The overall rate equation for this reaction is: Rate = k[R-R-OH][H2O] Where k is the rate constant and [R-R-OH] and [H2O] are the concentrations of the reactants R-R-OH and H2O, respectively.

1. Elimination Reaction
Elimination reactions involve the loss of elements from the starting material to form
a new π bond in the product.
Alkyl halides (RX) undergo elimination with bronsted bases. The elements of HX are
lost and alkene are formed.
X, OH, OR, N2
+, N3, H2O+, NR3
+ and SR2
+
Leaving groups
Classification
 α-elimination (1,1)
 β- elimination (1,2)
 γ- elimination (1,3)
β- elimination
E1
E2
E1CB

2. Substitution and Elimination in alkyl halides
Br
Br
fast
H2O or OH
OH
t-butyl bromide
t-butanol
 Substitution on t-butyl bromide invariably follow SN1 mechanism. The
following reaction can not be speeded up by 1) changing the nucleophile from
H2O to OH-, 2) increasing the concentration of OH-.
With conc. NaOH
rate = k [t-BuBr]
Br
+ OH HOH Br rate = k [t-BuBr] [OH-]
t-butyl bromide
isobutene
(2-methylpropene)
Br
OHH
r=k[t-BuBr][OH-] HOHBr
Elimination occurs when the nucleophile attacks H instead of carbon
E2 (elimination bimolecular)

3. Elimination of t-butyl happens because here the Nu- (OH) is basic. Hydrogen is
not acidic but proton removal occurs because Br- is a good leaving group.
Rate = k2[t-BuBr][OH-]
O H
t-b u ta n o l
O H 2
B r
fa s t
B r
t-b u ty l b ro m id e
Nucleophilic substitution of t-BuOH with HBr
H2SO4 H + S
O O
OHO
OH
H
OH2
H
t-butanol in H2SO4 does not undergo substitution but elimination. HSO4
- is a weak
nucleophile.

4. Role of nucleophile in Elimination vs Substitution
Basicity
H
X
Attackhereleadstosubstitution
Attackhereleadstoelimination
Cl OEt
Cl
SN1
HOEt
EtOH
pKaofEtOH2=-7
-H+
Weak base: substitution
C l
E tO -
C l
H
O E t-
p K a E tO H = 1 6
Strong base: elimination

5. Size of the nucleophile
Getting at a more exposed hydrogen atom in an elimination reaction is much easier,
which means if we use a bulky and basic nucleophile elimination becomes preferred
over substitution.
B r
B rH
O H
K O H O H
S N 2
B r K O t-B u
E 2
H
B r
H
H
O
Small nucleophile: substitution
large nucleophile: elimination
Large alkyl substituent makes it hard for oxygen to attack at carbon but there is no
problem in attacking hydrogen.

6. Temperature
2 molecules 3 molecules
Elimination
2 molecules 2 molecules
Substitution
ΔS elimination > ΔS substitution
ΔG = ΔH − 𝑇ΔS
This equation says that a reaction in which ΔS is positive becomes more
favourable (ΔG is more –ve) at higher temperature. Elimination should therefore
be favoured at high temperature.
H
R
H
H
R
X
H
R
H
H
R
Base
H
R R
H
rds
Rate=k[alkyl halide]
E1 elimination
E1 describes an elimination reaction E in which the rate-determining step is
unimolecular and does not involve the base. The leaving group leaves in first step
and the proton is removed in a separate step.

7. E2 elimination
H
H
R
R
H
X
base
H
RR
H
Rate = k[Base] [alkyl halide]
E 2 describes an elimination that has a bimolecular (2) rate determining step that must
involve the base . Loss of leaving group is simultaneous with removal of the proton by
the base.
Factors that affect the nature of elimination
High conc. of base favours E2
Strong base favours E2 over E1
Substrate structure for E1
If the starting material is tertiary alkyl halide it would substitute only by SN1. But it
eliminate either by E1 (with weak base) or E2(with strong bases).
E1 occurs with substrates which ionise to give stable carbocations. Ex: tertiary,
benzylic alkyl halides, allylic etc .
Base

8. tertiary
X
R
R
X
Ar
R
X
OR
X
H
R
R
H
H
H
H
R
Ar
RO
H
H
R
Ar
R
RO
allylic
benzylic
hetero
substituted
R
H
R R
Secondary
R
X
Rprimary
Substrates that eliminate by E1
Substrates that may eliminate by E1
Substrate that never eliminate by E1
Substrates that can not eliminate by either mechanism
CH3XPhX
X
HO
MayalsoeliminatebyE2
Cannot
eliminate
byE2

9. Role of leaving group
Any good leaving group will lead a faster E1 and E2 reaction.
OH is never a leaving group in elimination reaction. It is converted to tosylate or
mesylate for elimination.
H3C
S
O
O
Cl S
OO
H3C Cl
Tosyl chloride Mesyl chloride
O H
S
OO
C l
H 3 C
O T S
t-BuoK/E2

10. Stereoselectivity of elimination
Stereoselectivity- Whether the resulting alkene is cis or trans
Regioselectivity- Where is double bond is located in the product
Ph
OH
Ph
HOHH
Only one alkene possible
Two regioisomeric alkene possible
OH
and/or
H+
trisubstituted
alkene
disubstitutedalkene
Two stereoisomeric alkenes possible
Ph
OHH+
Ph Ph
and/or
trans cis

11. E alkenes are thermodynamically stable than Z alkenes.
Ph
OHH+
Ph Ph
and/or
95% E alkene 5% Z alkene
H2SO4
HO
Ph
CH3
H
H
H
Ph
CH3
H2O
H
H
H H
Ph
CH3
H
H
H2O
Ph
CH3
Geometry of the product
depends on conformation
about this bond
The new π bond can be formed if the vacant p orbital of the carbocation and the
breaking of C-H bond are aligned parallel.
Ph
H
H
HCH3
lowenergy
intermediate
Ph
H
H
H3C H
Highenergy
intermediate

12. Regioselectivity in E1
OH
HBr,H2O
Major Minor
More substituted alkenes are more stable.
π-system of the double bond is stabilized when the empty π* antibonding orbital can
interact with filled σ bonding orbital of C-H bond. More the C-C, C-H bonds, more stable
is the alkene.
H
H
H
CH3
H
HCH3
H3C
H
H
H
H
H
H
HH
H
H
C
H
H
H
H
H
H
Increasing substitution allows more and more C-H and C-C bonds to interact with π*.
(Saytzev Product)

13. E2 is highly sterioselective
In E2 the new π bond is formed by overlap of σ bonding orbital of C-H and σ* orbital of
C-X bond. Two orbitals have to lie in the same plane for the best overlap.
H
X HX
There are two possible overlaps where H and X are coplanar.
H
X HX
X HX
H
Antiperiplanner
(staggered)
Syn-periplanner
Eclipsed
The anti-periplanar conformation is more stable because it is staggered. E2
elimination takes place preferentially from anti-periplanar conformation.

14. 81%but2-ene
19%
Br
H
H CH3
Br
CH3H
H
H3C H
Br
CH3H
major
minor
2-Bromobutane has 2 conformers. But the one with less hindered leads to the
more of product. Hence E isomer predominates.
Stereospecificity of E2
When there is only one proton to take part in elimination, there is no choice of
antiperiplanar TS. In that case E2 leads to the production of a single isomer as a
direct result of mechanism of the reaction and stereochemistry of starting
material.

15. H
Ph
Ph
CH3
Br
Ph
H3C
H
Ph
Br
H
OH
Ph
PhH3C
NaOH
Ph
Ph
Br
H3C
Ph
H
Ph
Br
H
OH
H3C
PhPh
NaOH
H3C
thisisomerEalkenethis isomer Zalkene
Elimination reactions of substituted cyclohexanes
X
H
H
X
OR
equatorial X is antiperiplannar
only to C-C bonds and can not be
eliminated by E2
ring invesrion
For E2 elimination both H and X must be axial

16. CH3
Cl
CH3 CH3
ratio of 1:3
A
Cl
NaOEt
250 times slower
B
???Most substituted is the
major product
Cl
CH3
H
Cl
CH3
OEt
A has Cl axial all the time ready for E2 where as B has Cl axial only in minute proportion of the
molecule that happen not to be in lowest energy state. Concentration of reactive molecule
is low so rate is low.
Cl
H
CH3
H
H
H
H
CH3
Cl
OEt
Reactivity of diastereomers A and B are different when treated with NaOEt.
NaOC2H5
NaOC2H5

17. Hofmann Elimination

18. Hoffmann elimination product takes place in the following four cases.
 Bulky base
 When leaving group is poor such as F-, NR3
+ , SR2
+.
 Steric hinderance at β carbon
 When alkyl halide contains one or more double bond.
Poor leaving
group
F
H
F
H
H2C
H
OCH3
F
F
H
O C H 3
δ-
δ-
Base begins to abstract proton before
leaving group leaves . –Ve charge
develops on carbon. Transition state has
carbanion character.
NaOCH3 NaOCH3
More
stable
Less
stable

19. Br
Base
Major Minor
Conjugated alkenes are more stable than nonconjugated alekenes.
Hence Hofmann product.

20. Regiselectivity in E2
Cl
KOCEt3
OH
H3PO4
120C
E1, more substituted alkene Less substituted alkene
with hindered base
Br
NaOC2H5t-BuOK
69%31%73%26%
Base attacks methyl – H because
they are less hindered
More substituted product
with less hindered base
 E1 reaction gives more substituted alkene.
 E2 reactions may give more substituted alkene, but become more regioselective
for the less substituted alkene with more hindered bases.

21. E1CB Elimination unimolecular conjugate base
OOH
H
pKa=20
acidic
KOH
O
The base can remove –H before the leaving group departs. The resulting anion is stable.
OOH
H
KOH
O
OH
OH
OOH O
The leaving group is not lost from starting material it is lost from the conjugate base.
Hence E1CB mechanism. Although OH is not a leaving group in E2 it can be a leaving
group in E1CB. Establishment of conjugation in the product assists loss of OH-.
rds

22. EX-1
CO2Et
OH
CO2Et
90%yield
mixtureofE:Zalkene2:1
MsCl
Et3N
OH O
OEt
H
Et3N
OEt
OOMs
CO2Et
Mechanism
Ex 2
O H
O
C O 2 E t
M s C l
P y r i d i n e
O
C O 2 E t
O M s
N
H
O
C O 2 E t
O M s
O
C O 2 E t
O M s
O
C O 2 E t
OH
O
CO2Et
MsCl
Pyridine
100% yield
O
CO2Et
mechanism

23. Q. Write the rate equation for the following reaction.
R R
O O H
O H
R R
O
R R
O O H
O H H 2 O
R R
O O H
ra te
c o n s ta n t = k
R R
Oe q u illib riu m
c o n s ta n t = KA.
K
R R
O O H
R R
O O H
H 2O
O H
therefore
R R
O O H
=
K
H 2O R R
O O H
O H
rate = k
K
H 2O
R R
O O H
O H = constant x
R R
O O H
O H
=