≡CH) has a pKa value of 25, water has a pKa value of 15.7, and ammonia (NH3) has a pKa value of 36. Draw the equation, showing equilibrium arrows that indicate whether reactants or products are favored, for the acid–base reaction of ethyne with
b. −NH2.
c. Which would be a better base to use if you wanted to remove a proton from ethyne, HO− or -NH2?
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Textbook Question
For each of the acid–base reactions in [Section 2.3], compare the pKa values of the acids on either side of the equilibrium arrows to prove that the equilibrium lies in the direction indicated.
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4. <IMAGE>
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Textbook Question
Ethyne (HC≡CH) has a pKa value of 25, water has a pKa value of 15.7, and ammonia (NH3) has a pKa value of 36. Draw the equation, showing equilibrium arrows that indicate whether reactants or products are favored, for the acid–base reaction of ethyne with
a. HO-.
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Textbook Question
a. Which is a stronger base: CH3COO− or HCOO−? (The pKa of CH3COOH is 4.8; the pKa of HCOOH is 3.8.)
b. Which is a stronger base: HO− or -NH2? (The pKa of H2O is 15.7; the pKa of NH3 is 36.)
c. Which is a stronger base: H2O or CH3OH? (The pKa of H3O+ is −1.7; the pKa of CH3O+H2 is −2.5.)
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Textbook Question
Estimate the pKa values of the following compounds:
c. CH3CH2COOH
d. CH3CH2CH2N+H3
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