Textbook Question
Amino acids exist predominantly in one of the following forms. Which is it? Explain your answer.
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Ch. 4 - Acids and Bases: Electron Flow
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 3, Problem 70c
Using pKa values for the conjugate acids of the bases on each side of the reaction arrow, identify which side of the equilibrium would be favored in the following hypothetical reactions.
(c) 
Verified step by step guidance1
Step 1: Identify the bases and their conjugate acids on both sides of the equilibrium. On the left side, the base is the cyanide ion (N≡C⁻), and its conjugate acid is hydrogen cyanide (HC≡N). On the right side, the base is the deprotonated form of the imine (N⁻ with lone pairs), and its conjugate acid is the protonated imine.
Step 2: Look up the pKa values for the conjugate acids of the bases. The pKa of hydrogen cyanide (HC≡N) is approximately 9.2, and the pKa of the protonated imine is typically lower, around 4-5, depending on the specific structure.
Step 3: Compare the pKa values. The side of the equilibrium with the conjugate acid that has the higher pKa value will be favored because it indicates a weaker acid and a more stable conjugate base.
Step 4: Analyze the equilibrium direction. Since the pKa of hydrogen cyanide (9.2) is higher than the pKa of the protonated imine (4-5), the equilibrium will favor the side with the cyanide ion (N≡C⁻) and the neutral imine.
Step 5: Conclude that the equilibrium lies to the left, favoring the cyanide ion and the neutral imine, based on the relative stability of the conjugate bases and the pKa values of their conjugate acids.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
pKa and Acid-Base Equilibrium
pKa is a measure of the strength of an acid in solution; it is the negative logarithm of the acid dissociation constant (Ka). In acid-base reactions, the side with the weaker acid (higher pKa) is favored at equilibrium. Understanding pKa values allows chemists to predict the direction of equilibrium based on the relative strengths of acids and their conjugate bases.
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Determining Acid/Base Equilibrium
Conjugate Acids and Bases
A conjugate acid is formed when a base gains a proton (H+), while a conjugate base is what remains after an acid donates a proton. The strength of an acid is inversely related to the strength of its conjugate base; a strong acid has a weak conjugate base. Recognizing these relationships is crucial for determining which side of a reaction is favored based on pKa values.
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02:11Equilibrium constant and conjugates.
Le Chatelier's Principle
Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system shifts to counteract the change and restore equilibrium. In the context of acid-base reactions, this principle helps predict how changes in concentration, pressure, or temperature can affect the position of equilibrium, guiding the analysis of which side is favored based on pKa values.
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Related Practice
Textbook Question
Predict the product of the following Lewis acid–Lewis base reactions.
(c)
Textbook Question
A medicinal chemist needed to deprotonate acetylene ( HC≡CH ) for use in a coupling reaction. Among the options given, which base(s) could be used for this process?
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Textbook Question
Predict the product of the following Lewis acid–Lewis base reactions.
(d) [an intramolecular reaction]
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Textbook Question
Using pKa values for the conjugate acids of the bases on each side of the reaction arrow, identify which side of the equilibrium would be favored in the following hypothetical reactions.
(a)
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Textbook Question
Suggest an arrow-pushing mechanism for each of the following acid–base reactions
(c)
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