Nucleophilic Substitution
X
+
nucleophilic
substitution
Nu
Nu +
X
Nu
is the nucleophile
X
is the leaving group
nucleophilic
substitution
CH3
Br + Nu
Nucleophile
HO
CH3O
H3N
CH3
Nu +
Br
Product
CH3OH
CH3OCH3
CH3NH3
H
H2O
CH3
O
H
Base
CH3OH
OH-
Br
OH
NH3
Br
NH3
But where do alkyl bromides come from?
Br
HBr
But what about a primary bromide?
OH
HBr
Nucleophilic Substitution
works both ways!
Br
Mechanisms of Nucleophilic
Substitution Reaction
There is more than one.
The mechanisms are determined
by studying the rate of the reaction.
A rate of a reaction refers to the
change in concentration of the
reactants or products versus time.
First Case:
CH3
Br + Nu
Rate =
- d [CH3Br]
dt
CH3
Nu +
Br
= k [CH3Br] [Nu]
The rate that the CH3Br disappears
is proportional to the concentration of
the CH3Br and the concentration of the
nucleophile.
First Case:
CH3
Br + Nu
Rate =
- d [CH3Br]
dt
CH3
Nu +
Br
= k [CH3Br] [Nu]
The rate depends on the concentration
of two components so it is a bimolecular
reaction.
First Case:
CH3
Br + Nu
CH3
Nu +
Br
The reaction is called an:
SN2 reaction
S for substitution
N for nucleophilic
2 for bimolecular
Transition
State

HO


Br
Ea
HO- + CH3Br
H
HOCH3 + Br-

HO
H


Br
HH
Transition State. Exists only for a
very short time.
The key step in the reaction is
a collision between the two
reactants.
This means that an increase in
the concentration of either reactant
will result in a direct increase in the
rate.
Rate =
- d [CH3Br]
-]
=
k
[CH
Br]
[OH
3
dt
Second Case:
Br
(CH3)3CBr
Kind of Strange
HO
OH +
Br
No
substitution (CH3)3COH
H2O
OH
+
HBr
(CH3)3COH
No substitution with HO-, but reacts with H2O
No substitution with HO-, but reacts with H2O
HO- is much more basic than H2O.
It should be a better nucleophile.
Most interestingly the rate does not
depend upon how much H2O is present.
Rate =
- d [(CH3)3CBr]
dt
= k [(CH3)3CBr]
The rate does not depend upon
concentration of H2O
The key step in the reaction can not
be bimolecular. It must be unimolecular.
It turns out to be a multi step reaction:
Step one is ionization to give the
t-butyl carbocation and bromide.
+
Br
(CH3)3CBr
(CH3)3C
CH3
CH3
CH3
Br
+
Br
(CH3)3CBr
Step 1
Slow
Br
(CH3)3C
H
(CH3)3C
H2O
Step 2
Fast
H
O
H
(CH3)3COH2+
O
Step 3
Fast
(CH3)3COH
H+
The key is the relative rate of the steps.
+
Br
(CH3)3CBr
Step 1
Slow
Br
(CH3)3C
The slow step determines the
rate of the reaction. It is unimolecular.
Rate =
- d [(CH3)3CBr] = k [(CH3)3CBr]
dt
The reaction is called an:
SN1 reaction
S for substitution
N for nucleophilic
1 for unimolecular
Br
CH3
Br
(CH3)3CBr
Why does t-butylbromide react via
an SN1 reaction while methylbromide
reacts via an SN2 reaction?
Size Matters
Steric Hindrance
HO
Why does t-butylbromide react via
an SN1 reaction while methylbromide
reacts via an SN2 reaction?
1. Steric Hindrance
2. More stable carbocation
CH3
H
versus
CH3
CH3
H
H
Transition
State

Intermediate
Br
Transition
State
H
O
H
Ea
(CH3)3C
Br
+ H2O
(CH3)3C OH2
+ Br-
Transition States occur at a
maximum of a Energy Profile.
T.S.
T.S.
Int.
Intermediates occur at a minimum
of an Energy Profile. They are
potentially isolatable species.