Textbook Question
The nitrogen screened in purple in Figure 4.43 is not protonated at physiological pH. Why?
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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 52d
Identify the most acidic proton in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer. [Always start by drawing the conjugate base.]
(d) 
Verified step by step guidance1
Step 1: Begin by drawing the conjugate base for each molecule. For the benzyl proton (H attached to the CH2 group next to the benzene ring), remove the proton to form a benzyl carbanion. For the hydroxyl proton (OH group), remove the proton to form an alkoxide ion.
Step 2: Analyze the stability of the conjugate bases. The benzyl carbanion benefits from resonance stabilization because the negative charge can delocalize into the aromatic ring. This resonance effect significantly increases the stability of the benzyl carbanion.
Step 3: Compare this to the alkoxide ion formed from the hydroxyl group. The alkoxide ion does not have resonance stabilization, but it does have inductive effects from the surrounding alkyl groups. However, these effects are weaker compared to resonance stabilization.
Step 4: Consider the electronegativity of the atoms involved. Oxygen in the hydroxyl group is more electronegative than carbon in the benzyl group, which means the alkoxide ion can better stabilize the negative charge compared to a simple carbanion. However, resonance stabilization in the benzyl carbanion outweighs this factor.
Step 5: Conclude that the benzyl proton is more acidic because its conjugate base (benzyl carbanion) is more stable due to resonance effects. Structural features like resonance stabilization are weighted heavily in determining acidity.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acidity and pKa
Acidity in organic chemistry refers to the tendency of a compound to donate a proton (H+). The strength of an acid is often measured by its pKa value; lower pKa values indicate stronger acids. Understanding the relationship between acidity and pKa is crucial for identifying the most acidic proton in a given structure.
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Identifying pKa values
Conjugate Base Stability
The stability of the conjugate base formed after deprotonation is a key factor in determining acidity. A more stable conjugate base corresponds to a stronger acid. Factors influencing stability include resonance, electronegativity of atoms, and inductive effects, which can help predict which proton is more acidic.
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Structural Features Influencing Acidity
Various structural features can affect the acidity of protons in organic molecules. These include the presence of electronegative atoms, resonance structures, and hybridization of the atom bearing the proton. Analyzing these features allows for a systematic approach to determine which proton is the most acidic in a pair.
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Related Practice
Textbook Question
Identify the most stable conjugate base in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer.
(f)
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Textbook Question
Identify the most acidic proton in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer. [Always start by drawing the conjugate base.]
(b)
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Textbook Question
Identify the most acidic proton in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer. [Always start by drawing the conjugate base.]
(h)
1630
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Textbook Question
Identify the Lewis acid and the Lewis base in each of the following reactions.
(c)
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Textbook Question
Identify the most acidic proton in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer. [Always start by drawing the conjugate base.]
(a)
1410
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