Textbook Question
For each set of isomers, choose the isomer that you expect to be most stable and the isomer you expect to be least stable.
(c)
2
views
Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade9th EditionOrganic ChemistryISBN: 9780135213728
Not the one you use?Change textbookSelect a different textbook
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 7, Problem 17a
SN1 substitution and E1 elimination frequently compete in the same reaction.
a. Propose a mechanism and predict the products for the solvolysis of 2-bromo-2,3,3-trimethylbutane in methanol.
Verified step by step guidance1
Identify the substrate: The compound 2-bromo-2,3,3-trimethylbutane is a tertiary alkyl halide. Tertiary alkyl halides are prone to undergoing SN1 and E1 mechanisms due to the stability of the carbocation intermediate formed during the reaction.
Determine the reaction conditions: The reaction occurs in methanol, which is a polar protic solvent. Polar protic solvents stabilize carbocations and favor SN1 and E1 mechanisms. Methanol can also act as a nucleophile in the SN1 pathway.
Propose the first step of the mechanism: The reaction begins with the departure of the bromide ion (Br⁻), forming a tertiary carbocation at the 2-position. This step is the rate-determining step for both SN1 and E1 mechanisms. Represent this step as: .
Propose the SN1 pathway: In the SN1 mechanism, methanol acts as a nucleophile and attacks the carbocation, forming an intermediate oxonium ion. This intermediate is then deprotonated by another methanol molecule to yield the substitution product, 2-methoxy-2,3,3-trimethylbutane.
Propose the E1 pathway: In the E1 mechanism, a base (methanol or another molecule) abstracts a proton from a β-hydrogen (on the 3-position), leading to the formation of a double bond. This results in the elimination product, 2,3,3-trimethyl-2-butene. Note that the E1 product is favored at higher temperatures due to the entropy increase associated with elimination.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4mWas this helpful?
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 occurs in two steps. First, the leaving group departs, forming a carbocation intermediate. This is followed by the nucleophile attacking the carbocation. The rate of the reaction depends only on the concentration of the substrate, making it unimolecular.
Recommended video:
Guided course
10:49
Drawing the SN1 Mechanism
E1 Mechanism
The E1 mechanism is an elimination reaction that also proceeds in two steps. Similar to SN1, the first step involves the formation of a carbocation after the leaving group departs. In the second step, a base abstracts a proton from a neighboring carbon, resulting in the formation of a double bond. The E1 reaction competes with SN1 when conditions favor elimination.
Recommended video:
Guided course
08:09Drawing the E1 Mechanism.
Carbocation Stability
Carbocation stability is crucial in determining the pathway of reactions involving SN1 and E1 mechanisms. Tertiary carbocations, like the one formed from 2-bromo-2,3,3-trimethylbutane, are more stable due to hyperconjugation and inductive effects from surrounding alkyl groups. This stability influences the likelihood of substitution versus elimination reactions occurring.
Recommended video:
Guided course
05:58Determining Carbocation Stability
Related Practice
Textbook Question
[FIGURE: KEY MECHANISM 7-1]
Show what happens in step 2 of the example if the solvent acts as a nucleophile (forming a bond to carbon) rather than as a base (removing a proton).
2
views
Textbook Question
The solvolysis of 2-bromo-3-methylbutane potentially can give several products, including both E1 and products from both the unrearranged carbocation and the rearranged carbocation. Mechanisms 6-6 and 7-2 show the products from the rearranged carbocation. Summarize all the possible products, showing which carbocation they come from and whether they are the products of E1 or reactions.
<IMAGES>
1
views
Textbook Question
SN1 substitution and E1 elimination frequently compete in the same reaction.
b. Compare the function of the solvent (methanol) in the E1 and SN1 reactions.
2
views
Textbook Question
Finish Solved Problem 7-3 by showing how the rearranged carbocations give the four products shown in the problem. Be careful when using curved arrows to show deprotonation and/or nucleophilic attack by the solvent. The curved arrows always show movement of electrons, not movement of protons or other species.
1
views
Textbook Question
For each set of isomers, choose the isomer that you expect to be most stable and the isomer you expect to be least stable.
(b)
2
views