Substitution and Elimination
Reaction of Alkyl Halides
By: Ismiyarto, MSi
1
ALKIL HALIDA
1. Manfaat (Pestisida, Bahan
Sintesis Alkohol, Alkena)
2. Struktur (Metil, Primer,
Tersier, Benzil dan Vinil)
Dasar
Sekunder,
3. Reaksi (SN-2, SN-1, E-2 dan E-1)
2
PETA REAKSI ALKIL HALIDA
1.
2.
3.
4.
Metil Halida
Alkil halida Primer
Alkil Halida Sekunder
Alkil Halida Tersier
5.
6.
Alil Halida
Benzil Halida
SN-2
SN-2
SN-2, SN-1 dan E-2
SN-2, SN-1 dan E-2
SN-2, SN-1
SN-2, SN-1
7. Vinil Halida
8. Aril Halida
Dalam Pembahasan Tersendiri
3
Organic compounds with an
electronegative atom or an
electron-withdrawing group
bonded to a sp3 carbon
undergo substitution or
elimination reactions
Halide ions are good leaving
groups. Substitution reaction
on these compounds are easy
and are used to get a wide
variety of compounds
alkyl fluoride
-
alkyl chloride
alkyl bromide
alkyl iodide
4
Alkyl Halides in Nature
Synthesized by red algae
red algae
Synthesized by sea hare
a sea hare
5
Substitution Reaction with Halides
(1)
bromomethane
If concentration of (1) is
doubled, the rate of the
reaction is doubled.
If concentration of (2) is
doubled, the rate of the
reaction is doubled.
(2)
methanol
If concentration of (1) and
(2) is doubled, the rate of
the reaction quadruples.
6
Substitution Reaction with Halides
(1)
(2)
bromomethane
methanol
Rate law:
rate = k [bromoethane][OH-]
this reaction is an example of a SN2 reaction.
S stands for substitution
N stands for nucleophilic
2 stands for bimolecular
7
Mechanism of SN2 Reactions
Alkyl halide
The rate of reaction depends on the
concentrations of both reactants.
When the hydrogens of bromomethane
are replaced with methyl groups the
reaction rate slow down.
The reaction of an alkyl halide in which
the halogen is bonded to an asymetric
center leads to the formation of only
one stereoisomer
Relative rate
1200
40
1
≈0
8
Mechanism of SN2 Reactions
Hughes and Ingold proposed the following mechanism:
Transition state
Increasing the concentration of either of the
reactant makes their collision more probable.
9
Mechanism of SN2 Reactions
Steric effect
Energy
activation
energy: DG2
activation
energy: DG1
reaction coordinate
reaction coordinate
Inversion of configuration
(S)-2-bromobutane
(R)-2-butanol
10
Factor Affecting SN2 Reactions
The leaving group
- + RCH I
2
HO
HO
HO
HO
+ RCH2Br
+ RCH2Cl
+ RCH2F
-
RCH2OH + I
RCH2OH + Br
RCH2OH + Cl
RCH2OH + F
relative rates of reaction
30 000
10 000
200
1
pKa HX
-10
-9
-7
3.2
The nucleophile
In general, for halogen substitution the
strongest the base the better the
nucleophile.
pKa
Nuclephilicity
11
SN2 Reactions With Alkyl Halides
an alcohol
a thiol
an ether
a thioether
an amine
an alkyne
a nitrile
12
Substitution Reactions With Halides
1-bromo-1,1-dimethylethane
1,1-dimethylethanol
Rate law:
If concentration of (1) is
doubled, the rate of the
reaction is doubled.
If concentration of (2) is
doubled, the rate of the
reaction is not doubled.
rate = k [1-bromo-1,1-dimethylethane]
this reaction is an example of a SN1
reaction.
S stands for substitution
N stands for nucleophilic
1 stands for unimolecular
13
Mechanism of SN1 Reactions
Alkyl halide
Relative rate
The rate of reaction depends on the
concentrations of the alkyl halide only.
≈0*
When the methyl groups of 1-bromo1,1-dimethylethane are replaced with
hydrogens the reaction rate slow down.
≈0*
The reaction of an alkyl halide in which
the halogen is bonded to an asymetric
center leads to the formation of two
stereoisomers
12
1 200 000
* a small rate is actually observed as a result of a SN2
14
Mechanism of SN1 Reactions
nucleophile attacks the
carbocation
slow
C-Br bond breaks
fast
Proton dissociation
15
Mechanism of SN1 Reactions
Rate determining step
DG
Carbocation
intermediate
R++ X+
R-OH2
R-OH
16
Mechanism of SN1 Reactions
Inverted
configuration relative
the alkyl halide
Same configuration
as the alkyl halide
17
Factor Affecting SN1 reaction
Two factors affect the rate of a SN1 reaction:
• The ease with which the leaving group dissociate from the carbon
• The stability of the carbocation
The more the substituted the
carbocation is, the more
stable it is and therefore the
easier it is to form.
As in the case of SN2, the
weaker base is the leaving
group, the less tightly it is
bonded to the carbon and the
easier it is to break the bond
The reactivity of the
nucleophile has no effect on
the rate of a SN1 reaction
18
Comparison SN1 – SN2
SN1
SN2
A two-step mechanism
A one-step mechanism
A unimolecular rate-determining step
A bimolecular rate-determining step
Products have both retained and
inverted configuration relative to the
reactant
Product has inverted configuration
relative to the reactant
Reactivity order:
3o > 2o > 1o > methyl
Reactivity order:
methyl > 1o > 2o > 3o
19
Kestabilan Karbokation
H
H
+
H
H
H
propan-2-ylium
H
H
+H C
2
H
H
CH3+
H
Ethanylium
Methanylium
20
21
Elimination Reactions
1-bromo-1,1-dimethylethane
2-methylpropene
Rate law:
rate = k [1-bromo-1,1-dimethylethane][OH-]
this reaction is an example of a E2 reaction.
E stands for elimination
2 stands for bimolecular
22
The E2 Reaction
A proton is
removed
Br- is eliminated
The mechanism shows that an E2
reaction is a one-step reaction
23
Elimination Reactions
1-bromo-1,1-dimethylethane
2-methylpropene
Rate law:
If concentration of (1) is
doubled, the rate of the
reaction is doubled.
rate = k [1-bromo-1,1-dimethylethane]
If concentration of (2) is
doubled, the rate of the
reaction is not doubled.
this reaction is an example of a E1
reaction.
E stands for elimination
1 stands for unimolecular
24
The E1 Reaction
The base
removes a proton
The alkyl halide
dissociate, forming a
carbocation
The mechanism shows that an E1
reaction is a two-step reaction
25
Products of Elimination Reaction
30%
2-bromobutane
50%
80%
2-butene
20%
1-butene
The most stable alkene is the
major product of the reaction
for both E1 and E2 reaction
For both E1 and E2 reactions, tertiary alkyl halides
are the most reactive and primary alkyl halides
are the least reactive
The greater the number of
alkyl substituent the more
stable is the alkene
26
ELIMINATION REACTIONS:
ALKENES, ALKYNES
27
Elimination Reactions
Dehydrohalogenation (-HX) and
Dehydration (-H2O) are the main types of
elimination reactions.
C C
C C
+
X Y
X Y
28
Dehydrohalogenation (-HX)
H
C
C
strong
C
C
+
"H
X"
base
X
X = Cl, Br, I
29
The E2 mechanism
This reaction is done in strong base at high
concentration, such as 1 M NaOH in water.
H_
H
.. __
O:
..
..
O:
H
H
C
+
C
C
C
Br
_
+
Br
concerted mechanism
30
Kinetics
• The reaction in strong base at high concentration is
second order (bimolecular):
Rate law: rate = k[OH-]1[R-Br]1
31
The E1 mechanism
This reaction is done in strong base such as 0.01 M NaOH
in water!! Actually, the base solution is weak!
1)
H
C
H
slow
C
rate determining
step
Br
H
2)
H
C
C
+
_
Br
+
H
..
O:
fast
H
C
C
+
.. +
O
H
H
+
C
C
32
Kinetics
• The reaction in weak base or under neutral
conditions will be first order (unimolecular):
• Rate law: rate = k [R-Br]1
• The first step (slow step) is rate determining!
33
The E2 mechanism
•
•
•
•
•
•
Mechanism
Kinetics
Stereochemistry of reactants
Orientation of elimination (Zaitsev’s rule)
Stereochemistry of products
Competing reactions
34
E2 mechanism
This reaction is done in strong base at high
concentration, such as 1 M NaOH in water.
H
H
.. __
O:
..
..
O:
H
H
C
+
C
C
C
Br
_
+
Br
concerted mechanism
35
Kinetics of an E2 reaction
• The reactions are second order (bimolecular
reactions).
• Rate = k [R-Br]1[Base]1
second order reaction (1 + 1 = 2)
High powered math!!
36
Transition State
H
.. dO:
H
C
C
Br
d-
energy
H
.. __
O:
..
H O
H
C
C
Br
C
H
C
Br-
Reaction coordinate
37
Stereochemistry of reactants
• E2 reactions must go by an anti elimination
• This means that the hydrogen atom and
halogen atom must be 180o (coplanar) with
respect to each other!!
• Draw a Newman projection formula and place
the H and X on opposite sides.
38
Stereochemistry of E2 Reaction
CH3 C
CH3
CH3
Br
CH3
H
H
H
H
KOH
Alcohol
Solvent
CH3 C
CH3
H
H
H
H and Br are anti structure in conformation!!!!!!!!!
39
(S,S)-diastereomer
CH3
Br
CH3
C
H
t-butyl
C
H
KOH
ethanol
heat
???
CH3
CH3
C
H
???
H
C
CH3
C
t-butyl
(E)-isomer
CH3
C
t-butyl
(Z)-isomer
40
This one is formed!
CH3
CH3
C
H
C
T-butyl
(E)-isomer
41
(R,S)-diastereomer
CH3
Br
C
H
CH3
t-butyl
C
H
KOH
ethanol
heat
???
CH3
CH3
C
H
???
H
C
CH3
C
T-butyl
(E)-isomer
CH3
C
t-butyl
(Z)-isomer
42
This one is formed!
H
CH3
C
CH3
C
t-butyl
(Z)-isomer
43
Orientation of elimination:
regiochemistry/ Zaitsev’s Rule
• In reactions of removal of hydrogen halides from
alkyl halides or the removal of water from alcohols,
the hydrogen which is lost will come from the more
highly-branched b-carbon.
More branched
H H H

b 
H C C C
CH3
H
X
H
b
C H
Less branched
A. N. Zaitsev -- 1875
H
44
Product formed from previous slide
H
H
C
H
H
C
CH3
H
C
C
H
H
More substituted alkene is more stable!!!!!!!!
45
Typical bases used in E2 reactions
High concentration of the following >1M
If the concentration isn’t given, assume
that it is high concentration!
• Na+ -OH
• K+ -OH
• Na+ -OR
• Na+ -NH2
46
Orientation of elimination:
regiochemistry/ Zaitsev’s Rule
Explaination of Zaitsev’s rule:
When you remove a hydrogen atom from the
more branched position, you are forming a
more highly substituted alkene.
47
Stereochemistry of products
• The H and X must be anti with respect to each
other in an E2 reaction!
• You take what you get, especially with
diastereomers! See the previous slides of the
reaction of diastereomers.
48
Competing reactions
• The substitution reaction (SN2) competes with
the elimination reaction (E2).
• Both reactions follow second order kinetics!
49
The E1 mechanism
• Mechanism
• Kinetics
• Stereochemistry of reactants
• Orientation of elimination (Zaitsev’s rule)
• Stereochemistry of products
• Competing reactions
50
E1 mechanism
This reaction is done in strong base at low
concentration, such as 0.01 M NaOH in water)
1)
H
C
H
slow
C
water helps
to stabilize
carbocation
Br
H
2)
H
C
C
+
H
..
O:
H
fast
H
C
C
+
_
Br
+
..+
O
H
+
C
C
51
E1 Reactions
• These reactions proceed under neutral
conditions where a polar solvent helps to
stabilize the carbocation intermediate.
• This solvent also acts as a weak base and
removes a proton in the fast step.
• These types of reactions are referred to as
solvolysis reactions.
52
• tertiary substrates go by E1 in polar
solvents, with little or no base present!
• typical polar solvents are water, ethanol,
methanol and acetic acid
• These polar solvents help stabilize
carbocations
• E1 reactions also occur in a low
concentration of base (i.e. 0.01M NaOH).
53
However!!!!
•With strong base (i.e. >1M), goes by E2
54
Structure of the Carbocation
Intermediate
CH3
C
CH3
CH3
55
Carbocation stability order
Tertiary (3o) > secondary (2o) > primary (1o)
It is hard (but not impossible) to get primary
compounds to go by E1. The reason for this is
that primary carbocations are not stable!
56
Kinetics of an E1 reaction
• E1 reactions follow first order (unimolecular)
kinetics:
Rate = k [R-X]1
• The solvent helps to stabilize the carbocation,
but it doesn’t appear in the rate law!!
57
dBr
d+
C C
H d+
C
H
C
d+
+
C C
energy
H
intermediate
Br
C C
H
C
C
+ H+
Reaction coordinate
58
Stereochemistry of the reactants
• E1 reactions do not require an anti coplanar
orientation of H and X.
• Diastereomers give the same products with E1
reactions, including cis- and trans products.
• Remember, E2 reactions usually give different
products with diastereomers.
59
Orientation of elimination
• E1 reactions faithfully follow Zaitsev’s rule!
• This means that the major product should be
the product that is the most highly
substituted.
60
Stereochemistry of products
E1 reactions usually give the
thermodynamically most stable product as the
major product. This usually means that the
largest groups should be on opposite sides of
the double bond. Usually this means that the
trans product is obtained.
61
Competing reactions
• The substitution reaction (SN1) competes
with the elimination reaction (E1).
• Both reactions follow first order kinetics!
62
Whenever there are carbocations…
• They can undergo elimination (E1)
• They can undergo substitution (SN1)
• They can rearrange
– and then undergo elimination
– or substituion
63
Rearrangements
• Alkyl groups and hydrogen can migrate in
rearrangement reactions to give more stable
intermediate carbocations.
• You shouldn’t assume that rearrangements
always occur in all E1 reactions, otherwise
paranoia will set in!!
64
Comparison of E2 / E1
• E1 reactions occur under essentially neutral
conditions with polar solvents, such as water,
ethyl alcohol or acetic acid.
• E1 reactions can also occur with strong bases,
but only at low concentration, about 0.01 to
0.1 M or below.
• E2 reactions require strong base in high
concentration, about 1 M or above.
65
Comparison of E2 / E1
• E1 is a stepwise mechanism (two or more);
Carbocation intermediate!
• E2 is a concerted mechanism (one step)
No intermediate!
• E1 reactions may give rearranged products
• E2 reactions don’t give rearrangement
• Alcohol dehydration reactions are E1
66
Bulky leaving groups
Hofmann Elimination
This give the anti-Zaitsev product (least substituted
product is formed)!
b  b
_
CH3 CH2 CH2 CH CH3 OH
CH N+ CH
3
heat
CH3 CH2 CH CH CH3
6%
3
CH3
+
CH3 CH2 CH2 CH CH2
94%
67
Orientation of elimination:
regiochemistry/ Hofmann’s Rule
• In bimolecular elimination reactions in the presence
of either a bulky leaving group or a bulky base, the
hydrogen that is lost will come from the LEAST
highly-branched b-carbon.
More branched
H H H

b 
H C C C
CH3
H
X
Less branched
H
b
C H
H
68
Product from previous slide
H
H
C
H
H
C
CH3
H
C
C
H
H
69
Elimination with bulky bases
• Non-bulky bases, such as hydroxide and
ethoxide, give Zaitsev products.
• Bulky bases, such as potassium tertbutoxide, give larger amounts of the least
substituted alkene (Hoffmann) than with
simple bases.
70
Comparing Ordinary and Bulky
Bases
H
CH3 C
CH
CH3
NaOC2H5
C2H5OH
heat
CH3 Br
H
CH3 C
CH3
C
CH
CH3
Major
CH
CH2
Major
CH3
H
CH
CH3 Br
CH3
KOC(CH3)3
(CH3)3COH
heat
CH3 C
CH3
71
1-butene: watch out for
competing reactions!
H3C
CH2
CH2
CH2
O-CH3
Non-bulky
KOCH 3
H3C
CH2
CH2
CH2
SN2
Br
bulky base
KO-t-butyl
E2
H3C
CH2
CH
CH2
72
Highlights
•
•
•
•
•
Dehydrohalogenation -- E2 Mechanism
Zaitsev’s Rule
Dehydrohalogenation -- E1 Mechanism
Carbocation Rearrangements -- E1
Elimination with Bulky Leaving Groups and Bulky
Bases -- Hofmann Rule -- E2
73
Competition Between
SN2/E2 and SN1/E1
SN1
SN2
E1
E2
rate = k1[alkyl halide] + k2[alkyl halide][nucleo.] + k3[alkyl halide] + k2[alkyl halide][base]
• SN2 and E2 are favoured by a high concentration of a good
nucleophile/strong base
• SN1 and E1 are favoured by a poor nucleophile/weak base, because a
poor nucleophile/weak base disfavours SN2 and E2 reactions
74
Competition Between
Substitution and Elimination
• SN2/E2 conditions:
In a SN2 reaction: 1o > 2o > 3o
In a E2 reaction: 3o > 2o > 1o
10%
90%
75%
25%
100%
75
Competition Between
Substitution and Elimination
• SN1/E1 conditions:
All alkyl halides that react under SN1/E1 conditions will give
both substitution and elimination products (≈50%/50%)
76
Summary
• Alkyl halides undergo two kinds of nucleophilic subtitutions:
SN1 and SN2, and two kinds of elimination: E1 and E2.
• SN2 and E2 are bimolecular one-step reactions
• SN1 and E1 are unimolecular two step reactions
• SN1 lead to a mixture of stereoisomers
• SN2 inverts the configuration od an asymmetric carbon
• The major product of a elimination is the most stable alkene
• SN2 are E2 are favoured by strong nucleophile/strong base
• SN2 reactions are favoured by primary alkyl halides
• E2 reactions are favoured by tertiary alkyl halides
77
REAKSI ADISI ALKENA
78
Addition Reaction of Alkene
1. HX Addition
• Electrophilic Addition (Markovnikov Product)
• Free Radical Mechanism (Anti-Mark Product)
2. Hydration (+ H2O)
3. Halogenation/ Hydrohalogenation
4. Reduction or Hydrogenation (+ H2 )
5. Oxidation
6. Multi-step Synthesis
Addition of Halogens to Alkenes
C
C
+ X2
X
C
C
• electrophilic addition to double bond
• forms a vicinal dihalide
X2 = Cl2 or Br2
F2; explosive I2 ; endothermic
X
Example
Br2
CH3CH
CHCH(CH3)2
CH3CHCHCH(CH3)2
Br
Br
(100%)
Stereochemistry of Halogen Addition
•anti addition
H
Br2
H
Br
Br
H
H
trans-1,2-Dibromocyclopentane
80% yield; only product
Anti Addition ; Two Bromines add to opposite
sides of the ring
Example
H
H
Cl
Cl2
H
H
Cl
trans-1,2-Dichlorocyclooctane
73% yield; only product
Mechanism is electrophilic addition
•Br2 is not polar, but it is polarizable
•two steps
(1) formation of bromonium ion &
•
•
electrophilic attack
(2)
nucleophilic attack on bromo
ion by bromide
NET REACTION
CH2=CH2 + Br2 -> Br-CH2-CH2-Br
Step 1a: Formation of Bromonium
Ion
Br
Br
d–
Br
Br
d+
Mutual polarization
of electron distributions
of Br2 and alkene
Electrons flow
from alkene
toward Br2
d+
Step 1b; Electrophilic Addition to
form
Bromonium
Ion
Part i
H
H
C
H
C
+
Br
Br
H
Br
H
C
+
C
H
H
+ Br-
H
Part ii
Step 1b; Lone Pair on Bromine
Stabalizes Carbocation and
Forms Cyclic Bromonium Ion
H
C
+
Br
Br
+
C
H
H
C
C
H
+ Br-
H
H
H
+
Br
H
Step 2; Bromide Ion Must
Attack from Oppositte Side of
Cyclic Bromonium Ion (anti
addition)
+
Br
Br
C
C
H
H
H
+
H
H
Br
C
C
Br
H
H
+
Br
H
Br
Example
H
Br2
H
H
Br
Br
H
trans-1,2-Dibromocyclopentane
80% yield; only product
C
C
+ X2
C
X
C
X
alkenes react with X2 to form vicinal dihalides
alkenes react with X2 in water to give vicinal
halohydrins
C
C
+ X2
+ H2O
X
C
C
+ H—X
OH
Examples
H2C
CH2
+
H2O
Br2
BrCH2CH2OH
(70%)
H
Cl2
H
OH
H2O
H
Cl
H
anti addition: only product
Mechanism; 1) Cl2 is polarized and adds across double bond.
2) Ion formed is stabalized by lone pair of Cl.
3) Water attacks
chloronium ion
from side
opposite (anti
addition) carbonchlorine bond.
This gives trans
isomer
Regioselectivity
CH3
H3C
Br2
C
CH3
CH2
C
CH2Br
H2O
H3C
OH
(77%)
•Markovnikov's rule applied to
halohydrin formation: the halogen adds
to the carbon having the greater
number of hydrogens.
Hydrogenation (Reduction, +H2) of Ethylene
H
C
H

H

Metal
Catalyst

C
+
H
H—H
H
H
H
C
C
H
H

•exothermic DH° = –136 kJ/mol
•catalyzed by finely divided Pt, Pd, Rh, Ni
H
Two spatial (stereochemical) aspects of
alkene hydrogenation:
•(1) syn addition of both H atoms to double bond
•(2) hydrogenation is stereoselective, corresponding
to addition to less crowded face of double bond
H
CO2CH3
CO2CH3
H2, Pt
CO2CH3
CO2CH3
H
syn-Additon versus antiAddition
syn addition
anti addition
syn-Addition; Metal catalyst
breaks H-H bonds.
B
Y
C
C
A
H
X
H
H
H
syn-Addition; Addition of H2
across double bonds takes
place in two steps.
B
Y
A
X
H
H
C
H
C
H
H3C
Example of
Stereoselect
ive Reaction
CH3
H
H3C
H3C
H2, cat
H3C
CH3
H
H
H
H3C
H
Both products
correspond to
syn addition
of H2.
H
H
H3C
CH3
H3C
CH3
H
H3C
H3C
H2, cat
Top face of double
bond blocked by
this methyl group
CH3
H
H
H3C
H
But only this
one is formed.
H2 adds to
bottom face of
double bond.
Example of
Stereoselect
ive Reaction