Acidity Concepts-1
D. A. Evans, F. Michael
Chem 531
„ Definition of Ka
Activation of Organic Molecules
Let H–X be any Bronsted acid. In water ionization takes place:
H3O+ +
H–X + HOH
„ Base Activation
R1
C
R2
R1
base
H
- H-base
R3
Nucleophile
C
R2
R3
where
[H3O+] [X–]
Keq =
[H–X] [HOH]
X–
(A)
where [HOH] = 55.5 mol L-1
Since [HOH] is, for all practical purposes, a constant value, the acid
dissociation constant Ka is defined wilthout regard to this entity. e.g.
pKa describes quantitatively a molecule's propensity to act as an acid, i.e. to
release a proton.
Hence
- Medium effects
- Structural effects (influence of substituents R1)
H+ + X–
H–X
Ka =
where H+ = H3O+
[H+] [X–]
[H–X]
(B)
From the above definitions, Ka is related to Keq by the relation:
„ Acid Activation
Ka(H–X) = 55.5 Keq(H–X)
R1
acid (protic or Lewis acid)
R1
Electrophile
X
X
R2
R2
(C)
H (LA)
„ Autoionization of water
H3O+ +
HOH + HOH
HO–
–18
Keq = 3.3 X 10
X = e.g. O, NR ...
„ the aldol example
O
M
O
R
R
+
O
O
M
base catalysis
H
R
R
O
O
R
R
+
H
O
R
R
M
SiMe3
R
+
H
R
O
R
Ka = 1.8 X 10–16
Since pKa is defined in the following equation:
The pKa of HOH is +15.7
pKa = – log10 [Ka]
OH
R
O
Hence
The strongest base that can exist in water is HO–.
R
acid
O
Ka = 55.5 Keq = 55.5(3.3 X 10–18)
R
R
base
From Eq C:
H3O+
O
SiMe3
R
R
+
H2O
obviously:
H3O+
+
H2O
Keq = 1
Ka = [HOH] x Keq
acid catalysis
R
Lets now calculate the acid dissociation constant for hydronium ion.
hence Ka = 55.5
pKa = – log10 Ka = –1.7
The strongest acid that can exist in water is H3O+.
Acidity Trends for Carbonyl & Related Compounds
D. A. Evans, F. Michael
„ The Gibbs Relationship
or
„ Medium Effects on the pKa of HOH
∆ G° = - RT ln K
2.3 RT = 1.4
∆ G° = – 2.3 RT log10 K
at T = 298 K
in kcal ⋅ mol-1
∆ G°298 = - 1.4 log10 Keq
** The gas phase ionization of HOH is
endothermic by 391 kcal/mol !!!
with pK = – log10 K
0
10
-1
- 1.4
100
-2
- 2.8 kcal/mol
HOH
DMSO
Vacuum
DMSO
HOH
∆ pKa
HOH
31.2
15.7
15.5
HSH
14.7
7.0
7.7
MeOH
29.0
15.3
13.7
C6H5OH
18.0
9.9
8.1
O2N–CH3
17.2
10.0
7.2
24.6
17
7.6
A
Energy
0
15.7
32
Substrate
∆ G°
1
Medium
279 (est)**
Hence, pKa is proportional to the free energy change
pKeq
HOH pKa
„ Representative pKa Data
∆ G°298 = 1.4 pKeq ≈ 1.4 pKa
Keq
Chem 531
HA
∆ G°
Reaction coordinate
O
„ Medium Effects
H A
+
Ph
Consider the ionization process:
solvent
A: –
solvent(H+)
+
In the ionization of an acid in solution, two charged species are created from a neutral
species (usually). The ability of the medium to stabilize these charged species plays
an important role in the promotion of ionization. Let us consider two solvents, HOH
and DMSO and the performance of these solvents in the ionization process.
The Protonated Solvent
The change in pKa in going from water to DMSO is increasingly diminished as
the conjugate base becomes resonance stabilized (Internal solvation!).
Substrate
Conjug. Base Stabiliz.
O
H
Water
H
O+
H
O
H
A–
HO
S+
No H-bonding Capacity
Me
As shown above, although HOH can stabilize anions via H-bonding, DMSO cannot.
Hence, a given acid will show a greater propensity to dissociate in HOH. As
illustrated below, the acidity constants of water in HOH, DMSO and in a vacuum
dramatically reflect this trend.
OEt
O
Me
HOH
∆ pKa
18.1
16.0
2.1
16.4
13.3
3.1
13.3
8.9
4.5
11.1
11.2
O
Me
NC
DMSO
O
EtO
H
Me
DMSO
C CH3
CN
0
Acidity Trends for Carbonyl & Related Compounds
D. A. Evans, F. Michael
Chem 531
Inductive Effects
Substituent Effects
Electronegativity, Hybridization, Inductive Effects, and Resonance
„ Alkyl Substituents on Localized Carbanions are Destabilizilng:
A) inductive effect. B) steric hinderance of anion solvation
Electronegativity
e.g. Compare Carboxylic Acids vs. Ketones
(JACS 1975, 97, 190)
Compare:
H
H
C
H
O
C
H
H
O
C
H
O
C
CH2–H
H
R
pKA ≈ 19
pKA = 4.8
(DMSO)
Carboxylate ion
more stabile than
enolate because R
C
O more
O
electronegative than C
O–
O–
C
CH2
29
H
PhSO2-CH–Me
31
H
(DMSO)
- s-character of carbon hybridization
Hybridization
pKa (DMSO)
PhSO2-CH–H
pKa (DMSO)
PhSO2-CH-OCH3
2
sp -orbitals 33% s-character
sp-orbitals
H
S
H
S
Me
S
H
pKa (DMSO)
31.1
38.3
„ Heteroatom-Substituents: - 2nd row elements of periodic table
sp3-orbitals 25% s-character
Remember:
S
30.7
H
50% s-character
PhSO2-CH-OPh
Inductive Stabilization versus
Lone Pair Repulsion
(-I vs +M -Effect)
27.9
H
Carbon Acids
R
R
R
R
H
PhSO2-CH-NMe3
R
RR
RR
H
Inductive Stabilization
H
H
Hybridzation
sp
sp2
≈ sp2
sp3
Bond Angle
180°
120°
≈ 120
109°
pKa(DMSO)
19.4
H
25
40
≈ 39
50
„ Heteroatom-Substituents: - 3nd row elements of periodic table
Strong carbanion stabilizing effect
pKa (DMSO)
PhSO2-CH-H
H
PhSO2-CH-SPh
H
29
20.5
PhSO2-CH-SO2Ph
H
PhSO2-CH-PPh2
H
pKa (DMSO)
12.2
20.5
Acidity Trends for Carbonyl & Related Compounds
D. A. Evans, F. Michael
„ Carbanion Stabilization by 3rd–Row Atoms: SR, SO2R, PR3 etc
Chem 531
Resonance
„ Conjugative Stabilization of Conjugate Base
Me +
S CH3
O
H3C
Me
18.2 (DMSO)
Ph +
P CH3
Ph
Ph
22.5
S
O
S
H
S
H
CH3
H3C
31
pKA (DMSO)
O
S
31
CH3
H
H
C
35
NO2
H
The accepted explanation for carbanion stabilization in 3rd row
elements is delocalization into vicinal antibonding orbitals
CH3
H
σ*S–X
X
C
S
nC (filled)
This argument suggests a specific orientational requirement. This has been noted:
Me
He
S
Ha
31.5
N
pKA (DMSO)
Ph3C–H
5.2
31.5
O
H
He
Rates for deprotonation with n-BuLi
He : Ha = 8.6
(JACS 1974, 96, 1811)
H
O
He
He : Ha = 30
Ha
C
H
pKA
Ha
S
C
N
„ Impact of Stereoelectronic Requirement for Carbanion Overlap
O
S
C
For efficient conjugative stabilization, rehybridization of carbanion orbital
from nsp3 to np is required for efficient overlap with low-lying π*-orbital of
stabilizing group. However, the cost of rehybridization must be considered.
Anti (or syn) periplanar orientation of carbanion-orbital and σ* orbital mandatory
for efficient orbital overlap.
Me
CH3
H
H
σ*S–X (empty)
26.5
C
H
C
nC
O
C
C
H
E
O
H
O
C
17.2
N
H
H
(JACS 1976, 98, 7498; JACS 1977, 99, 5633; JACS 1978, 100, 200).
O
C
(JACS 1978, 100, 200)
47.7
C-H acidity not
detectable
O
H
Explain why 1 and 3 are ~4 pKa units more acidic than the corresponding
acyclic counterparts 3 and b. (Bordwell, J. Org. Chem. 1994, 59, 6456)
O
O
O
1
O
2 Me
O
Et
HN
O
O
Me
N
H
O
Et
Kinetic & Thermodynamic Acidity of Ketones
D. A. Evans, F. Michael
Consider enolization of the illustrated ketone under nonequilibrating conditions:
Kinetic Acidity
OLi
HB
HA
HB
Me
O
HA
LiNR2
HB
Me
LiNR2
HB
kB
B–
OLi
Me
O
O
(1) Me
(99)
(90) Me
(10)
A–
A‡
K
Kinetic & Equilibrium Ratios of Enolates Resulting from Enolization
with LDA & Subsequent Equilibration
HB
kA
Chem 531
Equilibrium
Ratios
Kinetic Ratios
B‡
∆G
‡
O
(98)
O
(2)
(66)
(34)
Kinetic Ratios
A
Equilibrium
Ratios
Energy
∆G‡B
O
K
LiNR2
B–
(13) H
(87)
Kinetic Ratios
Reaction Coordinate
Kinetic acidity refers to the rate of proton removal. e.g. kA vs kB. For
example, in reading the above energy diagram you would say that HA has a
lower kinetic acidity than HB. As such, the structure of the base (hindered vs
unhindered) employed plays a role in determining the magnitude of kA and kB.
For the case shown above, ∆G‡A will increase more than ∆G‡B as the base
becomes more hindered since the proton HA resides in a more sterically
hindered environment. The example shown below shows the high level of
selectivity which may be achieved with the sterically hindered base lithium
diisopropylamide (LDA).
H
Me
N
Li
H
O
H
OLi
Me
THF
-78 °C
H
H
H
Me
OLi
H
Me
Me
LDA
C3H7
Kinetic Ratio 99 : 1
Equilibrium Ratio 10 : 90
Me
H
Equilibrium
Ratios
(14)
(16)
CH3 C3H7
(84)
Kinetic Ratios
(87)
CH3
(13)
Equilibrium
Ratios
O
O
Ph
O
O
(53)
H
A–
Me
(47) H
O
CH3
(86)
Kinetic Ratios
Ph
(99)
CH3
(1)
Equilibrium
Ratios
Note that alkyl substitution stabilizes the enolate (Why??). This effect
shows up in the equilibrium ratios shown above.
Hence, enolization under kinetic control with LDA allows you to produce
the less-substituted enolate while subsequent equilibration by simply
heating the enolate mixture allows equilibration to the more substituted
enolate.
Kinetic Acidity: Carbon versus Oxygen Acids
Evans, Annis, Michael
„ Kinetic Acidity
Observation: The thermodynamic acidities (pKas) of phenol and
nitromethane are both approximately 10; however, using a common base,
phenol is deprotonated 10+6 times as fast.
Base
O
H
O
H3C
N
O
O
rel rate: 10+6
O
Base
H2C
rel rate: 1
N
O
Proton transfers from C-H bonds are slow.
„ Why???
Most carbon acids are stabilized by resonance. Hence significant
structural reorganization must accompany deprotonation.
Base
O
O
H
O-H electron density
is here.
H
H
O
N
H
O
C-H electron density
is here.
O-H electron density
is still here.
Base
O
H
N
H
O
C-H electron density
now resides here, and nuclei
have moved to accomodate
rehybridization.
The greater the structural reorganization during deprotonation, the lower
the kinetic acidity
Chem 531