Alkyl Halides.pdf

11-1
SUBSTITUTION AND ELIMINATION
REACTIONS OF ALKYL HALIDES
CHE 212
Sumani Jey
ORGANIC CHEMISTRY

11-2
Hybrid Orbitals
The Carbon Problem:
• Bonding by 2s and 2p atomic orbitals would give bond
angles of approximately 90°
• Instead we observe bond angles of approximately
109.5°, 120°, and 180°
A Solution to this problem is:
• Hybridization of atomic orbitals
s px py pz
Ground state of C 1s2 2s2 sp2

11-3
Hybrid Orbitals
Hybridization of orbitals (Linus Pauling)
• The combination of two or more atomic orbitals to
form a new set of atomic orbitals, called hybrid orbitals
We deal with three types of hybrid orbitals
sp3 (one s orbital + three p orbitals)
sp2 (one s orbital + two p orbitals)
sp (one s orbital + one p orbital)
Overlap of orbitals can form two types of bonds
depending on the geometry of overlap
 bonds are formed by “direct” overlap (hybrid and
unhybridized orbitals)
 bonds are formed by “parallel” overlap of unhybridized
p orbitals

11-4
sp3 Hybrid Orbitals
• Each sp3 hybrid orbital
has two lobes of
unequal size
• The sign of the wave
function is positive in
one lobe, negative in the
other, and zero at the
nucleus
• The four sp3 hybrid
orbitals are directed
toward the corners of a
regular tetrahedron at
angles of 109.5°

11-5
Hybridize at n=2 sp3 sp3 sp3 sp3
s px py pz
s px py pz
Hybridization
sp3 sp3 sp3 sp3
Four sp3 orbitals of equal length, energy and in
tetrahedral shape
Ground state of C 1s2 2s2 sp2
Promote electron at n=2 2s12p3

11-6
An energy-level diagram showing the
formation of four sp
3
orbitals

11-7
sp2 Hybrid Orbitals
When an s and two p orbitals are mixed to form a set of
three sp
2
orbitals, one p orbital remains unchanged and
is perpendicular to the plane of the hybrid orbitals.
The axes of the three sp2 hybrid orbitals lie in a plane and
are directed toward the corners of an equilateral triangle
The unhybridized 2p orbital lies perpendicular to the
plane of the three hybrid orbitals

11-9
An orbital energy-level diagram for sp
2
hybridization
Note that one p orbital remains unchanged

11-10
A sigma () bond centers along the internuclear axis.

11-11
(a) The orbitals used to form the bonds in ethylene. (b) The
Lewis structure for ethylene.
A pi () bond occupies the space above and below the
internuclear axis.

11-13
sp Hybrid Orbitals
When one s orbital and one p orbital are hybridized, a set
of two sp orbitals oriented at 180 degrees results
The unhybridized 2p orbitals are perpendicular to each
other and to the line created by the axes of the two sp
hybrid orbitals

11-15
The orbital energy-level diagram for the
formation of sp hybrid orbitals on carbon

11-17
Hybridization is about:
The mixing of atomic orbitals to form special
orbitals for bonding.
The atoms are responding as needed to give the
minimum energy for the molecule.
The number of atomic orbitals mixed is equal to
the number of hybrid orbitals formed

11-18
State the hybridization of O, N and S and indicate the location of the lone pairs of
electrons on these atoms

11-19
Halogenoalkane (Alkyl Halides)

11-20
Structure
Alkyl halide:
-Halogen, X, is directly bonded to an sp3 hybridized
carbon.
-Given the symbol RX
(where R is an alkyl
group)
a haloalkane
(alkyl halide)
X
R

11-21
Structures of other halides
X
Aryl Halide
X
Vinylic halide
-When a halogen is directly bonded to a phenyl
group then it is known as an aryl halide and vinyl
halide when a halogen is directly bonded to an
alkene functional group.

11-22
Nomenclature
Name the parent chain using rules previously
learned. Revisit your CHE 121 notes
In short : number the carbons of the parent chain
beginning at the end nearer the first substituent.
Assign each carbon a number.

11-23
Example
5-bromo-2,4-dimethylheptane
CH3CHCH2CHCHCH2CH3
CH3
CH3
Br

11-24
Example
2-bromo-4,5-dimethylheptane
CH3CHCH2CHCHCH2CH3
CH3
CH3
Br

11-25
Nomenclature
If more than one of the same kind of halogen is
present, number each and use the prefixes di-,
tri-, tetra-, and so on.
CH3CHCHCHCH2CH3
Cl Cl
CH3
2,3-dichloro-4-methylhexane

11-26
Nomenclature
If different halogens are present, number each
according to its position on the chain, but list
substituents alphabetically.
3-bromo-2,3-dichloro-4-methylhexane
CH3CHCCHCH2CH3
Cl Cl
CH3
Br

11-27
Nomenclature
If the parent chain can be numbered properly
from either end of the chain, begin at the end
nearer the substituent that has alphabetical
precedence.
2-bromo-5-methylhexane
CH3CHCH2CH2CHCH3
CH3 Br

11-28
Nomenclature
Cl
6-(2-chloropropyl)-2,3-dimethyldecane
Cl
1
10

11-29
primary alkyl halide secondary alkyl halide
tertiary alkyl halide
Primary, Secondary, Tertiary Alkyl
Halides
CH3CHCH2
CH3
Br CH3CH2CH
CH3
Br
CH3C
CH3
Br
CH3
*
*
*

11-30
Types of Dihalides
Geminal dihalide: two
halogen atoms are
bonded to the same
carbon.
Vicinal dihalide: two
halogen atoms are
bonded to adjacent
carbons.
CH3 CH
Br
Br
CH2CH2
Br Br
geminal dihalide
vicinal dihalide

11-31
Boiling Points
Among constitutional isomers, branched isomers
• Have a more compact shape,
• Decreased area of contact and decreased van der Waals
attractive forces between neighbors,
• And, therefore, lower boiling points
Br
Br
2-bromo-2-methylpropane
2-bromopentane
Br
1-bromopentane

11-32
Boiling Points
For an alkane and an alkyl halide of comparable
size and shape, the alkyl halide has the higher
boiling point
• The difference is due, almost entirely, to the greater
polarizability of the three unshared pairs of electrons
on halogen compared with the polarizability of shared
electron pairs of the hydrocarbon
Br
2-bromo-2-methylpropane 2,2-dimethylpropane

11-33
Uses of Alkyl Halides
Industrial and household cleaners.
Anesthetics:
• CHCl3 used originally as general anesthetic but it is
toxic and carcinogenic.
• CF3CHClBr is a mixed halide sold as Halothane®
Freons are used as refrigerants and foaming agents.
• Freons can harm the ozone layer so they have been
replaced by low-boiling hydrocarbons or carbon
dioxide.
Pesticides such as DDT are extremely toxic to insects
but not as toxic to mammals.
• Halogenoalkanes cannot be destroyed by bacteria so
they accumulate in the soil to a level which can be
toxic to mammals, especially, humans.

11-34
C C
Cl
Cl
H
Cl
trichloroethylene
(a solvent)
C C
Br
Cl
Cl
F
F
F
halothane
(an inhaled anesthetic)
C
F
Cl
Cl
F
dichlrodifluromethane
(a refrigerant)
C
H
H
H
Br
bromoethane
(a fumigate)

11-35
Preparation of alkyl halide
HX
HBr
X2
X
X CH3
H
CH3
Br
H H
CH3
Br
H
H X
CH3
H
H
CH3

11-36
Reactions of Alkyl Halides (R-X):
The -carbon in an alkyl halide is electrophilic (electron accepting)
for either or both of two reasons…
a) the C to X (F, Cl, Br) bond is polar making carbon d+ (partially positive)
(4.0 – 2.5) = 1.5
(3.0 – 2.5) = 0.5
(2.8 – 2.5) = 0.3
F
H3C
d+
d−
EN (F-C) =
EN (Cl-C) =
EN (Br-C) =
EN (I-C) = (2.5 – 2.5) = 0.0
b) X (Cl, Br, I) is a leaving group
pKb= 23 pKb= 22 pKb= 21 pKb= 11 pKb= -1.7
I-
Br -
Cl -
F-
HO -
30,000 10,000 200 1 0
decreasing basicity, increasing stability
increasing leaving ability
Br
H3C
d+
d−
Cl
H3C
d+
d−
I
H3C
The best
leaving
groups are
the weakest
bases.
The poorest
leaving
groups are
the strongest
bases.

11-37
Reactions of Alkyl Halides (R-X):
When a nucleophile (electron donor, e.g., -OH, -OCH3)
reacts with an alkyl halide, the halogen leaves as a
halide
There are two competing reactions of alkyl halides with nucleophiles
1) Substitution reaction
2) Elimination reaction
C C
H
X
Nu:-
+
C C
H
Nu
+ X-
+ C C
H
X
Nu:
- C C + X
- + Nu H
The Nu:- replaces the halogen on the -carbon.
The Nu:- removes an H+ from a b-carbon &
the halogen leaves forming an alkene.


b

11-38
Reaction Mechanism
Chemists use reaction mechanism to show how a
reaction proceeds in steps
• Use curved arrows to show the movement of electron
pairs
• Tail of arrow shows where the electron pair is before the
electrons move (lone pair or bond)
• Head of the arrow shows where the electron pair is
donated (its new position)
• Curved (curly) arrows show which bonds break and
which new ones form
Mechanism of the reaction—The events that are postulated
to take place at the molecular level as reactants are
transformed/converted into products

11-39
Examples of how curved arrows are
used in organic chemistry
CH3
C
CH3
CH3
H3C
C
H3C
H3C
Br HO HO
CH3
C
H3C CH3
CH3
C
H3C CH3
step 1 step 2
Br

11-40
CH2
C Br
H
H2C CH2
HO
H2O Br
+ +
H H
C
CH3
HO
O
Cl
d+
d−
C
CH3
HO
O
Cl
C
CH3
O
HO
C
CH3
HO
O
Cl
Cl
+

11-41
C
CH3
HO
O
Cl C
CH3
O
HO + Cl
or
C
CH3
HO
O
Cl C
CH3
O
HO + Cl
We will not be using this notation in this course
Consider the following uses of curved arrows:

11-42
H
C O
C
C
H
H
H
H
C O
C
C
H
H
H
R
R (a nucleophile)
H
Add acid at end
of reaction.
H
C O
C
C
H
H
H
R
H

11-43
H
C O
C
C
H
H
C
H
H
H
no reaction because C=C is not polar,
and the C=O is separated by a CH2 group.
Further resonance
of C=C bond is not possible
R

11-44
Unacceptable movement of electrons
C OH
S O
O
Cl
CH3
+ C O S
O
O
CH3
tosylate
C O S
O
O
CH3
Nu: C
Nu S
O
O
O-
H3C
+

11-45
Energy Diagrams
Energy diagram: a graph
showing the changes in
energy that occur during a
chemical reaction
Reaction coordinate
(Reaction progress): a
measure in the change in
positions of atoms during a
reaction
Reaction
coordinate
Energy

11-46
Terms used in Energy Diagrams
Enthalpy change, H0: the difference in total bond
energy between reactants and products
• a measure of bond making (exothermic) and bond
breaking (endothermic)
Heat of reaction: the difference in enthalpy
between reactants and products
• exothermic reaction: a reaction in which the enthalpy of
the products is lower than that of the reactants; a
reaction in which heat is released
• endothermic reaction: a reaction in which the enthalpy
of the products is higher than that of the reactants; a
reaction in which heat is absorbed

11-47
Gibbs Free Energy
Gibbs free energy change, G0: a thermodynamic
function relating enthalpy, entropy, and
temperature
• exergonic reaction: a reaction in which the Gibbs free
energy of the products is lower than that of the
reactants; the position of equilibrium for an exergonic
reaction favors products
• endergonic reaction: a reaction in which the Gibbs free
energy of the products is higher than that of the
reactants; the position of equilibrium for an
endergonic reaction favors starting materials
G0
= H0
–TS0

11-48
Activation Energy
Transition state:
• an unstable species of maximum energy formed
during the course of a reaction
• a maximum on an energy diagram
Activation Energy, G‡ (Ea): the difference in Gibbs
free energy between reactants (or any stable
intermediate) and a transition state
• if G‡ is large, few collisions occur with sufficient
energy to reach the transition state; reaction is slow
• if G‡ is small, many collisions occur with sufficient
energy to reach the transition state; reaction is fast

11-49
Exothermic energy diagram
A one-step reaction with no intermediate (one
step corresponds to one transition state)

11-50
Endothermic energy diagram
Energy
absorbed
in
reaction
Activation
Energy
Energy
used in
bond
breaking
Energy
released in
bond making

11-51
Exothermic Diagram
Energy
release
d in
bond
making
Activation
Energy
Energy used in
bond
breaking

11-52
Exothermic energy diagram
A two-step reaction with one intermediate has
two transition states

11-53
Consider the following Energy Diagram
Forming the
carbocation is an
endothermic step
(slow reaction step).
Step 2 is fast with a
low activation energy.

11-54
Hammond’s Postulate
“The structure of the transition state for an exothermic
reaction is reached early in the reaction, so it
resembles reactants more than products.
Conversely, the structure of the transition state for
an endothermic reaction step is reached relatively
late, so it resembles products more then reactants.”

11-55
Hammond’s Postulate
In reactions where the
starting material is
higher in energy (A),
the transition state
more closely
resembles the starting
material
In reactions where the
product is higher in
energy (C), the
transition state more
closely resembles the
product

11-56
Hammond’s Postulate Summary
In its simplest form the Hammond’s postulate
states that the transition state of the a reaction
step resembles higher energy component of
reaction i.e. the stable component which is
closest in energy to the transition state has the
same structure as the transition state.

11-57
Stereochemistry
• Although everything has a mirror image, mirror images
may or may not be superimposable.
• The mirror image of words like MOM is indistinguishable
whereas as the mirror image of DAD is distinguishable.
•Some molecules are like hands. Left and right hands are
mirror images, but they are not identical, or
superimposable.
Stereoisomers: molecules with the same connectivity but
different arrangement of atoms (groups) in space

11-58
Chiral: An object that cannot be superposed on
its mirror image
◼ Chiral objects don’t have
a plane of symmetry.
◼ Objects with a plane of
symmetry or a center of
symmetry are achiral.
Chiral and Achiral Molecules

11-59
Chiral Carbon
Tetrahedral carbon (sp3 hybridised carbon) with 4
different groups attached to it is chiral.
Chiral carbon is also called chiral centre or
stereogenic centre

11-60
Mirror Trick
Whenever two structures with the same molecular formula
and connectivity can be positioned around a plane of
symmetry and if they are not identical then those structures
are enantiomers.
Enantiomers are molecules that have the same molecular
formula and connectivity but different from their mirror
images

11-61
Properties of Enantiomers
-Same boiling point, melting point, density
-Same refractive index
-Different interaction with other chiral molecules
such as:
• Enzymes
• Taste buds, scent
In general:
Enantiomers: non-superimposable mirror image isomers
and are related to each other much like a right hand
is related to a left hand and have identical physical
properties, i.e., bp, mp, etc.
Enantiomers have no center of symmetry and are said to be
chiral.

11-62
A plane of symmetry divides an entire molecule into two
equal parts that are exactly the same
A molecule with a plane of symmetry is the same as its
mirror image and is said to be achiral

11-63
A substance is optically active if it rotates
the plane of a plain polarized light.
-Compounds which have ability to rotate
plane polarized light are said to be optically
active.
-Enantiomers rotate light in opposite
directions, but to the same degree i.e.
same number of degrees.
Optical Activity

11-64
Optical Activity
-In order for a substance to exhibit optical
activity, it must be chiral and one enantiomer
must be present in excess of the other.
-The optical rotation is dependent upon the
substance, the concentration, the path length
through the sample, and the wavelength of light.

11-65
Polarized light
Ordinary
(nonpolarized)
light consists of
many beams
vibrating in
different planes
Plane-polarized
light consists of
only those beams
that vibrate in the
same plane-Nicol prism

11-66
Rotation of plane-polarized light

11-67

11-68


11-69
Polarimeter
589 nm -
D-line of
a sodium
lamp

11-70
dextrorotatory – when the plane of polarized light is
rotated in a clockwise direction when viewed through a
polarimeter.
(+) or (d)
levorotatory – when the plane of polarized light is rotated
in a counter-clockwise direction when viewed through a
polarimeter.
(-) or (l)
The angle of rotation of plane polarized light by an
optically active substance is proportional to the number of
atoms in the path of the light.

11-71
Substitution Reactions
There are two kinds of substitution reactions,
called SN1 and SN2.
Typically these mechanisms are based on 3D and
not 2D structures. In other words: the majority of
the principles behind SN1 and SN2 mechanistic
pathways are meaningless when substrates are
presented in 2D format.

11-72
The Nucleophile
Neutral or negatively charged Lewis base
Reaction increases coordination at nucleophile
• Neutral nucleophile acquires positive charge
• Anionic nucleophile becomes neutral

11-73
Stereochemical Modes of Substitution
Substitution with inversion:
◼ Substitution with retention:
◼ Substitution with racemization: 50% - 50%

11-74
SN2 reactions
SN2 stands for Substitution, Nucleophilic,
bimolecular. Another word for bimolecular is
‘2nd order’.
Bimolecular (or 2nd order) means that the rate of
an SN2 reaction is directly proportional to the
molar concentration of two reacting molecules,
the alkyl halide ‘substrate’ and the nucleophile:
Nu:- + RX RNu + X:-
Rate = k [RX] [Nu:-] (This is a rate equation
and k is a rate constant).

11-75
Mechanism of SN2 reactions
C C
H
X
Nu:-
+
C C
H
Nu
+ X-

Note that the nucleophile must hit the back side of the -
carbon and steriochemistry at C is inverted.
The nucleophile to -C bond forms as the -C to X bond
breaks.
O H
..
..
: C Br
..
.. :
H
H
H
+
+
..
.. :
Br
:
C
H
H
H
O
H
..
..

11-76
The molecule inverts analogous to
an umbrella inverting in the wind

11-77
SN2 Mechanistic pathway
Nu CH L
HOMO LUMO BMO
CH
Nu L-
+
BMO
C6H13
H
Br
H3C
(R)-(-)-2-Bromooctane
[]D = -34o
ee = 100%
HO-
C6H13
H
Br
CH3
HO
C6H13
H
CH3
HO
(S)-(-)-2-Octanol
[]D = +10o
ee = 100%
H
CH3
H
Cl
OH-
H
CH3
OH
H
Cis-3-Methylchlorocyclopentane Trans-3-Methylcyclopentanol

11-79
The rate of SN2 reactions
The rate of SN2 reactions depends on 4 factors
❑ The nature of the substrate (the alkyl halide)
❑ The power of the nucleophile
❑ The ability of the leaving group to leave
❑ The nature of the solvent

11-80
1. Nature of substrate (Alkyl halide)
❑ Unhindered alkyl halides (those in which the
back side of the -carbon is not blocked), will
react fastest in SN2 reactions, that is:
Me° >> 1° >> 2° >> 3°
❑ While a methyl halide reacts quickly in SN2
reactions, a 3° alkyl halide does not react. The
back side of an -carbon in a 3° alkyl halide is
completely blocked.
O H
..
..
: C Br
..
.. :
H
H
H
+
transition state
C Br
..
.. :
H H
H
O
H
..
..
+
..
.. :
Br
:
C
H
H
H
O
H
..
..

11-81
Me° >> 1° >> 2° >> 3°
C
H3 Br C
H3 CH2 Br
CH Br
C
H3
C
H3
C Br
C
H3
C
H3
C
H3
t-butyl bromide
methyl bromide ethyl bromide isopropyl bromide
Back side of -C
of a methyl halide
is unhindered.
Back side of -C of a
1° alkyl halide is
slightly hindered.
2° alkyl halide is
mostly hindered.
3° alkyl halide is
completely blocked.
decreasing rate of SN2 reactions
SPACE FILLING MODELS SHOW ACTUAL SHAPES AND RELATIVE SIZES

11-83
C
H2 CH Br
H2C CH Br
Br
Br
vinyl bromide bromobenzene
Nu:-
Nu:-
The -carbon in vinyl and aryl halides, as in 3°
carbocations, is completely hindered and these alkyl
halides do not undergo SN2 reactions
The overlapping p-orbitals that form the -bonds in vinyl
and aryl halides completely block the access of a
nucleophile to the back side of the -carbon

11-84
The better the nucleophile, the faster the rate of SN2 rxns.
The best nucleophiles are the best electron donors.
The table below show the relative power or various Nu
Reactivity Nu:-
Relative Reactivity
very weak HSO4
-
, H2PO4
-
, RCOOH < 0.01
weak ROH 1
HOH, NO3
-
100
fair F-
500
Cl-
, RCOO-
20  103
NH3, CH3SCH3 300  103
good N3
-
, Br-
600  103
OH-
, CH3O-
2  106
very good CN-
, HS-
, RS-
, (CH3)3P:, NH2
-
,RMgX, I-
, H-
> 100  106
increasing
2.Effect of power of Nu:- on rate of SN2 Rxns

11-85
❑ The leaving group leaves together with electrons and has, therefore,
a negative charge
❑ Groups which best stabilize a negative charge are the best leaving
groups, i.e., the weakest bases are stable as anions and are the best
leaving groups.
❑ Weak bases are readily identified. They have high pKb values.
❑ Iodine (-I) is a good leaving group because iodide (I-) is non basic.
❑ The hydroxyl group (-OH) is a poor leaving group because hydroxide
(OH-) is a strong base.
pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21
I- Br - Cl- F- HO- RO- H2N-
30,000 10,000 200 1 0 0 0
Increasing leaving ability
3.Effect of ability of leaving group on rate of SN2
Rxns

11-86
There are 3 classes of organic solvents:
❑ Protic solvents, which contain –OH or –NH2 groups. Protic
solvents slow down SN2 reactions.
❑ Polar aprotic solvents like acetone, DMSO, DMF,DMA, HMPA,
which contain strong dipoles but no –OH or –NH2 groups. Polar
aprotic solvents speed up SN2 reactions.
❑ Non polar solvents, e.g., hydrocarbons. SN2 reactions are
relatively slow in non polar solvents.
4.Effect of solvent on rate of SN2 Rxns
CH3 C N :
acetonitrile
C
O
CH3
H3C
: :
acetone

11-87
Protic solvents (e.g., H2O, MeOH, EtOH,
CH3COOH, etc.) cluster around the Nu:- (solvate
it) and lower its energy (stabilize it) and reduce
its reactivity via H-bonding.
A solvated Nucleophile has difficulty hitting the
-carbon thereby slowing down SN2 reaction.
Effect of protic solvent on rate of SN2 Rxns
X:-
H
H
H
H OR
OR
OR
RO
d
+
d
+
d
+
d
+
-
d
-
d
-
d
-
d
A solvated anion (Nu:-) has reduced nucleophilicity,
reduced reactivity and increased stability

11-88
❑ Polar Aprotic Solvents solvate the cation counterion of the
nucleophile but not the nucleophile (“cage” the cation)
❑ Examples include acetonitrile (CH3CN), acetone (CH3COCH3),
dimethylformamide (DMF) [(CH3)2NC=O], dimethyl sulfoxide, DMSO
[(CH3)2SO], hexamethylphosphoramide, HMPA {[(CH3)2N]3PO} and
dimethylacetamide (DMA).
DMF
C
O
H N
CH3
CH3
C
O
N
CH3
CH3
DMSO
S
O
CH3
H3C
HMPA
[(CH3)2N]3P O H3C
DMA
: : : : : :
.. .. ..
..
..
..
Polar aprotic solvents solvate metal cations
leaving the anion counterion (Nu: -) bare and
thus more reactive
CH3C O
O
: :
..
..
:
_
Na+
Na+
N C CH3
N C CH3
N C CH3
N
C
H3C
-
d
-
d
-
d
-
d
d
+
d
+
d
+
d
+
+ CH3C O
O
: :
..
..
:
_
CH3CN
:
:
..
..
:
Effect of polar aprotic solvents

11-89
-Non polar solvents (benzene, carbon tetrachloride, hexane,
etc.) do not solvate or stabilize nucleophiles.
-SN2 reactions are relatively slow in non polar solvents
similar to that in protic solvents.
benzene
C
Cl
Cl
Cl Cl
carbon
tetrachloride
CH3CH2CH2CH2CH2CH3
n-hexane
Effect of non-polar solvents
-Polar Aprotic Solvents therefore accelerate the SN2
reaction

11-90
SN2 Energy Diagram
-The SN2 reaction is a one-step reaction. No reactive
intermediates
-Transition state is highest in energy. Only 1 TS for SN2
pathway

11-91
91
CH2CH3
Br
H3C
SN2
-
OH
CH3
C
H3C
CH3
H3C
SN2
-
OH
I
CH3
C
H3C
CH3
H3C
SN2
-
OH
I

11-92
92
SN2
-
OH
I
CH3
SN2
-
OH
H3C Br
SN2
-
OH
Br
SN2
-
OH
Cl
CH3
H3CH2C
Br SN2
-
OH
SN2
-
SH
Cl

11-93
Tertiary (3o) alkyl halides are not reactive in
substitutions that proceed by the SN2 mechanism.
Do they undergo nucleophilic substitution at all?
Yes. But by a mechanism different from SN2.
The most common examples are seen in
solvolysis reactions.
Solvolysis reaction is a reaction in which solvent
acts like a nucleophile
A question...
The SN1 Mechanism of Nucleophilic Substitution

11-94
The SN1 Reaction
The SN1 reaction is a unimolecular nucleophilic
substitution.
It is at the minimum a two step reaction with a
carbocation intermediate.
Rate is first order in the alkyl halide, zero order in
the nucleophile.
Rate = k*[RX].

11-95
first order kinetics: rate = k[RX]
•unimolecular rate-determining step
carbocation intermediate
•rate follows carbocation stability
•rearrangements sometimes observed
reaction is not stereospecific
•much racemization in reactions of
optically active alkyl halides
Characteristics of the SN1 mechanism

11-96
Comparison of a series of alkyl bromides under SN1 reaction
conditions (solvolysis) reveals that tertiary alkyl halides
react fastest.
The Alkyl Halide and the Rate of SN1 Reaction
In general, methyl and primary alkyl halides hardly react by
the SN1 mechanism and tertiary alkyl halides never react by
SN2.

11-97
The mechanism of an SN1 reaction occurs in a minimum of 2
steps:
Reaction Steps …
1. the slower, rate-limiting dissociation of the alkyl halide
forming a C+ intermediate (the C changes hybridization
from sp3 to sp2)
2. a rapid nucleophilic attack on the C+ intermediate
C
CH3
H3C
CH3
3°
Br
..
..
: + Na+
Br-
C
CH3
H3C
CH3
I
..
..
:
1.
Br -
-
C
CH3
H3C
CH3
+
3° C+
rapid
Na+
I -
..
..
: :
2.
Note that the nucleophile is not involved in the slower, rate-
limiting step.
Mechanism of SN1 reactions

11-98
Stereochemistry of SN1 Reaction
H3C
Pr
Et
H2O
H2O
H3C
Pr
Et
Br
Chiral Achiral
H3C
Pr
Et
OH2
H3C
Pr
Et
OH2
H2O
H2O
H3C
Pr
Et
OH
H3C
Pr
Et
OH
50%
50%
A 1:1 mixture of enantiomers is optically inactive since the
rotations cancel each other.
It is called a racemate or racemic mixture.
The process of converting an optically active compound
into a racemate is called racemization. It is common in SN1
reactions.

11-99
Stereochemistry of SN1 Reaction
-The planar intermediate (with an sp2 hybridized carbon)
leads to loss of chirality
-The intermediate carbocation is achiral
-Product is racemic (both inversion and retention of
configuration)

11-
• Again loss of the leaving group in Step [1] generates a planar
carbocation that is achiral. In Step [2], attack of the nucleophile
can occur on either side to afford two products which are a pair of
enantiomers.
• In most cases there is no preference for nucleophilic attack from
either direction and as such an equal amount of the two
enantiomers is formed - a racemic mixture. In that case we say that
racemization has occurred.

11-
Two examples of racemization in the SN1 reaction
There is a difference between a stable intermediate
and a product-refer to slide page 97

11-
Much as SN1 reactions proceed through the carbocation which
is planar the product is not always in a 50:50 ratio.
Consider the following reaction:
Expectation:

11-
The incomplete loss of stereochemistry is explained by a
partial shielding of one side of the carbocation by the halide
leaving group.
Stereochemistry of the SN1 Reaction

11-
(CH3)3CBr + 2 H2O → (CH3)3COH + H3O+ + Br-
Energy diagram
3o alkyl halide follows a first order rate law:
Rate = k[(CH3)3CBr]
And the reaction is termed SN1.
We cannot draw the energy diagram if we do not know the
step by step reaction equation (take note that we are not
talking about a mechanism here).
Step by step reaction is more less like a reaction mechanism
without curved arrows: All intermediates are included in a
step-by-step reaction equation.

11-
SN1 Mechanism of Nucleophilic Substitution
Step 1: Ionization to form a tertiary cation.
Step 2: Addition of a water molecule.
Step 3: Deprotonation.

11-
Potential energy graph of the a multi-step reaction

11-
Potential energy diagrams for multistep reactions:
The rate of oxonium ion
formation is very fast
The rate of carbocation
formation (dissociation
of the oxonium ion) is
slow
The rate of reaction
between the carbo-
cation and Cl- is fast
The overall rate is
dependent of the
slowest step (rate
limiting step)
rate = k [oxonium ion] , where k is the rate constant

11-
108
CH3
H3CH2C
Br SN1
OH2
CH3
Cl
H3C
SN1
OH2
CH3
Cl
C
N
SN1
KOH
CH3
Cl
H3C
SN1
CH3OH

11-
Rearrangements are evidence for carbocation
intermediates and to an extend serve to confirm the SN1
reaction mechanism.
Carbocation Rearrangements in SN1 Reactions

11-
i) Draw a step by step reaction equation showing how the given product is formed.
ii) Draw the structure(s) of the remaining substitution product(s)?
iii) Plot the energy diagram for the formation of given product and product(s) in question (ii)
above using the same margins (x- and y-axes). Use your knowledge of chemical reactions
to decide on the various energies involved in your diagram and no assumption is allowed.

11-111
Carbocation rearrangements in SN reactions
Wagner – Meerwein rearrangements
Rearrangement of a secondary carbocations -> more stable tertiary
carbocation

11-
OH
H2O/H2SO4
OH
OH
OH
H2O/H2SO4
H2O/H2SO4
CH3
HO CH3
OH

11-
The rate of SN1 reactions
The rate of SN1 reactions depends on 3 factors
❑ The nature of the substrate (the alkyl halide)
❑ The ability of the leaving group to leave
❑ The nature of the solvent

11-
C
H
H
H +
C
CH3
H
H +
C
CH3
H
H3C +
C
CH3
CH3
H3C +
tertiary
3º
secondary
2º
primary
1º
methyl
more
stable
less
stable
> > >
increasing rate of SN1 reactions
Highly substituted alkyl halides (substrates) form
a more stable C+.
Consider the Nature of substrate (Alkyl halide)

11-
-Alkyl groups are weak electron donors.
-They stabilise the carbocation by 2 ways
1. By donating electron density by induction (through 
bonds)
C
CH3
CH3
H3C +
Inductive effects:
Alkyl groups donate (shift) electron
density through sigma bonds to
electron deficient atoms.
This stabilizes the carbocation.
Stability of carbocations – alkyl groups

11-
Stability of carbocations
2. By hyperconjugation (by partial overlap of the
alkyl C-to-H bonds with the empty p-orbital of the
carbocation).

11-
Hyperconjugation
H3C
H3C
Vacant p-orbital
H
H
Occupied -orbitals
interacting with empty
p-orbital
C
H
H C
H
H
H
C
C
H C
H
H
H
H
H
H
C
C
C C
H
H
H
H
H
H
H
H
H
C
H
H H
0 3 6 9

11-
❑ Benzyl and allyl halides also react quickly by SN1
reactions even though the carbon holding the
halogen is a primary carbon. This is the case
because their carbocations are unusually stable
due to their resonance forms which delocalize
charge over an extended  system
Stability of carbocations

11-
Resonances
Benzyl bromide undergoes C-Br bond cleavage to generate the benzyl cation A.
Through resonance the positive charge can be delocalised throughout the ring.
Br
Br
A B C D
Benzyl Bromide Delocalisation of charges (positive or negative)
into an aromatic ring results in a relatively stable
ion.

11-
In general one can draw the following resonance
structures of benzylic carbocation:
2º benzylic
1º benzylic
C
H
R
+
C
H
H
+
C
H
H
C
H
H
C
H
H
+ +
+
H2C CH +
CH2
CH2
HC
H2C+
1º allyl carbocation
H2C CH +
CHR CHR
HC
H2C
+
2º allyl carbocation
The same applies for allylic carbocation:

11-
Br
OMe
Br
OMe
A B C
The resonance stabilisation is even more pronounced if the benzene ring has a
substituent which can supply electrons to the ring
Comment on why A undergoes C-Br cleavage more readily than the parent benzyl
bromide .

11-
Br
OMe
Br
OMe
A B C
OMe
The reason that the C-Br bond is cleaved so readily, is due to the positive charge
being able to delocalise through the ring and onto the oxygen atom. i.e. the charge
is spread out over many atoms leading to a stable electronic structure.

11-
-The nature of the leaving group has the same effect on both
SN1 and SN2 reactions.
-The better the leaving group, the faster a C+ can form and
hence the faster will be the SN1 reaction.
-The leaving group usually has a negative charge
pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21
I- Br - Cl- F- HO- RO- H2N-
30,000 10,000 200 1 0 0 0
Increasing leaving ability
❑ Iodine (-I) is a good leaving group because iodide (I-) is non
basic.
❑ The hydroxyl group (-OH) is a poor leaving group because
hydroxide (OH-) is a strong base.
Effect of ability of leaving group on rate of SN1 Rxns

11-
❑ For SN1 reactions, the solvent affects the rate only if it
influences the stability of the charged transition state, i.e.,
the steps leading to the formation of C+. The Nu:- is not
involved in the rate determining step so solvent effects on
the Nu:- do not affect the rate of SN1 reactions.
❑ Polar solvents, both protic and aprotic, will solvate and
stabilize the charged transition state (C+ intermediate),
lowering the activation energy and accelerating SN1
reactions.
❑ Nonpolar solvents do not lower the activation energy and
thus make SN1 reactions relatively slower
Effect of solvent on rate of SN1 Rxns

11-
reaction rate increases with polarity of solvent
The relative rates of an SN1 reaction due to
solvent effects are given
(CH3)3C-Cl + ROH → (CH3)3C-OR + HCl
H2O 20% EtOH (aq) 40% EtOH (aq) EtOH
100,000 14,000 100 1
Effect of solvent on rate of SN1 Rxns

11-
❑ Recall again that the nature of the nucleophile has no effect on
the rate of SN1 reactions because the slowest (rate-
determining) step of an SN1 reaction is the dissociation of the
leaving group and formation of the carbocation.
❑ All carbocations are very good electrophiles (electron
acceptors) and even weak nucleophiles, like H2O and
methanol, will react quickly with them.
❑ The two SN1 reactions will proceed at essentially the same
rate since the only difference is the nucleophile.
C
CH3
H3C
CH3
Br + Na+
I- C
CH3
H3C
CH3
I + Na+
Br-
3°
C
CH3
H3C
CH3
Br + C
CH3
H3C
CH3
F + K+
Br-
3° K+ F-
Effect of nucleophile on rate of SN1 Rxns

11-
127
• The strong nucleophile favors an SN2 mechanism.
• The weak nucleophile favors an SN1 mechanism.

11-
Elimination Reactions
Involve removal of some fragments from adjacent carbon
atoms of a reactant.
Converts an sp3 carbon to an sp2 or an sp2 carbon to an sp
therefore is important in the synthesis of alkenes and
alkynes.
There are two types of elimination reactions which are E2
and E1
Elimination and Nucleophilic Substitution reactions
competes with each other.
C C
Y Z
C C
sp3
sp3
+ Y-Z
Removed
fragments
C C
X
H
sp2
C C
sp
sp2
+ HX

11-
❑ E2 = Elimination, Bimolecular (2nd order).
❑ Rate = k [RX] [Nu:-]
❑ E2 reactions occur when a 2° or 3° alkyl halide is
treated with a strong base such as OH-, OR-, NH2
-, H-,
etc.
+ C C
H
Br
OH- C C + Br- + HO H
The Nu:- removes an H+ from a b-carbon
& the halogen leaves forming an alkene.

b
E2 Reaction

11-
E2: Stereochemistry of substrate
-Concerted mechanism in which abstraction of β-H,
formation of the π-bond and Elimination of Br- take place
simultaneously.
-Two TS are possible: Anti-coplanar and Syn-coplanar.
• Anti-coplanar is preferred since its TS is of low energy (more
stable).
• Syn-coplar is not possible under E2 mechanistic pathway even in
cases where it is the only conformation like in cases of rigid
system.
C C
H
H
CH3
Br
H
H
C C
H
H
CH3
Br
H
H
Anti-coplanar: prefered TS
Not possible under E2 mechanism
B-
B-

11-
E2 Reactions – Stereochemistry and Regiochemistry
❑ For SN2 reactions, you saw that the nucleophile had to attack
from the backside of the electrophilic site. In E2, since we are
concerned with bases and not nucleophiles, this restriction
reads ‘the proton removed must be anti-periplanar to the
leaving group.
❑ Consider the following E2 reactions:
KOH
KOH
CH3
Br
CH3
Br
CH3 CH3
CH3

11-
• Increasing the number of R groups on the carbon with the
leaving group forms more highly substituted, more stable
alkenes in E2 reactions.
• In the reactions below, since the disubstituted alkene is more
stable, the 3° alkyl halide reacts faster than the 10 alkyl halide.
Compounds with more than one β-H

11-
136
• The Zaitsev rule: the major product in b-elimination has the more
substituted double bond.
• A reaction is regioselective when it yields predominantly or
exclusively one constitutional isomer when more than one is
possible. Thus, the E2 reaction is regioselective.
The Zaitsev (Saytzeff) Rule:

11-
137
• Recall that when alkyl halides have two or more different
b -carbons, more than one alkene product is formed.
• When this happens, one of the products usually
predominates.
• The major product is the more stable product—the one
with the more substituted double bond.
• This phenomenon is called the Zaitsev rule.
The Zaitsev (Saytzeff) Rule:

11-
CH3 C CH CH3
Br
NaOC2H5
C2H5OH
heat
C CH
CH3 CH3
CH3 C CH CH3
Br
KOC(CH3)3
(CH3)3COH
heat
C CH
CH3 CH2
Major
H
CH3 CH3
CH3
H
CH3
Major
H
Non-bulky bases, such as hydroxide and ethoxide, give
Zaitsev products.
Bulky bases, such as potassium tert-butoxide, give larger
amounts of the least substituted alkene than with simple
bases-anti-Zaitsev product
EFFECT OF A BASE ON ELEMINATION PRODUCTS

11-
+
OH
_
heat
+
6%
94%
 b
b
CH3
CH2
CH2
CH CH3
N
CH3
CH3
CH3
CH3 CH2 CH CH CH3
CH3
CH2
CH2
CH CH2
-bulky leaving groups also gives the anti-
Zaitsev product (least substituted product is
formed)
EFFECT OF A LEAVING GROUP ON ELEMINATION PRODUCTS

11-
140
❑ When E2 reactions occur in open chain alkyl halides,
the Zaitsev product is usually the major product.
Single bonds can rotate to the proper alignment to
allow the anti-periplanar elimination.
❑ In cyclic structures, however, single bonds cannot
rotate. We need to be mindful of the stereochemistry in
cyclic alkyl halides undergoing E2 reactions. Example:
H
Cl
H3C
H
H
H
Na+ OH-
E2
H3C
H
H
H
Non Zaitsev product
is major product.
3-methylcyclopentene

11-
E2 Reactions and Cyclohexene Formation
-Abstracted proton and leaving group should align trans-
diaxial to be in anti-periplanar conformation in order to have
the required energy in the transition state
-Equatorial groups are not in proper alignment

11-
Energy diagram
energy
Reaction coordinate
C C
H O
H
Br-
..
:
..
__
H O
C C
Br
H
..
:
H O
C C
H
Br
d−
d−

11-
Cl
K OC(CH3)3
E2
NaOCH2CH2CH3
HOCH2CH2CH3
+
Br
15% 85%

11-
E1 Reaction
-Unimolecular elimination
-Two groups lost (usually X- and H+)
-Nucleophile acts as base
-Also have SN1 products (mixture)

11-
-Halide ion leaves first, forming a carbocation.
-Base removes H+ from adjacent carbon.
-Pi bond forms.
E1 reaction mechanism

11-
E1 Energy Diagram
Note that the first step is the same as in SN1
=>

11-
Comparing E1 and E2
-Strong base is needed for E2 but not for E1
-E2 is stereospecific, E1 is not
-E1 gives Zaitsev orientation

11-
Example of an E2
So the question is that do you expect the E1 reaction to
form the same product?

11-
C
C
H
Br
H3C
H
Na OCH2CH3
Examples of an E1 reactions

11-
CH3
Cl H2O
CH(CH3)2
Cl
H3C
OH
CH(CH3)2
Cl
H3C
OH

11-
A Closer Look
O
H
H
C
H
H
H
C CH3
CH3
C C
H
CH3
CH3
H
+ H3O
+

11-
E1 Energy Diagram
-Note: first step is same as in SN1
=>

11-
Bicyclic systems: Bredt’s Rule
No p orbitals on a bridgehead position in a rigid
bicyclic molecule can be formed. It means you
cannot form a carbocation at a bridgehead position.
It is not possible to have a double bond at a
bridgehead position because the p-orbitals do not
have right conformations to overlap
+
bridgehead
bridgehead

11-
Cl
AgNO3
Ethanol
No reaction!

11-
E1 or E2?
-Tertiary > Secondary
-Weak base
- Good ionizing solvent
-Rate = k[halide]
-Saytzeff product
-No required geometry
-Rearranged products
-Tertiary > Secondary
-Strong base required
-Solvent polarity not
important
-Rate = k[halide][base]
-Saytzeff product
-Coplanar leaving
groups (usually anti)
-No rearrangements

11-
Substitution or Elimination?
-Strength of the nucleophile determines order:
Strong nuc. will go SN2 or E2.
-Primary halide usually SN2.
-Tertiary halide mixture of SN1, E1 or E2
-High temperature favors elimination.
-Bulky bases favor elimination.
-Good nucleophiles, but weak bases, favor
substitution.

11-
Predicting Reaction Mechanisms
1. Non basic, good nucleophiles, like Br- and
I- will cause substitution not elimination.
2. In 3° substrates, only SN1 is possible. In
Me° and 1° substrates, SN2 is faster. For
2° substrates, the mechanism of
substitution depends upon the solvent.
3. Strong bases, like OH- and OR-, are also
good nucleophiles. Substitution and
elimination compete. In 3° and 2° alkyl
halides, E2 is faster. In 1° and Me° alkyl
halides, SN2 occurs.

11-
4. Weakly basic, weak nucleophiles, like H2O,
EtOH, CH3COOH, etc., cannot react unless a
C+ forms. This only occurs with 2° or 3°
substrates. Once the C+ forms, both SN1 and
E1 occur in competition. The substitution
product is usually predominant.
5. High temperatures increase the yield of
elimination product over substitution
product. (G = H –TS) Elimination
produces more products than substitution,
hence creates greater entropy (disorder).

11-
6. Polar solvents, both protic and aprotic, like
H2O and CH3CN, respectively, favor
unimolecular reactions (SN1 and E1) by
stabilizing the C+ intermediate.
7. Polar aprotic solvents enhance
bimolecular reactions (SN2 and E2) by
activating the nucleophile.

11-
Predicting Reaction Mechanisms
alkyl
halide
(substrate)
good Nu
-
nonbasic
e.g., bromide
Br
-
good Nu
-
strong base
e.g., ethoxide
C2H5
O
-
good Nu
-
strong bulky base
e.g., t-butoxide
(CH3
)3
CO
-
very poor Nu
-
nonbasic
e.g., acetic acid
CH3
COOH
Me
1°
2
3
SN1, E1
SN2
E2
SN2
SN2
SN1
SN2
SN2
E2
E2
SN2
E2 (SN2)
E2
no reaction
no reaction
SN1, E1
Strong bulky bases like t-butoxide are hindered.
They have difficulty hitting the -carbon in a 1°
alkyl halide. As a result, they favor E2 over SN2
products.

11-
174
Determining whether an alkyl halide reacts by an SN1, SN2, E1, or E2 mechanism
Alkyl Halides and Elimination Reactions
Predicting the Mechanism for SN1, SN2, E1 or E2.

11-
Alkyl Halides and Elimination Reactions

11-

11-
Carbocation rearrangements
Also 1,3- and other
shifts are possible
The driving force of rearrangements is -> to form a
more stable carbocation !!!
Happens often with secondary carbocations -> more
stable tertiary carbocation

11-
178
Carbocation rearrangements in SN + E reactions
Rearrangement

11-
Summary
SN1 SN2 E1 E2
Mechanism 2 or more steps
involving carbocation
intermediate
1 step bimolecular
process
2 or more steps
involving
carbocation
intermediate
1 step bimolecular process
Kinetics First order in
substrate
Second order, first in
substrate and
nucleophile
First order in
substrate
Second order, first in substrate
and base
Substrate
Dependence
Those substrates that
form stable
carbocations.
3°, allylic, benzylic
Those substrates that
are uncluttered at the
reaction site: 1°, 2°.
Good nucleophiles.
Those substrates
that form stable
carbocations.
3°, allylic, benzylic
Requires strong base and any
substrate with beta proton.
Stereochemistry Racemization. Stereospecific
inversion.
Usually mixtures. Stereospecific involving
antiperiplanar relationship of
beta-proton and leaving group.
Importance of
Base/nucleophile
Not involved in RDS,
but less basic form of
nucleophile will limit
E2.
Reactivity of
nucleophile is
important since it is
involved in RDS.
If a good, non-basic
nucleophile is
present (halides,
bisulfate) then SN1.
Strong, non-nucleophilic bases
(KOtBu, LDA) best to limit SN2.
Importance of
Leaving group
Involved in RDS so is
important.
Involved in RDS so is
important.
Involved in RDS so
is important.
Involved in RDS so is important.
Competes with.. E1 and E2 E2 when basic
nucleohiles employed.
SN1 SN2
Solvent Polar protic best Polar aprotic best Polar protic best Varies.

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