Elimination Reactions of Alkyl Halides
In addition to undergoing the nucleophilic substitution reactions, alkyl halides undergo
elimination reactions.
In an elimination reaction, groups are eliminated from a reactant.
For example, when an alkyl halide undergoes an elimination reaction, the halogen (X) is
removed from one carbon and a proton is removed from an adjacent carbon.
A double bond is formed between the two carbons from which the atoms are eliminated.
Therefore, the product of an elimination reaction is an alkene.
Just as there are two important nucleophilic substitution reactions— and are two
important elimination reactions: E1 and E2.
The E2 Reaction
“E” stands for elimination and “2” stands for bimolecular. The product of an elimination
reaction is an alkene.
An E2 reaction is a concerted, one-step reaction: The proton and the bromide ion
are removed in the same step, so no intermediate is formed.
The rate of an E2 reaction depends on the concentrations of both tert-butyl
bromide and hydroxide ion. It is, therefore, a second-order reaction.
Removal of a proton and a
halide ion is called
dehydrohalogenation.
The carbon to which the halogen is attached is called the α-carbon. A carbon adjacent to the
α-carbon is called a β-carbon. Because the elimination reaction is initiated by removing a
proton from a β-carbon, an E2 reaction is sometimes called a β-elimination reaction. It is also
called a 1,2-elimination reaction because the atoms being removed are on adjacent carbons.
The weaker the base, the better it is as a leaving group.
The Regioselectivity of the E2 Reaction
An alkyl halide such as 2-bromopropane has two β-carbons from which a proton can be
removed in an E2 reaction.
Because the two -carbons are identical, the proton can be removed with equal ease from
either one. The product of this elimination reaction is propene.
2-bromobutane has two structurally different β-carbons from which a proton can be
removed.
So when 2-bromobutane reacts with a base, two elimination products are formed: 2-butene
and 1-butene.
This E2 reaction is regioselective because more of one constitutional isomer is formed than
the other.
The major product of an E2 reaction is the most stable alkene.
The greater the number of substituents, the more stable is the alkene. (Zaitsev rule)
The more substituted alkene is not, however, always the more stable alkene.
In the following reactions, the conjugated alkene is the more stable alkene even though it is not the most
substituted alkene.
The major product of each reaction is, therefore, the conjugated alkene because, being more stable, it is
more easily formed.
Therefore, if there is a double bond or a benzene ring in the alkyl halide, do not use Zaitsev’s rule to predict
the major (most stable) product of an elimination reaction.
In some elimination reactions, the less stable alkene is the major product.
For example, if the base in an E2 reaction is sterically bulky and the approach to the alkyl
halide is sterically hindered, the base will preferentially remove the most accessible
hydrogen.
The percentage of the less substituted alkene increases as the size of the base increases.
If the alkyl halide is not sterically hindered and the base is only moderately hindered, the
major product will still be the more stable product.
Although the major product of an E2 dehydrohalogenation of alkyl chlorides, alkyl bromides,
and alkyl iodides is normally the more substituted alkene, the major product of the E2
dehydrohalogenation of alkyl fluorides is the less substituted alkene.
The fluoride ion, however, is the strongest base of the halide ions and, therefore, the
poorest leaving group. So when a base begins to remove a proton from an alkyl fluoride,
there is less tendency for the fluoride ion to leave than for the other halide ions to do so.
As a result, negative charge develops on the carbon that is losing the proton, causing the
transition state to resemble a carbanion rather than an alkene.
We can summarize by saying
that the major product of an
E2 elimination reaction is the
more stable alkene except if
the reactants are sterically
hindered or the leaving group
is poor (e.g., a fluoride ion), in
which case the major product
will be the less stable alkene.
The E1 Reaction
“E” stands for elimination and “1” stands for unimolecular.
The reaction of tert-butyl
bromide with water to form
2-methylpropene is an
example of an E1 reaction.
An E1 reaction is a first-order elimination
reaction because the rate of the reaction
depends only on the concentration of the alkyl
halide.
When two elimination products can be formed in an E1 reaction, the
major product is generally the more substituted alkene. Zaitsev’s rule.
Relative reactivities of alkyl halides in an E1 reaction = relative stabilities of carbocations
A tertiary alkyl halide and a weak base were chosen to illustrate the E1 reaction.
Whereas a tertiary alkyl halide and a strong base were used to illustrate the E2 reaction.
Tertiary alkyl halides are more reactive than secondary alkyl halides, which, in turn, are
more reactive than primary alkyl halides in both E1 and E2 reactions.
Because the E1 reaction forms a carbocation intermediate, the carbon skeleton can
rearrange before the proton is lost, if rearrangement leads to a more stable carbocation.
Competition Between E2 and E1 Reactions
Primary alkyl halides undergo only E2 elimination reactions. They cannot undergo E1
reactions because of the difficulty encountered in forming primary carbocations.
Secondary and tertiary alkyl halides undergo both E2 and E1 reactions.
For those alkyl halides that can undergo both E2 and E1 reactions, the E2 reaction is favored by
the same factors that favor an SN2 reaction and the E1 reaction is favored by the same factors
that favor an SN1 reaction.
Thus, an E2 reaction is favored by a high concentration of a strong base and an aprotic polar
solvent (e.g., DMSO or DMF), whereas an E1 reaction is favored by a weak base and a protic
polar solvent (e.g., H2O or ROH).
Competition Between Substitution and Elimination
A tertiary alkyl halide is the least reactive of the alkyl halides in an SN2 reaction and the
most reactive in an E2 reaction. Consequently, only the elimination product is formed
when a tertiary alkyl halide reacts with a nucleophile/base under conditions.
Substitution and Elimination Reactions in Synthesis
The alkoxide ion for the Williamson ether synthesis is prepared by using sodium
metal or sodium hydride (NaH) to remove a proton from an alcohol.
It is an SN2 reaction because it requires a high concentration of a good nucleophile.
You can use either a propyl halide and butoxide ion or a butyl halide and propoxide ion.
you want to synthesize tert-butyl ethyl ether, the starting materials must be an ethyl halide
and tert-butoxide ion.
Tertiary alkyl halide under SN2 / E2 conditions forms primarily the elimination product.
So, in carrying out a Williamson ether synthesis, the less hindered alkyl group should be
provided by the alkyl halide and the more hindered alkyl group should come from the
alkoxide.
This is an reaction (the alkyl halide reacts with a high concentration of a good nucleophile).
Best to use primary alkyl halides and methyl halides in the reaction.
A tertiary alkyl halide would form only the elimination product, and a secondary alkyl
halide would form mainly the elimination product because the acetylide ion is a very strong
base.
If you want to synthesize an alkene, you should choose the most hindered alkyl halide
in order to maximize the elimination product and minimize the substitution product.
Consecutive E2 Elimination Reactions
Alkyl dihalides can undergo two consecutive dehydrohalogenations, giving products that
contain two double bonds.
If the two halogens are on the same carbon (geminal dihalides) or on adjacent carbons
(vicinal dihalides), the two consecutive E2 dehydrohalogenations can result in the
formation of a triple bond. This is how alkynes are commonly synthesized.
The vinylic halide intermediates in the preceding reactions are relatively unreactive.
Consequently, a very strong base, such as –NH2 is needed for the second elimination.
If a weaker base, such as -OH is used at room temperature, the reaction will stop at the
vinylic halide and no alkyne will be formed.
How to convert a double bond into a triple bond.
Intermolecular Versus Intramolecular Reactions
A molecule with two functional groups is called a bifunctional molecule.
Which reaction is more likely to occur—an intermolecular reaction or an intramolecular
reaction?
Depends on
The concentration of the bifunctional molecule
And the size of the ring that will be formed in the intramolecular reaction
A low concentration of reactant favors an intramolecular reaction.
If the intramolecular reaction forms a five- or six-membered ring, it will be highly
favored over the intermolecular reaction.
Three- and four-membered rings are
strained, so they are less stable than
five- and six membered rings and,
therefore, are less easily formed.
Three-membered ring compounds
are formed more easily than four-
membered ring compounds.
To form a cyclic ether, the nucleophilic oxygen atom must be oriented so that it can attack
the back side of the carbon bonded to the halogen.
Rotation about a bond in the molecule forms conformations with the groups pointed away
from one another
The likelihood of the reacting groups finding each other decreases sharply when the
groups are in compounds that would form seven-membered and larger rings. Therefore,
the intramolecular reaction becomes less favored as the ring size increases beyond six
members.
Elimination from Cyclic Compounds
E2 Elimination from Cyclic Compounds
Elimination from cyclic compounds follows the same stereochemical rules.
To achieve the anti-periplanar geometry that is preferred for an E2 reaction, the two groups
that are being eliminated from a cyclic compound must be trans to one another.
In the case of six-membered rings, the groups being eliminated will be anti-periplanar only
if both are in axial positions.
More stable conformer of a monosubstituted cyclohexane is the one in which the
substituent is in an equatorial position does not undergo E2 reaction.
The less stable conformer, with the chloro substituent in the axial position, readily
undergoes an E2 reaction
Because one of the two conformers does not undergo an E2 reaction, the rate of an
elimination reaction is affected by the stability of the conformer that does undergo the
reaction.
The reaction is faster if elimination takes place by way of the more stable conformer.
If elimination has to take place by way of the less stable conformer, reaction will be slower.
Neomenthyl chloride undergoes
an E2 reaction with ethoxide ion
about 200 times faster than
menthyl chloride does.
The conformer of neomenthyl
chloride that undergoes
elimination is the more stable
conformer because when the Cl
and H are in the required axial
positions, the methyl and
isopropyl groups are in the
equatorial positions.
When menthyl chloride or trans-1-chloro-2-methylcyclohexane undergoes an E2 reaction, the
hydrogen that is eliminated is not removed from the -carbon bonded to the fewest hydrogens.
This may seem like a violation of Zaitsev’s rule.
In the preceding two reactions, the hydrogen that is removed has to be in an axial position and
only one -carbon has a hydrogen in an axial position. Therefore, that hydrogen is the one that is
removed even though it is not bonded to the α-carbon with the fewest hydrogens.
E1 Elimination from Cyclic Compounds:
When a substituted cyclohexane undergoes an E1 reaction, the two groups that are
eliminated do not have to both be in axial positions, because the elimination
reaction is not concerted.
In the following reaction, a carbocation is formed in the first step. It then loses a
proton from the adjacent carbon that is bonded to the fewest hydrogens— in other
words, Zaitsev’s rule is followed.
A carbocation is formed in an E1 reaction, you must check for the possibility of a
carbocation rearrangement before you use Zaitsev’s rule to determine the elimination
product.
Temperature:
Temperature has an important role to play in deciding whether a reaction is an elimination
or a substitution.
• In an elimination, two molecules become three.
• In a substitution, two molecules form two new molecules.
• The two reactions therefore differ in the change in entropy during the reaction: ΔS is
greater for elimination than for substitution.
∆G = ∆H − T∆S
• This equation says that a reaction in which ΔS is positive becomes more favorable (ΔG
becomes more negative) at higher temperature. Eliminations should therefore be
favored at high temperature, and this is indeed the case: most eliminations you will see
are conducted at room temperature or above.
E2 elimination from vinyl halides: how to make alkynes
An anti-periplanar arrangement of C–Br and C–H is attainable with a
vinylic bromide too, provided the Br and H are trans to one another.
E2 elimination from the Z isomer of a vinyl bromide gives an alkyne
rather faster than elimination from the E isomer because in the E isomer
the C–H and C–Br bonds are syn-periplanar.
The E1cB Mechanism
In the E1 mechanism, X leaves first and then H.
In the E2 mechanism, the two groups leave at the same time.
There is a third possibility:
The H leaves first, and then
the X. This is a two-step
process, called the E1cB
mechanism, or the
carbanion mechanism,
since the intermediate is a
carbanion.
The name E1cB comes from the fact that it is the conjugate base of the
substrate that is giving up the leaving group.
The greatest likelihood of finding the E1cB mechanism in substrates that have
(a) a poor nucleofuge
(b) an acidic hydrogen at β-carbon.
The second step is the rate-determining step of the
elimination—the elimination is unimolecular.
It’s important to note that, while HO− is never a leaving group in E2 reactions, it can
be a leaving group in E1cB reactions.
The combination of a poor leaving group and
an acidic proton means that E1cB elimination is
extremely easy.
With E1cB, the regioselectivity is straightforward: the location of the double bond is
defined by the position of (a) the acidic proton and (b) the leaving group.
E1cB reactions may be stereoselective—the one above, for example, gives mainly
the E alkene product (2:1 with Z).
The intermediate anion is planar, so the stereochemistry of the starting materials is
irrelevant, the less sterically hindered (usually E) product is preferred.
This double E1cB elimination, gives only the E,E product.
Formation of an amide in the presence of base
E1cB elimination
The product of the elimination is a substituted ketene—a highly reactive species.
It is the ketene that reacts with the amine to form the amide.
This deprotonation,
followed by loss of
chloride, is the first step in
the formation of a
mesylate ester. It is an
E1cB elimination and the
product is called a sulfene.
The sulfene is electrophilic: the alcohol acts as a nucleophile for sulfur
and generates an anion of carbon which undergoes a proton transfer
to give the mesylate.
The Hofmann rule states that the major product in Hofmann eliminations
and other similar elimination reactions is the less stable alkene (or the
alkene featuring a lesser substituted double bond).
Hofmann elimination is an elimination reaction of an amine to form alkenes.
The reaction starts with the formation of a quaternary ammonium iodide salt by
treatment of the amine with excess methyl iodide (exhaustive methylation), followed
by treatment with silver oxide and water to form a quaternary ammonium hydroxide.
When this salt is decomposed by heat, the Hofmann product is preferentially formed.
A positively charged leaving group increases the acidity of the β-protons.
A substituent at the β-position could hyperconjugatively decrease the acidity
of the β-proton.
Consequently, a terminal methyl group (this has no alkyl substituent) is more
acidic than the internal methine proton (bearing at least one alkyl
substituent).
The steric factors govern such elimination reactions. The charged leaving
groups are large compared to neutral leaving groups.
When the leaving group is a halogen, the stability of the developing double
bond becomes important and this leads to the thermodynamically more stable
product.
The larger leaving groups like –NR3+ and –SR2+ give more
Hoffmann product than smaller groups like halogens. The
bulkiness of the base also increases the Hoffmann product
at the cost of the Zaitsav product.
PYROLYTIC ELIMINATIONS
Several types of compound undergo elimination on heating, with no other reagent
present.
Reactions of this type are often run in the gas phase.
Mechanism for these reactions are different from those already discussed bescause
there is no base or solvent present in pyrolytic elimination.
Ei mechanism (Elimination Internal/Intramolecular), also known as a thermal syn
elimination or a pericyclic syn elimination.
Two mechanisms have been found to operate. One involves a cyclic transition state,
which may be four, five, or six membered.
Two vicinal (adjacent) substituents on an alkane framework leave simultaneously
via a cyclic transition state to form an alkene in a syn elimination.
The elimination must be syn and the
atoms coplanar for four and five-
membered transition states, but
coplanarity is not required for six-
membered transition states.
Exclusive syn elimination as demonstrated experimentally in various substrates
with the use of radio-labelling techniques.
For example, the pyrolysis of the erythro and threo isomers of 1-acetoxy-2-
deutero-1,2-diphenylethane gave in each case trans-stilbene but, the stilbene from
the erythro compound retained all the deuterium while the threo compound lost
all of it.
This must be due to syn elimination of hydrogen or the deuterium atom and the
acetoxy group
If two or more syn periplanar arrangements
are available, the one with least crowding
is preferred.
pyrolysis of 2-butyl acetate affords a mixture
of 1-butene, cis-2-butene, and
trans-2-butene, with the trans/cis ratio
around 2.
(since there is less steric interaction between
the two methyl groups in the transition
state leading to the trans-alkene).
Evidence for the existence of the Ei mechanism is
• The kinetics are first order, so only one molecule of the substrate is involved in the reaction.
• Free-radical inhibitors do not slow the reactions, so no free-radical mechanism is involved.
• The mechanism predicts exclusive syn elimination, and this behavior has been found in
many cases.
The second type of pyrolysis mechanism is completely different and
involves free radicals. Pyrolytic hemolytic cleavage initiates the process.
Free-radical mechanisms are mostly found in pyrolysis of polyhalides and of primary
monohalides, although they also have been postulated in pyrolysis of certain carboxylic
esters.
Orientation in Pyrolytic Eliminations
If a double bond is present, a conjugated system will be preferred, if sterically
possible.
Apart from these considerations, the following statements can be made for Ei
eliminations:
1. orientation is statistical and is determined by the number of β hydrogens
available.
3:2 distribution
predicted by the
number of hydrogens
available.
2. A cis β hydrogen is required. Therefore in cyclic systems, if there is a cis hydrogen on only
one side, the double bond will go that way. However, when there is a six-membered transition
state, this does not necessarily mean that the leaving groups must be cis to each other, since
such transition states need not be completely coplanar.
ethyl-2-acetoxycyclohexanecarboxylate in which the leaving group is axial, The double
bond in the direction of carbethoxyl group does not form even though it would be
conjugated, since there is no equatorial hydrogen on that side.
However, in the case of its diastereomer in which the leaving group is equatorial and the
hydrogen at the carbethoxyl group is axial, can adopt the necessary cyclic transition state
that leads to the conjugated alkene.
3. In some cases, especially with cyclic compounds, the
more stable alkene forms and Zaitsev’s rule applies.
For example, menthyl acetate gives 35% of the Hofmann
product and 65% of the Zaitsev, even although a cis β
hydrogen is present on both sides and the statistical
distribution is the other way.
A similar result was found for the pyrolysis of menthyl
chloride.
4. There are also steric effects. In some cases, the direction of elimination is determined
by the need to minimize steric interactions in the transition state or to relieve steric
interactions in the ground state.
Scope and synthetic importance of pyrolytic eliminations
Pyrolytic eliminations occur in a variety of substrates containing appropriate leaving
groups and are important from a synthetic perspective
It can be carried out by simple heating, if the boiling point of the ester is high enough or else
is carried out by passing the vapor through a heated tube.
Moreover, since these reactions do not necessarily require any solvent or base, hence, these
reactions are clean to perform and isolation of product is simplified (in cases when a single
product is formed).
Further, when a sensitive or reactive substrate is present, the lack of any added acidic or
basic reagents is an additional advantage.
Chugaev elimination of Xanthates and Thionocarbonate
The Russian chemist Lev Aleksandrovich Chugaev (1899) discovered a way to eliminate water
from alcohols to produce alkenes via the formation of an intermediate xanthate and this
methodology came to be known as Chugaev reaction.
Xanthate salts are easily produced by the reaction of an alcohol with sodium or potassium
hydroxide and carbon disulfide, and on treatment with iodomethane, it is transformed into a
xanthate.
At about 150-200 °C, the alkene is formed by a syn-elimination through a 6-membered
cyclic transition state. The side product decomposes to carbonyl sulfide (OCS) and
methanethiol.
Since these related thiocarbonates or xanthates eliminate at temperatures in the region of
150– 250oC, this reaction is of immense synthetic potential than ester pyrolysis as further
decomposition of the alkene product(s) can often be avoided.
However, on the flip side, separation of the alkene from sulfur-containing by-products can
sometimes be troublesome
The Cope Elimination of tert-amineoxides
The Cope elimination, developed by Arthur C. Cope, is an elimination reaction of the
N-oxide of a tertiary amine to form an alkene and a hydroxylamine at milder reaction
temperatures (100-200 oC).
The N-oxide can be easily prepared in situ from tertiary amines with an oxidant such
as m-CPBA or H2O2.
The Cope elimination is also a syn-elimination and involves a five-membered cyclic
transition state which must be planar (syn-coplanarity). The amine oxide (leaving
group) itself acts as a base. The net reaction is 1,2-elimination.
The Cope Elimination is extremely sensitive to
solvent effects, and a 106 -fold rate increase can be
obtained going from protic to aprotic solvents arising
from hydrogen bonding between the amine oxide
and the solvent. Within aprotic solvents, decreasing
polarity significantly increases the reaction rate.
The sterically demanding
amine-oxide reacts
preferentially with the more
easily accessible hydrogens,
and often gives good selectivity
favoring the less-substituted
alkene.
Sulfoxide containing substrates
Sulphur and selenium belong
to the same group (periodic
table) and show valences from
2 to 6.
Sulphoxides containing β-
hydrogen undergo syn-
elimination on heating to give
an olefin.
Oxidation of the sulphide,
followed by heating of the
sulphone introduces a double
bond. The reaction involves a
5-membered T.S. as in other
cases.
β-hydroxy phenyl sulfoxides were found to undergo thermal elimination through a 5-
membered cyclic transition state, yielding β-keto esters and methyl ketones after
tautomerization and a sulfenic acid.
• Pyrolytic eliminations are different from other types of elimination (E1, E2 or E1cB) and
do not require any added reagent (base or acid) but require the application of heat.
• Pyrolytic reactions often occur in the gas phase, without the addition of another
reagent.
• These may be α-eliminations, β-eliminations or 1,3/1,n-eliminations, though the β-
pyrolytic eliminations are the most common.
• These are exclusively syn-eliminations that undergo the Ei (Elimination internal)
mechanism involving a cyclic transition state, which may comprise of four-, five- or six-
membered transition states.
• In those versions of the Ei mechanism that involve a four- or five-membered cyclic
transition state, there is a requirement that all the atoms are co-planar.
• The other type of pyrolytic elimination reactions involves free radicals, for example in
polyhalides and primary monohalides.
• Free radical eliminations involve the normal sequence for radical reactions, namely:
initiation, propagation and termination steps and are frequent at 600-900 oC
temperatures.
• Carboxylic esters, xanthates, tert-amine oxides, sulfoxides and selenoxides are some
substrates which commonly undergo pyrolytic syn-eliminations (Ei mechanism).
• Of the various substrates for pyrolytic eliminations, selenoxide pyrolysis provides the
most milder conditions and is often carried-out in a one-pot reaction sequence (no
need to isolate the selenoxide) to produce the C=C in good yields.
• These pyrolytic eliminations are important from a synthetic perspective and these
methods have been used for the synthesis of natural products and compounds of
pharmaceutical significance
