Daley 13 eliminations sos

Richard F. Daley and Sally J. Daley
www.ochem4free.com
Organic
ChemistryChapter 13
Elimination Reactions
13.1 The Elimination Mechanisms 658
13.2 Stereochemistry of Eliminations 660
13.3 Direction of Elimination 663
13.4 E1 vs. E2 670
13.5 Elimination vs. Substitution 671
13.6 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 686
13.11 Hofmann Elimination 691
13.12 Oxidation of Alcohols 694
Synthesis of Citronellal 697
Key Ideas from Chapter 13 698

Organic Chemistry - Ch 13 656 Daley & Daley
Copyright 1996-2005 by Richard F. Daley & Sally J. Daley
All Rights Reserved.
No part of this publication may be reproduced, stored in a retrieval system, or
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www.ochem4free.com 5 July 2005

Chapter 13
Elimination Reactions
Chapter Outline
13.1 The Elimination Mechanisms
An overview of the two types of elimination mechanisms
13.2 Stereochemistry of Eliminations
A look at the stereochemical requirements of an elimination reaction
13.3 Direction of Elimination
Regioselectivity of the elimination mechanisms
13.4 Comparing E1 and E2 With SN1 and SN2 Reactions
Comparing the E1 with the SN1 reaction and the E2 with the SN2
mechanism
13.5 Elimination Versus Substitution
Factors that influence whether an elimination or substitution
reaction will occur
13.6 Summary of Elimination and Substitution
A summary of the factors that favor the various substitution and
elimination mechanisms
13.7 E and Z Nomenclature
Naming alkenes using E and Z nomenclature
13.8 Elimination of Organohalogens
Elimination reactions involving organohalogen substrates
13.9 Dehydration of Alcohols
Elimination reactions involving alcohol substrates
13.10 Pinacol Rearrangement
The elimination/rearrangement reaction involving a 1,2-diol
13.11 Hofmann Elimination
An elimination reaction involving a nitrogen substrate
13.12 Oxidation of Alcohols
Elimination reactions involving primary and secondary alcohols

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 elimination
reaction will occur with a particular substrate and set of reaction
conditions
✔ Apply the E and Z 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 aliphatic
nucleophilic substitution reactions. Recall, however, that a
competing elimination reaction takes place under some conditions
instead of the intended substitution reaction. In a β-elimination
reaction, the substrate loses two atoms or two groups of atoms to form
a π 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 β-, elimination
reactions and how they fit into the continuum with each other and
with nucleophilic substitution reactions. Organic chemists use various
methods to selectively control which of the possible pathways that a
particular elimination reaction follows. One other major elimination

reaction is the 1,1-, or α-, elimination reaction. Chapter 14 briefly
discusses the α-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 π 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 C
LE
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 reaction—
either 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 see
Section 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
H
C C
H
•
•

Figure 13.1. The E1 reaction mechanism. The first step is an ionization operation
and the second is a 1,3-electron pair abstraction operation.
Hyperconjugation stabilizes the carbocation in the second step
of 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
H
Overlap
+
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 the
leaving 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 reaction
is 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 13.3
summarizes the two steps in the E2 mechanism.
Transition State
C C
C C
H
L
B
B
C C
L
H
•
•
+ L + BH•
•
‡
-
-
Figure 13.3. The E2 mechanism is a 1,5-electron pair displacement operation.

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 less
activation 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 of
the 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 of
the 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. The
conformation 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 the
side 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, the
torsion angle between the electrophile and leaving group is 180o. In
the syn-coplanar conformation, the torsion angle between the
electrophile and the leaving group is 0o.
H
L H
LL
H
LH
Anti Syn
Coplanar conformations
When the leaving
group and the
in the same plane on
the same side of the
molecule in an eclipsed
conformation, they are
syn-coplanar.

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-coplanar
elimination 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 energy
eclipsed 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.

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 equatorial
position. 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)3C
OTs
(CH3)3C
OTs
Both compounds react with ethoxide ion in ethanol to eliminate the
tosyl group and a β 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 β hydrogen in an anti-coplanar conformation;
thus, allowing the base to easily remove the β 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)3C
OTs
H
H
C(CH3)3
H
OTs

Exercise 13.1
What is the approximate dihedral angle between the β hydrogens and
the tosylate group in trans-4-tert-butylcyclohexyl tosylate? Draw a
Newman 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 a
deuterium 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 β 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
Br
H
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

In an elimination reaction, the substrate loses the leaving
group and, in most cases, a hydrogen β to the leaving group.
Frequently, the leaving group has more than one type of β hydrogen
and, depending on a number of factors, any one of these β hydrogens
can leave with the leaving group. For example, in the elimination
reaction of HBr from 2-bromobutane a hydrogen from either carbon 1
or 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-Butene
19% 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 other
hand, produce several different isomers, but give one isomer in greater
quantity than the others. Whether a reaction is regiospecific or
regioselective depends on whether the reaction prefers to eliminate
only one particular β hydrogen or a mixture of β hydrogens. The
elimination of 2-bromobutane mentioned previously eliminates a
mixture of β hydrogens, so it is regioselective.
The “regio” prefix
means 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 elimination
reaction 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 leaving
group, 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 the
Saytzeff 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.

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 of
1-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 of
the 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

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-Butene
C 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 that
there 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 of cis- 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.

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 product
predominates. 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 β hydrogen from the less substituted carbon to
produce a less substituted alkene.
Further evidence for the steric sensitivity of the E2 elimination
is 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
Br
CH3CH2O
(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)

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 carbon—carbon double bond. 3-Methyl-1-butene
has only one substitutent.
b)
CH3
Br
KOH
Solution
In 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.

CH3
Br
H
H
This hydrogen
is anti to the Br.
This hydrogen cannot
be 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)3
H
H
d)
C C
CH3Br
H
H
Sample solution
b)

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 between
elimination 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 basic
conditions. 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 proceeds
best 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 β 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
ClC
H
CH3
O-
Carbon is difficult
to access for SN2
Protons are easy
to access for E2

Figure 13.6. The protons on a carbon adjacent to a tertiary leaving group are more
open to reaction than the backside of the carbon bearing the leaving group.
Both the E2 mechanism and the E1 mechanism work best on
tertiary 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 strongly
acidic 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 the
competition between elimination and substitution reactions. These
two 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 energetically
favorable. 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 and
concentration of the incoming base or nucleophile, the reaction
medium, 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 second
order 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 - OH ion to promote
elimination, in preference to using the softer - 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 occurs
because 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.

2-Methylpropenetert-Butyl methyl ether
36%64%
+
CH3OH
C
CH3
CH3
CH3
OCH3CCH3
CH3
CH3
Br C
CH3
CH2
CH3
50
o
C
However, adding methoxide ion, a stronger base, to the reaction
mixture 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
CH3
CH2C
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%
+
OCH3
CH3O
CH3OH, 50
o
C
OTs
100%0%
Cyclohexenetert-ButoxycyclohexaneCyclohexyl tosylate
+
OC(CH3)3
(CH3)3CO
(CH3)3COH, 50
o
C
OTs
Exercise 13.4
Explain why chemists seldom use the azide and cyanide anions as
bases in elimination reactions.

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 spreads
out, 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 elimination
reaction 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 product
would you expect to predominate at 70oC?
Although chemists have studied the E1 and SN1 mechanisms
extensively 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 SN1 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

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 the
outcome 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 or
bulky base
Mainly SN2 with weak
bases or E2 with
strong bases. Also SN1
or E1 in acid,
especially if the
substrate can
rearrange.
No SN2. In a
solvolysis
reaction, SN1 and
E1. 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 E2
Nucleophile
(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, lower
polarity than
for SN1
Polar-
aprotic
solvent
Polar-
aprotic
solvent,
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, and
reaction temperature on the mechanism of a particular substrate.

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)
CH3OH
refluxBr
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.

Exercise 13.6
Predict the product(s) of each of the following reactions. Indicate the
mechanism for each one and the major and minor product.
a)
reflux
CH3OH
CH3
Br
b)
reflux
KOH/ethanol
CH3
Br
c)
CH3OH, 0o
CH3O
CH3CH2CHCH2CH3
Br
d)
acetone
NaI
CH3CH2CHCH2CH3
Br
e)
70oC
CH3CH2OHCH2OTs
CH3
Sample Solution
c)
E2
SN2
minor
major
+
CH3CH2CHCH2CH3
OCH3
CH3CH2CH CHCH3
CH3OH, 0o
CH3O
CH3CH2CHCH2CH3
Br

13.7 E & Z Nomenclature
Section 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 Z and 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.
The E-Z method of naming is based on the Cahn-Ingold-Prelog
system of substituent priorities. To determine whether a compound
gets an E or a Z 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 group
gets which number, assign a 1 to the group with the atom having the
higher atomic number closest to the point of attachment to the double
bond.
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
➀
➀
➀➀
➁ ➁
➁
➁
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 π bond, the molecule is the Z
isomer. When the #1 are on opposite sides of the π 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 complex
alkenes use the E and Z terms.
Exercise 13.7
Name the following alkenes using the E-Z system of nomenclature.

a)
C C
Cl
CH3H
H
b)
C C
CCH3
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 forming
reactions that chemists investigated. You have already seen many
dehydrohalogenation examples in the illustrations of the elimination
reactions. 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)3CO
CH3CH2CH2CH CH2CH3CH2CH2CH2CH2Br
1-Pentene

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 a
nucleophile. 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 an
E2 reaction with potassium iodide in ethanol. The iodide ion is not
sufficiently basic to remove a proton. Thus, the double bond forms
between the two carbons bearing the bromine atoms.
Vicinal substituents
are substituents that
are on adjacent carbon
atoms.
I
Br
Br
••
••
••
••
•
•
•
•
•
•
•
•
•
•
•
•

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 the
same substrate.
(CH3)3CO
••
••
•
•
Br
H
••
•
•
•
•
(CH3)3CO
Br
Br
H
H
••
••
••
••
•
•
•
•
•
•
•
•
•
•
Iodide ion functions as the base in the dehydrohalogenation of vicinal
dibromides 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 of
preparing a vicinal dihalide is from an alkene. Unlike the
dehydrohalogenation, which forms a double bond at more than one
possible 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 and
geminal dihalides, as well as vinyl halides. This reaction follows the
same 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 substituents
are two substituents
that are on the same
carbon atom (1,1-
disubstituted).

leads to an alkyne because it does not have hydrogens on adjacent
carbons available for elimination to form a diene.
(68%)
KOH
CH3CH2OH
C C
Diphenylethyne
Br
Br
(73%)
CH2CHBr2 C CH
NaNH2
NH3
Ethynylbenzene
(79%)
C C
H
Cl
CH3
CH3CH2CH2 KOH
CH3CH2OH
CH3CH2CH2C 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 a
nucleophilic 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

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
NaNH2
CH3CCH2CH2CHBr2
CH3
CH3
b)
Ethanol,
KOH
CHCH3
Br
c)
(CH3)3CO
DMSO,
CHCH3
Br
d)
Ethanol
KI
CH3CH2CH2CHCH2Br
Br
e)
NH3
NaNH2
CH3C CCH3
CH3 CH3
Br Br
f)
Br
CH3 DBN
Sample Solution
d)
CH3CH2CH2CH CH2
Ethanol
KI
CH3CH2CH2CHCH2Br
Br

13.9 Dehydration of Alcohols
One of the most widely used elimination reactions is the
dehydration of an alcohol to produce an alkene.Dehydration means the
removal of water.
C C
OH H
C C + H2O
H
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 a
dehydration 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 than
primary 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, the
- OH group is a strong base and therefore a poor leaving group. As a
result, alcohol dehydration does not occur in a basic reaction mixture.

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 β hydrogen in an E2
mechanism. When the acid catalyst is sulfuric acid, the conjugate base
is the HSO4
- ion.
••
••
••
••
•
•
•
•
•
•
HO S
O
O
O
•
•
C CC C
OH
H
H
An alternate, and more common, reaction is a concerted
rearrangement 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
H
C C
OH
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.

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 the
reaction 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 slow—no 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?

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)
OH
H3PO4
c)
H3PO4
CH3
CH3
OH
d)
H2SO4
CH2OH
H
Sample Solution
b)
OH
H3PO4

13.10 Pinacol Rearrangement
An acid-catalyzed dehydration of a 1,2-diol is frequently
accompanied 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 from
the 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%)
H2SO4
CH3 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 tertiary
alcohol. 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.

••••
••
CH3 C C CH3
OH OH
CH3 CH3
H
H OSO3H
••••
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 pinacol
rearrangement 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 the
Baeyer-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 acid
plus a trace of H2SO4, the phenyl group migrates.

Phenyl migration
Methyl migration
CH3
CH3
O
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 higher
migratory 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.

•
••
•
••
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

Solution
This reaction has two possible migrating groups: a hydrogen from C1 and a
methyl group from C2.
OH
CH3
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)

H2SO4
(trace)
CH3COH
OOH
OH
Sample Solution
c)
(cold)
H2SO4
OH
CH2OH
H
O
13.11 Hofmann Elimination
The Hofmann elimination reaction has more historical
importance in the development of organic chemistry than it has in
modern 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
( - 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)3R
CH3I
I
Next they convert the tetraalkyl ammonium iodide to the hydroxide by
treating it with silver oxide in water.

N(CH3)3R I N(CH3)3R OH
Ag2O
H2O
If the quaternary ammonium hydroxide contains a hydrogen β 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 an
elimination 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 E2
mechanism 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 view
helps 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 C2—C3 bond.
The direction of elimination for the Saytzeff product is an elimination
at the C2—C3 bond. The Hofmann elimination reaction results in
elimination at the C1—C2 bond.

CH3H3C
N(CH3)3
CH3
HH
CH3H3C
N(CH3)3
Hydrogen not anti
to leaving group
Unstable gauche
steric 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 the
quaternary 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 methyl
groups 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.

+(CH3)3C
N(CH3)2
CH3OH
100%
(CH3)3C
N(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 oxidation
reaction, a molecule
typically loses two
hydrogens to form a
multiple bond.
RCH2OH RCH
O
Oxidation
(-2H)
RCHOH
R
RCR
O
Oxidation
(-2H)

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 oxidation
uses chromic acid or
chromium trioxide to
oxidize primary or
secondary alcohols to
carboxylic acids or
ketones. This reaction
is often called the
Jones oxidation.
(97%)
OH
CrO3
H2SO4, acetone
O
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
H Cr
O
OO
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 is
7.5, page 000.
H
O
H3O
OH
HHO
Like above
mechanism
OH
O

O
H
CrO3
H2SO4, acetone
O
OH
3-Methylbutanoic acid
(93%)
To oxidize primary alcohols to carboxylic acids, chemists use either
chromium trioxide or sodium dichromate.
OH
O
OH
H2SO4, 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 so
it 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
14.10, page 000.
Pyridinium
chlorochromate
(PCC)
Pyridine
••
ClCrO3
HCl
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

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.

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 β-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 π bond.

❏ 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, the
electrophile 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 and
electrophile 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 alkene
is 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 less
heat 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.

❏ Increasing solvent polarity favors substitution over
elimination.
❏ Increasing the temperature increases the rate of elimination
over 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.
❏ The E and Z 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 least
substituted alkene following the Hofmann orientation.
❏ Elimination can also take place with a primary or secondary
alcohol. This reaction produces an aldehyde or carboxylic acid
from a primary alcohol and a ketone from a secondary alcohol.

Daley 13 eliminations sos

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