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Ch. 11 - Properties and Synthesis of Alkyl Halides: Radical Reactions
Mullins - Organic Chemistry: A Learner Centered Approach 1st Edition
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All textbooksMullins 1st EditionCh. 11 - Properties and Synthesis of Alkyl Halides: Radical ReactionsProblem 16d
Chapter 10, Problem 16d

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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1
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.
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The radical stability trend.

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.
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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

Using the bond-dissociation energies in Table 5.6,

(a) predict whether or not an iodine radical would be selective for forming a single radical propane.

1
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Textbook Question

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

(c)

2
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Textbook Question

Using the bond-dissociation energies in Table 5.6,

(a) predict whether or not a fluorine radical would be selective for forming a single radical from propane.

2
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Textbook Question

Using the bond-dissociation energies in Table 5.6,

(b) Will the transition state be reactant-like or product-like? Explain your answer.

2
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Textbook Question

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

(a)

8
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Textbook Question

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

(b)

2
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