Textbook Question
There were actually two possible products in the solvolysis reaction from Figure 21.10. Show both products. Which would you expect to be more stable? Why?
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Ch. 21 - Conjugated Systems I: Stability and Addition Reactions
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
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Mullins 1st EditionCh. 21 - Conjugated Systems I: Stability and Addition ReactionsProblem 11
Mullins 1st EditionCh. 21 - Conjugated Systems I: Stability and Addition ReactionsProblem 11Chapter 20, Problem 11
p-Nitrophenol (pKa = 7.2) is ten times more acidic than m-nitrophenol (pKa = 8.4) Explain.
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Step 1: Understand the relationship between pKa and acidity. A lower pKa value indicates a stronger acid. Since p-nitrophenol has a pKa of 7.2 and m-nitrophenol has a pKa of 8.4, p-nitrophenol is more acidic.
Step 2: Analyze the structures of p-nitrophenol and m-nitrophenol. Both molecules contain a phenol group (-OH attached to a benzene ring) and a nitro group (-NO2). The position of the nitro group relative to the hydroxyl group differs: para (p) for p-nitrophenol and meta (m) for m-nitrophenol.
Step 3: Consider the electron-withdrawing effect of the nitro group. The nitro group is highly electronegative and withdraws electron density from the benzene ring through both inductive and resonance effects. In p-nitrophenol, the nitro group is positioned para to the hydroxyl group, allowing resonance stabilization of the phenoxide ion formed after deprotonation.
Step 4: Examine resonance stabilization in p-nitrophenol. When p-nitrophenol loses a proton, the negative charge on the oxygen can delocalize through the benzene ring and interact with the nitro group via resonance. This delocalization stabilizes the phenoxide ion, making p-nitrophenol more acidic.
Step 5: Compare with m-nitrophenol. In m-nitrophenol, the nitro group is not in a position to participate in resonance stabilization of the phenoxide ion. The electron-withdrawing effect is limited to inductive effects, which are weaker than the combined inductive and resonance effects in p-nitrophenol. This explains why m-nitrophenol is less acidic than p-nitrophenol.
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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 is often measured using the pKa value, which indicates the strength of an acid in solution. A lower pKa value corresponds to a stronger acid, meaning it more readily donates protons (H+). In this context, p-nitrophenol has a pKa of 7.2, making it a stronger acid than m-nitrophenol, which has a pKa of 8.4.
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Resonance Stabilization
Resonance stabilization occurs when a molecule can be represented by multiple valid Lewis structures, leading to a more stable overall structure. In p-nitrophenol, the nitro group is positioned para to the hydroxyl group, allowing for effective resonance that stabilizes the negative charge on the phenoxide ion formed after deprotonation. This stabilization is less effective in m-nitrophenol due to its meta positioning.
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03:43The radical stability trend.
Inductive Effect
The inductive effect refers to the electron-withdrawing or electron-donating effects of substituents on a molecule, influencing its reactivity and acidity. The nitro group in p-nitrophenol exerts a strong electron-withdrawing inductive effect, enhancing the acidity by stabilizing the negative charge on the conjugate base. In contrast, the inductive effect in m-nitrophenol is less pronounced, contributing to its lower acidity.
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Related Practice
Textbook Question
The molecular orbital picture of H2 can be represented by the following diagram. Label σ and σ* on the diagram—that is, which is (a) and which is (b)? Which is lower in energy? Why?
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Textbook Question
Hexa-1,3,5-triene uses six p orbitals, each containing a single electron, to make its three π bonds. How many total molecular orbitals are made by the six p orbitals?
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Textbook Question
Suppose a molecule is formed by the overlap of 16 atomic orbitals.
(a) How many molecular orbitals will be present?
(b) How many will be bonding?
(c) How many will be antibonding?
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Textbook Question
To this point, hydrogenation has always been an exothermic process. Using the numbers from Figure 21.6, calculate ∆Hohydr for each step of the reduction of benzene.
∆H1 + ∆H2 + ∆H3 = ∆Htot = ―49.5 kcal /mol (―208 kJ/mol)
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Textbook Question
Hydroboration, an electrophilic addition reaction like those studied in Section 21.4, only gives 1,2-addition to buta-1,3-diene, regardless of temperature. Why?
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