This power point presentation summarizes elimination reactions, specifically 1,2 elimination reactions. It defines elimination reactions as reactions where two atoms or groups are removed from a reactant to form an unsaturated product. 1,2 elimination reactions eliminate atoms or groups from the 1 and 2 positions on a molecule. Three possible mechanisms are discussed: E2, E1, and E1cb. Evidence for the E2 mechanism includes kinetic isotope effects, the element effect showing dependence on leaving group ability, and the lack of hydrogen exchange. The Saytzeff rule and factors influencing its application, like stability and transition state crowding, are also covered.

1. Power Point Presentation
0n
Elimination Reaction
T. Y. B. Sc.
By
Dr. G. D. Shirole
Assistant Professor
DEPARTMENT OF CHEMISTRY,
A. S. C. COLLEGE, RAHATA

2. Elimination Reaction
Defination –
A reaction in which two atoms or group of atoms are removed from the
reactant to form a product with high degree of unsaturation is known as
elimination reaction.
Generally the atoms or groups eliminated are from the neighboring
atom i.e.1,2 position hence such reaction is known as 1,2 elimination reaction.
Generally in 1,2 elimination the atom that is lost from C2 or Cβ is
hydrogen (X=H) and some good leaving group from C1 or Cα (Y=leaving Group).
Such 1,2 elimination reaction is termed as β elimination reaction
c c
1,2 elimination
1,2 elimination
c c
Alkynes
Alkene
c2 c1
Y
X
c2 c1
Y
X
+ X-Y
+ X-Y
c c
ß- elimination
Alkene
c2 c1
Y
H
+ H-Y

3. Mechanism of 1,2 elimination
The 1,2 elimination reactions which takes place in basic medium .
Three things must be happened here-
a.Breaking of Cα –Y Bond : The substrate has a good leaving group Y.
During the reaction C-Y Bond breaks heterolytically so that the
leaving group loss with bonding pair of electrons.
b.Breaking of Cβ –H Bond : The base used abstracts the β-hydrogen as a
proton. During the reaction Cβ –H bond also breaks
heterolytically so that the bonding pair of electron is retained by
the β-carbon.
c.Formation of the Л bond between Cα-Cβ: The pair of electron
retained by the β-carbon is used to form a Л bond bet Cα-Cβ.
c c
1,2 elimination
Alkene
cß ca
Y
H
+ H-B+
+ Y-
B:

4. Three Possible Mechanisms
1) The E2 Mechanism: (Elimination Bimolecular)
The breaking of Cα–Y bond, formation of the Л bond between
Cα-Cβ and breaking of Cβ–H bond takes place simultaneously.
c c
Alkene
cß ca
Y
H
+ H-B+
+ Y-
B:
E2,Base
Slow
c c
Y
H
B
T.S.
Substrate
Rate α [Substrate] [Base]
This mechanism has some similarities with SN2 mechanism-
a.It is bimolecular.
b.All bond formation and bond breakings are concerted.
c.Proceed through single T.S.
d.It is one step process.

5. 2) The E1 Mechanism:
(Elimination Unimolecular)
The breaking of Cα–Y bond taks place in the slow step to
form a carbonium ion, which is then followed by fast breaking
of Cβ–H bond and formation of Cα-Cβ Л bond .
Rate α [Substrate]
This mechanism has some similarities with SN1 mechanism-
a.It is Unimolecular.
b.Formation of carbonium ion occurs in the slow step.
c.It is two step process.
c c
Alkene
cß ca
Y
H
+ H-B
E1
Slow
c c
H
Substrate Carbonium ion
+ Y
c c
H
Base
Fast
Carbonium ion
B:
(a)
(b)

6. 3) The E1cB Mechanism:
(Elimination Proceeding through Conjugate base)
The breaking of Cβ–H bond taks place first to form a carbanion
or conjugate base of the substrate, followed by simultenious
formation of Cα-Cβ Л bond and breaking of Cα–Y bond.
Rate α [Substrate] [Base]
E2 and E1 mechanisms are very commonly observed but E1cB
mechanism is very rarely observed
c c
Alkene
cß ca
Y
H
+ H-B
E1cB
Slow
c c
Substrate Carbanion
+ Y
Base
Fast
B:
(a)
(b)
Y
Carbanion
c c
Y

7. The E2 Mechanism
Consider the following 1,2 elimination –
H3C CH
Br
CH3 + NaoMe H2C CH CH3 + NaBr + MeOH
a ß
ß
2-bromopropane Propene
Kinetics :- Rate α [Substrate] [Base]
Rate =K2 [2-bromopropane] [NaOMe]
Mechanism:- It is one step mechanism. The leaving group i.e. halogen atom
lossed with bonding e pair and β–H lossed without bonding e pair, to form
the double bond. All these operations occur simultaneously, in a single step
via single T.S.
c c
H
H3C H
H
Alkene
ca cß
H
H
H
Br
H
H3C + MeOH + Br-
MeO Base
Slow
c c
H
H
H3C
H
T.S.
2-bromopropane
Br H
OMe

8. Transition State:-
In the T.S. two bonds are broken (C-H and C-Br), the energy
required for this bond breaking comes from bond making process
i.e. formation of bond between proton and the base (B-H) and
formation of a Л bond.
As the base begins to pull β–H from the substrate, the β carbon
with electron pair begins to form the Л bond.
As the Л bond starts to form, the carbon halogen bond starts to
break. The halogen atom (L.G.) lossed with bonding pair of electron
Evidence for E2 Mechanism
1) KINETIC ISOTOPIC EFFECT:-
If the rate of reaction is changed due to
replacement of some atom in the molecule by its isotope, it
is known as Kinetic Isotopic Effect or Primary isotopic effect.

9. Consider the rates of the following two reactions-
The two reactions were carried out under identical conditions.
When the rate constants were determined it was observed that KH /KD
was close to 7. It means that comp. (A) reacts seven times faster than
comp. (B) which contains D in place of H.
This clearly indicates that in E2 elimination, breaking of Cβ–H
bond must occur in the slow step.
If breaking of Cβ–H bond were not in the slow step, both
compound would react at same rate i.e. there will be absence of
Kinetic Isotopic Effect
c c
D
Ph H
H
cß ca
H
Br
H
D
Ph + EtOD + NaBr
EtO
D
KD
B)
NaOEt
c c
H
Ph H
H
cß ca
H
Br
H
H
Ph + EtOH + NaBr
EtO
H
KH
A)
NaOEt

10. 2) THE ELEMENT EFFECT:- Nature of Leaving group
 As the leaving group also departs in the rate determining step of E2,
changing the leaving group should change the E2 rate.
 With extremely good leaving groups, the E2 rates are expected to
be very high.
 Consider the relative E2 rates in the following reactions:-
R-CH2-CH2-X R-CH=CH2
The relative rates are determined by changing the leaving group only
Substrate Relative E2 rate
R-CH2-CH2-F 1
R-CH2-CH2-Cl 70
R-CH2-CH2-Br 4 X 103
R-CH2-CH2-I 2.5 X 104
As leaving group ability is in the order I- > Br -> Cl- > F- . The relative rates of
E2 elimination are also in the same order.
This proves that the C-X bond breaks in the rate determining step. If this
bond had broken in fast step then all above substrate reacted at same rate.
Base
E2

11. 3) ABSENCE OF HYDROGEN EXCHANGE:-
 The E2 & E1cB mechanism are both follow second order Kinetics . So Kinetically
they are indistinguishable.
 The reaction follows E2 mech. and not E1cB, Hydrogen exchange expt. is
performed.
 Ex. 2-phenyl ethyl bromide is subjected to E2 with NaOEt in EtOD as solvent.
Reaction is allowed to proceed until half of the reactant is consumed and then
stopped.
 Suppose that the reaction follow E1cB mech.-
C C
H
Ph H
H
cß ca
H
Br
H
H
Ph
EtO
H
a) Fast ß
c ca
H
Br
H
Ph
H
ß
c ca
H
Br
H
Ph
H
b)
Slow
2-phenyl ethyl bromide Carbanion
Styrene
ß
c ca
H
Br
H
Ph
H
C)
Et-O-D
cß ca
H
Br
H
D
Ph
H
EtOD
Recovery of starting material
With D In Place of H

12. • If the reaction had really followed E1cB
Mechanism, the unreacted substrate should
contain deuterium.
• In actual experiment it was found that the
unreacted substrate did not contain
deuterium i.e. there was no hydrogen
exchange.
• Therefore this reaction is not following E1cB.
• All the reactions which follow E2 mechanism,
fail to show hydrogen exchange.

13. ORIENTATION & REACTIVITY IN E2
 If the substrate contains more than one type of β-hydrogens, more
than one product will be formed.
R C
H
CH2
H X
ß
a
only one type of ß-H
Elimination
R C CH2
a)
b)
R C
H
C
H Br
ß a
Two types of ß-Hydrogens
Elimination
R C C
CH2
H H
ß
+ C C CH2
R
H
H
Only one elimination Products
Two elimination Products
H
H
CH3
H
H
 In such cases one of the alkene is formed as major product and the
other as minor.

14. THE SAYTZEFF RULE
• In order to decide which of the alkene will be obtained as major
product, Alecxander Saytzeff praposed a rule in 1875 known as
Saytzeff rule.
• It states that- “In an elimination reaction the more substituted alkene
is obtained as the major product”.
a)
H3C C
H
C
H Br
ß' a
H3C C C
CH2
H H
ß
+ C C CH2
H3C
H
H H
H
CH3
H
KOH
Alcoholic
2-bromo butane 2-butene
Major
1-butene
Minor
b)
H3C-H2C C
H
C
H Br
ß' a
H3C-H2C C C
CH2
H H
ß
+ C C CH2
H3C-H2C
H
H H
H
CH3
H
NaOEt
2-bromo pentane 2-pentene
Major
1-pentene
Minor
H3C C
H
C
H Br
CH2
CH3 H
c)
ß' ß
ß
a
NaOEt
H3C C C
CH3
CH3
H
C C CH2
H3C
H
H CH3
2-methyl-2-butene
Major
2-methyl-1-butene
Minor
+
2-bromo-2-methyl-butane
Disubstituted Alkene
Disubstituted Alkene
Disubstituted Alkene
Monosubtituted Alkene
Monosubtituted Alkene
Trisubtituted Alkene

15. JUSTIFICATION OF SAYTZEFF RULE
 Why the more substituted alkene obtained as major product ?
Two major factors contribute to this-
a) Stability of the Alkene :-
As the No. of Alkyl Substituent Stability of the Alkene
The stability of the alkene is measured In terms of the
heat liberated when the double bond undergoes addition of
hydrogen i.e heat of hydrogenation.
Lower the heat of hydrogenation higher is the stability of alkene
General stability order-
Tetra-substituted > Tri-substituted > 1,1Di-substituted > 1,2Di-
substituted > Mono-substituted > Ethylene
With increase in the order of substitution –
No. of ‘R’ group attached to the olefinic carbon +I effect
No. of hyperconjugative structures also Stability of the Alkene

16. b) Lower energy of activation:-
Consider the E2 transition state-
c c
Alkene
cß ca
Y
H
B:
E2,Base
Slow
c c
Y
H
B
T.S.
Substrate
 As the Number of Alkyl Substituents attached to the olefinic carbon
atom increase, the energy of activation required to form T.S.
decreases & hence more substituted alkene is formed at faster rate.
 Thus the more substituted alkene obtained as major product
because it is more stable and formed at faster rate due to lower
energy of activation

17. MODIFIED SAYTZEFF RULE
• Modified Statement- “In an elimination reaction the more stable
alkene is obtained as the major product”.
• The stability of alkene could also be due to Resonance stabilisation.
e.g.
C
H
C
H Cl
ß a
C C
CH2
H H
ß
+
C C CH2
H
H H
H
CH3
H
KOH
2-Chloro- 1-Phenyl Propane (A) Conjugated
Major
(B) Non-conjugated
Minor
Alcoholic
In product (A) the double bond is in conjugation with the benzene
ring and hence resonance will stabilize the product.
C C
H
CH3
H
C C
H
CH3
H
etc
Such stabilisation is not possible in (B) because of non-conjugation

18. Hofmann Elimination
 In case of some E2 elimination reactions instead of formation of
more substituted alkene as major product, we get formation of less
substituted alkene as the major product. Such a elimination
reaction is kn as Hofmann elimination.
 In all the E2 elimination reactions which leads to the formation of
more than one product-
If more substituted alkene is obtained as major product it is termed
as Saytzeff elimination and
If less substituted alkene is obtained as major product it is termed
as Hofmann elimination.
 Whether a given elimination reaction will follow the Saytzeff rule or
Hofmann rule depands upon-
1) the nature of leaving group and
2) the nature of base used.

19. The Effect of Leaving Group(X)
 Consider the following elimination reaction-
R C
H
C
H X
ß a
R C C
CH2
H H
ß'
+ C C CH2
R
H
H H
H
CH3
H
Base
 Effect of leaving group on elimination-
Leaving Group (X) % of Saytzeff product % of Hofmann product
i) –Br 80 20
ii) -OTs 60 40
iii) –S+(CH3)3 25 75
iv) –N+(CH3)3 05 95
 These results indicates that as the size of leaving group increases,
% of Hofmann product increases.
With smaller L.G. Saytzeff product always predominates and with
highly bulky L.G. Hofmann product predominates.

20. The Effect of Attacking Base
 Effect of Size of Attacking Base on elimination-
Base % of Saytzeff product % of Hofmann product
i) C2H5O- 70 30
ii) (CH3)3C- O- 28 72
iii) (CH3)2(C2H5)C- O- 22 78
iv) (C2H5)3C-O- 20 80
 As the size of Attacking Base increases, the T.S. for the Saytzeff
elimination becomes more crowded than the Hoffmann Elimination and
hence less substituted alkene is obtained as major product i.e. Hofmann
product predominates.
R C
H
C
H X
ß' a R C C
CH2
H H
ß
C C CH2
R
H
H H
H
CH3
H
B:
-ß'H R C
H
C
H X
CH2
H H
B
More crowded T.S.
Requires more energy
Saytzeff Product
Hofmann Product
R C
H
C
H X
CH2
H H
ß' a ß
B:
-ßH
Less crowded T.S.
Requires less energy
R C
H
C
H X
CH2
H H B
Saytzeff Elimination
Hofmann Elimination

21. The E1 Mechanism
 Consider the following elimination reaction-
H3C C
H
C
H Br
ß' a
CH2
CH3 H
ß
H3C C C
H
ß
2-bromo-2methyl butane
NaOEt
EtOH
CH3
CH3
2-methyl-2-butene
+ NaBr + EtOH
 Kinetics :- Rate α [Substrate]
Rate = K1 [2-bromo-2-methyl-butane]
 Mechanism :-
 The First step involves hetrolysis of Cα-Br bond to form carbocation
which is well stabilized by +I effect of alkyl groups.
This step is slow step and contains only one molecule and hence it
is unimolecular reaction.
 The Second Step is a fast step and involves abstraction of β-
hydrogen by base to form an alkene.

22. H3C C
H
C
H Br
ß' a
CH2
CH3 H
ß
H3C C C
H
ß
2-bromo-2methyl butane
NaOEt
+ EtOH
CH3
CH3
2-methyl-2-butene
H3C C
H
C
H
CH2
CH3 H
Carbocation
H3C C
H
C
H
CH2
CH3 H
Carbocation
Slow
Fast
+ Br
ß' a
Step-I
Step-II
EtO
 The carbocation formed in the first step may undergo following
reactions in the second step-
a. Combine with a nucleophile (NaOEt) to yield Sub. product by SN1.
b. Rearrange to a more stable carbocation (if possible).
c. Eliminate a β-hydrogen to form an alkene by E1 reaction .
H3C C C
H
SN1
CH3
CH3
H3C C
H
C
H
CH2
CH3 H
Carbocation
Fast
ß' a
EtO
H
OEt
Substitution product
+

23. Evidence of E1Mechanism
1) No Kinetic Isotopic Effect :-
• The isotope effect is not expected in E1, as the Cβ-H is lost in the fast
step and not in the slow step.
• When following expts. carried out, kinetic isotope effect was not
observed.
H3C C
Br
CH3
CH3
D2C C
NaOEt
CD3
CD3
C CH3
CH3
D3C C CD3
CD3
(I)
(II)
NaOEt
KH
H2C
Br
KD
• Experimentally it is shown that KH=KD This clearly indicates that ,
breaking of C-H/C-D bond does not takes place in the slow step or
rate determining step.

24. 2) Structural effect :-
• The first step in E1 is same as SN1 i.e. formation of carbocation. Therefore
the order of reactivity of alkyl halides must be the same as in SN1 i.e.
Reactivity in E1 is – 3o>2o>1o
• The rate of E1 as well as SN1 decreases as the stability of carbocation
decreases in the order- 3o>2o>1o
H3C C
Br
CH3
CH3
C CH3
CH3
(I)
H3C
H3C C
Br
CH3
H
C CH3
H
(II)
H3C
H3C C
H
Br
H
C H
H
(III)
H3C
3o
halide
2o
halide
1o
halide
3o
Carbocation
2o
Carbocation
1o
Carbocation
Rate of E1/SN1
Decreases
As Stability of
carbocation
Decreases

25. 3) Rearrangement :-
 If the structure of the substrate permits, the first order elimination
(E1) are accompanied by rearrangement.
 The primary carbocation rearranges to the 2o or 3o whereas the 2o
carbocation will rearrange to 3o carbocation
 Example-
H3C C
CH3
C
H OTs
ß' a
CH3
H
C C C
H
3-methyl-2-butyl tosylate H
H
2-methyl-2-butene
H3C C
CH3
C
H
CH3
H
Carbocation (2o
)
Slow
ß nBuOH ß
ß' a
H3C C C
H
CH3
CH3
-ß'H
-ßH
3-methyl-1-butene
CH3
H3C
H
A)Saytzeff product
B)Hofmann product

26. Conclusion:-Formation of rearranged product (C) clearly indicates
that the reaction did not follow E2 but proceed through formation of
carbocation i.e. E1 route.
2-methyl-2-butene
H3C C
CH3
C
H
CH3
H
Carbocation (2o
)
Less stable
H3C C C
H
CH3
CH3
-ßH
-ß'H
2-methyl-1-butene
A)Saytzeff product
C)Hofmann product
Hydride shift
H3C C
CH3
C CH3
H
H
Carbocation (3o
)
More stable
ß' ß
ß'
H2C C
CH3
CH2-CH3
 In addition to above elimination products A & B we also get a
rearranged elimination product by hydride shift as-

27. 4) Absence of β-H :-
If the substrate does not contain β-H atom, E2 elimination is not
possible at all. Formation of alkene takes place through
carbocation i.e. by E1 route only .
 Example- Neopentyl bromide does not react under E2 conditions
as it has no β-H. However, under E1 condition it does undergo
elimination as shown
2-methyl-2-butene
H3C C
CH3
C
CH3
H
H
Carbocation (1o
)
Less stable
H3C C C
H
CH3
CH3
-ßH
-ß'H
2-methyl-1-butene
A)Saytzeff product
B)Hofmann product
Methyl shift
H3C C
CH3
C CH3
H
H
Carbocation (3o
)
More stable
ß' ß
ß'
H2C C
CH3
CH2-CH3
H3C C
CH3
C
CH3
Br
H
H
EtOH
Neopentyl bromide

28. ORIENTATION
• The elimination by E1 always show strong Saytzeff orientation, i.e. when
more than one alkene can be formed, the highly branched or more stable
alkene is the preferred product.
• The reactivity of the substrate in E1 depends upon slow step & orientation
depends upon which β-H lost faster from the carbocation.
 Factors affecting E2 & E1
a)Structure of the substrate- As we proceed along the series 1o,2o,3o for
substrate, the reactivity by both E1 & E2 increases although for different
reasons.-
i) Reactivity by E2 increases because of the greater stability of the highly
branched alkene being formed.
ii) Reactivity by E1 increases because of the greater stability of
carbocation being formed in the rate determining step.
b)Nature of Base-We have seen that rate of E2 reaction depends upon
concentration as well as nature of base. In general E2 is favored by strong
and higher conc. of base.
The rate of E1 is independent of concentration as well as nature of base.

29. A) SN2 vs E2

30. THANK YOU