Chemistry 125: Lecture 44
January 26, 2011
This
Nucleophilic Substitution
and Mechanistic Tools:
Stereochemistry, Rate Law,
Substrate, Nucleophile
For copyright
notice see final
page of this file
SN2 Nucleophilic
Substitution
Generality
of Nucleophilic
Substitution
Solvent
Nu:
R-L
Nucleophile
Substrate
(+)
(-)
Nu-R
L
Product
Leaving
Group
But there are different mechanisms!
the Pragmatic Logic
of Proving a Mechanism
with Experiment & Theory
(mostly by disproving all
alternative mechanisms)
"It is an old maxim of
mine that when you have
excluded the impossible,
whatever remains,
however improbable,
must be the truth."
The Adventure of the Beryl Coronet
SN2 Nucleophilic Substitution
Nu:
R-L
(+)
Nu-R
(-)
L
Break bond Make bond
the Pragmatic
Logic
(Dissociation)
(Association)
Simultaneous
of
Proving
a
Mechanism
D then A
A then D
“Concerted”
with Experiment &(make-as-you-break)
Theory
(mostly by disproving all
alternative mechanisms)
Concerted
Nu
C
L
Transition
State
A/D
Nu
C
L
Pentavalent
Intermediate
D/A
C
Trivalent
Intermediate
Concerted
a
Nu
A/D
b
C
a
L
c
Transition
State
Nu
enantiomers
a b
b
C
D/A
L
c
Pentavalent
Intermediate
Nu C Nu
c
Trivalent
Intermediate
chiralfor
chiral
achiral
Unlikely
Which
is it
very
exothermic
normally?
processStereochemical Implications!
(
(Hammond implausibility)
Tools for Testing
(i.e. Excluding) Mechanisms:
Stereochemistry (J&F sec 7.4b)
Rate Law (J&F sec 7.4a)
Rate Constant (J&F sec 7.4cdefg)
Structure
X-Ray and Quantum Mechanics
Nucleophilic Substitution
Walden
N Inversion
+ RL (1898) - “the most
L +astounding
RN
discovery in stereochemistry since the
groundbreaking work of van’t Hoff.” E. Fischer
N +
C
L
L
N
+C
C+ L
N
Displacement
Replacement
STEREOCHEMISTRY
Kenyon and Phillips (1923)
nucleophilic substitution at S
O H
PhCH2 CH
+33° CH3
(A/D, A favored by vacant d orbital of S)
PhCH CH
CH3
-OH
CH3
O SO2
PhCH2 CH
+31°
Why
notBackside
avoid AttackHin CH
3
nucleophilic
acetate
steps by substitution
nucleophilic
O
substitution
at
C=O
using OH?
Because
at
saturated
O C CH3
(A/D, A favored by C.
*)
C
it
attacks
H
(the only step involving chiral C)
Same as
O
O
O
starting
O C CH3
material? O H
O C CH3
HO OH
PhCH2 CH
PhCH2 CH OH
PhCH2 CH
CH3
CH3
CH3
Inversion!
-7°
-32° (R)  (S)
Proves nothing
Cl SO2
CH3
Concerted
Nu
C
L
Pentavalent
Transition State
A/D
Nu
C
L
Pentavalent
Intermediate
D/A
C
Trivalent
Intermediate
Trivalent intermediate could be attacked from
either face  racemization, not inversion.
Tools for Testing
(i.e. Excluding) Mechanisms:
Stereochemistry
Rate Law
Rate Constant
Structure
X-Ray and Quantum Mechanics
NaOEt + EtBr
EtOEt + NaBr
rate
d[EtO-]
dt
Second Order (SN2)
= k2 [EtO-] [EtBr]
0
[NaOEt]
( fixed [EtBr] )
Concerted
A/D
D/A
Nu enters
Nu
C
L
Pentavalent
Transition State
Nu
C
Nu enters
L
Pentavalent
Intermediate
C
Trivalent
Intermediate
Initial rate-limiting dissociation in D/A would
give a rate independent of [Nu], not SN2.
 Not D/A
NaOEt + EtBr
Analogy
EtO- + EtBr
H+  EtOH
EtOEt
H
H
pKa
15.7
-1.7
EtOEt + NaBr
1017.4
at equilibrium
d[EtO-]
rate
EtO: +
H
++
+
EtBr
H EtOH
EtOEt
EtOH
dt
Ratio should be
much less drastic
at early SN2
transition state.
Second Order (SN2)
= k2 [EtO-] [EtBr]
+ k1 [EtBr]
[EtOH] [EtBr]
~ const
Pseudo
First Order
First(D/A?)
Order
k2 = 20,000  k
0
[NaOEt]
Is it reasonable to
be so different?
Tools for Testing
(i.e. Excluding) Mechanisms:
Stereochemistry
Rate Law
Rate Constant
Structure
X-Ray and Quantum Mechanics
Rate Constant Dependance on
Nu:
R-L
Nucleophile
Substrate
23x
Leaving
Group
R
krel
CH3
145
CH3CH2
[1]
CH3CH2CH2
0.82

1.2x

(CH3)2CH
(CH3)2CHCH2
3000x
Solvent
Nu-R
L
Product
RBr + Iacetone / 25°C
C-L
antibonding
node
128x
~same
H
0.036
>15x
0.000012
LUMO
145x
0.0078
(CH3)3C ~ 0.0005 ?
(CH3)3CCH2
(-)
(+)
Surface Potential
+26 to -25 kcal/mole
Something else happens
e.g. J&F Table 7.1 p. 275
Methyl
Ethyl
iso-Propyl
Steric Hindrance
Total Density (vdW)
-Methylation
t-Butyl
Methyl
Ethyl
iso-Propyl
LUMO
at
0.06
LUMO
at
0.04
Total Density (vdW)
-Methylation
t-Butyl
Methyl
Ethyl
iso-Propyl
Surface Potential
+26 to -25 kcal/mole
-Methylation
t-Butyl
Ethyl [1]
n-Propyl 0.82
iso-Butyl 0.036
-Methylation
Neopentyl
0.000012
No way to avoid the third -CH3
Nu
C
L
C
Transition
State
Backside Attack
Planar
Trivalent
Intermediate
Might it be possible to
have frontside attack?
or formation of a
non-planar cation?
(remember planar BH3)
C
Nu
L
Transition
State
Frontside Attack
C+
Nonplanar
Trivalent
Intermediate
“In 1939 Bartlett and Knox published
the account of their work on the bridgehead chloride, apocamphyl chloride. I
believed then, and I believe now, that this
was a fantastically influential paper. For
thirty years afterwards, no one really
accepted any mechanism unless it had
been tested out on a bridgehead case.
Indeed, the Bartlett-Knox paper shaped
the interests and viewpoint of many
chemists about the kind of physical
organic they wanted to do.”
John D. Roberts
Caltech
1975
Molecule specifically
designed and prepared
to test these
mechanistic questions
Bartlett and Knox *
(J.Am.Chem.Soc. - 1939)
“bridgehead” chloride
Cl
boat c-hexane
with a bridge
bicyclo[2.2.1]heptane
Flattening would generate highly
strained angles (estimated >23 kcal/mole).
Cation would not be planar.
Backside of s*C-Cl is inaccessible,
and inversion would be impossible.
Attack would have to be frontside.
Bartlett and Knox *
(J.Am.Chem.Soc. - 1939)
“C=C bonds cannot originate
from such a bridgehead.”
(Bredt’s Rule)
Although there are -H
atoms, they are not in the
anti position necessary to
allow sCH - s*C-X overlap
during elimination of H-X
to form C=C.
Horrid
Overlap!
gauche
H
H
H
Would competition from loss
of HCl make it impossible to
measure the expected really
slow rate of substitution?
Bartlett and Knox *
(J.Am.Chem.Soc. - 1939)
C
Nu
L
C+
>>106 slower than
typical backside attack
>109 slower than from
Et(CH3)2C-Cl
60°cooler and without Ag+
pull on Cl instead of pushing at C
R-Cl: + Ag+
R+ + AgCl ( )
Bartlett and Knox *
(J.Am.Chem.Soc. - 1939)
Cycloalkyl Halides (e.g. J&F Table 7.2)
krelative
~109°
60
strain°
in starting material
90
°
109
°
[1]
<0.0001
???
Br
C
120
C
°
C
H sp2
I
increased strain in transition state
0.008
1.6
0.01
OK
bent
Rate Constant Dependance on
Nu:
Nucleophile
Nu
Solvent
R-L
(+)
Nu-R
Leaving
Substrate Group
krel
(-)
L
Product
pKa (NuH+)
H2O
[1]
-1.7
F-
80
3.2
Cl-
1,000
-8
Br-
10,000
-9
HO-
16,000
15.7
I-
80,000
-10
HS-
126,000
7
For first-row elements
nucleophilicity (attack sC-L )
parallels basicity (attack H+).
Both require high HOMO.
But as atoms get bigger, they
get better at attacking sC-L
(compared to attacking H+)
e.g. J&F Sec. 7.4d, Table 7.3
Rate Constant Dependance on
Nucleophile
Nu
R-L
Leaving
Substrate Group
krel
pKa
[1]
-1.7
F-
80
3.2
Cl-
1,000
-8
Br-
10,000
-9
HO-
16,000
15.7
I-
80,000
-10
HS-
126,000
7
Nu-R
L
Polar solvents accelerate reactions
that generate (or concentrate) charge,
and vice versa.
(NuH+)
krel
CH3I
in H2O
harder [1]
to break 14
H-bonds
to smaller
ions 160
krel
CH3Br
in Acetone
Backwards
H2O
(-)
(+)
11
5
Sensible
Nu:
Solvent
[1]
e.g. J&F Sec. 7.4dg
End of Lecture 44
Jan. 26, 2011
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J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0