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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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 23e,f
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
Verified step by step guidance1
Step 1: Understand the SN1 mechanism. The SN1 reaction proceeds via a two-step process: (1) formation of a carbocation intermediate after the leaving group departs, and (2) nucleophilic attack on the carbocation. The rate of the reaction depends on the stability of the carbocation formed and the nature of the leaving group.
Step 2: Compare the leaving groups in each pair. The leaving group ability is determined by how easily it can depart and stabilize itself. Iodide (I⁻) is a better leaving group than bromide (Br⁻) or chloride (Cl⁻) because it is larger and more polarizable.
Step 3: Evaluate the carbocation stability for each compound. Carbocation stability increases in the order: primary < secondary < tertiary. Tertiary carbocations are the most stable due to hyperconjugation and inductive effects from surrounding alkyl groups.
Step 4: Analyze pair (e): 2-iodo-2-methylbutane vs. tert-butyl chloride. Both compounds form tertiary carbocations upon losing their leaving groups, but iodide is a better leaving group than chloride, making 2-iodo-2-methylbutane react faster via the SN1 mechanism.
Step 5: Analyze pair (f): 2-bromo-2-methylbutane vs. ethyl iodide. 2-bromo-2-methylbutane forms a tertiary carbocation, while ethyl iodide forms a primary carbocation. Since tertiary carbocations are more stable, 2-bromo-2-methylbutane will react faster via the SN1 mechanism despite bromide being a slightly weaker leaving group than iodide.
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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 is a type of nucleophilic substitution reaction that involves two main steps: the formation of a carbocation intermediate and the subsequent attack by a nucleophile. The rate of the reaction depends primarily on the stability of the carbocation formed, which is influenced by factors such as the degree of substitution and the presence of electron-donating groups.
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Drawing the SN1 Mechanism
Carbocation Stability
Carbocation stability is a crucial factor in determining the rate of SN1 reactions. Tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects from surrounding alkyl groups. Therefore, substrates that can form more stable carbocations will react faster via the SN1 mechanism.
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Leaving Groups
The quality of the leaving group significantly affects the rate of SN1 reactions. Good leaving groups, such as iodide and bromide, can stabilize the transition state and facilitate the formation of the carbocation. In general, the better the leaving group, the faster the reaction will proceed, as it can depart more easily during the first step of the SN1 mechanism.
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Related Practice
Textbook Question
Propose an SN1 mechanism for the solvolysis of 3-bromo-2,3-dimethylpentane in ethanol.
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Textbook Question
Choose the member of each pair that will react faster by the SN1 mechanism.
c. n-propyl bromide or allyl bromide
d. 1-bromo-2,2-dimethylpropane or 2-bromopropane
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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
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.
2000
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Textbook Question
Choose the member of each pair that will react faster by the SN1 mechanism.
a. 1-bromopropane or 2-bromopropane
b. 2-bromo-2-methylbutane or 2-bromo-3-methylbutane
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