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(3,4)
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 16a(3,4)Chapter 3, Problem 16a(3,4)
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 on both sides of the equilibrium. In the first image, CH3NH2 acts as the acid (donating a proton), and H-O⁻ acts as the base (accepting a proton). In the second image, CH3NH2 acts as the base (accepting a proton), and H3O⁺ acts as the acid (donating a proton).
Step 2: Compare the pKa values of the acids on both sides of the equilibrium. For the first reaction, the pKa of CH3NH2 (acid) is approximately 36, and the pKa of H2O (conjugate acid of H-O⁻) is 15.7. For the second reaction, the pKa of H3O⁺ (acid) is -1.7, and the pKa of CH3NH3⁺ (conjugate acid of CH3NH2) is approximately 10.6.
Step 3: Recall that equilibrium favors the side with the weaker acid (higher pKa value). For the first reaction, the equilibrium lies toward the side with H2O (pKa 15.7) and CH3NH2 (pKa 36). For the second reaction, the equilibrium lies toward the side with CH3NH3⁺ (pKa 10.6) and H2O (pKa 15.7).
Step 4: Analyze the bond-breaking and bond-forming events. In the first image, the bond between CH3NH2 and H breaks, and a new bond forms between H and O⁻. In the second image, the bond between H3O⁺ and H breaks, and a new bond forms between CH3NH2 and H.
Step 5: Summarize the direction of equilibrium based on pKa values. For the first reaction, equilibrium lies to the right (formation of CH3NH2 and H2O). For the second reaction, equilibrium lies to the left (formation of CH3NH3⁺ and H2O).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acid-Base Equilibrium
In acid-base chemistry, equilibrium refers to the state where the rates of the forward and reverse reactions are equal. The position of this 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, which has a higher pK<sub>a</sub> value.
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Determining Acid/Base Equilibrium
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>). Lower pK<sub>a</sub> values indicate stronger acids, as they dissociate more completely in solution. This concept is crucial for predicting the direction of acid-base reactions.
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Proton Transfer Mechanism
Acid-base reactions often involve the transfer of a proton (H<sup>+</sup>) from the acid to the base. This mechanism can be visualized through reaction arrows indicating the breaking and forming of bonds. Understanding this process is essential for analyzing the reaction and determining the equilibrium position based on the relative strengths of the acids and bases involved.
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Related Practice
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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2. <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
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
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
Which of the following bases can remove a proton from acetic acid in a reaction that favors products?
HO− CH3NH2 HC≡C− CH3OH H2O Cl−
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