Textbook Question
In the presence of a small amount of bromine, cyclohexene undergoes the following light-promoted reaction:
d. Explain why cyclohexene reacts with bromine much faster than cyclohexane, which must be heated to react.
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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 26c
Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate.
Hint: Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations.
(c) 
Verified step by step guidance1
Step 1: Begin by identifying the leaving group in the substrate. In this case, the iodine atom (I) attached to the cyclohexene ring is the leaving group. Under solvolysis conditions, the bond between the carbon and iodine breaks, forming a carbocation intermediate.
Step 2: Analyze the stability of the initially formed carbocation. The carbocation formed directly after the iodine leaves is a secondary carbocation. Secondary carbocations are less stable than tertiary carbocations or resonance-stabilized carbocations.
Step 3: Propose a rearrangement mechanism to stabilize the carbocation. A hydride shift or an alkyl shift can occur to convert the secondary carbocation into a more stable tertiary carbocation. In this case, a hydride shift from an adjacent carbon atom can occur, moving a hydrogen atom along with its bonding electrons to the carbocation center.
Step 4: After the hydride shift, the carbocation is now tertiary, which is more stable due to increased hyperconjugation and inductive effects. This rearranged carbocation is the key intermediate for the subsequent reaction.
Step 5: The rearranged carbocation reacts with acetic acid (CH3COOH) to form the final product. The nucleophilic oxygen atom in acetic acid attacks the carbocation, leading to the formation of the ester products shown in the reaction.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Carbocation Stability
Carbocations are positively charged carbon species that can undergo rearrangements to achieve greater stability. The stability of carbocations increases in the order of primary < secondary < tertiary due to hyperconjugation and inductive effects from adjacent alkyl groups. In solvolysis reactions, rearrangements often convert less stable carbocations into more stable ones, such as transforming a secondary carbocation into a tertiary one.
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Determining Carbocation Stability
Hydride and Alkyl Shifts
Hydride shifts involve the migration of a hydrogen atom with its bonding electrons from one carbon to an adjacent positively charged carbon, while alkyl shifts involve the movement of an alkyl group. These shifts are key mechanisms in carbocation rearrangements, allowing the molecule to form a more stable carbocation intermediate. Such shifts are often driven by the need to minimize the energy of the transition state and stabilize the resulting carbocation.
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SN1 Mechanism
The SN1 mechanism is a type of nucleophilic substitution reaction characterized by a two-step process: the formation of a carbocation intermediate followed by nucleophilic attack. The first step is the rate-determining step, where the leaving group departs, forming a carbocation. The stability of this intermediate is crucial, as more stable carbocations lead to faster reaction rates and a higher likelihood of rearrangements, such as hydride or alkyl shifts, to achieve stability.
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Related Practice
Textbook Question
For each reaction, give the expected substitution product, and predict whether the mechanism will be predominantly first order (SN1) or second order (SN2).
a. 2-chloro-2-methylbutane + CH3COOH
b. isobutylbromide + sodium methoxide
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Textbook Question
Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate.
Hint: Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations.
(b)
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Textbook Question
Give the SN1 mechanism for the formation of 2-ethoxy-3-methylbutane, the unrearranged product in this reaction.
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Textbook Question
Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate.
Hint: Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations.
(d)
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Textbook Question
For each reaction, give the expected substitution product, and predict whether the mechanism will be predominantly first order (SN1) or second order (SN2).
c. 1-iodo-1-methylcyclohexane + ethanol
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