Textbook Question
Show the products you expect when each compound reacts with NBS with light shining on the reaction.
(c)
2
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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 59a
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.
Verified step by step guidance1
Step 1: Understand the reaction mechanism. NBS (N-Bromosuccinimide) is commonly used for allylic bromination in the presence of light or heat. The reaction proceeds via a free radical mechanism, where the allylic hydrogen is abstracted, forming an allylic radical intermediate.
Step 2: Identify the allylic positions in 1-methylcyclohexene. The allylic position is the carbon atom adjacent to the double bond. In 1-methylcyclohexene, there are two distinct allylic positions: one on the methyl group and one on the cyclohexene ring.
Step 3: Draw the first allylic radical intermediate. When the hydrogen from the methyl group is abstracted, a resonance-stabilized allylic radical is formed. Use curved arrows to show the delocalization of the unpaired electron across the π-system of the double bond.
Step 4: Draw the second allylic radical intermediate. When the hydrogen from the cyclohexene ring is abstracted, another resonance-stabilized allylic radical is formed. Again, use curved arrows to show the delocalization of the unpaired electron across the π-system.
Step 5: Draw the third allylic radical intermediate. Consider the possibility of resonance structures for the second allylic radical. This involves shifting the double bond and the unpaired electron to generate an additional resonance form. Ensure all resonance forms are properly drawn and labeled.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Allylic Free Radicals
Allylic free radicals are reactive intermediates formed when a radical is located on a carbon atom adjacent to a double bond. These radicals are stabilized by resonance, allowing the unpaired electron to be delocalized over the π system of the alkene. This delocalization increases the stability of the radical, making it more favorable for reactions such as halogenation.
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Resonance Structures
Resonance structures are different Lewis structures for the same molecule that illustrate the delocalization of electrons. In the case of allylic free radicals, resonance forms show how the radical can be represented at different positions along the carbon chain. These structures help predict the stability and reactivity of the radical, as well as the possible products formed during reactions.
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N-Bromosuccinimide (NBS) and its Role
N-Bromosuccinimide (NBS) is a reagent commonly used in organic chemistry for bromination reactions, particularly in allylic positions. When NBS is added to alkenes, it generates bromine radicals under light irradiation, which can abstract hydrogen atoms from the allylic position, forming allylic free radicals. This process leads to the formation of brominated products, which are crucial for understanding the reaction mechanism and product distribution.
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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)
1
views
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.
b. Rank these three intermediates from most stable to least stable.
2
views
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.
3
views