10 Lecture

© 2014 Pearson Education, Inc.

Elimination Reactions of Alkyl Halides

Competition Between Substitution and

Elimination

Paula Yurkanis BruiceUniversity of California,

Santa Barbara

Chapter 10


Alkyl Halides UndergoSubstitution and Elimination Reactions

In an elimination reaction, a halogen is removed from one carbonand a hydrogen is removed from an adjacent carbon.

A double bond is formed between the two carbonsfrom which the atoms were removed.


An E2 Reaction


Mechanism for an E2 Reaction


The Halogen Comes off the Alpha Carbon;the Hydrogen Comes off the Beta Carbon

dehydrohalogenation


The Weakest Base Is the Best Leaving Group


An E2 Reaction is Regioselective

The most stable alkene is (generally) obtained by removing a hydrogen from the beta carbon that is bonded to the fewest hydrogens.

The major product is the most stable alkene.


The More Stable AlkeneHas the More Stable Transition State


Alkene-Like Transition State


More E2 Reactions


Relative Reactivities of Alkyl Halides in an E2 Reaction


The More Stable Alkene is the Major Product

The conjugated alkene is the more stable alkene (an exception to Zaitsev’s rule).


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).


Need a Lot of Steric Hindrance

forms the more stable alkene


Fluoride Ion is a Poor Leaving Group

This is another exception to Zaitsev’s rule.


The Transition State is Carbanion-Like


Relative Stabilities of Carbocations


Relative Stabilities of Carbanions


How the Leaving Group Affects the Product Distribution


An E1 Reaction


The Mechanism for an E1 Reaction


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.


The most stable alkene is obtained by removing a hydrogen from the beta carbon that is bonded to the fewest hydrogens.

An E1 Reaction is Regioselective

The major product is the more stable alkene.


The More Stable Alkene is the Major Product


The Weakest Base is the Best Leaving Group


Benzylic and Allylic Halides Undergo E2 Reactions


Benzylic and Allylic Halides Undergo E1 Reactions


The E1 Reaction of Allylic Halides Can Form Two Products


The Reactivity of Alkyl Halides in Elimination Reactions


The Bonds to Be Broken Must Be in the Same Plane


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.


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.


The More Stable Product is Easier to Form


E2 and E1 Reactions are Regioselective


The Major Product of an E2 Reaction (Largest Groups Are on Opposite Sides)


When Only One Hydrogen is Bonded to the β-Carbon, the Major Product of an E2 Reaction Depends on the

Structure of the Alkene


When Only One Hydrogen is Bonded to theβ-Carbon, the Major Product is

Still the More Stable Alkene


Summary of Stereochemistry


E2 Elimination from Six-Membered Rings

Both groups being eliminated must be in axial positions.


H and Cl Must Both Be Axial


The Value of Keq Depends on Whether the ReactionTakes Place Through the More Stable

Conformer or Through the Less Stable Conformer

Keq is large Keq is small


Neomenthyl Chloride is Faster

Elimination occurs through the more stable conformer.


Menthyl Chloride is Slower

Elimination occurs through the less stable conformer.


E1 Elimination from Six-Membered Rings

The H and Cl do not have to be in axial positions because the reaction is not concerted.


Proof That the E2 Reaction is Concerted

The deuterium kinetic isotope effect = 7.1. Therefore, the C—D bond is broken in the rate-limiting step.

A carbon deuterium bond (C—D) is stronger than a carbon hydrogen bond (C—H).


Alkyl Halides in SN2 and E2 Reactions


Under SN2/E2 Conditions Primary Alkyl Halide = Primarily Substitution


Steric Hindrance Favors Elimination


Under SN2/E2 Conditions Secondary Alkyl Halide = Substitution and Elimination

Substitution is favored by a weak base.Elimination is favored by a strong base.


Bulky Bases are Used to Encourage Elimination


Although They are Neutral Bases, They are Strong Bases


A High Temperature Favors Elimination

Why? Because elimination has a greater ∆S‡.


Under SN2/E2 ConditionsTertiary Alkyl Halide = Only Elimination


Under SN1/E1 Conditions Tertiary Alkyl Halides Undergo Substitution and Elimination


Tertiary (SN1/E1): Substitution is FavoredTertiary (SN2/E2): Only Elimination


Summary of the Products Obtained From Substitution and Elimination Reactions


William Ether Synthesis:an SN2 Reaction


Forming an Alkoxide Ion


Synthesizing Butyl Propyl Ether


The less hindered group should be provided by the alkyl halide.

Synthesizing tert-Butyl Ethyl Ether


Synthesizing an Alkene

The more hindered group should be provided by the alkyl halide.


If the reactant is a tertiary alkyl halide, use SN2/E2 conditions because it gives only elimination.

Synthesizing an Alkene


Synthesizing an Alkyne


Converting an Alkene to an Alkyne


Designing a Synthesis

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