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

11. Radical Reactions

Radical Stability

Organic Chemistry

11. Radical Reactions

Radical Stability

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Problem 16d

Textbook Question

Based on the stability of the radicals produced, predict which bond in each pair would have the higher bond-dissociation energy.
(d)

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Identify the two molecules in the image: one is a bromocyclohexane and the other is an allylic bromide.

Understand that bond-dissociation energy is related to the stability of the radicals formed when the bond is broken. More stable radicals result in lower bond-dissociation energy.

Consider the stability of the radicals formed: breaking the C-Br bond in bromocyclohexane forms a cyclohexyl radical, while breaking the C-Br bond in the allylic bromide forms an allylic radical.

Recall that allylic radicals are stabilized by resonance, which allows the unpaired electron to be delocalized over the π system, increasing stability.

Conclude that the allylic radical is more stable than the cyclohexyl radical due to resonance stabilization, suggesting that the C-Br bond in bromocyclohexane has a higher bond-dissociation energy.

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Key Concepts

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

Radical Stability

Radical stability is crucial in determining bond-dissociation energy. More stable radicals form from weaker bonds, as they require less energy to break. Stability is influenced by factors like hyperconjugation and resonance. In the given image, the radical formed from the allylic bromine bond benefits from resonance stabilization, making it more stable than the cyclohexyl radical.

Bond-Dissociation Energy

Bond-dissociation energy is the energy required to break a bond homolytically, forming radicals. It is higher for bonds that produce less stable radicals upon dissociation. In the image, the cyclohexyl bromine bond likely has a higher bond-dissociation energy because the resulting radical is less stable compared to the allylic radical, which is resonance-stabilized.

Resonance Stabilization

Resonance stabilization occurs when a radical can delocalize its unpaired electron over multiple atoms, increasing stability. This is common in allylic positions, where the radical can be spread across a conjugated system. In the image, the allylic radical formed from the second structure is resonance-stabilized, making it more stable than the cyclohexyl radical, which lacks such delocalization.

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Related Practice

Textbook Question

Based on the stability of the radicals produced, predict which bond in each pair would have the higher bond-dissociation energy.

(c)

Textbook Question

Why is a 2° carbon radical more stable than a 1° carbon radical?

Textbook Question

The following polyunsaturated fat, which does not exist in nature, is oxidized at a rate similar to oleic acid (Table 11.14), despite having two cis-alkenes. Why?

Textbook Question

Ethers can be converted into radicals, some more easily than others. Which of the following radicals is more stable, and thus, more likely to form?

Textbook Question

Methyl t-butyl ether (MTBE) is used preferentially over diethyl ether because it is less prone to form peroxides. Explain this observation in terms of the two structures.

Textbook Question

a. Which of the hydrogens in the structure in the margin is the easiest for a chlorine radical to remove?

b. How many secondary hydrogens does the structure have?

Textbook Question

Rank the following radicals in decreasing order of stability. Classify each as primary, secondary, or tertiary.

a. The isopentyl radical,

b. The 3-methyl-2-butyl radical,

c. The 2-methyl-2-butyl radical,

Textbook Question

Use the information in Table 4-2 (p. 167) to rank the following radicals in decreasing order of stability.

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Based on the stability of the radicals produced, predict which bond in each pair would have the higher bond-dissociation energy.(d)

Key Concepts

Radical Stability

Bond-Dissociation Energy

Resonance Stabilization

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Based on the stability of the radicals produced, predict which bond in each pair would have the higher bond-dissociation energy.
(d)