ELIMINATION REACTIONS

An elimination reaction is one where starting

material loses the elements of a small molecule
such as HCl or H2O or Cl2 during the course of
the reaction to form the product.
ELIMINATION REACTIONS
- HCl
C C C C
H Cl
TWO EXAMPLES FOLLOW

ELIMINATION REACTIONS ELIMINATION IS THE REVERSE OF ADDITION

E2 TWO EXAMPLES
interconversions of alkyl halides and alkenes
Alkyl halide + strong base and heat LOSS OF HCl basic conditions + heat

NaOH NaOH + heat

CH3CH2CH2CH2 Cl CH3CH2 CH CH2
∆ - HCl
CH3CH2 CH CH3 CH3 CH CH CH3
NaOH
CH3CH2 CH CH3 CH3 CH CH CH3 Cl + HCl
∆
Cl ALKYL HALIDE ALKENE
E1 conc. HCl
acidic conditions

Alcohol + strong acid and heat LOSS OF H2O

H2 SO4 H Cl
CH3CH2CH2CH2 OH CH3CH2 CH CH2 C C C C
∆
 ELIMINATION IS THE REVERSE OF ADDITION
interconversions of alcohols and alkenes

strong acid conditions + heat

conc. H2SO4

- H2O
CH3CH2 CH CH3
CH3 CH CH CH3
OH
ALCOHOL
+ H2O
ALKENE
STRONG BASES
2-6 M H2SO4
dilute aqueous acid conditions
FIRST WE MUST LEARN WHAT IS A STRONG BASE

H OH
C C C C

WHAT ARE STRONG BASES ? STRONG BASES (continued)

NaOH sodium hydroxide .. NaOR sodium alkoxides
used with H2O solvent, .. :
H-O always used with the parent alcohol as solvent,
most alkyl halides are soluble
but most RCl insoluble in H2O

2 Na + 2 H2O 2 NaOH + H2 (g) 2 Na + 2 ROH 2 NaOR + H2 (g)

examples: sodium methoxide

NaOCH3 (NaOMe)
KOH potassium hydroxide .. ..
.. :
R-O sodium ethoxide
soluble in H2O, methanol or ethanol,
most RCl soluble in methanol and ethanol
.. :
H-O
NaOCH2CH3 (NaOEt)

2 K + 2 H2O 2 KOH + H2 (g) sodium t-butoxide

Stronger bases
than hydroxides NaOC(CH3)3 (NaOtBu)
….. why?
 STRONG BASES (continued) SUMMARY OF STRONG BASES
USED FOR ELIMINATION REACTIONS

.. Base Solvents Allowed

NaNH2 sodium amide :NH2
always used with liquid ammonia NaOH water
solvent

2 Na + 2 NH3
FeCl3
2 NaNH2 + H2 KOH water, MeOH, EtOH
NH3 (liq)
-33 oC
NaOR ROH (same R group)
A stronger base than
A gas at room temp the hydroxides or the NaNH2 NH3 (liq) -33o C
alkoxides …. why?
ammonia
NH3 bp -33.4 oC liquifies Halides (RX) are not soluble in water, but are
mp -77.7 oC solidifies soluble in most alcohols, therefore, KOH or
sodium alkoxides in alcohol are most often used.

THE REACTION IS A β-ELIMINATION

LEARN THIS ! IT IS THE FORMULA
FOR A DEHYDROHALOGENATION The β-hydrogen
REACTION is attached to the
H
β-carbon. α-carbon
C C
ALKYL HALIDE + STRONG BASE β-carbon Cl The functional
+ HEAT group is attached
to the α-carbon.

shorthand designation
Since the β-hydrogen is lost this reaction is
E2 for this type of reaction
called a β-elimination.
defined later...

Reagent = a strong base

 MECHANISM ALKYL HALIDE + STRONG BASE (E2)

THE BASE TAKES THE β-HYDROGEN

B:
B H
H
C C C C REGIOSELECTIVITY
:Cl
.. :
..
: Cl
.. :

ELIMINATION IS REGIOSELECTIVE
WHAT HAPPENS IF THERE IS MORE
THAN ONE β-HYDROGEN ? β β’

CH3CH2 CH CH3 CH3 CH CH CH3

Br major product 2-butene
β-H
β β’ 2-bromobutane
81 %

H H CH3CH2 CH CH2

C C C β’-H
minor product 1-butene
The major product is 19 %
Cl the one which has the
lowest energy.

See the next slide.

Which one do we lose ?

 HAMMOND POSTULATE BUTENE ISOMERS - HEATS OF HYDROGENATION

lowest RECALL
energy Usually the pathway section 4.13
pathway
leading to the lowest
energy product is
the lowest energy +H2 +H2 +H2
pathway (lower TS). ∆H
-30.3 -28.6 -27.6 kcal / mole

…. but not always

(exceptions later)
CH3CH2CH2CH3
CH3CH2-CH=CH2 1-butene butane

All are hydrogenated to the

lowest
energy CH3-CH=CH-CH3 2-butene same product therefore their
energies may be compared.
product you get more trans than cis

MORE REGIOSELECTIVITY METHYLCYCLOHEXENE ISOMERS

1-chloro-1-methylcyclohexane methylenecyclohexane
1-methylcyclohexene CH2 RECALL
β’’ β’ CH3
β β’
CH3 CH3 CH2
NaOCH3
Cl + +H2 +H2
kcal
β
CH3 OH / ∆ ∆H mole -27.8 -25.4
major product minor product
three possibilities
to lose β -hydrogens
β’’
CH3
1-methylcyclohexene Both are hydrogenated
to the same product CH3
identical to therefore their energies
β’’= β β-product may be compared.
methylcyclohexane
 Different positions
ALKENE ISOMERS of the double bond.
Zaitsev
THE MORE SUBSTITUTED ISOMER IS MORE STABLE
Tsayseff
etc. decreasing energy
R H
1,1-
R H

SAYTZEV RULE R H R R R R R R
cis
H H H H 1,2- R H R R
R H
monosubstituted trans trisubstituted
1,2-
H R tetrasubstituted
The reaction gives the most highly-substituted
(lowest energy) alkene as the major product. disubstituted

increasing substitution

APPLICATIONS OF THE ZAITZEV RULE

ALKYL HALIDE + STRONG BASE (E2)
2-bromobutane

CH3CH2 CH CH3 CH3 CH CH CH3 +

Br
major product
81 % CH3CH2 CH CH2
DISUBSTITUTED
minor product
19 %
MONOSUBSTITUTED
STEREOCHEMISTRY
1-chloro-1-methylcyclohexane
OF THE REACTION
CH3 CH3 CH2
NaOCH3
Cl +
CH3 OH / ∆

major product minor product

TRISUBSTITUTED DISUBSTITUTED
 STEREOCHEMISTRY
TWO EXTREME POSSIBILITIES FOR THE ELIMINATION PROCESS

STEREOCHEMISTRY
C C syn elimination
H Cl not common
HCl
ACYCLIC HALIDES
H H
anti elimination
C C
observed
Cl most often
Cl anti-coplanar

ACYCLIC MOLECULES MAY HAVE TO ROTATE 2-Bromo-3-phenylbutane

IN ORDER TO REACT
two stereocenters RS SS enantiomers
SR RR
* * NaOEt
H CH CH CH3
EtOH diastereomers
CH3 Br

C CH CH3
*CH CH CH
+ 2

CH3 CH3
Cl H Cl anti-coplanar
Major Product Minor Product
Zaitsev

cis or trans ?
Is stereochemistry important?
 2S,3R-DIASTEREOMER MAKING THE 2S,3S-DIASTEREOMER
R S R S S S
CH3 CH3 CH3 CH3 Ph
CH3 CH3 CH3
C C C C C C R S C C H
Ph H Ph H CH3
Ph H
H Br H Br H Br
make
not observed
rotate diastereomer
H ( change one
H
CH3 NaOMe Ph CH3
Ph C C H
CH3 stereocenter )
? cis or
Ph C C H C C trans?
trans CH3 Br
CH3 MeOH CH3 H
Br (Z)
observed WHAT DO YOU THINK?
anti-coplanar major product Ph CH3
C C Will the 2S,3S-diastereomer
trans-2-phenyl-2-butene
CH3 H give the same product as
(plus some 1-butene
due to β-H on the CH3)
trans (Z) its 2S,3R diastereomer?

2S,3S-DIASTEREOMER CONVERTING THE ALKYL HALIDE TO AN ALKENE

VISUALIZING THE PRODUCT THAT FORMS
S S
Ph Ph CH3
CH3
CH3 C C H C C H (top)

CH3 H H
H Br CH3 CH3 CH3 CH3
rotate
not observed Ph H
Ph H
H CH3 CH3 Cl
CH3 NaOMe
CH3 C C H C C cis alkyl halide alkene Cl (bottom)
Ph MeOH Ph H
Br (E)
observed
anti-coplanar major product
The methyl groups (blue) are in back in both structures.
cis-2-phenyl-2-butene
The phenyl and the hydrogen (black) are in front in both.
NO A DIFFERENT PRODUCT IS (plus some 1-butene)
FORMED THAN WITH 2S,3R !
 ANOTHER VISUALIZATION OF THE REACTION
2S,3R back CH3 H
H carbon C
CH3 H
C
Ph CH3 front
carbon Ph CH3
Cl same same
side side

2S,3S CYCLIC HALIDES

H CH3 H
CH3 H C
STEREOCHEMISTRY REARS ITS UGLY HEAD AGAIN !
CH3 Ph C
Cl same CH3 Ph
side same
side

1-Bromo-2-methylcyclohexane THE CIS STEREOISOMER

The elimination needs to have H and Br anti-coplanar
The expected result (naïve) :

H major product (Zaitsev)

CH3 CH3 H
CH3
CH3
+ CH3
H
Br plus a small
Br amount of
major product minor cis The other chair CH3
drawn flat without Zaitsev won’t work. Why?
stereochemistry H
CH3

The result actually depends on the stereochemistry Br

of the starting material ( cis or trans ). trans shown on next slide
 THE TRANS STEREOISOMER THE RING MAY HAVE TO INVERT FOR REACTION

this ring
cannot react H anti-
CH3 CH3 Cl
H coplanar
invert
CH3
H H Cl H
H H
H H H
Br H
only product CH3
trans chlorine is not
The other chair
anti-coplanar KOH / EtOH
won’t work. Why?
to any hydrogen
CH3
CH3

Br CH3 H
no methylcyclohexene
is formed CH3
Cl

REACTION DOES NOT OCCUR EASILY IF

ANTI-COPLANAR GEOMETRY CANNOT BE ACHIEVED
A CASE THAT FOLLOWS THE
1 HAMMOND POSTULATE
trans e,e Br is not axial,
H no anti-coplanar H
CH3 NaOEt
Br no reaction
CH3 C EtOH HAMMOND POSTULATE
CH3
Br
NaOEt
CH3 CH3 The activation energy leading to the product
H EtOH of lower energy will be lower than the activation
CH3 C CH3 C
H energy leading to the product of higher energy
CH3 H hydrogens CH3
equivalent ….. unless stereochemistry or some other
cis e,a important factor interferes.
THESE RINGS WILL NOT INVERT
( WHYNOT ? )
REGIOSELECTIVE FOLLOWS HAMMOND POSTULATE *
AND
ZAITSEV A CASE THAT CAN NOT FOLLOW THE
H H H RULE 2 HAMMOND POSTULATE

CH3
Br
CH3 HAMMOND POSTULATE
E2 NaOEt / EtOH

OR….STEREOCHEMISTRY REARS
CH3 CH3 ITS UGLY HEAD !
major product
* The activation energy leading to the product
of lower energy will be lower than the activation
energy leading to the product of higher energy.

REGIOSPECIFIC
DOES NOT FOLLOW
incorrect HAMMOND
stereochemistry
H H CH3 raises Ea POSTULATE
OR
ZAITSEV RULE *

H
Br

E2 NaOEt / EtOH
CH3

CH3
only product CH3

* Due to stereochemical complications.

 SUMMARY TO DATE
E2 ELIMINATION REACTIONS OF ALKYL HALIDES
STRONG BASE Required HEAT Usually
required
REGIOSELECTIVE
ALKYL HALIDES + STRONG BASE + HEAT
Follows Zaitsev Rule (Unless Stereochemistry Prevents)
....... continued - favors most substituted alkene

STEREOSPECIFIC
β-H and X must be ANTI-COPLANAR
- acyclics may have to rotate
- rings may have to invert

RATE EXPRESSIONS
exponents
n m
RATE = K [A] [B]

concentrations of reactants
rate “constant”
A and B in moles / liter
Actually will change with
KINETICS temperature and solvent,
the specific molecule, etc.

REACTION ORDER = SUM OF EXPONENTS = n + m

Examination of the rate expression often helps
to understand the mechanism.
Not all reactants will necessarily show up in the rate expression.
The rate expression is determined by experiment. 3/2 2
Fractional rate orders are possible : RATE = K [A] [B]
 KINETICS MECHANISM
CH3CH2CH2 CH CH3 + NaOCH3 Elimination
35oC
Bimolecular
Br
CH3CH2 CH CHCH3 + CH3OH strong
B:
base E2
Rate = + d (CH3CH2 CH CHCH3) H
REACTION RATE DATA dt
Rate [RBr] [OCH3]
CH CH
This equation explains the alkyl
x1 x1 x1 rate behavior : Br halide
x2 x2 x1
x2 x1 x2 Rate = K [RBr] [OCH3]
x4 x2 x2 CONCERTED = only one step
x8 x4 x2 All bonds are broken and formed without
x9 x3 x3 second order rate the formation of any intermediates.

Reaction Order
Sum of the exponents of the concentration terms
in the rate expression.
Molecularity
E2
Number of species that come together in the
rate-determining step.

Rate-determining Step
CONCERTED REACTION
The slowest step in the reaction sequence.

Transition State
An energy highpoint in the energy profile
of a reaction. One Step - No Intermediates

Activated Complex
The species that exists at the transition state.
 E2 ELIMINATION Concerted (one step) reaction

δ− transition
B: B state TS

H E
H N activation
E energy Ea
R
CH CH CH CH G
Y heat of
Br starting reaction
δ− material ∆H
activated Br
mechanism
complex product
Concerted : everything
happens at once with- This is what E2 looks like.
out any intermediates.

ORBITAL ALIGNMENT .. When these electrons enter the

back lobe of the adjacent orbital
CH3 O : they “push” the bonding pair out
IN MOST CONCERTED REACTIONS THE
ORBITALS BECOME PREALIGNED FOR .. the other end (along with Br).

A SMOOTH PROGRESSION OF EVENTS

H
The anti-coplanar arrangement of the β-H and The critical event is sp 3
.. R
the halide leaving group X places the orbitals the removal of the β -H.
that undergo change in a perfect alignment.
H C C H
The coplanar arrangement allows a continuous
movement of electrons from one end of the system R
to the other, much like a stack of dominoes each
sp3
..
pushing the next one over. :Br
.. :
The two orbitals that will form the pi bond are The attack of the base on the
already parallel (anti-coplanar) so that the double β-hydrogen starts the reaction.
Notice the parallel aligment
can form easily. of the two sp3 orbitals.
 Note the parallel orbitals

.. in the pi bond.

CH3 O
.. H
2p
. . 2p

H C C R
FRONTIER MO THEORY
R H
.. The LUMO is present on the β-H only in the
The formation of the double bond
and the loss of bromide finish it.
:Br
.. : anti -coplanar arrangement.

ANTI CONFORMATION SYN CONFORMATION

LUMO has RECALL :
- density on H Frontier theory
B: -
LUMO
requires a base
B: HOMO or nucleophile
to add to the
LUMO.

-
B:

LUMO has
no density
on any H DENSITY-ELPOT
LUMO
negative end
of molecule DENSITY-ELPOT LUMO
 ISOTOPES OF HYDROGEN
NAME SYMBOL MASS COMPOSITION

PROTIUM H 1 1 proton + 1 electron

KINETIC ISOTOPE EFFECT 99.985%

DEUTERIUM D 2 1 proton + 1 neutron

+ 1 electron
0.015%
The reaction slows if the β-H is replaced by D.

This “kinetic isotope effect” shows that breaking TRITIUM

radioactive
T 3 1 proton + 2 neutrons
+ 1 electron
the β C-H bond is a part of the rate-determining ( β- ) 12.26 yrs
step.

ISOTOPE EFFECT
B B
H C-H D C-D

CH CH CH CH

ORIGIN OF THE ISOTOPE EFFECT

Br Br
C-D bond is stronger than C-H
kH Slows the reaction The effect is due to differences
approx. 5-8 if breaking this bond in C-H and C-D bond strengths.
kD is part of the
for an isotope effect rate-determining step.
 BONDING CURVE (D) 1s C-D BONDS ARE STRONGER THAN C-H BONDS
C(sp3) (H)1s
++ vibrational
energy + +
levels bond breaks
here
E r E
N N
E E
R + + R
G G
Y Y
+ = nucleus
bond dissociation
Since D is heavier than H the
zero-point CH CH energies (CD > CH)
C-D bond vibrates slower over
energy
CD a shorter distance. CD
oo oo
r (bond distance) r (bond distance)
D is heavier than H and the
average bond + + CD bond vibrates more slowly
length
over a shorter distance than CH.

Three types of elimination reactions are conceivable

B:
H
concerted
C C C C E2
X just studied

B: carbocation
H H
OTHER ELIMINATION MECHANISMS C C C C C C E1
+
X halogen proton
first second
B:
carbanion
H
C C C C C C E1cb
X proton X halogen
first second
 Three types of elimination reactions are conceivable

B:
H just
concerted studied
C C C C E2
X

carbocation
ELIMINATION H
B:
H
E1
OTHER POSSIBLE MECHANISMS C C C C C C
halogen + proton
X first second

B: carbanion
H
Do some elimination reactions E1cb
occur in a different fashion? C C C C C C
proton halogen
X first X second

The E1 Elimination Reaction (two steps)

E1 weak
base B: carbocation

H H
slow
C C C C + :X
ALKYL HALIDES + WEAK BASE X
step one +
(SOLVOLYSIS) 3o > 2o > 1o
also favored
unimolecular step
two fast if a resonance
stabilized
rate = k [RX] carbocation
is formed
The removal of a β-hydrogen becomes difficult without
a strong base and a different mechanism (ionization) C C
begins to take place
Works best in a
….. if the substrate is capable. IONS
polar solvent.
FORMED
 ENERGY PROFILE
two step reaction
carbocation
E1
intermediate
TS1
E TS2
N
E Ea2
R
G Ea1
STEREOSPECIFICITY
Y

starting step 1 step 2

slow ∆H
material

product

THE E1 REACTION IS NOT STEREOSPECIFIC

( THE OPEN CARBOCATION IS PLANAR AND CAN ROTATE)

H +
Carbocation is

anti H sp2 hybridized

( planar )

C C
and can react

C C rotation
from either side.

X These two REGIOSELECTIVITY

carbocations
syn are equivalent
by rotation and

C C C C
by symmetry.

H X H + rate of C-C rotation

= 1010 to 1012 / sec
Elimination can be either syn or anti.
 E1 REACTION IS REGIOSELECTIVE

THE ZAITSEV RULE IS FOLLOWED

(stereochemistry is not a problem as in E2)

very dilute base

CH3
0.001 M CH3 CH2 DIFFERENCES BETWEEN
Br KOH / EtOH + E1 AND E2
major minor
tertiary trisubstituted disubstituted
Zaitsev

BEHAVIOR OF THE RATE EFFECT OF BASE CONCENTRATION ON E1/E2 REACTIONS

WITH INCREASING BASE CONCENTRATION
k2 [RX] [B]
E2
rate = k2 [RX] [B] At high base concentration E2
second order E1 never has a chance.

E1 dominates
at low base At low base concentration
concentration E2 dominates E2 is nonexistent
at higher base
k1 [RX]
concentration tertiary RX, k’’
1

rate = k1 [RX] secondary RX, k’1 E1

first order Rate
Rate E1 primary RX, k 1

[RX] constant, [Base] increasing

[RX] constant, [B] increasing
For E1 elimination : k’’ (tertiary) > k’ (secondary) > k (primary).
EFFECT OF BASE CONCENTRATION ON E1/E2 REACTIONS
tertiary
STRUCTURE OF SUBSTRATE
secondary
k’’2
R H H
E2
k’2
R-C-X R-C-X R-C-X
k2
k2 [RX] [B] primary R primary
R secondary
H tertiary

Obviously for E1 which forms a carbocation intermediate

k1 [RX] rate : tertiary > secondary > primary > methyl
tertiary RX, k’’
1

Rate
secondary RX, k’1 E1 But this same order holds for E2 also.
primary RX, k
H
H H
tertiary has more H C more opportunites
[RX] constant, [B]
β -hydrogens H C C Br for reaction
For E2 elimination : line slopes k2 differ for 1o,2o,3o . H C
H H
Different substrates react at different rates, EtO- H

WHEN THE E1 MECHANISM OCCURS ALKYL HALIDE + BASE

E1 occurs only strong base weak base

high base conc. low base conc.
1) at zero or low base concentration or
2) with solvolysis (the solvent is the base) solvolysis
3) with tertiary and resonance capable (solvent is base)
substrates (alkyl halides)
E2 mechanism E1 mechanism
If a strong base is present in moderate anti-coplanar must be able to make
to high concentration, or the substrate requirement “good” carbocation
is a primary halide, the E2 reaction stereospecific not stereospecific
dominates. regioselective regioselective
EXAMPLES
rate = k [RBr] [OEt]
..
:OEt
..
6M
E2
NaOEt
CH CH3 CH CH2
EtOH
Br
SOLVOLYSIS
0.01 M E1
KOH
CH CH3 CH CH3
EtOH +
Br
The solvent is the thing !
rate = k [RBr]

SOLVOLYSIS
MANY E1 REACTIONS ARE SOLVOLYSIS REACTIONS

SOLVOLYSIS = THE SOLVENT IS THE REAGENT (BASE)

CH3
EtOH CH3 CH3
Cl
∆ + OEt

E1
competing product SOMETIMES E1 AND E2 RESULTS DIFFER
EtOH adds to the
CH3 carbocation
CH3
+ O Et + O Et
EtOH solvent
H
H H acts as base - H
H H
no other base
is present
 A COMPARISON OF E1 AND E2
major product
E2
syn
H Anti-Zaitsev
NaOEt
Br H H stereospecific
EtOH / ∆ anti
H
H CH3 CH3

EtOH / ∆
E1
anti
Zaitsev
not
CH3 stereospecific
E1 doesn’t require
anti-coplanarity