Textbook Question
Show the products you expect when each compound reacts with NBS with light shining on the reaction.
(c)
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Ch.6 - Alkyl Halides; Nucleophilic Substitution
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 6, Problem 57b
Propose mechanisms to account for the observed products in the following reactions.
(b) 
Verified step by step guidance1
Step 1: Recognize that the reaction involves allylic bromination using NBS (N-Bromosuccinimide) under light (hv). NBS is commonly used to selectively brominate allylic or benzylic positions in the presence of a double bond.
Step 2: Initiation phase: Light (hv) causes homolytic cleavage of the N-Br bond in NBS, generating a bromine radical (Br•). This radical is highly reactive and initiates the reaction.
Step 3: Propagation phase: The bromine radical abstracts a hydrogen atom from the allylic position of the cyclohexene molecule, forming an allylic radical. The allylic radical is stabilized by resonance, creating two possible resonance structures.
Step 4: Bromine from NBS reacts with the allylic radical to form the brominated products. The resonance structures of the allylic radical lead to two different products: (1) 3-bromocyclohexene and (2) 1-bromocyclohexene.
Step 5: Termination phase: Any remaining radicals combine to form stable molecules, completing the reaction. The observed products are the result of the selective bromination at the allylic positions due to the resonance stabilization of the intermediate radical.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Allylic Bromination
Allylic bromination is a reaction where bromine is added to the allylic position of an alkene or alkane. This process typically involves the formation of a radical intermediate, allowing for the substitution of a hydrogen atom with a bromine atom. The reaction is often facilitated by reagents like N-bromosuccinimide (NBS) in the presence of light, which generates bromine radicals that can abstract hydrogen atoms from the allylic position.
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Mechanism of Allylic Bromination.
Radical Mechanism
The radical mechanism involves the formation and reaction of free radicals, which are highly reactive species with unpaired electrons. In the context of allylic bromination, the mechanism proceeds through three main steps: initiation (formation of bromine radicals), propagation (reaction of radicals with the substrate), and termination (recombination of radicals). Understanding this mechanism is crucial for predicting the products of the reaction and the regioselectivity of bromination.
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Regioselectivity
Regioselectivity refers to the preference of a chemical reaction to yield one structural isomer over others when multiple possibilities exist. In allylic bromination, the regioselectivity is influenced by the stability of the radical intermediates formed during the reaction. More stable radicals, such as those formed at more substituted carbon centers, are favored, leading to specific products. This concept is essential for predicting the distribution of products in the reaction.
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Related Practice
Textbook Question
Show the products you expect when each compound reacts with NBS with light shining on the reaction.
(a)
(b)
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Textbook Question
The following reaction takes place under second-order conditions (strong nucleophile), yet the structure of the product shows rearrangement. Also, the rate of this reaction is several thousand times faster than the rate of substitution of hydroxide ion on 2-chlorobutane under similar conditions. Propose a mechanism to explain the enhanced rate and rearrangement observed in this unusual reaction. (“Et” is the abbreviation for ethyl.)
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Textbook Question
Propose mechanisms to account for the observed products in the following reactions.
(a)
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Textbook Question
A student adds NBS to a solution of 1-methylcyclohexene and irradiates the mixture with a sunlamp until all the NBS has reacted. After a careful distillation, the product mixture contains two major products of formula C7H11Br.
(a) Draw the resonance forms of the three possible allylic free radical intermediates.
(b) Rank these three intermediates from most stable to least stable.
(c) Draw the products obtained from each free-radical intermediate.
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Textbook Question
Furfuryl chloride can undergo substitution by both SN2 and SN1 mechanisms. Because it is a 1° alkyl halide, we expect SN2 but not SN1 reactions. Draw a mechanism for the SN1 reaction shown below, paying careful attention to the structure of the intermediate. How can this primary halide undergo SN1 reactions?
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