Chapter 6
Ionic Reactions-Nucleophilic Substitution and
Elimination Reactions of Alkyl Halides
Chapter 6
Ionic Reactions: Nucleophilic Substitution
and Elimination Reactions of Alkyl Halides
Alkyl Halides: R-X
The carbon center is sp3 hybridized in alkyl halides and the C-X bond is
polarized as shown because of the greater electronegativity of the
halogen.
 
C X
Vinylic and Phenyl Halides
In these compounds, the carbon is sp2 hybridized which results in a
stronger C-X bond. The nucleophilic substitution reactions discussed
in this chapter do not occur with these types of halides.
Cl
X
a vinylic halide
X
a phenyl halide
R
"vinyl chloride"
(chloroethene)
Br
bromobenzene
the vinyl group
R
the phenyl group
Introduction
• The polarity of a carbon-halogen bond leads to the
carbon having a partial positive charge
– In alkyl halides this polarity causes the carbon to become
activated to substitution reactions with nucleophiles
• Carbon-halogen bonds get less polar, longer and
weaker in going from fluorine to iodine
4
Nucleophilic Substitution Reactions
• In this reaction a nucleophile is a species with an
unshared electron pair which reacts with an
electron deficient carbon
• A leaving group is substituted by a nucleophile
• Examples of nucleophilic substitution
5
Timing of Bond Breaking and Bond Making Processes
The timing of the bond breaking and bond making steps is fundamental
to a discussion of the mechanism of a reaction. In nucleophilic
substitution, there are two distinguishable mechanisms.
(A) Simultaneous Bond Breaking and Bond Making
Nu:R +
: X:
: :


Nu---R---X:
: :
: :
Nu: + R-X:
Bond to incoming nucleophile is
developing as bond to halogen is
breaking.
(B) Bond Breaking precedes Bond Making
:
R+ + Nu:-
:
: :
R-X:
-
R+ + :X:
R:Nu
The operational mechanism
depends on the structure of
R-X, the nucleophile and
other factors.
Nucleophile
• The nucleophile reacts at the electron deficient
carbon
• A nucleophile may be any molecule with an
unshared electron pair
7
Leaving Group
• A leaving group is a substituent that can leave as a
relatively stable entity
• It can leave as an anion or a neutral species
8
Nucleophiles
A nucleophile has an unshared pair of electrons available
for bonding to a positive center.
Nucleophiles may be negatively charged:
:
: :
: : ::
: :
: :
:
:
: :
or neutral:
: :
- HO , CH3O , I , NH2
-
H2O , H3N, CH3OH
The polarity of the

C
X 
carbon-halogen bond
determines the
reactivity pattern: Nucleophiles attack
electropositive center.
Halide ion
is the leaving
group.
Examples
H
(1)
HO
-
+
nucleophile
C
Cl
H
H3C
substrate
H
O
(2) H
H+
nucleophile
C
Cl
H
H3C
substrate
H
H
H3C
C
OH
product
+
Cl
leaving group
H
C O H
H
H3C
H
ethyloxonium ion
+
leaving group
H2O
H
H
H3C
C
product
Cl
OH + H3O
Leaving Groups
The halogen is only one of many leaving groups, "L". A more general
description of nucleophilic substitution is :
-
Nu:
+
R-Nu
R-L
+
L:leaving group
A good leaving group produces a stable anion or neutral molecule.
Generally, the anions (conjugate bases) of strong acids are good
leaving groups.
H-A
strong acid
+ H2O
+
H3O
+
A:anion
very stable
A good
leaving group
in R-A .
Neutral Molecules as Leaving Groups
Poor leaving groups can
be turned into good
CH3-OH
leaving groups by
protonation.
Hydroxide ion is a
poor leaving group
because it is the anion
of a weak acid, H2O.
In the presence of a strong acid,
: :
CH3OH
H2SO4
+
+
CH3OH +
H
HSO4
a nucleophilic substitution reaction occurs:
CH3OH
nucleophile
+
+
CH3OH
H
good leaving group
+
CH3OCH3
H
+
H 2O
leaving group
Kinetics of a Nucleophilic Substitution Reaction: An SN2
Reaction
• The initial rate of the following reaction is measured
• The rate is directly proportional to the initial
concentrations of both methyl chloride and hydroxide
• The rate equation reflects this dependence
• SN2 reaction: substitution, nucleophilic, 2nd order
(bimolecular)
13
Thermodynamics and Kinetics of Chemical Reactions
A chemical reaction can be analyzed in terms of:
(1) The extent to which it takes place---how completely the reactants
are converted into products after equilibrium is reached.
This analysis is based on thermodynamic parameters.
(2) The rate of the reaction---how fast (moles per liter per second)
the reactants are converted to products. The kinetics of the
reaction means the rate of the reaction.
Thermodynamics
The extent to which a reaction spontaneously proceeds is given by the
equilibrium constant, Keq. The Keq is related to the Gibbs standard
free energy change by :
Go = -2.303 RT Log Keq
A negative value for Go means the product state is lower in free
energy than the reactant state, and the Keq > 1, while a positive value
for Go means the product state is above the reactant state, and the
Keq < 1.
The Rate of a Chemical Reaction: Kinetics
Even if a reaction has a very favorable Go (negative), the rate of the
reaction may still be very slow.
Example
The reaction
2H2(g) + O2(g)
2H2O(l)
has a very favorable free energy change: Go = -237 kJ/mol
Yet when hydrogen and oxygen gases are mixed, no reaction occurs.
If there is a high energy source (flame or spark), an explosive reaction
occurs.
The rate of a chemical reaction depends on a particular path from
reactants to products---a specific mechanism involving defined
chemical intermediates. In the reaction of hydrogen and oxygen to
produce water, radical intermediates must be formed to carry the
reaction along .
Sometimes, the product of a chemical reaction is not the one with the
most favorableGo (the thermodynamic product). Rather, it is the
one formed fastest because of a highly favorable reaction pathway.
This product is called the kinetic product.
Kinetics of Nucleophilic Substitution Reactions
The SN2 Reaction
Methyl chloride reacts
with hydroxide ion in
CH3Cl
a nucleophilic
substitution reaction: substrate
+
HO-
CH3OH
nucleophile
product
+
Cl-
leaving group
The rate of the reaction, which depends on experimental conditions such as
temperature, solvent, and concentrations, can be measured by following
the consumption of the reactants (CH3Cl or HO-) or the appearance of the
products (CH3OH or Cl-) over time.
The time dependence of the concentration of a reactant or
product is determined by periodically analyzing the reaction
mixture. A plot of these concentrations versus time gives
either a growth or decay curve.
Graphical Determination of Reaction Rates
The progress of the chemical reaction is followed graphically.
Decrease in reactant
concentration.
Increase in product
concentration.
concentration, M
At time = T, the concentration
of a reactant is measured..
The rate of the reaction can be
expressed as the amount of
reactant (moles/L) consumed
per second:
time
 [Reactant] = Rate = T
[Reactant]T=T - [Reactant]T=0
Time in seconds
where the notation  means the difference.
Method of Initial Rates
The mathematical form of the rate expression can be experimentally
determined by the method of initial rates. The rate of the reaction is
measured just after the reaction begins.
This initial rate is calculated directly from the slope.
concentration, M
initial conc.
T=O
conc. at
T=T
time
Initial Rate = -  [Reactant]
T
(from slope)
An Example of the Method of Initial Rates
For the nucleophilic substitution reaction:
o
CH3-Cl
+
HO-
60 C
H2O
CH3-OH +
Cl-
The following initial rates were determined as the
concentrations of the two reactants were changed.
Initial Rate Study at 60o C
initial
[CH3Cl]
initial
[HO ]
initial rate
mole L-1, s-1
0.0010 M
0.0020
0.0010
0.0020
1.0 M
1.0
2.0
2.0
4.9 x 10 -7
9.8 x 10-7
9.8 x 10-7
19.6 x 10-7
result
doubled
doubled
quadrupled
Conclusion: The rate of the reaction doubles when the concentration of
either reactant is doubled.
The Kinetic Rate Expression
The method of initial rates indicates that the rate of the
reaction depends on the concentrations of both reactants:
Rate  [CH3Cl] [HO-]
where  means "proportional to."
The more exact rate expression
includes the rate constant k:
Rate = k [CH3Cl] [HO-]
Evaluation of the Rate Constant
The data in the table allow k to be calculated:
k=
Initial Rate (experimentally measured)
[CH3Cl] [HO-]
k = 4.9 x 10-4 L mol-1 s-1 (note units)
A reaction rate that shows a kinetic dependence on the concentrations
of two reactants, each to the first power, is called second order. The kinetic
order implies that the two reactants must collide for reaction to occur and
the reaction is termed bimolecular and given the symbol:
SN2 Substitution Nucleophilic Bimolecular
Example: The reaction of methyl bromide and sodium hydroxide
CH3Br + NaOH
CH3OH + NaBr
The reaction of methyl bromide and sodium hydroxide in a solvent of
80% ethanol and 20% water at 550 C has a k = 0.0214 M-1 s-1. What is
the rate of formation of methyl alcohol if [CH3Br] = 0.5 M and [HO-] =
0.5 M ?
Rate = k[CH3Br][HO-] = (0.0214 M-1s-1) (0.5 M) (0.5 M)
Rate = 0.00535 mol per liter per second
Note that k is defined for a solvent and temperature.
Quiz Chapter 6 Section 5
The nucleophilic substitution reaction of methyl chloride and
hydroxide ion follows second order kinetics.
CH3OH + Cl-
CH3Cl + HO-
What is the Hughes-Ingold symbol for this reaction?
SN2
What is the exact rate expression for this reaction?
rate = k [CH3Cl] [HO-]
The method of initial rates is often used to determine the form of the
rate expression using the initial concentrations of reactants. As the
reaction proceeds, what happens to the rate of the reaction and the
rate constant k?
The observed rate of the reaction decreases because the
concentrations of the reactants decrease over time, but the rate
constant k remains the same.
A Mechanism for the SN2 Reaction
In 1937 Edward Hughes and Sir Christopher Ingold proposed a
mechanism to explain the second order kinetics and other important
features of this nucleophilic substitution reaction that were known at
that time.
The Hughes-Ingold Mechanism for the SN2 Reaction
In their mechanism, the nucleophile attacks the carbon center on the side
opposite the leaving group. As overlap develops between the orbital with
the electron pair of the nucleophile and the antibonding orbital of the
substrate, the bond between the carbon and the leaving group weakens.
Nu
-
:
C :
L
Nu
:
developing
bond
C
:
L
weakening
bond
A Mechanism for the SN2 Reaction
• A transition state is the high energy state of the reaction
– It is an unstable entity with a very brief existence (10-12 s)
• In the transition state of this reaction bonds are partially
formed and broken
– Both chloromethane and hydroxide are involved in the transition
state and this explains why the reaction is second order
24
Transition State Theory: Free-Energy Diagrams
• Exergonic reaction: negative Go (products favored)
• Endergonic reaction: positive Go (products not
favored)
• The reaction of chloromethane with hydroxide is
highly exergonic
• The equilibrium constant is very large
25
Transition State Theory: Free Energy Diagrams
The standard free energy change (Go) in a chemical reaction may be
negative (exergonic, energy released), or positive (endergonic, energy
taken on).
CH3Cl + HO-
CH3OH + Cl-
At 60oC in water, this reaction is
both exergonic and exothermic.
Go = -100.5 kJ/mol
Ho = -75.4 kJ/mol
Spontaneity of the Reaction
Since the standard free energy change is
related to the equilibrium constant by
Go = -2.3RTLog Keq
at 60oC,
Keq = 5.6 x 1015
Such a large value for Keq means that this nucleophilic substitution
reaction spontaneously goes to completion.
However, this thermodynamic analysis only examines the relative free
energies of the starting and product states. The rate of the reaction
depends on the detailed bonding changes as the reactants proceed up an
energy hill to the transition state (energy barrier) before moving down in
energy to the product state.
A Free Energy Diagram of the Bonding Changes
CH3-Cl + HO-

HO
H
TS
CH3-OH +
Cl-

Cl
C
HH
G
G
measures the energy barrier and
is related to the reaction rate
Free Energy
CH3-Cl + HO
o
G
o
G is a thermodynamic
parameter related to Keq
Progress of Reaction
CH3-OH +
Cl-
Thermodynamic Parameters Related to the Transition State
The changes in the standard thermodynamic parameters between the
starting and product states describe the thermodynamics of a chemical
change.
A + B
C + D
starting state
product state
Go = Ho - TSo
According to Transition State Theory, the transition state (even though
it only exists for the duration of a bond vibration) may be described by
the thermodynamic parameters of free energy, enthalpy and entropy.
G = H
-
TS
[TS]
The G is the free energy of activation which measures the change in
free energy between the starting state and the transition state. The
enthalpy of activation, H , and the entropy of activation, S ,
measure the changes in those parameters in going from the starting
state to the transition state.
• An energy diagram of a typical SN2 reaction
• An energy barrier is evident because a bond is being broken in going
to the transition state (which is the top of the energy barrier)
• The difference in energy between starting material and the
transition state is the free energy of activation (G‡ )
• The difference in energy between starting molecules and products is
the free energy change of the reaction, Go
The General Case for an Exergonic Reaction
29
• In a highly endergonic reaction of the same type the
energy barrier will be even higher (G‡ is very large)
The General Case for an Endergonic Reaction
30
• There is a direct relationship between G‡ and the
temperature of a reaction
– The higher the temperature, the faster the rate
– Near room temperature, a 10oC increase in temperature
causes a doubling of rate
– Higher temperatures cause more molecules to collide with
enough energy to reach the transition state and react
31
• The energy diagram for the reaction of chloromethane
with hydroxide:
– A reaction with G‡ above 84 kJ mol-1 will require heating to
proceed at a reasonable rate
– This reaction has G‡ = 103 kJ mol-1 so it will require heating
32
Enthalpy and Entropy of Activation
While the overall rate of a chemical reaction depends on the G , the
components of the free energy of activation (the enthalpy and entropy
of activations) often provide mechanistic information.
G = H - TS
H measures the enthalpy change as the reactants move towards the
transition state. Bonding changes make important contributions
to this parameter. Solvation effects are also important..
S
measures the change in entropy as the reactants move towards
the transition state. In most reactions, the transition state is more
ordered than the reactant state leading to a negative value for the
entropy of activation, and an unfavorable energy contribution to
the free energy of activation..
An Example: The Reaction of Methyl Iodide and Pyridine
Pyridine reacts rapidly with methyl iodide in a nucleophilic
substitution reaction yielding N-methylpyridinium iodide:
25 oC
N: + CH3I
pyridine
+
N CH3 + I-
acetonitrile
N-methylpyridinium iodide
o
G = -50.2 kJ/mol Exergonic
A kinetic study of this reaction provides the overall rate of this
reaction, and a breakdown of the free energy of activation into its
components of the enthalpy and entropy of activation.
G = +92.1 kJ/mol H = +53.6 kJ/mol S = -129.8 J K-1 mol-1
The relative contributions to the free energy of activation are:
G = H - TS
at 25 oC, 92.1 kJ/mol = 53.6 kJ/mol + 38.5 kJ/mol
Both the enthalpy and entropy of activation terms make significant
contributions to the free energy of activation and the rate of the reaction.
The Energy Barrier: A Closer Look
When reactants collide in solution, there are numerous possible orientations
with paths leading to products. The operating transition state is the most
energetically favorable path through an "energy mountain range."
The mountains on either side
of the pass represent
energetically less favorable
reaction pathways, described
in terms of bonding changes,
solvation effects, etc.
A very wide pass means many possible orientations of reactants will
lead to products...a less restrictive transition state structure with a less
negative entropy of activation term.
Quiz Chapter 6 Section 7
Define the following terms:
Exergonic Reaction
A reaction where G is negative.
Endergonic Reaction
A reaction where G is positive.
Transition State
The transitory state at the top of the free energy barrier between a
starting state and a product state. According to transition state theory,
the transition state is described by the state functions of free energy,
enthalpy and entropy.
Free Energy of Activation
The change in free energy between the starting state and the transition
state is the free energy of activation, G .
The Role of Kinetic Energy
The Boltzmann Distribution
fraction of molecules
At a given temperature, molecules will have a distribution of kinetic
energies (1/2 mv2) called a Boltzmann distribution. The shape of the curve
changes with temperature as shown below.
The distribution of kinetic
energies moves to higher
values and the curve
broadens with increasing
temperature.
T1
T2 > T1
T2
kinetic energy (1/2 mv2)
Kinetic Energy and the Energy Barrier
When reactants collide, their kinetic energies are available to move
them towards the top of the barrier in the free energy diagram. As
molecules surmount the energy barrier, kinetic energy is being
transformed into chemical potential energy.
fraction of molecules
Distribution of Kinetic Energies
of all Colliding Molecules
G
Only those colliding
molecules with sufficient
kinetic energy to reach
the free energy barrier,
G , will react.
fraction of effective = Area A
collisions
Total Area
under curve
A
energy
Note that the fraction of effective collisions increases in going
from lower to higher temperature because of the shift in the
Boltzmann distribution with increasing temperature.
Relationship Between the Free Energy of Activation
and the Rate Constant
The rate of a chemical reaction is given by:
Rate (mol L-1s-1) = k [concentration of reactant]
where k, the rate constant, is specific for the
reaction and the reaction conditions.
The rate constant is related to G by:
k = ko e - G/RT
where ko is the collision frequency at temperature T.
At 25o C, ko = 6.2 x 1012 collisions/s.
e is the base of the natural logarithm (2.718), and
e - G/RT is the fraction of collisions sufficiently energized
to reach the top of the reaction barrier. It will be
called the Energy Factor.
G/RT
Some Consequences of the Energy Factor: e (1) The energy factor (fraction of effective collisions) decreases as the
G increases at constant T.
Example: An increase in G from 10 to 20 kcal/mol at 25o C
decreases the energy factor from 4.36 x 10-8 to
1.90 x 10-15, a factor of almost 10-8.
(2) The energy factor for a particular reaction ( G is constant)
increases as the temperature increases.
Example: For a reaction, G = 20 kcal/mol. A change in T
from 25oC to 50oC increases the energy factor from
1.90 x 10-15 to 2.62 x 10-14, an increase by a factor of 14.
Changes in the energy factor directly change the rate of the
reaction by changing the magnitude of the rate constant:
Rate (mol L-1s-1)
k [concentration of reactant]
and k = ko e - G/RT
Origin of these Rate Outcomes
fraction of molecules
The sources of these changes in the energy factor are found from an
examination of the Boltzmann distribution curves.
T2 > T1
T1
For a specific reaction with a particular
free energy of activation, an increase in
temperature from T1 to T2 leads to a
greater fraction of effective collisions
because of the shift in the distribution
curve with change in temperature .
T2
G1 G2
kinetic energy
At a constant temperature (T1 or T2), an increase in the free energy
of activation leads to a smaller fraction of effective collisions. For
either curve, there is less area under the curve beyond G2.
Quiz Chapter 6 Section 7
In transition state theory, the rate constant for a reaction is given by
k = ko e - G/RT
Describe the effect on the magnitude of k, and the rate of a reaction:
when the free energy of activation is decreased.
The magnitude of k increases and the rate of a reaction increases.
when the temperature is increased.
The magnitude of k increases and the rate of a reaction increases.
The Stereochemistry of SN2 Reactions
• Backside attack of nucleophile results in an
inversion of configuration
• In cyclic systems a cis compound can react and
become trans product
43
The Stereochemistry of the SN2 Reaction
The proposed mechanism of the SN2 reaction
is based on two different methodologies:
(1) Kinetics: Rate = k[Nu:][substrate]
(2) Stereochemistry: An inversion of configuration is observed
at reacting stereocenters.
OH-

H
C
H
H
Cl
HO
H
C
HH
TS

Cl
H
HO
C
+ Cl
H
H
The stereochemistry of the SN2 reaction was elucidated
in studies on substrates with stereocenters.
-
Stereochemical Studies
Numerous studies on substrates with well-defined stereochemical
features reveal that the SN2 reaction proceeds with inversion at the
reacting carbon center.
CH3
Cl
H
H
+ HO-
cis-1-chloro-3-methylcyclopentane
A backside attack by the nucleophile explains
the formation of only the trans diastereomer
in this nucleophilic substitution reaction.
H
CH3
H
+ Cl-
OH
trans-3-methylcyclopentanol
Cl 
CH3
H
via
H
OH 
A Second Example
Studies on chiral substrates where the absolute configurations and
rotatory properties are known are also widely used as stereochemical
probes of mechanisms.
C6H13
Br
C
H
CH3
HO
NaOH
inversion
CH3
(R)-(-)-2-bromooctane
(S)-(+)-2-octanol
25
[a]
= -34.25o
D
enantiomeric purity 100%
C6H13
H
C
[a]
25
= +9.90o
D
enantiomeric purity 100%
The Hughes-Ingold mechanism explains
the stereochemical outcome of 100% inversion.
C 6H 13 H
HO
+
Br
-
CH 3
via this TS
Reaction of t-Butyl Chloride with Hydroxide: SN1 Reaction
The reaction of t-butyl chloride with sodium hydroxide in a mixture of
water and acetone (to help dissolve the RCl) shows the following rate
expression:
CH3
CH3-C-Cl
CH3
+ HO-
H2O
acetone
CH3
CH3-C-OH
CH3
+ Cl-
Rate = k[t-butyl chloride]
The reaction rate depends on the concentration of t-butyl chloride, but
shows no dependence on the concentration of hydroxide ion.
A reaction rate that depends on
the concentration of only one
The symbol is SN1.
reactant (to the first power) is
called first-order or unimolecular.
An Example
tert-Butyl chloride reacts with sodium hydroxide in a solvent of 80%
ethanol-20% water at 55oC with k = 0.0105 s-1. What is the rate of
the reaction when [t-BuCl] = 0.25 M and [HO-] = 1.0 M?
Solution
This reaction proceeds by the SN1 mechanism so the rate expression
does not include the concentration of the nucleophile.
Rate = k[t-BuCl] = (0.0105 s-1) (0.25 M) = 0.0025 Mol L-1 s-1
Multi-Step Reactions and the Rate-Determining Step
Many reactions in organic chemistry involve a sequence of steps,
each with an energy barrier linking starting and product states.
Each step proceeds with a unique rate constant.
A
k1
B
k2
C
a reaction
intermediate
Note: B is an intermediate. It is not a
transition state. This important difference is
shown in the free energy diagrams below.
Two general cases for the above two-step reaction sequence will be
examined. The relative magnitudes of rate constants k1 and k2 are
determined by the relative magnitudes of the free energies of
activation (the "energy barriers") for each step.
Two General Cases
Case A: k1 < k2
A
k1
slower
B
k2
faster
C
Step 1 (A
B) is slower
than step 2 (B
C).
Free Energy
The free energy of activation for
step 1 is larger leading to a smaller
rate constant.
slow step
C
The overall rate of A
is controlled by step 1 which
TS 1
is the rate-determing step.
TS 2
B
A
C
Progress of Reaction
Note: A chemical intermediate, such as B, is an energy minimum in the free energy
diagram. It persists for a finite time. A transition state is an energy maximum that is
transitory, existing for perhaps the time of a bond vibration, ~10-12 s.
Case B: k1 > k 2
A
k1
B
faster
In this scheme, the first step
is fast and reversible: A
k2
C
slower
B (fast).
When B is formed, it returns
to A faster than it proceeds to C.
The rate determining (slow) step is B
C.
slower
TS 2
Free Energy
faster
TS 1
The rate of formation of product C is
Rate = k2 [B]
where the concentration of B is
determined by the equilibrium
A
B.
B
A
Progress of Reaction
C
The Reaction of tert-Butyl Chloride with Hydroxide Ion: An
SN1 Reaction
• tert-Butyl chloride undergoes substitution with hydroxide
• The rate is independent of hydroxide concentration and
depends only on concentration of tert-butyl chloride
• SN1 reaction: Substitution, nucleophilic, 1st order
(unimolecular)
– The rate depends only on the concentration of the alkyl halide
– Only the alkyl halide (and not the nucleophile) is involved in the
transition state of the step that controls the rate
52
Multistep Reactions and the Rate-Determining Step
• In multistep reactions, the rate of the slowest step
will be the rate of the entire reaction
• This is called the rate determining step
• In the case below k1<<k2 or k3 and the first step is
rate determining
53
A Mechanism for the SN1 Reaction (next slide)
• Step 1 is rate determining (slow) because it requires
the formation of unstable ionic products
• In step 1 water molecules help stabilize the ionic
products
54
A Mechanism for the SN1 Reaction of t-Butyl Chloride in Water
The overall stoichiometry of this reaction is:
CH3
CH3-C-Cl + 2 H2O
CH3
CH3
CH3-C-OH
CH3
The reaction of tert-butyl chloride
with water to produce tert-butyl
alcohol shows first-order kinetics:
+ H3O+ + Cl-
Rate = k [t-butyl chloride]
This SN1 reaction proceeds by a multi-step mechanism of the Case A type
described previously with a slow (rate-determining) first step.
56
A Proposed Mechanism with a Carbocation Intermediate
bond heterolysis:
(1)
RDS
slow step
CH3
CH3-C-Cl
CH3
CH3
CH3-C + + ClCH3
t-butyl carbocation
a high energy intermediate
nucleophilic addition:
CH3 +
CH3-C O-H
CH3 H
fast
+ :O-H
H
:
CH3
CH3-C +
CH3
:
(2)
nucleophile
t-butyloxonium ion
proton exchange:
:
:
(3)
CH3 +
CH3-C O-H
CH3 H
+ :O-H
H
base
fast
CH3
CH3-C O-H + H3O+
CH3
The Free Energy Diagram
slow step
Bond Heterolysis
TS 1
CH3
CH3-C + + 2H2O + ClCH3
key intermediate
fast
Free Energy
TS 2
G2
G1
RDS
fast
TS 3
(CH3)3C-Cl + 2H2O
(CH3)3COH2+
+ H2O + Cl-
Progress of Reaction
(CH3)3COH
+ H3O+ + Cl-
The Carbocation Intermediate: The Importance of Solvation
Carbocations are high energy chemical species because they are ions.
In the gas phase, in the absence of solvation, the free energy of
activation for the bond heterolysis of an R-Cl compound is hundreds of
kilocalories per mole.
R-Cl
gas phase
R+ + Cl-
The ion state is very high in
energy relative to the starting state .
Solvation
In solution phase, solvation stabilizes both the carbocation and
chloride anion through charge-dipole interactions. The stabilizing
influence of solvation lowers the energy of the ions by hundreds of
kilocalories per mole, to the extent that the bond heterolysis step may
occur at ordinary temperatures.
These solvation forces already
operate in the transition state
for the bond heterolysis step
as the solvent dipoles interact
with the developing ions.
H
CH3
H O
H
  O
Cl H
CH3 C
CH3 O H
H
Quiz Chapter 6 Section 10
The free energy diagram for the multi-step reaction
k1
k2
A
B
C
is shown below.
RDS
TS A B
TS B
C
B
free energy
G
A
C
progress of reaction
In the diagram: Show the locations of A, B and C.
Identify the transition states for each step.
Indicate the transition state of the rate-determining step.
Show the free energy of activation for the RDS.
Carbocations
Early in the last century, carbocations began to be postulated as
intermediates in many organic chemical reactions. By the midtwentieth century, the indirect mechanistic evidence for their existence
was overwhelming. In the early 1960s, George Olah and his coworkers
and others developed methods to directly observe carbocations by
nuclear magnetic resonance spectroscopy. George Olah received the
Nobel Prize in Chemistry in 1994 for this and other discoveries.
The Structure of Carbocations
Carbocations are trivalent carbon
species with a positive charge.
Carbocations are locally planar
with the three valences in a plane.
R3C +
R
+
C
R
R
A trigonal planar geometry
is predicted by VSEPR theory.
Carbocations
• A carbocation has only 6 electrons, is sp2 hybridized
and has an empty p orbital
• The more highly substituted a carbocation is, the
more stable it is
– The more stable a carbocation is, the easier it is to form
63
Molecular orbital theory also predicts a planar geometry that
utilizes sp2 hybrid orbitals for the 6 bonding electrons.
a planar geometry energetically
p
better
+ z
C
sp2
a nonplanar geometry
3
sp
+
C
sp3
sp2 hybridization
The six valence level electrons
reside in atomic orbitals
that use 100% of the lower
energy 2s orbital.
sp3 hybridization
The six valence level electrons
reside in atomic orbitals that use
75% of the lower energy 2s orbital.
Relative Stabilities of Carbocations
Many experiments indicate
the stability order of
carbocations is:
R
R C+
R
3o
>
R
R C+
H
2o
>
H
R C+
H
1o
H
H C+
H
>
methyl
least stable
most stable
Gas phase experiments (in the absence of solvation) provide
H values for simple alkanes undergoing the heterolytic process :
R-H
cation
H (kJ/mol)
R+ + H:+
CH3
1305
+
CH3CH2
1142
The larger the value for H,
the less stable the cation.
+
CH3CH
CH3
CH3
CH3C+
CH3
1035
961
increasing stability
Source of Carbocation Stability Order
Inductive Effect
Replacement of a hydrogen with an alkyl group around the carbocation
center stabilizes the positive charge by the inductive effect, electron
donation through the sigma bonds. The alkyl group is electron-donating
relative to hydrogen.

H
+
C
H3C
H
H
less stable


C

electron-donation
CH3 disperses positive
H3C
more stable
charge
• Hyperconjugation stabilizes the carbocation by donation
of electrons from an adjacent carbon-hydrogen or
carbon-carbon s bond into the empty p orbital
– More substitution provides more opportunity for
hyperconjugation
67
The Stereochemistry of SN1 Reactions
• When the leaving group leaves from a stereogenic
center of an optically active compound in an SN1
reaction, racemization will occur
– This is because an achiral carbocation intermediate is
formed
• Racemization: transformation of an optically active
compound to a racemic mixture
68
69
Stereochemistry of the SN1 Reaction
Because carbocations have a trigonal planar geometry, they
possess a plane of symmetry and are achiral. Therefore, a racemic
form of a chiral product is predicted in reactions that involve
carbocation intermediates.
R1
C
A General Scheme
R2
X
R3
chiral
SN1 -X
H
:
R2
R3
H
O:
C
:O
R2 R3
achiral
(plane of symmetry)
H
R1
H
C
R2
O
H
:
C
H
:
O
R1
:
H
R1
H
R3
chiral
There are equal probabilities of reaction at the two faces of the
carbocation. The two oxonium ions formed in equal amounts are
enantiomers. A racemic form of the product is necessarily produced.
Racemization
The reaction of an optically active starting material that yields a racemic
form of a chiral product is said to proceed with racemization.
The SN1 reaction of an optically active substrate necessarily produces a
racemic form of a chiral product, so the stereochemical course of such a
reaction is racemization.
Example: The Hydrolysis of (S)-3-Bromo-3-methylhexane
H3CH2CH2C
H3C C
H3CH2C
Br
H2O
acetone
(S)-3-bromo-3-methylhexane
H3CH2CH2C
CH2CH2CH3
H3C C OH + HO C CH3
(S)
(R) CH CH
H3CH2C
2
3
(50%)
(50%)
3-methyl-3-hexanol
Racemic Form
CH2CH2CH3
C
via SN2
H3C CH2CH3
achiral
There is an equal probability of
attack by H2O at each face.
Stereoselective Reactions
Many SN1 reactions yield a mixture of diastereomeric products because
there is more than one stereocenter in the structure. Such reactions
typically produce the products in unequal amounts because the
competing reaction pathways are diastereomerically related. Such a
reaction is stereoselective.
Example
Hydrolysis of either cis or trans-4-tert-butyl-1-methylcyclohexyl iodide
yields the same unequal mixture of cis and trans-4-tert-butyl-1methylcyclohexanol. This reaction is stereoselective .
H3C
I
H3C
(H3C)3C
H
(H3C)3C
cis or trans
OH
H
same mixture of diastereomeric
alcohols from either starting iodide
Reaction Scheme
I
CH3
I
CH3
(CH3)3C
(CH3)3C
H
H2O
acetone
trans
H
H2O
acetone
SN1
cis
SN1
Path A
+ CH3
(CH3)3C
H
OH
CH3
(CH3)3C
Path B
A common intermediate reacts
with H2O via competing paths A
and B. The two paths are
diastereomerically related.
same product mixture
from either source
CH3
OH
(CH3)3C
H
H
trans
cis
Solvolysis
A nucleophilic substitution reaction where the solvent is the
nucleophile is called solvolysis.
When water is the solvent, the term hydrolysis is used.
Methanolysis means methanol is the nucleophile/solvent.
RX
+ H2O
RX
+ CH3OH
RX
+ CH3CH2OH
ethanol
acetic acid
hydrolysis
ROCH3 + HX
methanolysis
ROCH2CH3 + HX
ethanolysis
O
ROCCH3 + HX
=
O
CH3COH
methanol
=
RX +
ROH + HX
water
acetolysis
Solvolysis
• A molecule of the solvent is the nucleophile in a
substitution reaction
– If the solvent is water the reaction is a hydrolysis
76
Factors Influencing the Rates of the
SN1 and SN2 Reactions
The SN1 and SN2 are always potentially competitive pathways for any
nucleophilic substitution reaction. However, because of structural,
electronic, and other factors, one pathway usually dominates.
Effect of Structure on the SN1 and SN2 Reactions
A change in alkyl group substitution around the reacting carbon
influences nucleophilic substitution by the SN1 and SN2 mechanisms
in opposite ways.
The SN2 Mechanism
Increasing alkyl substitution slows down this reaction:
R-X + Nu:-
R-Nu + X:-
R = methyl > primary > secondary >> tertiary
fastest
A special case of a primary alkyl
bromide that does not react readily
by the SN2 mechanism is neopentyl
bromide.
CH3
CH3-C-CH2Br
CH3
"neopentyl bromide"
1-bromo-2,2-dimethylpropane
An Example: Incorporation of isotopic bromine in RBr
Relative second-order rate constants in acetone
at 25oC for the SN2 reaction:
*Br- +
R-Br
isotope label
R = CH3
relative
rate
76
acetone
CH3CH2
CH3CH
CH3
(1.0)
0.011
*Br-R
+
Br-
CH3
CH3
CH3C
CH3-C-CH2
CH3
CH3
0.003
0.000015
faster
This study shows that methyl bromide reacts 25,000
times faster than tert-butyl bromide, and 5 million times
faster than neopentyl bromide, in this SN2 reaction.
The Steric Effect
This slowdown in the rate of the SN2 reaction with increasing
alkyl substitution around the reaction center is attributed to a
steric effect, increasing steric hindrance in approaching the
reaction center by the incoming nucleophile.
The incoming nucleophile approaches
the carbon center on an axis opposite
the leaving group. Groups larger than
H attached to the carbon center screen
this line of approach.
-
H
H
ethyl
-
Nu:
C X
-
Nu:
C X
H
H
methyl
CH3
CH3
Nu:
H
H
C X
CH3
-
Nu:
CH3
isopropyl
increasing steric hindrance
CH3
C
CH3
tert-butyl
X
Factors Affecting the Rate of SN1 and SN2 Reactions
• The Effects of the Structure of the Substrate
• SN2 Reactions
• In SN2 reactions alkyl halides show the following general
order of reactivity
• Steric hinderance: the spatial arrangement of the atoms or
groups at or near a reacting site hinders or retards a
reaction
– In tertiary and neopentyl halides, the reacting carbon is too
sterically hindered to react
81
The Special Case of Neopentyl
Space-filling models indicate hydrogen atoms on -methyl groups,
as in neopentyl, actually screen the backside line of approach by
the nucleophile more effectively than hydrogens on an a-methyl
group as in tert-butyl.
CH3
CH3
-
Nu:
CH3
C
X
-
Nu:
CH3
tert-butyl
hindered, slow
CH3
H
CH3
C
C X
H
neopentyl
extremely hindered, very slow
When tert-butyl and neopentyl RX compounds undergo nucleophilic
substitution, it is not by the SN2 mechanism. Rather, alternative
reaction pathways operate.
The SN1 Mechanism
The primary factor that determines the reactivity of RX compounds
in the SN1 reaction is the stability of the carbocation intermediate.
Structural and electronic factors that stabilize a carbocation promote
the SN1 mechanism in nucleophilic substitution.
The reactivity order of RX compounds in nucleophilic
substitution reactions by the SN1 mechanism is:
tertiary-RX > secondary-RX > primary-RX > methyl-RX
which reflects the stability of the carbocation intermediate.
Analysis of the SN1 Mechanism
Because the rate of a chemical reaction depends on the free energy of
activation of the rate-determining step, it is more accurate to examine
influences on the stability of the transition state structure of the ratedetermining step. But there is a close relationship between the carbocation
intermediate and the transition-state structure in SN1 reactions.
The Structure of the Transition State
Because bond heterolysis to produce a carbocation
is a highly endergonic and endothermic process,
R-X
R+ + X-
both Go and Ho
are very positive
the reactants proceed well along the reaction coordinate before
reaching the transition state.
This postulate by George Hammond and J.E. Leffler states
that the transition state resembles the product state of that
step for energetically uphill reactions .
• SN1 reactions
• Generally only tertiary halides undergo SN1 reactions because
only they can form relatively stabilized carbocations
• The Hammond-Leffler Postulate
• The transition state for an exergonic reaction looks very much
like starting material
• The transition state for an endergonic reaction looks very much
like product
• Generally the transition state looks most like the species it is
closest to in energy
85
• In the first step of the SN1 reaction the transition
state looks very much like carbocation
• The carbocation-like transition state is stabilized by
all the factors that stabilize carbocations
• The transition state leading to tertiary carbocations
is much more stable and lower in energy than
transition states leading to other carbocations
86
Effect of the Concentration and Strength of the Nucleophile
A change in the concentration of the nucleophile will affect the
rate of an SN2 reaction but will have no effect on an SN1 reaction.
SN1
Rate = k [R-X]
SN2
R-X
R+ + Nu:R-X + Nu:-
slow step
R+ +
fast step
R-Nu
single step
X-
R-Nu + X-
Rate = k [R-X] [Nu:-]
Since the concentration of nucleophile appears only in the rate
expression for the SN2 reaction, the concentration of nucleophile
only affects that reaction rate.
• The Effects of the Concentration and Strength of
Nucleophile
• SN1 Reaction
• Rate does not depend on the identity or concentration of
nucleophile
• SN2 Reaction
• Rate is directly proportional to the concentration of nucleophile
• Stronger nucleophiles also react faster
– A negatively charged nucleophile is always more reactive than its
neutral conjugate acid
– When comparing nucleophiles with the same nucleophilic atom,
nucleophilicities parallel basicities
• Methoxide is a much better nucleophile than methanol
88
Nucleophile Strength
Nucleophile strength means "reactivity." A good nucleophile reacts
rapidly with a substrate.
Some General Observations
(1) A negatively charged nucleophile always reacts faster with a
substrate than its conjugate acid.
relative
reactivities
RO- > ROH
HO->HOH
NH2- > NH3
(2) In a group of nucleophiles with the same nucleophilic atom,
nucleophile strength parallels base strength.
nitrogen nucleophiles
base strength
:
=
oxygen nucleophiles
base strength
:
RO- > HO- > RCO2-
O
RNH2 > RCNH2
Influence of Nucleophile on Competition
between SN1 and SN2 Reactions
Example: Conversion of Isopropyl Bromide to Isopropyl Alcohol
H
H
H3C
Br
CH3
H3C
OH
CH3
The conversion of isopropyl bromide to isopropyl alcohol may occur by
either an SN1 or SN2 mechanism. This secondary alkyl bromide is a
"borderline" system.
In water the SN1 mechanism dominates. But with added HO-, the SN2
mechanism becomes increasingly competitive.
A Rate Analysis of the Competing Reactions
In 80% ethanol-20% water, at 55o C, isopropyl bromide reacts
according to the comprehensive rate expression:
Rate =
k2 [RX] [HO-]
SN2 mechanism
Rate =
+
k1 [RX]
SN1 mechanism
(4.7 x 10-5) [RX] [HO-] + (0.24 x 10-5) [RX]
At [HO-] = 0.001 M, the percentage of the nucleophilic substitution
going by the SN2 reaction is
% SN2 =
(4.7 x 10-5) [RX] [0.001 M]
(4.7 x
10-5)
[RX] [0.001 M] + (0.24 x
% SN2 = 1.95%
But at [HO-] = 0.10 M, % SN2 = 66.1%
x 100
10-5)
[RX]
Quiz Chapter 6 Section 13
The nucleophilic substitution reaction
R-X
+ Nu:-
R-Nu + X:-
may proceed by either the SN1 or SN2 mechanism.
With which mechanism is each of the following observations associated?
Addition of alkyl groups around the reacting
center slows down the rate of the reaction.
SN2
When the reaction takes place at a stereocenter, racemization is
observed with optically active substrates.
SN1
Increasing the concentration of the nucleophile results in
an increase in the rate of the reaction.
SN2
• Solvent Effects on SN2 Reactions: Polar Protic and
Aprotic Solvents
• Polar Protic Solvents
– Polar solvents have a hydrogen atom attached to strongly
electronegative atoms
– They solvate nucleophiles and make them less reactive
– Larger nucleophilic atoms are less solvated and therefore
more reactive in polar protic solvents
– Larger nucleophiles are also more polarizable and can
donate more electron density
– Relative nucleophilicity in polar solvents:
93
• Polar Aprotic Solvents
– Polar aprotic solvents do not have a hydrogen attached to an
electronegative atom
– They solvate cations well but leave anions unsolvated because positive
centers in the solvent are sterically hindered
– Polar protic solvents lead to generation of “naked” and very reactive
nucleophiles
– Trends for nucleophilicity are the same as for basicity
– They are excellent solvents for SN2 reactions
94
Nucleophile Stability and Reactivity
The strength of the H-bonding increases with the density of the negative
charge on X-. Fluoride ion, the smallest halide, has the largest negative
charge density and strongest H-bonding interaction with polar protic
solvents. This interaction stabilizes the ion and decreases its nucleophile
strength in polar protic solvents.
Reactivity Order of Halide Ions in Polar Protic Solvents
I- > Br- > Cl- > Fweakest solvation
strongest solvation
Gas Phase Reactivitiy
Reactivity order of halide
ions in the gas phase:
F- > Cl- > Br- > I-
In the absence of solvent, the reactivity order of the halide
ions parallels their base strength.
Nucleophile Strength and Base Strength
A strong correlation is often seen between base strength and
nucleophile strength: stronger bases are better nucleophiles.
Exceptions exist, however. For example, hydroxide ion (HO-) is a
stronger base than cyanide ion N C , but cyanide is a better
nucleophile towards RX compounds.
Polar Aprotic Solvents: Enhanced Reactivity of
Nucleophiles in SN2 Reactions
Aprotic solvents do not have hydrogen atoms attached to highly
electronegative atoms, and do not participate in hydrogen bonding.
Many aprotic solvents are nonpolar, or have low polarity and do
not dissolve ionic materials. The polar aprotic solvents dissolve
ionic materials and selectively interact with cations through dipolecharge interactions.
Examples of Polar Aprotic Solvents
N,N-dimethylformamide
dimethyl sulfoxide
(DMF)
(DMSO)
=
O
CH3-S-CH3
=
O CH
4
H-C-N
CH3
H3C
O
S
CH3
Na
CH3
O
S
S
CH3
O
H3C
N
acetonitrile
O
H3C
Polar aprotic solvents
S
selectively solvate cations
through dipole-charge
H3C
interactions.
CH3C
CH3
While polar aprotic solvents strongly interact with cations,
they do not solvate anions (nucleophiles) very well.
"Naked Anions" in Polar Aprotic Solvents
Because anions are weakly solvated , and because there is little cationanion association in polar aprotic solvents, the term "naked anions" is
often used to describe the state of anions in these solvents.
Naked anions are very reactive as nucleophiles and bases.
The Reactivity of Naked Anions
The reactivity order of the halides
in DMSO is the same as in the gas
phase where there is no solvation:
F- > Cl- > Br- > I-
The reactivity of nucleophiles towards a single substrate increases
dramatically in polar aprotic solvents:
CH3I + Clsolvent
CH3OH
H 2O
HCONH2
CH3CN
HCON(CH3)2
CH3Cl
+
I-
relative rate
0.9
protic
1.0
solvents
14.1
35,800
708,000
polar aprotic
solvents
For the proton transfer (acid-base) reaction:
H
C6H5CCN + CH3OCH3
Rate in DMSO
Rate in CH3OH
-
C6H5CCN + CH3OH
CH3
= 109
• Solvent Effects on SN1 Reactions: The Ionizing
Ability of the Solvent
• Polar protic solvents are excellent solvents for SN1
reactions
• Polar protic solvents stabilize the carbocation-like
transition state leading to the carbocation thus
lowering G‡
• Water-ethanol and water-methanol mixtures are
most common
100
Change from a Less Polar to a More Polar Solvent
When there is a change from a less polar to a more polar solvent,
the rate of an SN1 reaction generally increases because the energy
of the more polar transition state decreases more than the energy
of the less polar starting state .
R-Cl

R

Cl
less polar
more polar
R+ + Cl-
Free Energy Changes for the Rate-Determining Step of an SN1
Reaction in Nonpolar and Polar Solvents
R-Cl

R

Cl
R+ + Cl-
less polar
more polar
more polar
TS and ion state
large stabilization
effect by polar
solvent
Free Energy
Gpolar< Gnonpolar
less polar
starting state
nonpolar
solvent
polar
solvent
Gnonpolar
Gpolar
modest stabilization effect
by polar solvent
Reaction Coordinate
The reaction rate
increases in going
from a nonpolar to
a polar solvent.
A Measure of Solvent Polarity: Dielectric Constant
One measure of solvent polarity is the dielectric constant, e, that measures
the ability of a solvent to screen the electrostatic attraction between
opposite charges. As a solvent interacts with and stabilizes charges, the
attractive forces decrease. The higher the value of e, the more stable the
ions, the more the attractive forces are screened, and the easier it is to
separate the charges as in the TS for the SN1 reaction.
Example: The Solvolysis of tert-Butyl Chloride
CH3
CH3-C-Cl
CH3
solvent
e
relative rate
H2O/CH3CH2OH
100%
ethanol
24
25o
C
CH3
CH3-C-OR + HCl
CH3
90% 80% 70% 60% 50% 40% ....100%
water
increasing polarity
80
1.0
5.3
23.5 73.6 214 756
increasing rate of reaction
Specific Solvent Effects in the SN1 Reaction: Hydrogen Bonding
Polar protic solvents typically are very rate enhancing in SN1
reactions because they specifically interact with the leaving group by
hydrogen bonding. This interaction provides an additional mode for
charge dispersal and stabilization of the transition state.
Example: The Methanolysis of tert-Butyl Chloride
Hydrogen bonding by methanol molecules helps
disperse developing charge in the transition state.
CH 3
H 3C
C
Cl
CH 3
SN 1
CH 3OH
H3C
CH3
H3C
CH3
H3C
CH3

C
C+
CH3
+
H3C

 O CH3
 H
Cl 
H 
O CH3

O 
H
-

 O CH3
H
 Cl 
H 
 H
O
O CH3
Nature of the Leaving Group
Good leaving groups in both the SN1 and SN2 reactions produce
stable anions. Stable anions are weak bases----the anions of strong
acids.
H3O+ + Astable anion
H-A + H2O
strong acid
SN1
R-A
good leaving group
Nu:-
+
R-A
good leaving group
SN2
R+
+ Astable anion
Nu-R + Astable anion
• The Nature of the Leaving Group
• The best leaving groups are weak bases which are
relatively stable
– The leaving group can be an anion or a neutral molecule
• Leaving group ability of halides:
• This trend is opposite to basicity:
• Other very weak bases which are good leaving groups:
• The poor leaving group hydroxide can be changed into
the good leaving group water by protonation
107
Leaving Group Ability
Leaving group ability influences the rates of both the SN1 and SN2
reactions through the developing anion character of the leaving
group in the transition states.
SN1
C
X
RDS

C

X
The stability of X- influences the energy
of both the SN1 and SN2 transition states.
SN2
Nu:
+
C
X
RDS

Nu
C

X
Example: The Halides as Leaving Groups
leaving group ability
I > Br > Cl > F
This order of leaving group ability of the halides in R-X
compounds is consistent with the acidity order of the hydrogen
halides, and the related base strength of the halide ions.
order of acid strength
H-F << H-Cl < H-Br < H-I
weakest
pKa
order of base strength
3.2
-2.2
-4.7
F >> Cl > Br > I
strongest
-5.2
Influence of the Leaving Group on Chemical Reactivity
The typical relative reactivities of RX compounds in SN2
reactions differ greatly because of the different leaving group
abilities of the halides.
R-I
30,000
R-Br
10,000
R-Cl
200
R-F
1
often a good choice
for organic synthesis
Other Good Leaving Groups
Another important class of good leaving groups are the
sulfonate esters, R-O3SR', that produce stable sulfonate anions
in nucleophilic substitution reactions.
=
=
=
O
=
O
trifluoromethanesulfonic acid
"triflic acid" a very strong acid
O
a "super "
CF3-S-O- leaving group
=
=
O
CF3-S-OH
p-toluenesulfonate anion
=
=
p-toluenesulfonic acid
Ka ~ 10
CH3
O
O
CH3
O
-S-O-
=
=
O
-S-OH
methanesulfonate anion
=
methanesulfonic acid
Ka = 16
O
=
O
CH3-S-OH
Stable Anions
O
CH3-S-O-
O
Acids
trifluoromethanesulfonate anion
"triflate" anion
Acid-Promoted Leaving Group
Poor leaving groups sometimes can be chemically modified to
turn them into better leaving groups. An example is the
protonation of the hydroxyl function of an alcohol.
: :
-
:X:
+ R-OH
X
No Reaction
Hydroxide is a poor leaving group.
-
: :
But,
:X:
R-X
+ R-OH
H
+ H2O
Water is a much better leaving group than hydroxide ion.
Note: Very strong bases such as simple carbanions
(R:-) and hydride (H:-) are never leaving groups
because of the instability of the anionic leaving
groups that results in very unfavorable energetics.
Nu:- +
R-H
X
No Reaction
Nu:- +
R-R'
X
No Reaction
Quiz Chapter 6 Section 13
The rate of a nucleophilic substitution reaction
R-Nu + X:RX + Nu:by either the SN1 or SN2 mechanism is influenced by several factors.
Which mechanism is associated with each of the
following observations?
The rate of the reaction typically increases by many
powers of 10 with a change from a polar protic to a
polar aprotic solvent.
S N2
The rate of the reaction in protic media increases with
increasing polar character of the solvent.
S N1
The reactivity of RX depends on the halide leaving group as
follows: I> Br> Cl > F
S N1
and
SN2
The rate of the reaction increases as the strength of the
nucleophile increases.
SN2
Summary of SN1 and SN2 Reactions
These two modes of nucleophilic substitution of alkyl halides are
always potential competing reactions. Often, one path dominates.
The SN1 reaction is favored by:
substrates that form relatively stable carbocations,
such as tertiary halides
weak nucleophiles
solvents of high ionizing ability (high e)
The SN2 reaction is favored by:
use of relatively unhindered substrates such as
methyl and primary RX
strong nucleophiles (usually strong bases)
polar aprotic solvents
Summary of SN1 and SN2 Reactions
Overall for R-X:
CH3X
RCH2X
SN2
R
RCHX
R
RCX
R
SN1
The effect of the leaving group
is the same for the
SN1 and SN2 reactions.
R-I > R-Br > R-Cl > R-F
• Summary SN1 vs. SN2
• In both types of reaction alkyl iodides react the
fastest because of superior leaving group ability
116
Organic Synthesis: Functional Group Transformations
Using SN2 Reactions
• Stereochemistry can be controlled in SN2 reactions
117
Stereochemistry of SN2 Synthetic Reactions
As in all SN2 reactions, these syntheses proceed
with inversion at a stereocenter.
N
C
+
CH3
H3C
C
N
Br
H
H3CH2C
(R)-2-bromobutane
C
C H
CH2CH3
(S)-2-methylbutanenitrile
Note: These synthetic reactions are limited to alkyl halides. Vinylic
and aryl halides do not react readily by either the SN1 or SN2
mechanisms.
X
C C
X
vinylic halide
aryl halide
Elimination Reactions of Alkyl Halides
• Dehydrohalogenation
• Used for the synthesis of alkenes
– Elimination competes with substitution reaction
– Strong bases such as alkoxides favor elimination
119
Elimination Reactions of Alkyl Halides
In an elimination reaction, the atoms or groups X and Y are lost from
adjacent carbons forming a multiple bond.
C C
X Y
C C
(-XY)
The Dehydrohalogenation Reaction
A standard synthesis of
alkenes is the
dehydrohalogenation
reaction of alkyl halides.
C C
X H
(-HX)
alkyl halide
C C
alkene
Example: The Dehydrobromination of tert-Butyl Bromide
CH3
CH3C-Br + NaOCH2CH3
CH3
tert-butyl bromide
CH3
CH2=C
CH3
isobutene
+ HOCH2CH3
+ Na+ Br-
A Beta- or 1,2-Elimination Reaction
This reaction is described as a beta-elimination or 1,2-elimination
indicating the positions of the lost atoms or groups.
H
H
C
 position H
CH3
C
(-HBr)
Br
 CH3
or
CH2 C
CH3  position
2
CH2
CH3
1 CH3
C
CH3
Theor 1 position is the carbon with the halogen leaving group.
The Role of Base in Dehydrohalogenation Reactions
A number of different bases may be used in the dehydrohalogenation
reaction. Typical bases are potassium hydroxide in ethanol (to solubilize
the alkyl halide) or sodium ethoxide in ethanol. Potassium tert-butoxide
is another oxygen base that is often used in dehydrohalogenation
reactions.
Some Oxygen Bases
KOH
NaOEt
KOBu-t
Preparation of Alkoxide Bases
Although most alkoxide bases are available from commercial sources,
they may be "freshly" prepared from the rapid redox reaction of the
"dry" (no water) alcohol with an alkali metal.
2 CH3CH2OH + 2 Na
2 NaOCH2CH3 + H2
sodium ethoxide
CH3
2 CH3COH
CH3
+ 2K
CH3
2 CH3CO-K+
CH3
+ H2
potassium tert-butoxide
Sodium alkoxides also may be prepared by reacting alcohols with the
strong base sodium hydride:
RO-H
pKa ~ 16
+
Na:H
sodium hydride
RO-Na+
a sodium alkoxide
+
H2
pKa = 35
Mechanism of Dehydrohalogenation: The E2 Reaction
The reaction of isopropyl bromide with sodium ethoxide in ethanol to
give propene:
CH3CH=CH2 +
Br
propene
CH3CHCH3 + NaOCH2CH3
isopropyl bromide sodium ethoxide ethanol HOCH2CH3 + Na+Br-
follows the rate expression: Rate = k [CH3CHBrCH3][NaOCH2CH3]
second-order kinetics
The rate expression implies that both the alkyl halide and base are
involved in the transition state of the rate-determining step.
The notation for the bimolecular reaction is E2.
A Mechanism for the E2 Reaction
The mechanism proposed for the E2 reaction is based on the observed
second order rate expression, as well as the stereochemical outcome
observed in alkyl halides with multiple stereocenters. The E2 like the
SN2 is a single step mechanism.
CH3CH2O-
H CH3
H
C
H
H

CH3CH2O
C
Br
C
The alkyl bromide reacts
from a conformation where
the leaving groups are
anti-coplanar.
H
+
H
C
H
H
C
Br

As the base removes the H+,the
double bond begins to develop
and the Br- begins to depart.
H
Br + CH3CH2OH
H CH3
H
C
CH3
The E2 Reaction: A Closer Look
The stereoelectronic requirements of the transition state for the E2
reaction are more easily seen in a Newman projection.
CH3CH2O-
Newman
H
CH3 Projection
H
C C
H
Br
H
CH3CH2O-
H
H
CH3
H
H
Br
The alkyl bromide reacts
from a conformation where
the leaving groups are
anti-coplanar.
Overlap of Developing p-Orbitals in the Transition State
The anti-coplanar conformation allows immediate overlap of the porbitals of the developing double bond in the transition state.

CH3CH2O
CH3CH2OH
H
H C
H
H
C CH3
H CH CH H
H
CH3
Br 
The required change in hybridization
from sp3 to sp2 is underway in the TS
with the p-orbitals "growing in."
The stabilization energy of this developing
bonding  MO drives the stereoelectronic
requirements of the E2 transition state.
Br-
• The alkoxide bases are made from the
corresponding alcohols
127
The E2 Reaction
• E2 reaction involves concerted removal of the proton,
formation of the double bond, and departure of the leaving
group
• Both alkyl halide and base concentrations affect rate and
therefore the reaction is 2nd order
128
The E1 Reaction
• The E1 reaction competes with the SN1 reaction and
likewise goes through a carbocation intermediate
129
The E1 Reaction
The reaction of tert-butyl chloride in the mixed solvent of 80%
ethanol-20% water at 25o C yields a product mixture from two
competing reaction paths: substitution and elimination.
substitution
CH3
CH3CCl
CH3
83%
17%
elimination
CH3
CH3C-OR R = H, CH3CH2CH3
CH3
CH2=C
CH3
The rate of formation of isobutene is given by:
Rate = ke [tert-butyl chloride]
first-order kinetics
The notation for this
reaction that shows first
order kinetics is E1.
Note that the rate expression for the E1 reaction has the same
kinetic form as the SN1 reaction:
ke/ks = 17/83,
Rate = ks [tert-butyl chloride]
the ratio of the
two products.
first-order kinetics
A Mechanism for the Competing E1/SN1 Reactions
These two competing reactions have the same rate-determining step:
bond heterolysis to produce the tert-butyl carbocation:
bond heterolysis
CH3
RDS
CH3CCl
slow step
CH3
CH3
CH3C + +
CH3
Cl-
tert-butyl carbocation
After the rate-determining step, two competing modes of reaction
between the tert-butyl carbocation and water/ethanol (acting as
nucleophile/base) lead to substitution and elimination products.
: :
CH3
CH3C + + R-O-H
CH4 ethanol
or water
S N1
fast
E1
CH3
CH3C-OR ROH as
CH3
nucleophile
CH3
CH2=C
CH3
ROH as
base
The E1 Path: Deprotonation Step
The substitution pathway follows the usual course for an SN1
reaction. In a fast step, the carbocation intermediate reacts with a
nucleophile (water or ethanol) to yield the substitution product.
: :
CH3
CH3C + ROH
CH3
fast
fast
(-H+)
nucleophile
CH3
CH3COR
CH3
substitution product
Along the elimination pathway, in a fast step, a base (water or
ethanol) removes a beta proton from the carbocation to produce
the alkene product.
:
:
R-O
H
H
H C

H
+
C
CH3
CH3
fast
ROH2
+
H
H
C
C
CH3
CH3
In the above conformation, the bonding orbital of the beta-H is
aligned with the empty p-orbital. This stereoelectronic requirement
allows immediate overlap of the developing p-orbitals of the 
bonding molecular orbital.
Free Energy Diagram of Competing E1 and SN1 Pathways
(CH3)3CCl
(CH3)3C+
Free Energy
RDS
TS
G
E1
CH2=C(CH3)2
SN1
(CH3)3COR
The RDS leads to the carbocation intermediate
that sits in a "saddle point" where a partitioning
occurs between the two product forming paths.
Free Energy Diagram of the Product Forming Steps
The competition between the E1 and SN1 paths, and the resulting
product mixture, is determined by the relative magnitudes of the
activation energies, Gelim and Gsub , for the fast steps.
partitioning factor
83:17
Free Energy
Gsub (CH3)3C+Gelim
(CH3)3COR
SN1
CH2=C(CH3)2
E1
Substitution versus Elimination
• SN2 versus E2
• Primary substrate
– If the base is small, SN2 competes strongly because approach at
carbon is unhindered
• Secondary substrate
– Approach to carbon is sterically hindered and E2 elimination is
favored
135
• Tertiary substrate
– Approach to carbon is extremely hindered and
elimination predominates especially at high temperatures
136
• Temperature
– Increasing temperature favors elimination over
substitution
• Size of the Base/Nucleophile
– Large sterically hindered bases favor elimination because
they cannot directly approach the carbon closely enough
to react in a substitution
– Potassium tert-butoxide is an extremely bulky base and is
routinely used to favor E2 reaction
137
The SN2 versus the E2 Reaction
Because strong nucleophiles are usually also strong bases, the SN2 and E2
reactions are usually competing pathways as are the SN1 and E1.
SN2
E2
H C C X
(a)
(a)
(b)
(b)
Nu: -
C=C
Rate = kelim [RX] [Nu-]
H C C Nu
Rate = ksub[RX][Nu-]
Because the rate expressions for the substitution and elimination
reactions both depend on the concentrations of the substrate and the
nucleophile/base, the competition between the paths cannot be changed
by varying the concentration of either reactant.
However, this competition does very
much depend on the structure of RX
and choice of base.
Some Examples
The reaction of primary halides with unhindered alkoxide bases favors
substitution.
CH3CH2Br + CH3CH2ONa
ethanol
55o C
CH3CH2OCH2CH3 + CH2=CH2
SN2
(90%)
E2
(10%)
Secondary alkyl halides react with alkoxide bases to give mostly alkenes.
CH3CHCH3 + CH3CH2ONa
Br
ethanol
55o C
+ CH2=CHCH3
CH3CHCH3
OCH2CH3
SN2
(21%)
E2
(79%)
Tertiary halides do not react by the SN2 reaction. The observed
substitution product probably comes from a competing SN1 reaction.
CH3
CH3C-Br
CH3
+ CH3CH2ONa
ethanol
25o C
. 3
CH
CH3
CH3C-OCH2CH3 + CH2=C
CH3
CH3
SN1
(9%)
E2
(91%)
Sterically Hindered Bases Promote Elimination
While a sterically unhindered nucleophile/base gives:
CH3CH2CH2Br + CH3CH2ONa
ethanol
CH3CH=CH2 + CH3CH2CH2OCH2CH3
(9%)
(91%)
a sterically hindered nucleophile/base gives:
CH3
t-butyl alcohol
CH3CH2CH2Br + CH3COK
CH3
potassium tert-butoxide
(a hindered base)
CH3CH=CH2 + CH3CH2CH2OC(CH3)3
(85%)
(15%)
Steric constraints are much more severe in nucleophilic
substitution where the incoming nucleophile/base must
approach the carbon center.
Weak Bases Promote Substitution
CH3CHCH3 + CH3CH2ONa
Br
ethanol
55o C
CH3CHCH3 + CH2=CHCH3
E2
OCH2CH3
SN2
(21%)
=
O
CH3CONa
sodium acetate
(a weaker base)
CH3
CH3CHCH2Br + CH3CH2ONa
isobutyl bromide
CH3
CH3CHCH2Br
+ I-
CH3CHCH3 (~100%)
OCCH3
=
+
CH3CHCH3
Br
SN2
O
But,
(79%)
ethanol
55o C
CH3
CH3
CH3CHCH2OCH2CH3 + CH2=C
CH3
(40%)
(60%)
branching slows down SN2
acetone
CH3
CH3CHCH2I
+ Br-
(100%)
Iodide ion is a very weak base and also is much more
polarizable than alkoxide ions. Substitution is favored
by nucleophiles with high polarizability.
Quiz Chapter 6 Section 18
For the following reactions, which mechanistic pathway will
dominate among SN1, SN2, E1 and E2?
CH3
CH3CHCH2Br + KOBu-t
t-BuOH
E2
Br
CH3CCH3
CH3
CH3OH
SN1
acetone
SN2
Br
CH3CH2CHCH3 +
NaI
Overall Summary
143
Overall Summary of Substitution/Elimination
Reactions of Alkyl Halides
SN1
RCH2X
R'
RCX
R''
E1
E2
yes,
very fast
CH3X
R'
RCHX
SN2
but hindered
bases give
mostly alkenes
mostly
very
little
very
favorable
mostly SN2
with weak
bases
none
very
little
strong bases
promote E2
always
strong bases
competes promote E2 path
with SN1