Transcript Slide 1

Chapter 10
Elimination
Reactions of Alkyl
Halides
Competition Between
Substitution and
Elimination
Paula Yurkanis Bruice
University of California,
Santa Barbara
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Alkyl Halides Undergo
Substitution and Elimination Reactions
In an elimination reaction, a halogen is removed from one carbon
and a hydrogen is removed from an adjacent carbon.
A double bond is formed between the two carbons
from which the atoms were removed.
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An E2 Reaction
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Mechanism for an E2 Reaction
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The Halogen Comes off the Alpha Carbon;
the Hydrogen Comes off the Beta Carbon
dehydrohalogenation
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The Weakest Base Is the Best Leaving Group
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An E2 Reaction is Regioselective
The major product is the most stable alkene.
The most stable alkene is (generally) obtained by removing a hydrogen
from the beta carbon that is bonded to the fewest hydrogens.
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The More Stable Alkene
Has the More Stable Transition State
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Alkene-Like Transition State
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More E2 Reactions
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Relative Reactivities of Alkyl Halides in an
E2 Reaction
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The More Stable Alkene
is the Major Product
The conjugated alkene is the more stable alkene (an
exception to Zaitsev’s rule).
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The More Stable Alkene
is Not the Major Product
A sterically hindered alkyl halide and a sterically hindered base
forms the less stable alkene (another exception to Zaitsev’s rule).
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Need a Lot of Steric Hindrance
forms the more stable alkene
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Fluoride Ion is a Poor Leaving Group
This is another exception to Zaitsev’s rule.
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The Transition State is
Carbanion-Like
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Relative Stabilities of Carbocations
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Relative Stabilities of Carbanions
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How the Leaving Group Affects
the Product Distribution
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An E1 Reaction
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The Mechanism for an E1 Reaction
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How Does a Weak Base (Like Water)
Remove a Proton From an sp3 Carbon?
1. The presence of the positive charge greatly reduces the pKa.
2. Hyperconjugation weakens the C—H bond by draining electron
density.
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An E1 Reaction is Regioselective
The major product is the more stable alkene.
The most stable alkene is obtained by removing a hydrogen
from the beta carbon that is bonded to the fewest hydrogens.
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The More Stable Alkene is the
Major Product
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The Weakest Base is the Best Leaving Group
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Benzylic and Allylic Halides
Undergo E2 Reactions
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Benzylic and Allylic Halides
Undergo E1 Reactions
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The E1 Reaction of Allylic Halides
Can Form Two Products
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The Reactivity of Alkyl Halides
in Elimination Reactions
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The Bonds to Be Broken
Must Be in the Same Plane
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Anti Elimination is Preferred
1. Anti requires the molecule to be in a staggered conformation.
2. Back-side attack achieves the best overlap of interacting orbitals (see Figure 9.1).
3. It avoids repulsion of the electron-rich base with the electron-rich leaving group.
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Anti Elimination
The alkene with the bulkiest groups on opposite sides
of the double bond will be formed in greater yield,
because it is the more stable alkene.
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The More Stable Product is
Easier to Form
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E2 and E1 Reactions are Regioselective
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E2 and E1 Reactions are Regioselective
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The Major Product of an E2 Reaction
(Largest Groups Are on Opposite Sides)
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When Only One Hydrogen is Bonded to the β-Carbon,
the Major Product of an E2 Reaction Depends on the
Structure of the Alkene
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When Only One Hydrogen is Bonded to the
β-Carbon, the Major Product is
Still the More Stable Alkene
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Summary of Stereochemistry
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E2 Elimination from Six-Membered Rings
Both groups being eliminated must be in axial positions.
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H and Cl Must Both Be Axial
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The Value of Keq Depends on Whether the Reaction
Takes Place Through the More Stable
Conformer or Through the Less Stable Conformer
Keq is large
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Keq is small
Neomenthyl Chloride is Faster
Elimination occurs through the more stable conformer.
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Menthyl Chloride is Slower
Elimination occurs through the less stable conformer.
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E1 Elimination from Six-Membered Rings
The H and Cl do not have to be in axial positions
because the reaction is not concerted.
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Proof That the E2 Reaction is Concerted
A carbon deuterium bond (C—D) is stronger
than a carbon hydrogen bond (C—H).
The deuterium kinetic isotope effect = 7.1.
Therefore, the C—D bond is broken in the rate-limiting step.
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Alkyl Halides in SN2 and E2 Reactions
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Under SN2/E2 Conditions
Primary Alkyl Halide = Primarily Substitution
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Steric Hindrance Favors Elimination
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Under SN2/E2 Conditions
Secondary Alkyl Halide = Substitution and Elimination
Substitution is favored by a weak base.
Elimination is favored by a strong base.
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Bulky Bases are Used to Encourage Elimination
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Although They are Neutral Bases,
They are Strong Bases
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A High Temperature Favors Elimination
Why? Because elimination has a greater ∆S‡.
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Under SN2/E2 Conditions
Tertiary Alkyl Halide = Only Elimination
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Under SN1/E1 Conditions Tertiary Alkyl Halides
Undergo Substitution and Elimination
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Tertiary (SN1/E1): Substitution is Favored
Tertiary (SN2/E2): Only Elimination
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Summary of the Products Obtained
From Substitution and Elimination Reactions
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William Ether Synthesis:
an SN2 Reaction
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Forming an Alkoxide Ion
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Synthesizing Butyl Propyl Ether
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Synthesizing tert-Butyl Ethyl Ether
The less hindered group should be provided by the alkyl halide.
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Synthesizing an Alkene
The more hindered group should be provided by the alkyl halide.
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Synthesizing an Alkene
If the reactant is a tertiary alkyl halide,
use SN2/E2 conditions because it gives only elimination.
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Synthesizing an Alkyne
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Converting an Alkene to an Alkyne
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Designing a Synthesis
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Designing a Synthesis
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Designing a Synthesis
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Designing a Synthesis
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