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Ch.6 - Alkyl Halides; Nucleophilic Substitution
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh.6 - Alkyl Halides; Nucleophilic SubstitutionProblem 10a,b
Chapter 6, Problem 10a,b

Show how free-radical halogenation might be used to synthesize the following ­compounds. In each case, explain why we expect to get a single major product.
(a) 1-chloro-2,2-dimethylpropane (neopentyl chloride)
(b) 2-bromo-2-methylbutane

Verified step by step guidance
1
Identify the starting alkane for each compound. For neopentyl chloride (1-chloro-2,2-dimethylpropane), the starting alkane is neopentane (2,2-dimethylpropane). For 2-bromo-2-methylbutane, the starting alkane is 2-methylbutane.
Understand the mechanism of free-radical halogenation. It involves three steps: initiation (generation of halogen radicals), propagation (reaction of halogen radicals with the alkane to form alkyl radicals), and termination (recombination of radicals).
For neopentyl chloride, consider the structure of neopentane. The tertiary hydrogen on the central carbon is the most reactive due to the stability of the resulting tertiary radical. This selectivity ensures that chlorination occurs at this position, leading to a single major product.
For 2-bromo-2-methylbutane, analyze the structure of 2-methylbutane. The tertiary hydrogen on the second carbon is the most reactive, as the tertiary radical formed during the propagation step is highly stable. Bromination at this position will yield the single major product.
Explain why a single major product is expected in each case. The regioselectivity of free-radical halogenation is influenced by the stability of the intermediate radical. Tertiary radicals are more stable than secondary or primary radicals, so halogenation preferentially occurs at the tertiary carbon in both cases.

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Key Concepts

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

Free-Radical Halogenation

Free-radical halogenation is a reaction where alkanes react with halogens (like Cl2 or Br2) in the presence of heat or light to form alkyl halides. This process involves the formation of free radicals, which are highly reactive species with unpaired electrons. The reaction proceeds through three main steps: initiation, propagation, and termination, leading to the substitution of hydrogen atoms with halogen atoms.
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Selectivity in Free-Radical Reactions

Selectivity in free-radical reactions refers to the preference for the formation of certain products over others. In the case of halogenation, the stability of the resulting radical influences which hydrogen is replaced. More stable radicals (tertiary > secondary > primary) are formed preferentially, leading to a single major product when the structure of the alkane allows for this selectivity.
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Stereochemistry and Product Distribution

Stereochemistry plays a crucial role in determining the product distribution in free-radical halogenation. In cases where the starting material has symmetrical or similar substituents, the reaction tends to yield a single major product due to the uniformity in the environment around the reactive site. This is particularly evident in compounds like 1-chloro-2,2-dimethylpropane and 2-bromo-2-methylbutane, where the structure leads to minimal competition for radical formation.
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Related Practice
Textbook Question

For each pair of compounds, predict which compound has the higher boiling point. Check [TABLE 6-2] to see if your prediction was right; then explain why that compound has the higher boiling point.

c. 1-bromobutane and 1-chlorobutane

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Textbook Question

Show how free-radical halogenation might be used to synthesize the following ­compounds. In each case, explain why we expect to get a single major product.

(d)

Textbook Question

Show how free-radical halogenation might be used to synthesize the following ­compounds. In each case, explain why we expect to get a single major product.

(c)

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Textbook Question

The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products:

a. Give a mechanism for this reaction, showing how the two products arise as a ­consequence of the resonance-stabilized intermediate.

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Textbook Question

Classify each reaction as a substitution, an elimination, or neither. Identify the leaving group in each reaction, and the nucleophile in substitutions.

a.

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Textbook Question

The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products:

b. The bromination of cyclohexene using NBS gives only one major product, as shown on the previous page. Explain why there is no second product from an allylic shift.

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