Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

6. Thermodynamics and Kinetics

Enthalpy

Organic Chemistry

6. Thermodynamics and Kinetics

Enthalpy

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Problem 38c

Textbook Question

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.
(c)

Verified step by step guidance

Step 1: Understand the concept of radical stability. Radicals are stabilized by factors such as resonance, hyperconjugation, and inductive effects. Resonance is particularly significant because it delocalizes the unpaired electron over multiple atoms, reducing its energy.

Step 2: Analyze the structure of the radicals. The first radical is an alkyl radical (ethyl radical), while the second radical is a benzyl radical. The benzyl radical has the unpaired electron adjacent to a benzene ring, which allows for resonance stabilization.

Step 3: Evaluate resonance stabilization in the benzyl radical. The unpaired electron on the benzyl radical can delocalize into the π-system of the benzene ring, forming multiple resonance structures. This delocalization significantly increases the stability of the benzyl radical.

Step 4: Compare the ethyl radical. The ethyl radical lacks resonance stabilization because the unpaired electron is localized on the carbon atom. It can only be stabilized through hyperconjugation with adjacent C-H bonds, which is less effective than resonance.

Step 5: Justify the difference in stability using bond-dissociation energies. Bond-dissociation energy is lower for the bond that forms the benzyl radical compared to the ethyl radical, indicating that the benzyl radical is more stable. This difference arises due to the resonance stabilization in the benzyl radical.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Bond-Dissociation Energy

Bond-dissociation energy (BDE) is the energy required to break a bond in a molecule, resulting in the formation of radicals. Higher BDE values indicate stronger bonds, which typically correlate with lower stability of the resulting radicals. Understanding BDE is crucial for predicting the stability of radicals, as radicals formed from weaker bonds are generally more stable due to lower energy requirements for bond cleavage.

Radical Stability

Radical stability refers to the relative stability of radical species, which can be influenced by factors such as hybridization, resonance, and steric effects. Generally, tertiary radicals are more stable than secondary, which are more stable than primary radicals. This stability is often due to the ability of surrounding groups to donate electron density to the unpaired electron, thus stabilizing the radical.

Resonance Effects

Resonance effects occur when a molecule can be represented by multiple valid Lewis structures, allowing for the delocalization of electrons. In the context of radicals, resonance can significantly enhance stability by spreading out the unpaired electron over multiple atoms. This delocalization reduces the energy of the radical, making it more stable compared to localized radicals that lack such resonance stabilization.

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Related Practice

Textbook Question

Calculate ∆H° for the following reactions.

(b) CH₃Br + HCl → CH₃Cl + HBr

Textbook Question

Calculate ∆H° for the following reactions.

Textbook Question

Calculate ∆H° for the following reactions.

(a)

Textbook Question

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(a)

Textbook Question

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(d) I• vs •OH

Textbook Question

In the following reactions, identify the bonds formed and the bonds broken.

(a)

Textbook Question

In the following reactions, identify the bonds formed and the bonds broken.

(c)

Textbook Question

Teflon-coated frying pans routinely endure temperatures that would cause polyethylene or polypropylene to oxidize and decompose. Decomposition of polyethylene is initiated by free-radical abstraction of a hydrogen atom by O₂. Bond-dissociation energies of C—H bonds are about 400 kJ/mol, and C—F bonds are about 460 kJ/mol. The BDE of the H—OO bond is about 192 kJ/mol, and the F—OO bond is about 63 kJ/mol. Show why Teflon (Figure 7-5) is much more resistant to oxidation than polyethylene is.

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Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.(c)

Key Concepts

Bond-Dissociation Energy

Radical Stability

Resonance Effects

Watch next

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.
(c)