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 25
Give the SN1 mechanism for the formation of 2-ethoxy-3-methylbutane, the unrearranged product in this reaction.
Verified step by step guidance1
Step 1: Identify the substrate and the conditions for the reaction. The substrate is 3-methyl-2-butanol, and the reaction occurs under acidic conditions, which favor the SN1 mechanism. SN1 reactions proceed via a two-step process involving the formation of a carbocation intermediate.
Step 2: Protonation of the alcohol group (-OH) occurs under acidic conditions. The hydroxyl group is converted into a better leaving group, such as water (H₂O), by reacting with H⁺ from the acid.
Step 3: The leaving group (H₂O) departs, forming a carbocation intermediate. The carbocation is formed at the 2-position of the 3-methylbutane structure, resulting in a secondary carbocation. This step is the rate-determining step of the SN1 mechanism.
Step 4: Nucleophilic attack occurs. The ethoxide ion (CH₃CH₂O⁻) acts as the nucleophile and attacks the carbocation at the 2-position, forming a new bond between the ethoxy group and the carbocation.
Step 5: The final product, 2-ethoxy-3-methylbutane, is formed. This is the unrearranged product because no carbocation rearrangement occurs during the reaction.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
SN1 Mechanism
The SN1 mechanism, or unimolecular nucleophilic substitution, involves two main steps: the formation of a carbocation intermediate and the nucleophilic attack. The rate-determining step is the first step, where the leaving group departs, creating a positively charged carbocation. This mechanism is favored in tertiary substrates due to their ability to stabilize the carbocation through hyperconjugation and inductive effects.
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Drawing the SN1 Mechanism
Carbocation Stability
Carbocation stability is crucial in SN1 reactions, as more stable carbocations form more readily. Stability increases with the degree of substitution: tertiary > secondary > primary. Factors contributing to stability include hyperconjugation and resonance, which help to delocalize the positive charge. Understanding carbocation stability is essential for predicting the outcome of SN1 reactions.
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Nucleophiles and Electrophiles
In organic chemistry, nucleophiles are species that donate an electron pair to form a chemical bond, while electrophiles are electron-deficient species that accept electron pairs. In the context of the SN1 mechanism, the nucleophile attacks the carbocation formed after the leaving group departs. The choice of nucleophile can significantly influence the reaction pathway and the final product.
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Related Practice
Textbook Question
Benzyl bromide is a primary halide. It undergoes SN1 substitution about as fast as most tertiary halides. Use resonance structures to explain this enhanced reactivity.
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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
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)
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Textbook Question
3-Bromocyclohexene is a secondary halide. It undergoes SN1 substitution about as fast as most tertiary halides. Use resonance structures to explain this enhanced reactivity.
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Textbook Question
Choose the member of each pair that will react faster by the SN1 mechanism.
e. 2-iodo-2-methylbutane or tert-butyl chloride
f. 2-bromo-2-methylbutane or ethyl iodide
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