Organic Chemistry 6 th Edition Chapter 9 Paula

In addition to substitution, an alkyl halide can undergo an elimination reaction 2 ©

The E 2 Reaction Hydroxide cannot act as a nucleophile in this reaction because

The removal of a proton and a halide ion is called dehydrohalogenation: 4 ©

An E 2 reaction is also called a b-elimination or a 1, 2 -elimination

The Regioselectivity of the E 2 Reaction The major product of an E 2

Reaction coordinate diagram for the E 2 reaction of 2 -bromobutane and methoxide ion

The Zaitsev Rule The more substituted alkene product is obtained when a proton is

Conjugated alkene products are preferred over the more substituted alkene product: Do not use

Bulky bases affect the product distribution resulting in the Hofmann product, the least substituted

Another exception to Zaitsev’s rule 14 © 2011 Pearson Education, Inc.

Consider the elimination of 2 -fluoropentane… A carbanion-like transition state 15 © 2011 Pearson

Therefore, the major product of an E 2 elimination reaction is the more stable

The E 1 Reaction A unimolecular reaction: 17 © 2011 Pearson Education, Inc.

The E 1 reaction, like the SN 1 reaction, involves a carbocation intermediate: 18

How does a weak base like water remove a proton from an sp 3

The major product in an E 1 reaction is generally the more substituted alkene

Reaction coordinate diagram for the E 1 reaction of 2 -chloro-2 -methylbutane 21 ©

Because the first step is the rate-determining step, the rate of an E 1

Because the E 1 reaction forms a carbocation intermediate, we need to consider carbocation

Competition Between E 2 and E 1 Reactions An E 2 is favored by

Stereochemistry of the E 2 Reaction The bonds to the eliminated groups (H and

Another view of the E 2 reaction mechanism The best overlap of the interacting

Consider the stereoselectivity of the E 2 reaction The alkene with the bulkiest groups

Reaction coordinate diagram for the E 2 reaction of 2 -bromopentane and ethoxide ion

When only one hydrogen is bonded to the b-carbon, the major product of an

Stereochemistry of the E 1 Reaction The major stereoisomer obtained from an E 1

Elimination from Cyclic Compounds In an E 2 reaction, groups to be eliminated must

The hydrogen that is removed and the leaving group both have to be in

E 1 Elimination from Cyclic Compounds No axial orientation required for the E 1

Utilization of kinetic isotope effect to study a reaction mechanism A carbon deuterium bond

Competition Between Substitution and Elimination Alkyl halides can undergo SN 2, SN 1, E

Consider the SN 2/E 2 conditions 44 © 2011 Pearson Education, Inc.

A bulky alkyl halide or a sterically hindered nucleophile encourages elimination over substitution 45

A strong or a bulky base encourages elimination over substitution 46 © 2011 Pearson

High temperature favors elimination over substitution: Why? Because elimination is entropically favorable. 47 ©

Tertiary alkyl halides undergo only elimination under SN 2/E 2 conditions: 48 © 2011

Consider SN 1/E 1 conditions: The elimination reaction favored at higher temperatures. Primary alkyl

Substitution and Elimination Reactions in Synthesis 51 © 2011 Pearson Education, Inc.

In synthesizing an ether, the less hindered group should be provided by the alkyl

However, the most hindered alkyl halide should be used if you want to synthesize

The E 2 reaction should be used if you want to synthesize an alkene:

Consecutive E 2 Elimination Reactions Requires a strong base Na. NH 2. Reaction carried

Synthesis Example 56 © 2011 Pearson Education, Inc.

Synthesis Example Racemic 57 © 2011 Pearson Education, Inc.

Synthesis Example 58 © 2011 Pearson Education, Inc.

Synthesis Example 59 © 2011 Pearson Education, Inc.

Slides: 59

Download presentation

The Zaitsev Rule The more substituted alkene product is obtained when a proton is removed from the b-carbon that is bonded to the fewest hydrogens The most stable alkene is generally (but not always) the most substituted alkene 9 © 2011 Pearson Education, Inc.

Bulky bases affect the product distribution resulting in the Hofmann product, the least substituted alkene: Hofmann product Zaitsev product However, it takes a lot of steric hindrance for the less stable product to be the major product: 12 © 2011 Pearson Education, Inc.

How does a weak base like water remove a proton from an sp 3 carbon? 1) The presence of a positive charge greatly reduces the p. Ka 2) Hyperconjugation weakens the C-H bond by electron density 19 © 2011 Pearson Education, Inc.

Because the first step is the rate-determining step, the rate of an E 1 reaction depends both on the ease with which the carbocation is formed and how readily the leaving group leaves 22 © 2011 Pearson Education, Inc.

Competition Between E 2 and E 1 Reactions An E 2 is favored by a high concentration of strong base and an aprotic polar solvent An E 1 is favored by a weak base and a protic polar solvent Review Section 8. 10 24 © 2011 Pearson Education, Inc.

Another view of the E 2 reaction mechanism The best overlap of the interacting orbitals is achieved through back-side attack Anti elimination avoids repulsion of the electron-rich base © 2011 Pearson Education, Inc. 26

Consider the stereoselectivity of the E 2 reaction 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 27 © 2011 Pearson Education, Inc.

Stereochemistry of the E 1 Reaction The major stereoisomer obtained from an E 1 reaction is the alkene in which the bulkiest substituents are on opposite sides of the double bond Both syn and anti elimination can occur in an E 1 reaction 32 © 2011 Pearson Education, Inc.

Utilization of kinetic isotope effect to study a reaction mechanism A carbon deuterium bond (C–D) is stronger than a carbon hydrogen bond (C–H) The deuterium kinetic isotope effect is 7. 1, indicating that a C–H (or C–D) bond must be broken in the rate-determining step 42 © 2011 Pearson Education, Inc.

Competition Between Substitution and Elimination Alkyl halides can undergo SN 2, SN 1, E 2, and E 1 1) decide whether the reaction conditions favor SN 2/E 2 or SN 1/E 1 • SN 2/E 2 reactions are favored by a high concentration of a good nucleophile/strong base • SN 1/E 1 reactions are favored by a poor nucleophile/weak base 2) decide how much of the product will be the substitution product and how much of the product will be the elimination product 43 © 2011 Pearson Education, Inc.

Consider SN 1/E 1 conditions: The elimination reaction favored at higher temperatures. Primary alkyl halides do not form carbocations; therefore they cannot undergo SN 1 and E 1 reactions. 49 © 2011 Pearson Education, Inc.