13 eliminations

7/31/2019 13 Eliminations

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Richard F. Daley and Sally J. Daleywww.ochem4free.com

Organic

ChemistryChapter 13

Elimination Reactions

13.1 The Elimination Mechanisms 65813.2 Stereochemistry of Eliminations 660

13.3 Direction of Elimination 663

13.4 E1 vs. E2 670

13.5 Elimination vs. Substitution 67113.6 Summary of Elimination and Substitution 673

13.7 E & Z Nomenclature 677

13.8 Elimination of Organohalogens 678

13.9 Dehydration of Alcohols 682

Synthesis of Cyclohexene 685

13.10 Pinacol Rearrangement 68613.11 Hofmann Elimination 691

13.12 Oxidation of Alcohols 694Synthesis of Citronellal 697

Key Ideas from Chapter 13 698


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Organic Chemistry - Ch 13 656 Daley & Daley

Copyright 1996-2005 by Richard F. Daley & Sally J. DaleyAll Rights Reserved.

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www.ochem4free.com 5 July 2005


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Chapter 13

Elimination Reactions

Chapter Outline

13.1 The Elimination MechanismsAn overview of the two types of elimination mechanisms

13.2 Stereochemistry of EliminationsA look at the stereochemical requirements of an elimination reaction13.3 Direction of Elimination

Regioselectivity of the elimination mechanisms

13.4 Comparing E1 and E2 With SN1 and SN2 ReactionsComparing the E1 with the SN1 reaction and the E2 with the SN2

mechanism

13.5 Elimination Versus SubstitutionFactors that influence whether an elimination or substitution

reaction will occur

13.6 Summary of Elimination and SubstitutionA summary of the factors that favor the various substitution and

elimination mechanisms

13.7 E and ZNomenclatureNaming alkenes usingE andZ nomenclature

13.8 Elimination of OrganohalogensElimination reactions involving organohalogen substrates

13.9 Dehydration of AlcoholsElimination reactions involving alcohol substrates

13.10 Pinacol RearrangementThe elimination/rearrangement reaction involving a 1,2-diol

13.11 Hofmann EliminationAn elimination reaction involving a nitrogen substrate

13.12 Oxidation of AlcoholsElimination reactions involving primary and secondary alcohols



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Objectives

Be able to write E1 and E2 mechanisms

Recognize the stereochemistry required for the substrate to react

via an E1 or E2 mechanism

Know the regiochemistry of the product of an elimination reaction

Know how to predict whether a substitution or eliminationreaction will occur with a particular substrate and set of reaction

conditions

Apply theE andZ descriptors to naming alkenes

Predict the products of elimination reactions involving

organohalides, alcohols, and vicinal diols

Recognize the regiochemistry of the Hofmann elimination reaction

Know the various pyrolytic eliminations

Know the elimination reaction involving alcohols that form

carbonyl compounds

Nothing in his life becomes him like the

leaving it.Shakespeare

Chapter 12 covered the continuum of SN1 and SN2 aliphaticnucleophilic substitution reactions. Recall, however, that acompeting elimination reaction takes place under some conditions

instead of the intended substitution reaction. In a -elimination

reaction, the substrate loses two atoms or two groups of atoms to form

a bond. This new double bond forms after, or concurrent with, the

loss of the leaving groups. Usually these atoms, or groups of atoms,leave from adjacent atoms.

This chapter covers two types of 1,2-, or -, elimination

reactions and how they fit into the continuum with each other and

with nucleophilic substitution reactions. Organic chemists use variousmethods to selectively control which of the possible pathways that a

particular elimination reaction follows. One other major elimination



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reaction is the 1,1-, or -, elimination reaction. Chapter 14 briefly

discusses the -elimination.

13.1 The Elimination Mechanisms

1,2-Elimination reactions follow two mechanistic pathways.

Both pathways involve the loss of the elements E and L from adjacent

atoms in the substrate to form a bond. E is some electrophile, often a

proton, and L is the leaving group. L has the same characteristics of

the leaving groups studied in Chapter 12.

C CC CLE

A 1,2-elimination

The two mechanistic pathways that 1,2-elimination reactions

take are designated as E1 and E2. The E stands for the elimination

pathway, and the number describes the kinetics of the reactioneither unimolecular or bimolecular. E1 and E2 reactions relate closely

to SN1 and SN2 reactions, so both a substitution reaction and an

elimination reaction frequently take place competitively on a

particular substrate.For a discussion of

nucleophiles seeSection 12.6, page 000. The E1 mechanism proceeds best on a tertiary substrate or

some other substrate that forms a stabilized carbocation intermediate.The energy profile for the E1 mechanism is very similar to that for the

SN1 mechanism. The first step of an E1 mechanism is identical to the

ionization operation in the first step of an SN1 mechanism. The

leaving group departs, and the substrate forms a carbocation. The

second step, the 1,3-electron pair abstraction operation, is the fast

step. In it, the carbocation loses a proton and forms the double bond.Figure 13.1 summarizes the two steps of the E1 mechanism.

See Figure 12.2, page

000 for the energy

profile of the SN1

reaction mechanism.

H+C CC C

H

L+C C

L

HC C

H



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Figure 13.1. The E1 reaction mechanism. The first step is an ionization operationand the second is a 1,3-electron pair abstraction operation.

Hyperconjugation stabilizes the carbocation in the second stepof the E1 mechanism and actually makes the electrophile depart more

readily. Because the overlap between the empty p orbital on the

carbocation and the bond on an adjacent carbon lowers the electron

density of the bond, the bond breaks more readily. Figure 13.2

illustrates this process.

C C

HOverlap

+

C C

Figure 13.2. The orbital interactions of the second step in the E1 mechanism show

how the hyperconjugation of the carbocation leads to loss of a hydrogen electrophile.

The transition state of an E2 mechanism involves both theleaving group and the base, so reactions that follow the E2 mechanism

are concerted. The base removes the electrophile, often a proton, at

the same time that the leaving group departs. Also, while the reactionis in the transition state, the overlap of the orbitals for the double

bond begins forming. The reaction rate of an E2 mechanism depends

on the concentrations of both the substrate and the base. Figure 13.3summarizes the two steps in the E2 mechanism.

Transition State

C C

C C

H

L

BB

C C

L

H

+ L + BH

-

-

Figure 13.3. The E2 mechanism is a 1,5-electron pair displacement operation.



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The E2 reaction has a very similar energy profile to the SN2

reaction. Because the E2 reaction is concerted, it usually requires less

activation energy to proceed than does an E1 reaction. That is,

simultaneous bond breaking and bond formation requires lessactivation energy than does sequential bond breaking and bond

formation.

13.2 Stereochemistry of Eliminations

This section describes the stereochemistry of the E1 and E2

reaction mechanisms and compares them with the stereochemistry ofthe SN1 and SN2 reaction mechanisms, which are covered in Chapter

12. The E1 reaction mechanism is a two-step process that, as with the

SN1 mechanism, usually loses all the stereochemical information ofthe substrate as the reaction proceeds. The E2 mechanism, similar to

the SN2 mechanism, is a concerted mechanism. A concerted reaction

usually requires that the substrate have a specific conformation. Theconformation must allow the orbitals of the bonds being broken to

overlap the bonds being formed. Only when the orbitals overlap do the

electrons flow smoothly from the breaking bonds to the forming bonds.

Without this smooth flow of electrons, a concerted reaction does not

take place. An SN2 reaction requires a geometric arrangement that

leaves the rear of the electrophilic carbon open for reaction from theside opposite from the leaving group. An elimination reaction

following the E2 mechanism requires that the leaving group and the

electrophilic group be coplanar.

When the leaving

group and the

electrophilic group are

in the same plane, they

are coplanar.

When the leaving

group and the


in a plane on opposite

sides of the molecule in

a staggered

conformation, they are

anti-coplanar.

Coplanar groups exist in two conformations called anti-

coplanar and syn-coplanar. In the anti-coplanar conformation, thetorsion angle between the electrophile and leaving group is 180o. Inthe syn-coplanar conformation, the torsion angle between the

electrophile and the leaving group is 0o.

H

L HLL

H

LH

Anti Syn

Coplanar conformations

When the leaving

group and the


in the same plane on

the same side of themolecule in an eclipsed

conformation, they are

syn-coplanar.



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Of these two conformations, the anti-coplanar conformation is

generally more stable. Thus, most E2 eliminations occur when the

substrate is in the anti-coplanar conformation. In some cases, where

the anti-coplanar conformation is not possible, the syn-coplanarelimination does occur, but at a much slower rate.

Review conformational

stabilities in Section

3.4, page 000.

The anti-coplanar conformation puts the electron-rich base as

far away from the electron-rich leaving group as is possible, as shown

in Figure 13.4a. In comparison, the syn-coplanar conformation puts

the leaving group and the electrophilic group in the higher energyeclipsed conformation shown in Figure 13.4b. Because the base comes

close to the leaving group in the syn-coplanar conformation, if the

leaving group is large (which many are) then the leaving group stops

the base and repels it. Thus, the syn-coplanar conformation is less

stable than the anti-coplanar conformation.

B

L Anti-coplanar E2 elimination

C C

H

C C

(a)

C C

H

C C

Syn-coplanar E2 elimination

L

B

Repulsion

(b)

Figure 13.4. Orbital picture of E2 elimination reactions in the (a) anti- and (b) syn-coplanar conformations.



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The difference in elimination pathways between cis- and trans-

4-tert-butylcyclohexyl tosylate illustrates the importance of molecular

conformation. Because of its size, the tert-butyl group effectively locks

the cyclohexane ring with the tert-butyl group in the equatorialposition. In the cis isomer, the smaller tosylate group is in the axial

position; and in the trans isomer, it is in the equatorial position.

Section 3.10, Page 000,

discusses the stability

of various substituents

on a cyclohexane ring.

trans-4-tert-Butylcyclohexyl tosylatecis-4-tert-Butylcyclohexyl tosylate

(CH3)3COTs

(CH3)3C

OTs

Both compounds react with ethoxide ion in ethanol to eliminate thetosyl group and a hydrogen to form 4-tert-butylcyclohexene.

However, the reaction of the cis isomer follows E2 kinetics, whereas

the reaction of the trans isomer follows E1 kinetics. The cis isomer has

the tosyl group and the hydrogen in an anti-coplanar conformation;

thus, allowing the base to easily remove the hydrogen.

(CH3)3C

OCH2CH3

(CH3)3C

OTs

H

4-tert-Butylcyclohexene

Because of the size of the tert-butyl group, the trans isomer does not

easily get into either the anti- or syn-coplanar conformation. Thus, the

rate of reaction is quite slow.

(CH3)3C

OCH2CH3

(CH3)3COTs

H

H

C(CH3)3

H

OTs



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Exercise 13.1

What is the approximate dihedral angle between the hydrogens and

the tosylate group in trans-4-tert-butylcyclohexyl tosylate? Draw aNewman projection looking from carbon 1 towards carbon 2.

The structure of a molecule such as 2-bromo-3-

deuterobicyclo[2.2.1]heptane is so rigid that it exists only in the syn-

coplanar conformation. Chemists labeled this compound with adeuterium atom trans and a hydrogen atom cis to a bromine atom and

followed the course of the E2 reaction.

Syn-coplanar

D

H

H

Br

2-Bromo-3-deuterobicyclo[2.2.1]heptane

They knew that if the elimination product contained the deuterium

atom, then syn-coplanar elimination took place. If the product

contained the hydrogen, then another process took place because the

molecule was so rigid that the deuterium atom could not move back

and forth between the syn- and anti-coplanar conformations. The

experimental results showed that the product contained deuterium,

thus, the reaction followed a syn-coplanar E2 mechanism.

(87%)

CH3OH

CH3O

D

H

H

BrH

D

2-Deuterobicyclo[2.2.1]-2-hexene

Exercise 13.2

The threo isomer of 1-bromo-1,2-diphenylpropane undergoes

elimination at a faster rate than does the erythro isomer. Explain the

rate difference.

13.3 The Direction of Elimination



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In an elimination reaction, the substrate loses the leaving

group and, in most cases, a hydrogen to the leaving group.

Frequently, the leaving group has more than one type of hydrogenand, depending on a number of factors, any one of these hydrogens

can leave with the leaving group. For example, in the elimination

reaction of HBr from 2-bromobutane a hydrogen from either carbon 1or carbon 3 leaves with the bromine. Experimental results indicate

that the reaction mixture contains 1-butene (19%), cis-2-butene (22%),

and trans-2-butene (59%).

CH3CH2CHCH3

Br

+CH3CH2O

CH3CH2OH

2-Bromobutane cis-2-Butene1-Butene19% 22%

CH3CH2CH CH2 C C

HH

CH3H3C

C C

CH3H

HH3C

59%

trans-2-Butene

+

Elimination reactions may be either regiospecific or

regioselective. Regiospecific elimination reactions produce only one

isomer of an alkene. Regioselective elimination reactions, on the otherhand, produce several different isomers, but give one isomer in greater

quantity than the others. Whether a reaction is regiospecific orregioselective depends on whether the reaction prefers to eliminate

only one particular hydrogen or a mixture of hydrogens. The

elimination of 2-bromobutane mentioned previously eliminates a

mixture of hydrogens, so it is regioselective.

The regio prefixmeans a preference for

the location of the

functional group in the

product. Specific

means only one

product. Selective

means one major

product with a minor

product(s).When an elimination reaction produces the more highly

substituted alkene, for example, 2-butene from 2-bromobutene, the

reaction is said to follow the Saytzeff orientation. If the eliminationreaction forms the less substituted alkene, for example, 1-butene from

2-bromobutene, the reaction is said to follow the Hofmann

orientation. A number of factors including the substrate, the leavinggroup, and the reaction conditions determine whether a reaction

follows the Saytzeff or the Hofmann orientation.

A Saytzeff orientation

produces the more

substituted alkene.

The Hofmann

orientation produces

the less substituted

alkene.Because the more stable isomer of a particular alkene is the

most substituted isomer, most elimination reactions follow theSaytzeff orientation. In comparison to a hydrogen, alkyl groups help to

stabilize a double bond. Alkyl groups are slightly electron donating,

whereas hydrogens are slightly electron withdrawing.



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One way to measure stabilization is to measure the amount of

energy given off when hydrogen adds to a double bond to form an

alkane.

C C C C

H H

H2

Pt

This method is also convenient for comparing the energy content of

various alkenes. For example, the energy released by hydrogenation of1-butene is 30.3 kcal/mole. For cis-2-butene, it is 28.6 kcal/mole; and

for trans-2-butene, it is 27.6 kcal/mole. All three compounds

hydrogenate to produce butane.

H2/Pt

30.3 kcal/mole

28.6 kcal/mole

27.6 kcal/mole

Figure 13.5 shows the relationship between the Gibbs free energy of

the three butene isomers.

Goterminal

Gotrans Gocis

Figure 13.5. The reaction energy diagram for the free energies of hydrogenation ofthe three butene isomers.

The data in Table 13.1 show that the energy needed to

hydrogenate a double bond decreases and the stability of a double

bond increases, as the number of alkyl substituents bonded to a



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double bond increases. For example, the hydrogenation of 2,3-

dimethyl-2-butene requires 6.1 kcal/mole less energy than the

hydrogenation of ethene (26.6 (32.7) = 6.1). So 2,3-dimethyl-2-

butene is 6.1 kcal/mole more stable than ethene.

Name Structure Gibbs Free Energy

of Hydrogenation

(kcal/mole)

Ethene CH2 CH2 32.7

1-Butene CH3CH2CH CH2 30.3

cis-2-ButeneC C

H

CH3

H

CH3

28.6

trans-2-Butene

C C

HCH3

H C 3H

27.6

2-Methyl-2-butene

C C

CH3CH3

H C 3H

26.9

2,3-Dimethyl-2-

butene C C

CH3

CH3

CH3

CH3

26.6

Table 13.1. Gibbs free energies of hydrogenation for various alkenes.

When the pathway of an elimination reaction proceeds along

the E1 mechanism, the direction of elimination usually follows the

Saytzeff orientation. It does so because the rapid second step of the E1

mechanism loses the electrophile with the lowest bond energy. An E2

reaction mechanism shows less discrimination between the Saytzeff

and Hofmann products. With an E2 reaction, the energy difference

between the various double bond positions is often small enough thatthere is not a strong preference for the Saytzeff product. An example is

the loss of HBr from 2-bromobutane with ethoxide ion. The product

mixture contains 19% 1-butene and 81% of a mixture ofcis- and trans-

2-butene. Although the difference between 19% and 81% may seem

large, for a chemist it may not be large enough to make the reaction

synthetically useful. Additionally, the difficulty in separating the cis

and trans isomers may make other synthetic methods more attractive.



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Although the bond energy of a double bond is not a significant

influence in determining whether the course of the E2 process follows

the Saytzeff orientation or the Hofmann orientation, steric factors are

significant. If the leaving group is large, then the Hofmann productpredominates. For example, a quaternary ammonium salt has a large

trialkylamine as the leaving group, so an E2 reaction produces the

less substituted alkene in the greater amount. The difference here

between 95% and 5% is synthetically significant.

ammonium hydroxide

N,N,N-Trimethyl-2-butyl-

5%95%

1-Butene 2-Butene

CH3CH CHCH3CH3CH2CH CH2 +CH3CH2CHCH3

N(CH3)3 OH

The quaternary ammonium salt in this reaction is large, so the base

causes crowding within the substrate as it approaches. This crowding

increases the energy of the transition state and reduces the rate of

reaction compared to a smaller leaving group. Thus, the base more

easily abstracts a hydrogen from the less substituted carbon to

produce a less substituted alkene.

Further evidence for the steric sensitivity of the E2 eliminationis the observation that, in reactions using the same substrate, the

amount of Hofmann elimination product increases with increasing

base size.

Hofmann eliminations

are discussed in more

detail in Section 13.11,page 000.

81%19%CH3CH2OH

+ CH3CH CHCH3CH3CH2CH CH2CH3CH2CHCH3

BrCH3CH2O

(CH3)3COH

(CH3)3CO

77% 23%

+ CH3CH CHCH3CH3CH2CH CH2CH3CH2CHCH3

Br

Solved Exercise 13.1

Draw the structure for the elimination product of each of the following

reactions.

a)



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Br

KOH

Solution

There are two possible elimination products from this reaction: 3-methyl-1-

butene and 2-methyl-2-butene.

Br

KOH

3-Methyl-1-butene 2-Methyl-2-butene

The major product is 2-methyl-2-butene because there are three alkyl

substituents attached to the carboncarbon double bond. 3-Methyl-1-butene

has only one substitutent.

b)

CH3

Br

KOH

SolutionIn this reaction there are two possible products: 1-methylcyclohexene and 3-

methylcyclohexene.

CH3

Br

KOH

+

1-Methylcyclohexene 3-Methylcyclohexene

Because 1-methylcyclohexene has three alkyl groups attached to the double

bond and 3-methylcyclohexene has only two, you would expect 1-

methylcyclohexene to be the major product. However, the anti conformation

of 1-bromo-2-methylcyclohexane required for the E2 elimination cannot exist.



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CH3

Br

HH

This hydrogenis anti to the Br.

This hydrogen cannotbe anti to the Br.

Thus, the 3-methylcyclohexene is the major product formed in this reaction.

Exercise 13.3

Give the structure of the major elimination product for each of the

following substrates with sodium ethoxide in ethanol.

a)

Br

CH3

b)

CH3CH2CH2C CHCH3

CH3

CH3

Br

c)

Br

C(CH3)3H

H

d)

C CCH3Br

H

H

Sample solution

b)



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CH3CH2CH2C CH CH2

CH3

CH3

13.4 Comparing E1 and E2 with SN1 and SN2

Much of the information covered in Chapter 12 about the SN1

and SN2 reaction mechanisms also applies to the E1 and E2 reaction

mechanisms. The two pairs of reaction mechanisms are very similar

and very competitive with each other. This section looks at the

similarities of the substitution and the elimination mechanisms, as

well as pointing out the differences between the two elimination

mechanisms. Section 13.5 begins outlining the competition betweenelimination reactions and substitution reactions.

Similar to the SN1 reaction, an E1 reaction occurs best on a

tertiary substrate under reaction conditions that enable the formation

of a carbocation. Both reaction types require a neutral or acidic

reaction mixture because carbocations do not form under basicconditions. An E1 reaction also depends on the presence of a polar-

protic solvent to promote ionization. The more polar solvents allow a

secondary substrate to undergo E1 eliminations, but primary

substrates usually do not.

The E2 mechanism and SN2 mechanism are very similar, but

there are some differences in the reactions. An SN2 reaction proceedsbest with primary substrates and most poorly with tertiary substrates,

whereas an E2 reaction is just the opposite. E2 reactions proceed best

with tertiary substrates and most poorly with primary substrates.

This preference for tertiary substrates is because they have less steric

hindrance to the approach of the base to the hydrogens than to the

approach of the nucleophile to the carbon bearing the leaving group in

an SN2 reaction. Figure 13.6 illustrates this concept.

CH3

CH3

ClCH

CH3

O-

Carbon is difficultto access for SN2

Protons are easyto access for E2



18/48


Figure 13.6. The protons on a carbon adjacent to a tertiary leaving group are moreopen to reaction than the backside of the carbon bearing the leaving group.

Both the E2 mechanism and the E1 mechanism work best ontertiary substrates and more poorly on primary substrates. Generally,

the difference is that the E1 reaction requires acidic reaction

conditions and higher temperatures to proceed because the stronglyacidic carbocation intermediate is formed. A carbocation cannot be

formed in a base. Most, but not all, E2 reactions require basic reaction

conditions to proceed.

13.5 Elimination versus Substitution

A problem you will face when planning a synthesis is thecompetition between elimination and substitution reactions. Thesetwo important reaction types are closely related mechanistically, but

they lead to very different products. Except with a tertiary substrate,

the substitution pathway predominates in the competition between

elimination and substitution reactions because substitution reactions

require fewer bond changes and have fewer conformational

requirements. Therefore, they are usually more energeticallyfavorable. By choosing reaction conditions that favor one pathway

compared to the other pathway, you can control the outcome of the

reaction.

Three variables that affect whether a particular substrate

follows the elimination or substitution pathway are: the strength andconcentration of the incoming base or nucleophile, the reactionmedium, and the reaction temperature. The first variable is the most

important of the three because changes in the strength or

concentration of the base have the greatest effect on the course of the

reaction. The other two variables have a significant, though smaller,

effect on the course of the reaction.

The presence of a strong nucleophile usually forces a secondorder kinetics, either SN2 or E2. When the nucleophile is a soft base, it

promotes substitution; when it is a hard base, elimination results. For

example, chemists often use the hard base c- OH ion to promote

elimination, in preference to using the softer c- SH ion.

Hard and soft

acid/base theory is

discussed in more

detail in Section 5.3,

page 000.

In the presence of low concentrations of base, or with weaker

bases, the amount of E1 reaction increases. This increase occursbecause both the E1 and SN1 mechanisms begin with an identical

rate-determining step that forms the carbocation. The formation of a

carbocation is followed by a fast substitution or elimination step. For

example, solvolysis of tert-butyl bromide in methanol at 50oC yields

about a 2:1 ratio of substitution to elimination.



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2-Methylpropenetert-Butyl methyl ether

36%64%

+CH3OH

C

CH3

CH3

CH3

OCH3CCH3

CH3

CH3

Br C

CH3

CH2

CH3

50oC

However, adding methoxide ion, a stronger base, to the reactionmixture changes the yield to a 1:24 ratio of substitution to

elimination. The elimination is an E2 mechanism, but substitution on

a tertiary substrate does not readily occur via an SN2 mechanism and

SN1 does not occur readily in base. Thus, there is only a small yield of

the substitution product.

CH3O

4% 96%

+CH3OH

C

CH3

CH3CH2C

CH3

CH3

CH3

OCH3CCH3

CH3

CH3

Br

Base size also affects the outcome of the elimination and

substitution competition. The larger the base, the more the

elimination reaction increases compared to the substitution reaction.

A comparison of the reactions of methoxide and tert-butoxide ions with

the tosylate of cyclohexanol at 50oC shows the effects of size.

CyclohexeneMethoxycyclohexaneCyclohexyl tosylate

46%54%

+

OCH3CH3O

CH3OH, 50oC

OTs

100%0%

Cyclohexenetert-ButoxycyclohexaneCyclohexyl tosylate

+

OC(CH3)3(CH3)3CO

(CH3)3COH, 50oC

OTs

Exercise 13.4

Explain why chemists seldom use the azide and cyanide anions as

bases in elimination reactions.



20/48


Elimination reactions involve two carbons, whereas

substitution reactions involve only one. Thus, in the transition states

of both the E1 and E2 mechanisms, the charge disperses, or spreadsout, more than it does in the transition states of the substitution

reactions. When chemists decrease the solvent polarity of the reaction

mixture, the elimination mechanism is favored compared to the

substitution mechanism.

As the temperature increases, the rate for an eliminationreaction increases faster than the rate for a substitution reaction. This

increase is true for both the unimolecular and bimolecular processes.

C CH2

H3C

H3C

C

CH3

CH3

BrH3C C

CH3

CH3

OCH2CH3H3C

tert-Butyl ethyl ether 2-Methylpropene

25oC

83% 17%

CH3CH2OH+

CH3CH2OH+

70oC

(39%)(61%)

C

CH3

CH3

BrH3C C

CH3

CH3

OCH2CH3H3C C CH2

H3C

H3C

Exercise 13.5

What are the possible products from the reaction of 2-chloro-2-

methylbutane with sodium hydroxide in aqueous ethanol? Which

product would you expect to predominate at 25oC? Which productwould you expect to predominate at 70oC?

Although chemists have studied the E1 and SN1 mechanismsextensively for a number of years, most reactions of these types are of

little practical use in organic synthesis. The problem is the inability to

precisely control the outcome of the reactions. The lack of control is

made worse by the competition between SN

1 and E1 mechanisms

under the same reaction conditions. In general, chemists try to avoid

using a reaction with a carbocation intermediate in organic synthesis.

13.6 Summary of Elimination and Substitution



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This chapter and Chapter 12 present a number of factors that

help place a particular reaction on the continuum of elimination and

substitution reactions. Using tables, this section summarizes these

factors. Table 13.2 compares the effects of various substrates on theoutcome of the reaction. Table 13.3 shows how the nucleophile/base,

the leaving group, the solvent and the temperature determine which

mechanism a substrate will follow.

LCH3

CH2R L CH

R

LR

CR

R

L

R

Methyl 1o 2o 3o

Bimolecular reactions only. Bimolecular and unimolecular reactions.

SN2 only. Mainly SN2,

but E2 with

very strong orbulky base

Mainly SN2 with weak

bases or E2 with

strong bases. Also SN1or E1 in acid,

especially if the

substrate can

rearrange.

No SN2. In a

solvolysis

reaction, SN1 andE1. With a strong

base, E2

predominates.

Table 13.2. Summary of reaction pathways of elimination and substitution reactions

by substrate type.

Factor SN1 E1 SN2 E2Nucleophile

(base)

Poor

Nucleophile

Very poor

nucleophile or

very low

concentration

Nucleophile

is soft base

Hard base

Leaving

group.

Weaker base than the nucleophile (base)

Solvent Polar-protic

solvent

Polar-protic

solvent, lowerpolarity than

for SN1

Polar-

aproticsolvent

Polar-

aproticsolvent,

lower

polarity

than for

SN2

Reaction

temperature

Raising the temperature favors the elimination reaction

Table 13.3. Summary of effects of nucleophile (or base), leaving group, solvent, andreaction temperature on the mechanism of a particular substrate.



22/48



Predict all reasonable products for each of the following reactions and

determine the major product. Explain your choice.

a)

CH3O

CH3OH, 0o

Br

Solution

The reaction conditions are basic; thus, a carbocation cannot form. The

reaction forms two possible products: one from an SN2 reaction and the other

from an E2 reaction.

CH3O

CH3OH, 0o

Br OCH3

+

Because the substrate is a relatively open secondary substrate, an SN2

reaction proceeds readily. However, the nucleophile is a sufficiently strong

base that an E2 reaction also proceeds readily. At 0oC, however, the amount

of elimination is reduced significantly.

b)

CH3OHrefluxBr

Solution

A tertiary substrate does not readily undergo an SN2 reaction. Thus, this

reaction must proceed via an SN1, E1, or E2 reaction mechanism. The SN1

and E1 are preferred because of the high temperature and weak base. The

reaction forms three possible products.

CH3OH

refluxBr

OCH3

+ +

The ether is the major product because this is a solvolysis reaction. When a

carbocation is produced, many solvent molecules are available to react with

it.



23/48


Exercise 13.6

Predict the product(s) of each of the following reactions. Indicate themechanism for each one and the major and minor product.

a)

reflux

CH3OH

CH3

Br

b)

reflux

KOH/ethanol

CH3

Br

c)

CH3OH, 0o

CH3OCH3CH2CHCH2CH3

Br

d)

acetone

NaI

CH3CH2CHCH2CH3

Br

e)

70oC

CH3CH2OHCH2OTs

CH3

Sample Solution

c)

E2

SN2

minor

major

+

CH3CH2CHCH2CH3

OCH3

CH3CH2CH CHCH3

CH3OH, 0oCH3OCH3CH2CHCH2CH3

Br



24/48


13.7E

&Z

NomenclatureSection 3.9 introduces the terms cis and trans to describe the

geometric relationship between groups on the double bond. However,

in some cases these terms are insufficient and even confusing for

describing the geometric relationships. A more general system uses

the letters Zand E to describe cis and trans relationships.

Section 3.9 begins on

page 000.

The Z comes from the

German word for

together,zusammen,

and the E comes from

the German word for

opposite,entgegen.The Cahn-Ingold-

Prelog priority system

is discussed in Section

11.2, page 000.

TheE-Z method of naming is based on the Cahn-Ingold-Prelog

system of substituent priorities. To determine whether a compound

gets anE or aZ designation, assign each group bonded to the double

bonded carbons a priority number. Using the condensed structural

drawing of a molecule, assign the groups on the left a 1 and a 2, and

assign the groups on the right a 1 and a 2. To determine which groupgets which number, assign a 1 to the group with the atom having the

higher atomic number closest to the point of attachment to the doublebond.

For example, consider cis- and trans-2-butene. Each carbon of

the two double bonds has a methyl group and a hydrogen attached to

it. The methyl group has the higher priority because carbon has a

higher atomic number than hydrogen.

(trans-2-Butene)(cis-2-Butene)

E-2-ButeneZ-2-Butene

C C

H

H

CH3

H3C

C C

H H

CH3H3C

Number the methyl groups as #1 and the hydrogens as #2. When the

#1 groups are on the same side of the bond, the molecule is the Z

isomer. When the #1 are on opposite sides of the bond, the molecule

is the E isomer. Both E and Z as well as cis and trans nomenclature

are used interchangeably in disubstituted alkenes. The more complexalkenes use theE andZ terms.

Exercise 13.7

Name the following alkenes using theE-Z system of nomenclature.



25/48


a)

C C

Cl

CH3H

H

b)

C CCCH3

O

HCH3CH2CH2

CH3CH

CH3

c)

C C

CH2OH

CH2CH3

CH3CH

CH3

H

d)

C C

Br

CH3H

CH2CH2CH3

Sample solution

a)E-1-Chloro-1-propene

13.8 Elimination of Organohalogens

Dehydrohalogenation was one of the earliest alkene formingreactions that chemists investigated. You have already seen many

dehydrohalogenation examples in the illustrations of the eliminationreactions. Most dehydrohalogenation reactions take place with

alkoxide bases and at elevated temperatures. These conditions, as

Section 13.5 discusses, favor the E2 elimination reaction mechanism.

Dehydrohalogenation

is the loss of the

elements H and X. H is

a hydrogen, and X is a

halogen. The result is a

double bond.

(87%)

Cl

CH3CH2O

CH3CH2OH

Ethenylcyclohexane

(Vinyl cyclohexane)

(85%)

(CH3)3COH

(CH3)3COCH3CH2CH2CH CH2CH3CH2CH2CH2CH2Br

1-Pentene



26/48


1,3-Cyclohexadiene

(57%)

Br

Br

(CH3)2CHO

Diglyme

125oC

Amines also function well as bases in elimination reactions.

Amines are readily available in a variety of structures. The low

basicity, or poor nucleophilicity, of some amines minimizes competing

side reactions. In addition, amines are much more soluble in organic

solvents than are the alkoxide bases. Both amines used in the

following examples are nonnucleophilic.Nonnucleophilic

amines are sufficiently

basic to act as bases

but do not act as anucleophile. Generally,

nonnucleophilic

amines are sterically

blocked or are tertiary.

(93%)

Br

+N

N

1,5-Diazabicyclo[4.3.0]-

5-nonene (DBN)

Cyclohexene

(84%)

CHCH3

Br

CH CH2

N

+

Quinoline Ethenylbenzene

(Styrene)

A 1,2-dibromoalkane, which is a vicinal dibromide, gives anE2 reaction with potassium iodide in ethanol. The iodide ion is not

sufficiently basic to remove a proton. Thus, the double bond formsbetween the two carbons bearing the bromine atoms.

Vicinal substituents

are substituents that

are on adjacent carbon

atoms.

I

Br

Br



27/48


This reaction is one of the few reactions you will encounter in which

the electrophile is not a proton. The reaction is also specific to the use

of iodide ion. None of the other halide ions brings about this reaction.

A stronger base, such as the tert-butoxide ion, forms a diene from thesame substrate.

(CH3)3CO

Br

H

(CH3)3CO

Br

BrH

H

Iodide ion functions as the base in the dehydrohalogenation of vicinaldibromides because iodine, like bromine, is soft. Thus, the bromine is

polarizible and as the iodine approaches a bromine atom, the electrons

shift easily toward the carbon. The use of the iodide ion is specific to

vicinal dibromide reactions.

The reaction of a 1,2-dibromoalkane with potassium iodide has

limited value as a synthesis of an alkene because the major method ofpreparing a vicinal dihalide is from an alkene. Unlike the

dehydrohalogenation, which forms a double bond at more than onepossible position, the reaction of vicinal dihalides with potassium

iodide forms the double bond only between the two carbons bearing

the bromine.

Exercise 13.8

Some chemists think that an alternate mechanism for the

dehalogenation reaction with potassium includes a substitution step

followed by an elimination step. Write a mechanism for this process.

The reaction for the dehalogenation reaction of cis-1,2-

dibromocyclohexane with potassium iodide in ethanol is much slower

than the reaction with the trans isomer. Does this experimental result

fit with the alternate mechanism?

1,2-Eliminations also form alkynes from both vicinal andgeminal dihalides, as well as vinyl halides. This reaction follows thesame kinetics and stereochemistry as the E2 reaction. A competing

reaction is the formation of a diene. The reaction of 1,2-

dibromocyclohexane, shown at the beginning of this section to

illustrate the dehydrohalogenation reaction, forms a diene because it

has hydrogens available for elimination on the carbons adjacent to the

bromines. The first reaction below is a very similar reaction, but it

Geminal substituentsare two substituents

that are on the same

carbon atom (1,1-

disubstituted).



28/48


leads to an alkyne because it does not have hydrogens on adjacent

carbons available for elimination to form a diene.

(68%)

KOH

CH3CH2OHC C

Diphenylethyne

Br

Br

(73%)

CH2CHBr2 C CHNaNH2

NH3

Ethynylbenzene

(79%)

C C

H

Cl

CH3

CH3CH2CH2 KOH

CH3CH2OHCH3CH2CH2C CCH3

2-Hexyne

Elimination reactions forming alkynes are usually slower than

elimination reactions forming alkenes. In the same way that internal

double bonds give more stable alkenes than terminal double bonds,internal triple bonds give more stable alkynes than do terminal triple

bonds. Because the reaction requires stronger bases and higher

temperatures, the triple bond has a tendency to migrate. Thus, a 1-alkyne product frequently isomerizes to a 2-alkyne. This tendency to

isomerize limits the utility of the reaction. However, the use of NaNH2

in liquid NH3 generally does not cause isomerization of the product

because of low reaction temperature and the fact that the base reacts

with the terminal alkyne forming its conjugate base. Nevertheless,

chemists normally prefer to synthesize a terminal alkyne via anucleophilic substitution instead of an elimination reaction.

Under ideal reaction conditions, the E1 reaction, like the SN1

reaction, produces an excellent yield of a single product. However,ideal conditions rarely occur. Often the intermediate carbocations

rearrange or give a mixture of products. Thus, chemists rarely use

organohalogens as substrates in E1 reactions.

Exercise 13.9



29/48


Predict the product(s) of each of the following reactions. If more than

one product is possible, indicate which is the major product.

a)

NH3

NaNH2CH3CCH2CH2CHBr2

CH3

CH3

b)

Ethanol,

KOHCHCH3

Br

c)

(CH3)3CO

DMSO,

CHCH3

Br

d)

Ethanol

KICH3CH2CH2CHCH2Br

Br

e)

NH3

NaNH2CH3C CCH3

CH3 CH3

Br Br

f)

Br

CH3 DBN

Sample Solution

d)

CH3CH2CH2CH CH2Ethanol

KICH3CH2CH2CHCH2Br

Br



30/48


13.9 Dehydration of Alcohols

One of the most widely used elimination reactions is thedehydration of an alcohol to produce an alkene.Dehydration means theremoval of water.

C C

OH H

C C + H2OH

A dehydration reaction is reversible and usually has a small

equilibrium constant. This means that the equilibrium favors the

substrate and not the product. In fact, the reverse reaction is a method

often used to synthesize alcohols. To improve the yields of adehydration reaction, chemists heat the reaction mixture in the

presence of a catalytic quantity of acid, to distill off the lower boiling

product, either the water or the alkene, as it forms. This is a viable

laboratory technique because the alkene has a lower boiling point

than the alcohol.

Tertiary and secondary alcohols react much faster thanprimary alcohols, suggesting that carbocation formation is the rate-

controlling step in the reaction. Primary carbocations are less stable

than tertiary carbocations. Thus, the rate of reaction depends on the

stability of the carbocation. The mechanism that dehydration

reactions of tertiary and secondary substrates follow is E1.

C C

H

C C

OH H

HHH2O

C C

C C

OH H

H2O

Unless the carbocation can rearrange to become a more stable

tertiary or secondary carbocation, primary carbocations do not form

from the dehydration of primary alcohols. With a primary alcohol, thec- OH group is a strong base and therefore a poor leaving group. As aresult, alcohol dehydration does not occur in a basic reaction mixture.



31/48


The first step in the dehydration of a primary alcohol is

protonation of the OH group to form a good leaving group, OH2. At

this point, the reaction follows a typical elimination reaction, and the

conjugate base of the acid catalyst removes a hydrogen in an E2mechanism. When the acid catalyst is sulfuric acid, the conjugate base

is the HSO4c- ion.

HO S

O

O

O

C CC C

OH

H

H

An alternate, and more common, reaction is a concertedrearrangement reaction. In a concerted rearrangement, the OH

group is protonated as the hydride moves to the carbon atom bearing

the protonated oxygen. This migration produces a more highly

substituted, and thus more stable, carbocation than that which forms

directly from the primary alcohol.

C CH

H

HC COH

H

H H

H

The products of dehydration reactions using either 1-butanol or 2-

butanol as the substrate always contain the same mixture of alkenes.

This indicates that the two reactions form the same carbocation

intermediate. The following illustration shows how the dehydration

reactions of two different substrates can give the same products.



32/48


E- and Z-2-Butene

1-Butene

+

70%

30%

CH3CH CHCH3

CH3CH2CH CH2

CH3CH2CHCH3

CH3CH2CHCH3

OH

CH3CH2CH2CH2OH

The best acid catalysts for the alcohol dehydration reaction are

concentrated sulfuric acid and phosphoric acid because they have a

high affinity for water. Chemists usually choose phosphoric acid

because it is less reactive and causes less decomposition of thereaction product. Using phosphoric acid is especially important when

the boiling point of the alkene being formed is too high to readily

distill from the reaction mixture.

Acid-catalyzed dehydration reactions generally follow the

Saytzeff rule in which the most substituted alkene predominates if

two or more alkenes can form from the carbocation intermediate.Because carbocation rearrangements are common, the synthetic

utility of alcohol dehydrations are limited to cases where no

rearrangement is likely or where only one dehydration product is

possible.

Synthesis of Cyclohexene

OH

(90%)

H3PO4/H2SO4

Place 5.2 mL (0.05 mol) of cyclohexanol in a reaction flask. Add 1.0 mL of phosphoric

acid along with 0.3 mL of sulfuric acid. Mix thoroughly and add a magnetic stir bar or

a boiling stone. Heat until the product mixture begins to distill. Adjust the heat so

that the rate of distillation is slowno more than 5-10 drops per minute. Continue

heating until no more liquid distills off. Separate the water from the distillate and dry

the organic layer over anhydrous sodium sulfate. Remove the drying agent and distill

the product. Yield of cyclohexene is 3.7 g (90%), b.p. 82-84oC.

Discussion Questions

1. What is the purpose of having a slow rate of distillation of product from the

reaction mixture?



33/48


Exercise 13.10

Propose a mechanism for the following reaction. Indicate the major

product.

++

CH3CH2

H3PO4

CH2OH

Exercise 13.11

Predict the major products for each of the following reactions.

a)

CH

CH3

CH3 CHCH3

OH

H3PO4

b)

OHH3PO4

c)

H3PO4

CH3

CH3

OH

d)

H2SO4CH2OH

H

Sample Solution

b)

OHH3PO4



34/48


13.10 Pinacol Rearrangement

An acid-catalyzed dehydration of a 1,2-diol is frequentlyaccompanied by a rearrangement of the carbon skeleton to form an

aldehyde or ketone. This reaction is called the pinacol

rearrangement. The pinacol rearrangement receives its name fromthe common name of the simplest compound that undergoes this

rearrangement.

The pinacol

rearrangement is a

dehydration reaction of

a vicinal diol

accompanied by a

rearrangement to form

an aldehyde or ketone.

C C CH3

CH3

OHOH

CH3

CH3

(82%)

H2SO4CH3 C

CH3

CH3

C CH3

O

"Pinacol" "Pinacolone"

2,3-Dimethyl-2,3-butanediol 3,3-Dimethyl-2-butanone

Because the substrate loses a molecule of water, the pinacol

rearrangement is a form of dehydration reaction. The reaction follows

these steps: 1) A proton from the acid protonates one of the OH

groups. 2) The compound loses a molecule of water and forms a

tertiary carbocation, much as would be expected from any tertiaryalcohol. 3) A methyl group migrates from the carbon bearing the OH

group to the carbon with the charge forming a resonance-stabilized

carbocation that is even more stable than the original carbocation. 4)

This carbocation loses a proton and forms the product.



35/48


CH3 C C CH3

OH OH

CH3 CH3

H

H OSO3

H

HSO4

CH3 C

CH3

CH3

C CH3

O

CH3 C

CH3

CH3

C CH3

O H

CH3 C C CH3

OH OH

CH3 CH3

CH3 C C CH3

OH

CH3CH3

CH3 C

CH3

CH3

C CH3

O H

The pinacol rearrangement has long been used as a model for

the study of the rearrangement of a carbocation. Studies of the pinacolrearrangement show that the migratory aptitude for groups generally

is H > aryl > alkyl. The best migrating groups are the most stable

cations. Thus, a tertiary group migrates better than a primary group.

Pinacol migratory aptitudes are different from those seen in theBaeyer-Villiger reaction. The migration in the pinacol rearrangement

is stereospecific with the migrating group retaining its configuration.

The Baeyer-Villiger

reaction is discussed in

Section 8.8, page 000.

The actual migration depends on the substrate and the

reaction conditions. For example, consider the reaction of 1,1-

diphenyl-2-methyl-1,2-propanediol. If it reacts with cold, concentrated

H2SO4, the methyl group migrates. When it reacts with acetic acidplus a trace of H2SO4, the phenyl group migrates.



36/48


Phenyl migration

Methyl migration

CH3

CH3O

CH3

CH3O

CH3COH

O

(cold)

H2SO4(trace)

H2SO4

OH

CH3

OH

CH3(83%)

(77%)

The difference is in the acid strength of the reaction medium. In

stronger acid (H2SO4), the methyl-migration mechanism

predominates. The driving force for the reaction is the stability of the

carbocations in the solution. However, in weaker acid (CH3COOH), a

concerted reaction occurs such that the group with the highermigratory aptitude (the phenyl group) is the one that actually moves.

The mechanism for the concerted reaction begins with protonation of

one of the OH groups. As the water molecule leaves, the migrating

group simultaneously moves to form the protonated ketone.



37/48


H OCCH3

O

CH3CO

O

CH3 C

CH3

C

O

CH3 C

CH3

C

O H

CH3 C C

OH OH

CH3

CH3 C C

OH OH

CH3

H


Predict the major product from the following pinacol reactions.

a)OH

HO

H2SO4

Solution

Four equivalent alkyl groups can migrate in this reaction. Thus, the product

is a spiro bicyclic product with a ketone on the larger ring.

OH

HO

H2SO4

O

b)

OH

OH

H2SO4



38/48


Solution

This reaction has two possible migrating groups: a hydrogen from C1 and a

methyl group from C2.

OHCH3

CH3 OH

H H

Possible migrating groups

or

The migratory aptitude of hydrogen is greater than for a methyl group, so the

hydrogen migrates to form an aldehyde.

OH

OH

H2SO4

O

H

Exercise 13.12

Predict the major products for the following reactions.

a)

H2SO4

(cold)H

OH

OH

H

b)

H2SO4

(cold)

OH

OH

c)

OH

CH2OH H2SO4

(cold)

d)



39/48


H2SO4(trace)

CH3COH

OOH

OH

Sample Solution

c)

(cold)

H2SO4

OH

CH2OH

H

O

13.11 Hofmann Elimination

The Hofmann elimination reaction has more historicalimportance in the development of organic chemistry than it has inmodern synthesis. Chemists now prefer other reactions, such as the

Wittig reaction, to the Hofmann elimination reaction. However, the

mechanism of the Hofmann is still important to consider because it

gives insight into the E2 mechanism. A. W. von Hofmann conducted

the original experiments in 1851. Based on the understanding he

gained from this experiment, he proposed what became known as the

Hofmann orientation rule.

The Hofmann

elimination reaction

eliminates a

quaternary ammonium

hydroxide generally to

form the least

substituted alkene.

The Wittig reaction is

discussed in Section

7.10, page 000. The Hofmann reaction forms a quaternary ammonium salt

from an amine (RNH2) and methyl iodide, then it eliminates

trimethylamine and a proton to form an alkene. The amide ion itself(c- NH2) is not eliminated because, similar to the OH group, the

NH2 group is not a very good leaving group. However, exhaustive

methylation to form a quaternary ammonium group puts a positive

charge on the nitrogen of the amine group making the neutral

trialkylamine a very good leaving group. A Hofmann elimination

reaction follows E2 kinetics.

The Hofmann

orientation rule states

that the least

substituted alkene is

synthesized. It is the

opposite of the

Markovnikov rule.

To accomplish the first step in a Hofmann elimination reaction,

the conversion of the amine to the quaternary ammonium salt,

chemists usually react the amine with an excess of methyl iodide.

Formation of

quaternary ammonium

salts is discussed in

Section 12.10, page

000.

R NH2 N(CH3)3RCH3I

I

Next they convert the tetraalkyl ammonium iodide to the hydroxide by

treating it with silver oxide in water.



40/48


N(CH3)3R I N(CH3)3R OHAg2O

H2O

If the quaternary ammonium hydroxide contains a hydrogen to the

quaternary amine, then heating the compound to 150oC forms the

alkene by loss of the neutral trimethylamine.

C CCC

H

N(CH3)3

HO

The Hofmann reaction is an exception to the rule that anelimination reaction forms the more substituted alkene, the Saytzeff

product. Instead, the Hofmann elimination of the quaternary

ammonium hydroxide produces the least substituted alkene because

the trimethylammonium group is too large to form the more

substituted alkene. Its size interferes with the ability of the molecule

to achieve the anti-coplanar arrangement required for the E2mechanism to form the Saytzeff product.

OHCH3CH2C

CH3

CH3

N(CH3)3 C CH2

CH3

CH3CH2

C C

CH3

CH3

CH3

H

+

93% 7%

Looking at the above example from a three-dimensional viewhelps you to see how the size of the trimethylamino group prevents

significant production of the more substituted alkene. Figure 13.7

compares the conformation of the quaternary ammonium hydroxide

required for an E2 elimination with a more stable conformation of the

quaternary ammonium hydroxide by looking along the C2C3 bond.

The direction of elimination for the Saytzeff product is an eliminationat the C2C3 bond. The Hofmann elimination reaction results in

elimination at the C1C2 bond.



41/48


CH3H3C

N(CH3)3

CH3

HH

CH3H3C

N(CH3)3

Hydrogen not antito leaving group

Unstable gauchesteric interaction

More stableNeeded for Saytzeff E2 product

CH3

H

H

(a)

H

H

H

N(CH3)3

H3C CH2CH3

Conformation for Hofmann E2 product

(b)

Figure 13.7. Hofmann elimination yields the least substituted alkene because thequaternary ammonium group is too large to allow formation of the conformation

needed for Saytzeff elimination. (a) Shows the conformations needed for the Saytzeff

product. (b) Shows the conformation for the Hofmann elimination.

Nucleophilic substitution is a significant side reaction in many

Hofmann eliminations. In the previous reaction, nearly 20% of the

starting material undergoes an SN2 reaction on one of the methylgroups attached to the nitrogen to produce methanol and N,N,2-

trimethyl-2-butanamine.

MethanolN,N,2-Trimethyl-2-butanamine

CH3OH+CH3CH2C

CH3

CH3

N(CH3)2CH3CH2C

CH3

CH3

N(CH3)3 OH

Exercise 13.13

Explain the differences between the reactions of the following isomers.



42/48


+

(CH3)3CN(CH3)2

CH3OH

100%

(CH3)3CN(CH3)3 OH

+

(CH3)3C

N(CH3)3 OH

(CH3)3C

N(CH3)2

CH3OH

+

(CH3)3C

91%

9%

13.12 Oxidation of Alcohols

With the exception of the pinacol rearrangement (Section 13.9),all the reactions discussed so far in this chapter are elimination

reactions that form alkenes or alkynes. This section discusses another

exception: the elimination of two hydrogen atoms from a primary or

secondary alcohol to form an aldehyde or ketone. This elimination

reaction is called an oxidation reaction.In an oxidationreaction, a molecule

typically loses two

hydrogens to form a

multiple bond.RCH2OH RCH

OOxidation

(-2H)

RCHOH

R

RCR

O

Oxidation

(-2H)



43/48


Many procedures are available to accomplish an oxidation

reaction, but this book covers only a few. A common one is the

chromate oxidation. A chromate oxidation uses chromium trioxide

or chromic acid to oxidize secondary alcohols to ketones.A chromate oxidationuses chromic acid orchromium trioxide to

oxidize primary or

secondary alcohols to

carboxylic acids or

ketones. This reaction

is often called the

Jones oxidation.

(97%)

OHCrO3

H2SO4, acetoneO

Cycloheptanone

(91%)

OOH

H2SO4, H2O

Na2Cr2O7

2-Pentanone

The mechanism for a chromate oxidation involves the initial formation

of a chromate ester. The chromate ester then undergoes an E2 type

elimination to form the carbonyl group.

C

O

OSO3H

Cr

O

O O

HOH

H

O CrO3H

H

HCr

O

O

O

H

OH

When starting from a secondary alcohol, the product is a ketone.

When starting from primary alcohol, the product is an aldehyde. With

the aqueous acid in the reaction mixture, the aldehyde forms a

hydrate. This hydrate can react further to form a carboxylic acid.Hydrate formation isdiscussed in Section

7.5, page 000.

H

O

H3O

OH

HHO

Like above

mechanism OH

O



44/48


O

H

CrO3

H2SO4, acetone

O

OH

3-Methylbutanoic acid

(93%)

To oxidize primary alcohols to carboxylic acids, chemists use eitherchromium trioxide or sodium dichromate.

OH

O

OHH2SO4, acetone

CrO3

(91%)Pentanoic acid

A drawback of chromate oxidation is the reaction conditions

can oxidize other functional groups such as double or triple bonds.

Pyridinium chlorochromate (PCC), however, is a much milder

oxidizing agent than chromic acid. In addition, no water is present soit cannot oxidize an aldehyde to a carboxylic acid. PCC is synthesized

by reacting equal amounts of pyridine, chromium trioxide, and HCl.

Oxidation of double

and triple bonds is

discussed in Section

14.10, page 000.

Pyridiniumchlorochromate

(PCC)

Pyridine

ClCrO3HCl

CrO3

N

H

N

PCC is quite soluble in low polarity solvents. The most commonly used

solvent is methylene chloride (CH2Cl2), which is specific for the

oxidation of primary and secondary alcohols to aldehydes and ketones.

(81%)

H

O

OH CH2Cl2

PCC

Hexanal



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PCC

CH2Cl2OH H

O

(93%)

3,7-Dimethyl-6-octenal

Synthesis of Citronellal

(93%)

H3C H

O

H

Alumina

PCC

H3C H

OH

Pyridinium chlorochromate on alumina

Dissolve 0.6 g of chromium trioxide in 1.1 mL of 6 M hydrochloric acid. Slowly add

0.47 g of pyridine maintaining the temperature of the reaction mixture at 45oC. Cool

the mixture to 10oC to initiate crystal formation, then reheat to 40oC to dissolve the

crystals formed. To the resulting solution, add 5 g of alumina with stirring. Remove

the solvent on a rotary evaporator and dry the orange solid for 2 hours in a vacuum at

room temperature. Store this reagent under vacuum in the dark until use. Yield about

6.3 g of PCC on alumina.

Citronellal

Place the PCC on alumina synthesized above in a 50 mL Erlenmeyer flask and add

546 mg (3.5 mmol) of citronellol and 10 mL of hexane. Stir this mixture for up to 3

hours. Follow the course of the reaction by using thin layer chromatography (TLC)

plates. Determine when the reaction is complete by the disappearance of the

citronellol spot in a TLC plate using methylene chloride as eluant. Filter off the solid

and wash it three times with 5 mL portions of ether. Combine the ether washes with

the hexane filtrate. Evaporate the combined solvents using a rotary evaporator. The

yield of product is 500 mg (93%); the b.p. is 204-207oC.

Discussion Questions

1. TLC is a convenient way to follow the course of the synthesis of citronellal, but can

you think other methods for following the course of the reaction? If so, what do you

see that would help you know when the reaction is complete?

Exercise 13.14

Predict the major products for the following reactions.



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a)

CrO3

H2SO4, acetone

OH

b)

OH

PCC

CH2Cl2

c)

PCC

CH2Cl2

OH

O

d)

Na2Cr2O7

H2SO4, H2O

OH

Sample Solution

b)

OH

PCC

CH2Cl2

H

O

Key Ideas from Chapter 13

A 1,2-elimination, or -elimination, reaction involves the loss of

a leaving group and an electrophile from adjacent atoms in a

substrate. This loss results in the formation of a bond.



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There are two major elimination mechanisms: E1 and E2. In

the E1 mechanism, the leaving group departs from the

substrate, then the electrophile leaves. The reaction proceeds

via a carbocation intermediate. In the E2 mechanism, theelectrophile and leaving groups depart at the same time in a

concerted reaction.

Both E1 and E2 mechanisms proceed best on tertiary

substrates.

An E1 mechanism loses all the stereochemistry at the reaction

site due to the formation of a symmetrical carbocation

intermediate.

An E2 mechanism requires that the leaving group andelectrophile be in the same plane. The rate of reaction for the

anti-coplanar conformation is generally much higher than for

the syn-coplanar conformation.

Most E1 and E2 elimination reactions follow the Saytzeff rule

for forming a double bond. The most highly substituted alkeneis generally the most stable product, so it is the major product.

A reaction following the Saytzeff rule is a regioselective

reaction.

A measurement of alkene stability is the amount of heat

evolved when a particular alkene is hydrogenated. The lessheat given off when adding hydrogen to a double bond, the

more stable the double bond.

E1 and SN1 reactions work best in acidic conditions with a

polar-protic solvent and a tertiary substrate.

An E2 reaction usually works best in strongly basic conditions

with a large base, a less polar solvent, and a tertiary substrate.

E1 takes preference over SN1 when the base is very weak or is

present in very low concentrations. Both pathways form the

same reactive carbocation.

SN1 takes preference over E1 when a good nucleophile is

present or is present in moderate to high concentrations.

A large or hard base increases the rate of E2 over SN2.



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Increasing solvent polarity favors substitution over

elimination.

Increasing the temperature increases the rate of eliminationover substitution.

Alkyl halides readily eliminate via an E2 mechanism to form

an alkene. Chemists often use vicinal dihalides to form an

alkene regiospecifically.

A strong base reacts with a vicinal dihalide to form a diene or

an alkyne, depending on the stereochemistry of the substrate.

A geminal dihalide also forms an alkyne.

TheE andZ prefixes are used in preference to the trans and cis

prefixes in the IUPAC nomenclature rules.

Alcohols dehydrate to form alkenes. Most dehydration

reactions follow the E1 mechanism. The carbon skeleton

commonly rearranges.

The Hofmann elimination forms an alkene from a quaternary

ammonium hydroxide. The product is usually the leastsubstituted alkene following the Hofmann orientation.

Elimination can also take place with a primary or secondary

alcohol. This reaction produces an aldehyde or carboxylic acidfrom a primary alcohol and a ketone from a secondary alcohol.

13 eliminations

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