CHAPTER 1 SUBSTITUENT EFFECTS ON
ORGANIC RATES AND EQUILIBRIA
I QUALITATIVE DESCRIPTION OF SUBSTITUENT EFFECTS
References:
1.
Carey & Sundberg Part A, 3rd. ed. p.196 (4th Ed, p. 204) (5th Ed. pp. 297-318)
2.
Isaacs, p. 135
3.
Carroll, p. 366
4.
J. Hine, Substituent Effects on Equilibria in Organic Chemistry, WileyInterscience, 1975, Chapt. 2. QD 503.H56
5.
R.D. Topson. Acc. Chem. Res. 16, 292(1983)
[Lowry and Richardson and Miller texts have no non-quantitative analysis of
substituents.]
A INTRODUCTION
Substituent effects on the rates and equilibria of organic reactions are of two types:
a) Steric Effects. Steric substituent effects arise from the size (i.e., space requirements)
of the substituent and the fact that strong forces of repulsion result when two atoms are
forced to be closer than the sum of their van der Waals radii.
b) Electronic Effects. Electronic substituent effects are a result of changes in the
electronic distribution within a substrate, caused by the substituents on that substrate.
Electronic effects have been ascribed to several different processes, or mechanisms of
electron displacement, the principle ones being the following:
i) field effect: an electrostatic effect of a charge or dipole transmitted through space
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ii) inductive effect: the effect from differences in electronegativity between a substituent
X and the carbon to which it is attached, transmitted through σ bonds. For a
substituent such as Cl (i.e., more electronegative than C), the effect is often pictured
in the following way:
−
δ
Cl
+
δ
CH2
+
δδ
CH2
+
δδδ
CH2
CH3
iii) resonance (or mesomeric) effects: delocalization of the π electrons in a conjugated
system, affecting alternate positions.
Field effects and inductive effects are usually difficult to separate because they operate
in the same direction. The combined field and inductive effects are often termed polar
effects. Quantum-mechanical calculations indicate that inductive effects through σbonds should be negligible beyond the first two carbons. In a few specifically
constructed rigid systems (e.g., page 1-5), inductive and field mechanisms operate in
opposite directions so that their relative importance of such systems can be assessed.
These studies suggest that inductive effects through σ-bonds should be negligible
beyond the first two carbons.
B CLASSIFICATION OF SUBSTITUENTS (C.K. Ingold)
-substituents can be classified as + or - I and + or - R.
“-“ means electron withdrawing
“+“ means electron donating (or supplying or releasing)
“I” means polar effects (through σ bonds and space)
“R” means resonance effects (delocalization of π electrons affecting alternate
positions)
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To summarize:
+R π-donating;
-R π-withdrawing
+I σ-donating
-I σ-withdrawing
Often the types of substituents are grouped together:
+R, +I
+R, -I
-R, -I
alkyl,
-NH2, -OH, -X, -SH
O
CH3-,
-NR2, -OR, -SR
C
CH3CH2-,
-N
CH3
C
trialkylsilyl-,
examples:
H
O
-O
C
CH3
O
O
C
R
O
OR
C N
SO3H
NO2
SO2R
C
+R
O
CH3
+ CH3
O
etc.
-
-R
-
O
N+
O
-
O
-
O
N+
etc.
+
C EXAMPLES OF SUBSTITUENT EFFECTS ON BRÖNSTED ACIDITY
a) Polar Effects
i) pKa’s of aliphatic carboxylic acids (H2O, 25 °C)
Effect of Multiple Substitution on pKa.
Compound
pKa
single ∆pKa
total ∆pKa
CH3CO2H
4.74
--
--
ClCH2CO2H
2.87
1.87
1.87
Cl2CHCO2H
1.29
1.58
3.45
CCl3CO2H
0.65
0.64
4.09
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OH
pKa’s of Monosubstituted Carboxylic Acids, X-CH2-C(O)OH
X
pKa
X
pKa
-NO2
1.68
-I
3.16
+
N(CH3)3
1.83
C CH
1.84
-OCH3
3.53
-C(O)CH3
3.58
-SO2CH3
2.36
-SCH3
3.72
-CN
2.47
-C6H5
4.31
-F
2.66
-CH=CH2
4.26
-C(O)OH
2.83
-H
4.74
-Cl
2.87
-CO2-
5.69
-Br
2.90
Effect of Distance of Substituent on pKa
Compound
pKa
∆pKa
CH3 CH2 CH2 CO2 H
4.82
--
Cl
CH3CH2CHCO2H
2.84
1.98
Cl
CH3CHCH2CO2H
4.06
0.76
Cl
CH2CH2CH2CO2H
4.52
0.30
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ii)
angular dependence of substituent effects
Grubbs and Firzgerald, Tetrahedron Letters, 4901(1968)
(50% aq. ethanol, 25 °C)
Cl
Cl
Cl
Cl
CO2 H
pKa:
6.26
CO2 H
6.08
CO2 H
5.68
b) Resonance Effects
i) pKa’s of benzoic acids and phenols (H2O, 25 °C)
-I, +R group:
CO2H
CO2H
CO2H
+R effect of CH3O- group
puts negative charge on the
carboxyl group
OCH3
-I effect only
pKa:
4.18
OCH3
4.09
4.47
OH
OH
-I, -R group:
OH
NO2
-I effect only
pKa:
9.92
8.40
-R effect of nitro group
stabilizes negative charge
of the phenolate
NO2
7.15
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c) Hydrogen Bonding
CO2H
CO2H
CO2H
CO2H
CO2H
OH
HO
OH
OH
OH
pKa:
4.18
4.08
CO2H
OH
pka:
2.98
4.58
2.98
2.30
O - O
C
H
O
CO2H
OMe
4.09
d) Steric Inhibition of Resonance
i) pKa’s of benzoic acids and phenols (H2O, 25 °C)
CO2H
CO2H
CO2H
CO2H
CO2H
CH3 H3C
CH3
CH3
CH3
pKa:
4.18
4.24
CO2H
4.34
CO2H
CH3
pKa:
3.91
3.91
CO2H
Et
3.79
3.25
CO2H
iPr
3.64
tBu
3.46
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OH
OH
H3C
CH3
OH
H3C
OH
CH3
H3C
CH3
H3C
NO2
NO2
pKa:
10.60
CH3
7.16
10.17
∆pKa = 3.44
8.25
∆pKa = 1.79
ii) pKa’s of ammonium ions (H2O, 25 °C)
H
+
N(CH3)2
pKa:
5.06
H
+
N
7.79
∆ = 2.73
H
+
N
10.58
∆ = 2.79
e) Hyperconjugation
Another effect that will be called upon through the course should be introduced here.
Whereas conjugation is the resonance shifting or distribution of π electrons by way of π
bonds and lone pairs, hyperconjugation is the movement of electrons through σ bonds.
As with normal resonance, it too requires alignment of the orbitals involved. The general
concept is introduced here and will be explained in more detail in later sections. When
negative charge is involved, the effect is sometimes called negative hyperconjugation,
whereas positive charge is stabilized by positive hyperconjugation. See King, Li, Cheng,
& Dave, Heteroatom Chemistry, 2002, 13, 397 and Alabugin & Zeidan, J. Am. Chem.
Soc. 2002, 124, 3175.
The MO energy level diagram shows why electronegative atoms as more willing
participants in negative hyperconjugation.
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Orbital interaction diagrams comparing negative hyperconjugation in -C-C-X with (a) X =
carbon and (b) X = oxygen. The left side of each diagram shows the interaction of the
filled p-orbital of the carbanion with each σ* orbital. The resulting energy lowering is
given by ∆ε CC for X = C and ∆ε CO for X = O; ∆ε CO > ∆ε CC, the difference being due to
the lower energy of σ*CO vs σ*CC, the origin of which in turn is shown on the right side of
each diagram.
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II QUANTITATIVE CORRELATION OF RATES AND EQUILIBRIA
References:
1.
Carey and Sundberg, Part A, pp. 196-209 (3rd ed.) (4th Ed. pp. 204-215) (5th Ed.
pp. 335-344)
2.
Carroll, p. 371
3.
Miller, p.122
4.
L&R pp. 143-158
5.
Isaacs Chapter 4
6.
Anslyn/Dougherty, pp 445-451
6.
J. Hine, Structural Effects on Equilibria in Organic Chemistry (see p.1)
7.
C.D. Johnson, The Hammett Equation, Cambridge, Univ. Press, (QD 502 J63).
8.
Hansch, Leo & Taft Chem. Rev. 91, 165(1991).
9.
For a short biography on Louis Hammett, see J. Shorter, Prog. Phys. Org. Chem.
17, 1(1990)
A THE HAMMETT EQUATION
The Hammett equation is one of several important linear free energy relationships. It was
developed as a correlation of reactivities (rates) and equilibria in reactions of meta- and
para-substituted benzene derivatives.
As early as 1912 it was noted that in reactions of m- and p-substituted benzene derivatives
there were regularities in substituent effects. For example, whatever effect a p-NO2 group
had on a rate or equilibrium constant, a m-Cl had an effect in the same direction but
smaller. Hammett (1935) found that for a number of reactions involving a series of m- and
p-substituted benzene derivatives, a plot of the log of the rate constant (k) or equilibrium
constant (K) for one reaction vs log k or log K for another reaction gave a fairly straight line.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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Examples
O
O
O
1
Et
+
-
OH
-
k2
X
O
+ EtOH
X
(85% aq. ethanol, 25 °C)
O
O
O
2
H
+
H2O
O
K
X
-
+
H3O+
X
(H2O, 25 °C)
CH3
N
CH3
3
X
+
CH3I
CH3
+
N
CH +
CH3 3
k2
I-
X
(95% aq. acetone, 35 °C
A plot of log k2 for reaction 1 vs. log K for reaction 2 gives a good straight line (slope ca.
2.6) if the points for the ortho substituents are omitted. Similarly, a plot of log k2 for reaction
1 vs. log k2 for reaction 3 gives a reasonable straight line of slope ca -1.0 (see graphs,
starting on page 1-11).
In all such correlations the rate and equilibrium constants within a series must all be
measured under a single set of conditions (same temperature, solvent). Satisfactory linear
correlations are generally obtained as long as substituents are restricted to the meta and
para positions. Ortho substituents give poor correlations. For example, in the plot of log k2
(reaction 1) vs. log Ka, the ortho-substituted esters are consistently less reactive than
predicted by the correlation line for the meta and para substituents. Similarly, the same
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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procedure applied to aliphatic systems generally does not give a good linear correlation. (A
plot of log k2 for basic hydrolysis of XCH2CO2Et vs. log Ka for XCH2CO2H is shown.)
To put all such correlations on the same basis, log Ka for benzoic acids in H2O at
25°C was taken as the standard reaction. For all other reactions we have a relationship of
the type
log k = ρlog Ka (benzoic acids, H2O, 25°) + C
or
log K = ρlog Ka + C
(1)
For no substituent on the benzene ring,
log ko or log Ko = ρlog Ka° + C
(2)
Subtracting (2) for (1),
log (k/ko) or log (K/Ko) = ρlog(Ka/Ka°)
log k for
2
O
OEt
X
+
-OH
85% aq. ethanol, 25 °C
log Ka, benzoic acids, H2O, 25 °C
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(3)
log k for
2
O
OEt
X
+
-OH
85% aq. ethanol, 25 °C
log k2 for
Me
N
X
Me
+ Me-I
90% aqueous acetone, 35 °C (k2 in M-1 min-1)
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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log(k/ko) for
O
X
OEt
+ -OH
85% aq. ethanol, 25 °C
log (K/Ko), ionization of X-CH2-CO2H, H2O, 25 °C
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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log k2
Me
N
X
Me
+ Me-
90% aq. ethanol, 30 °C
log Ka, benzoic acids, H2O, 25 °C
Equation (3) (p. 1-11)
is a linear free energy (LFE) relationship.
Multiplying both sides of eqn. (3) by -2.303 RT, we obtain
-RTln(k/ko) = -ρRTln(Ka/Ka°)
∆G‡ - ∆G‡o = ρ (∆G° - ∆Go°)
or
∆∆G‡ = ρ (∆∆G°)
Stated in words, the change in the standard free energy of activation produced by a
substituent X is proportional to the change in the standard free energy produced by the
same substituent in ionization of benzoic acids in H2O at 25°C.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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Though the effect of a substituent (e.g., a m-Cl) on the Ka of benzoic acid is exceedingly
complex and something that cannot be calculated from first principles, it is a good model
for the effect of the same substitution on the rate of basic hydrolysis of ethyl benzoates
(equally complex).
LFE relationships, then, amount to using one process that is too
complex to understand completely as an experimental model for some other process that
is too complex to understand completely.
define: (log Ka - log Ka°) or log (Ka/Ka°) ≡ σ
(4)
(Ka and Ka° measured in H2O at 25 °C)
Substituting into (3): log(k/ko) = ρσ (rate)
(5)
or log (K/Ko) = ρσ (equilibrium)
Equation (5) is known as the Hammett equation.
σ is fixed by the substituent (substituent constant) and measures the effect of a substituent
on the Ka of benzoic acid (H2O, 25 °C). Since electron withdrawing substituents increase
the acidity of benzoic acids, substituents having a net electron-withdrawing effect
(compared to H) have positive σ values, and those with a net electron donating effect
(compared to H) have negative σ values.
Some σ values (from Hine) are tabulated on page 1-16. Some of these values were
obtained directly from the Ka values of the corresponding substituted benzoic acids in H2O
at 25 °C using equation (4). These are called primary values. Sometimes the σ values
cannot be obtained in this way, either because the substituent reacts with water, or (more
commonly) because the substituted benzoic acid is not soluble enough in water. In such
cases a secondary reaction of known ρ values can be used, with the expression σ = 1/ρ log
(K/Ko) or σ = 1/ρ log (k/ko). For example, the Ka's of benzoic acids in 50% aqueous
ethanol at 25 °C (ρ = 1.52) is a secondary reaction. Values obtained in this way are
secondary values. Still other σ values are statistical values that are averages of values
obtained from two or more secondary reactions.
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Substituent Constants, σ
substituent
σ
σ
substituent
Me
meta
-0.07
para
-0.17
meta
para
NHCO2Et
0.09c
0.13c
Et
iPr
tBu
-0.07
-0.07b
-0.10
-0.15
-0.15
-0.20
NHAc
NHCHO
NHC(O)CF3
0.21
0.22c
0.35c
0.00
0.05c
0.14c
CH2Ph
-0.18c
-0.11c
NHSO2Me
0.21c
0.05c
C≡CH
Ph
Picryl
CH2SiMe3
0.20d
0.06
0.27e
-0.16
0.23d
-0.01
0.31e
-0.21
OH
OMe
OEt
OPh
0.12
0.12
0.10
0.25
-0.37
-0.27
-0.24
CH2OMe
0.02c
0.03c
OCF3
0.40h
0.35h
CH2OPh
0.03c
0.07c
OAc
0.39
0.31
CHO
0.36f
0.44f
OSO2Me
0.39c
0.37c
C(O)Ph
CO2-
0.36c
-0.10
0.44c
-0.00
F
SiMe3
0.34
-0.04
0.06
-0.07
CO2Et
0.37
0.45
PO3H-
0.20
0.26
CN
CH2CN
0.56
0.15c
0.66
0.17c
SH
SMe
0.25
015
0.15
0.00
CH2I
0.07c
0.09c
SCF3
0.0h
0.50h
CH2Br
0.11c
0.12c
SAc
0.39
0.44
CH2Cl
0.09c
0.12c
S(O)Me
0.52
0.49
CF3
0.43
0.54
SO2Me
0.60
0.72
CF(CF3)2
0.37h
0.53h
SO2CF3
0.79h
0.93h
NH2
-0.16
-0.66
SO2NH2
0.46
0.57
NMe2
-0.15I
-0.83
SO3-
0.05
0.09
N(CF3)2
0.40j
0.53j
SMe2+
1.00
0.90
NMe3+
0.88
0.82
SF5
0.61
0.68
N3
0.37k
0.08k
Cl
0.37
0.23
N=NPh
N2+
0.30l
1.76m
0.35l
1.91m
Br
I
0.39
0.35
0.23
0.28
NO2
0.71
0.78
Ac
0.38
0.50
-0.32n
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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These substituent constants are based on the ionization constants of benzoic acids taken from the compilation of McDaniel
and Brown and rounded off, unless otherwise stated. D.H. McDaniel and H.C. Brown, J. Org. Chem., 23, 420 (1958).
b H. van Bekkum, P.E. Verkade, and B.M. Wepster, Rec. Trav. Chim. Pays-Bas, 78, 815 (1959).
c Based on a pK value in 50% aqueous ethanol and the corresponding ρ value (1.52).
d J.A. Landgrebe and R.H. Rynbrandt, J. Org. Chem., 31, 2585 (1966).
e D.J. Glover, J. Org. Chem., 31, 1660 (1966).
f A.A. Humffray, J.J. Ryan, J.P. Warren, and Y.H. Yung, Chem. Commun., 610 (1965).
g Values of σ for electrically charged groups are relatively unreliable.
h W.A. Sheppard, J. Am. Chem. Soc., 85, 1314 (1963); 87, 2410 (1965).
i J.C. Howard and J.P. Lewis, J. Org. Chem., 31, 2005 (1966).
j F.S. Fawcett and W.A. Sheppard, J. Am. Chem. Soc., 87, 4341 (1965).
k P.A.S. Smith, J.H. Hall, and R.O. Kan, J. Am. Chem. Soc., 84, 485 (1962).
l M. Syz and H. Zollinger, Helv. Chim. Acta, 48, 383 (1965).
m E.S. Lewis and M.D. Johnson, J. Am. Chem. Soc., 81, 2070 (1959).
n This value is almost implausibly small, much smaller than the value 0.14 reported in 50% ethanol. See footnote c.
Table taken from J. Hine, Structural Effects on Equilibria in Organic Chemistry, Wiley-Interscience, 1974, p.66.
a
In examining the fit of a set of experimental rate or equilibrium constants to a
Hammett relationship, log k or log K is plotted vs. σ, and the best fit slope and intercept
obtained by statistical analysis. The ρ value is the statistical slope, while ko (or Ko) is the
statistical intercept approximately k (or K) for the unsubstituted compound. A recent paper
indicates how easily the relative data can be obtained (J. Am. Chem. Soc. 2000, 122,
6357) although most researchers are much more rigorous, and garner more precise data.
ρ is fixed by the reaction (reaction constant) and the reaction conditions (temperature,
solvent). ρ measures the sensitivity of the rate or equilibrium constant to substitution
(relative to the sensitivity of the rate or equilibrium constant to substitution (relative to the
sensitivity of the Ka of benzoic acids in water at 25 °C). The definition σ ≡ log (Ka/Ka°) for
benzoic acids (H2O, 25 °C) fixes ρ = 1.00 for this reaction. Reactions with a positive ρ
value respond qualitatively to substitution in the same way as the Ka's of benzoic acids do,
i.e., electron withdrawing groups increase k or K. If ⏐ρ⏐ > 1.00, the rate or equilibrium
constant is more sensitive to substitution than Ka.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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a) Qualitative interpretation of the sign of ρ
positive ρ
equilibrium:
The product has a higher electron density at the reaction site than the
reactant.
rate:
The transition state has a higher electron density at the reaction site than the
reactant.
(By reactant we mean the substituted benzene derivative.)
negative ρ
equilibrium:
The product has a lower electron density at the reaction site than the
reactant.
rate:
The transition state has a lower electron density at the reaction site than the
reactant.
(The reactant again means the substituted benzene derivative.)
b) Interpretation of σ Values
Let us examine the σ values in the large table (p. 1-16) in the light of our qualitative ideas
about the electrical effects of substituents, based on polar and resonance effects (earlier).
R.W. Taft and others have attempted to separate the total electronic effect of a substituent,
as measured by its σ value, into polar (inductive and field) and resonance contributions.
We will not discuss these attempts in detail at this point, but we will use his ideas and
conclusions to examine some values.
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Assumptions
(i)
σ measures the sum of polar and resonance contributions, and can be expressed
as a weighted sum of two position-independent parameters, σI and σR, which
measure the polar and resonance properties of the substituent, respectively.
(ii)
The polar effect of a substituent is about the same in the meta and para positions.
As partial justification for this assumption we note that for the +N(CH3)3 group,
which should not exert an appreciable resonance effect, σm/σp = 0.88/0.82 ≈ 1.
More recent experimental evidence puts this ratio near 1. (JOC 2007, 72, 5327)
(iii)
The resonance effect of a substituent in the meta position is some constant fraction,
α, of its effect in the para position.
σp = σI + σR
That is:
σm = σI + ασR
where σI & σR depend on
the substituent only and
α < 1.
It should be noted that it has been determined (JCS, Perkin Trans. II, 133 (1988)) that this
assumption is not a good one).
The reason that the resonance effect is not zero in the meta position is because resonance
can supply or withdraw electrons to or from the positions ortho to the reacting group, from
which the effect can be relayed by the inductive and field mechanisms.
e.g.
G
G
G
NH2
+
NH2
+
NH2
σI values can be estimated from substituent effects in some rigid saturated systems. Taft
found that by using a value of 0.33 for α, a σR value could be obtained for each substituent
that reproduced each σm and σp value fairly well.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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Thus, σR ≈ 3/2(σp - σm) and σI ≈ 1/2 (3σm - σp)
We can now examine the σ values for some selected substituents using this separation of
polar and resonance effects to see if the results are in reasonable agreement with our
qualitative ideas.
(i)
(ii)
NH2 group (-I, +R) (very strong +R)
σm = -0.16
The negative value of σm shows that the +R effect
σp = -0.66
is the dominant effect, even in the meta position.
σR ≈ 3/2 (-0.66 + 0.16) = -0.75
σm = 0.09 - 0.75/3 = -0.16
σI ≈ 1/2(-0.48 + 0.66) = 0.09
σp = 0.09 - 0.75 = -0.66
Other -I, +R groups
group
σp
σm
σp - σm (≈ 2/3 σR )
approx. σI
approx. σR
-OH
-0.37
0.12
-0.49
0.37
-0.74
-OMe
-0.27
0.12
-0.39
0.32
-0.59
With the OH and OR groups the -I effect dominates in the meta position and the +R effect
dominates in the para position.
Halogens: -I effect: F > Cl > Br > I
+R effect: F > Cl > Br > I
halogen
σp
σm
σp - σm (≈ 2/3 σR)
approx. σI
approx. σR
F
0.06
0.34
-0.28
0.48
-0.42
Cl
0.23
0.37
-0.14
0.44
-0.21
Br
0.23
0.39
-0.16
0.47
-0.24
I
0.28
0.35
-0.07
0.39
-0.11
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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(iii)
-I, -R groups
group
σp
σm
σp - σm (≈ 2/3 σR)
approx. σI
approx. σR
-CO2Et
0.45
0.37
0.08
0.33
0.12
-C(O)Me
0.38
0.50
0.12
0.32
0.18
-CN
0.56
0.66
0.10
0.51
0.15
-NO2
0.71
0.78
0.07
0.67
0.11
(iv)
Alkyl Groups (+I, +R)
group
σp
σm
σp - σm (≈ 2/3 σR)
approx. σI
approx. σR
-CH3
-0.07
-0.17
-0.10
-0.02
-0.15
PRACTICE PROBLEMS 1
1. a) Using the table provided on p. 1-16 of your notes, determine the σΙ and σR
values for i) the diazonium group and ii) the trifluoromethyl group
b) Having performed the exercise in part a) classify those substituents as +R, +I, -R
and/or -I.
2. Provide two reasons why compound 2 is a much stronger base than N,Ndimethylaniline (1).
H3C
N
CH3
Et2N
CH3O
1
pKa of conjugate acid = 5.1
NEt2
OCH3
2
pKa of conjugate acid = 16.3
3. Rank the following functional groups according to the strength of their +R effect.
Rank from strongest to weakest. Provide explanations for your rankings.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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a)
b)
O
N
N
CH3
CH3
N
CH3
c)
CH3
CH3 + CH3
N
d)
CH2CH3
N
CH3
N
CH3
CH3
CH3
4. Please explain the difference in acidity of the underlined H of the two isomers
drawn.
N
N
H H
pKa = 26.7
H H
pKa = 30.1
OH
5. (From a previous midterm) Please indicate which
of compounds 1 or 2 has the lower pKa and indicate
the reason for your choice.
MeO
OH
O
MeO
1
O
2
6. A number of N-protonated amides have pKa’s in the range of 0-1 as indicated by
these examples:
O
CH3
+H
N CH3
O
CH3
+H
N
CH3
0.1
1.0
O
O
N+
H H
0.4
N+
H CH3
1.0
N+
H
A
O
Explain why compound A has a pKa of 5.33.
7. The pKa values for methine hydrogen of triphenylmethane (4) and the substituted
analogs are as indicated: 4: 30.8; 5, 16.8; 6, 14.4; 7, 12.7. Note that the introduction
of one para-nitro group enhances the acidity by about 14 pKa units, but the
introduction of more para-nitro groups offers substantially less effect (about 2 units).
Please provide an explanation; use structures liberally.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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H
H
C
C
H
C
NO2
4
NO2
5
H
C
O2N
6
NO2
7
O2N
O2N
8. Compounds 9 and 10 are similar acetic acids in that they both bear the
deprotonated form of an acid at the α-position. Nevertheless one has a pKa greater
than 4.74 and the other has a pKa that is less than 4.74. Please provide an
explanation for the relatively large difference in pKa values, despite the similarities in
structure.
O
O
-
CO2 Na
HO
SO3-Na+
+
HO
9, pKa = 5.59
10, pKa = 4.05
9. Compound 1 is quite acidic, with an experimental pKa of about –1.
a) Given the extensive functionality within the molecule and given that more than
one of the various atoms or groups may contribute independent –I and –R
modes of stabilization of anion 2, identify eight stabilization factors that lower the
pKa to –1, a value significant lower than 16. Draw useful resonance structures for
any resonance efforts (i.e., -R) you invoke.
b) Identify the significance of 16.
OH
S
Cl
O
1
O
O-
pKa = -1
S
CH3
Cl
O
CH3
O
2
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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10. Amidine 3 is a strong organic base with two basic nitrogens. One of the nitrogen
accepts and hold protons preferentially.
a) Identify which nitrogen is more basic and justify your
answer with the aid of a drawing(s).
b) Does the pKa of the conjugate acid of 3 increase or
decrease if a nitro group is added to the para
position of one of the aryl groups?
c) Given your answer in b) part, will the effect of
adding a p-nitro group be more significant on the Naryl group or the C-aryl group? Justify your answer.
C aryl
H3C N
NH
N aryl
3
SOLUTIONS TO PRACTICE PROBLEMS 1
1.
σΙ = ½ (3σm-σp)
σR = 3/2 (σp-σm)
for diazonium: (-I, -R)
σΙ = ½ (3σm-σp)
σR = 3/2 (σp-σm)
= ½ (3(1.76) - 1.91)
= 3/2 (1.91 - 1.76)
= 1.69
= 0.23
for the trifluoromethyl group: (-I, -R)
σΙ = ½ (3σm-σp)
σR = 3/2 (σp-σm)
= ½ (3(0.43) - 0.54)
= 3/2 (0.54 - 0.43)
= 0.38
= 0.16
2. Compound 1 is a simple aromatic
amine. It is a weak base and its
conjugate acid is comparatively
strong for an ammonium ion. The
reason is that the lone pair will share
its electrons, by resonance, with the
aromatic ring when those electrons
are free. To do this the C-N-C atoms
must be coplanar with the ring.
H3C
N
CH3
H3C + CH3
N
-
1
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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In 2 the two methoxy groups are offering minimal electronic forces but are acting
as a steric barrier. These prevent the C-N-C atoms of the Et2N group from getting
coplanar with the rings and thus when the lone pair is free on the nitrogens of 2,
there is essentially no +R donation of electron density into the ring. Thus the lone
pairs are more available to accept a hydrogen (and thus be a stronger base and
weaker acid). This is one of the reasons and is an example of steric inhibition of
resonance.
Since the lone pair are not in line with the
aromatic p orbitals, they can pick up an H
and in one certain arrangement, that H is
directed right at the other nitrogen lone
pair, creating a perfect situation for Hbonding. So the second reason for the
high basicity of 2 is the ability for Hbonding, whereby the H is simply captured
perfectly between the nitrogens and the
nitrogens share the positive charge.
Et
Et
+N
δ
H
Et
Et
N+
δ
CH3O
OCH3
protonated 2
3. A measure of the +R strength of these compounds is tied to the availability of the
lone pair of electrons on the N which is common to all the groups. Below is a table
which ranks the compounds and describes the role played by each N’s lone pair.
Rank
strongest
Compound
d)
N
CH2CH3
Explanation
N lone pair is fully free to donate and the alkyl groups
enhance this effect a little with their +I character
CH3
2nd
b)
N
N
CH3
CH3
CH3
3rd
a)
O
N
CH3
CH3
NCH3
NCH3
N
CH3
CH3
N+ CH3
CH3
N
N+ CH3
CH3
The N’s lone
pair participates in resonance similar to normal amide
resonance except a nitrogen, slightly less
electronegative than an O, is not quite as good at
accepting electron density.
O
O-
CH3
CH3
The N’s lone
pair participates in normal amide resonance and
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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therefore is less available than for a normal amine.
Since the oxygen is more electronegative than N, this
places this molecule 3rd instead of 2nd.
weakest
c)
CH3 + CH3
N
N
CH3 + CH3
N
CH3
N
CH3
CH3
N
+
N
CH3
CH3
CH3
CH3
CH3
The N’s lone
pair is rather tied up trying to assist the other N with the
burden of sharing a positive charge. In doing so, that
lone pair is not available to donate elsewhere. The
positive charge on the other N makes it a particularly
good -R group.
4. Deprotonation makes the anions shown. We have assumed that inductive effects
are about the same whether the influencing group is in the meta or the para position.
Hence the pKa difference rests on resonance. Looking at all of the resonance
structures below, only one offers particular stabilization over the others, and this one
caused the enhanced acidity of the particular isomer.
N
-H+
N
N
H H
pKa = 30.1
H
H
N
N
H
H
N
N
N
-H+
H H
pKa = 26.7
H
H
N
N
H
H
particular stabilization
of the -ve charge on the
electronegative atom
5. a) Compound 1
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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b)
O-
O
MeO
MeO
O-
O-
O
MeO
1
O
O
MeO
O-
2
steric hindrance prevents
full planarity and hence
full conjugation
6.
O
+H
N CH3
CH3
O
+H
N
CH3
1.0
N+
N+
H CH3
N+
H H
CH3
0.1
O
O
0.4
H
1.0
A
O
Loss of H+ from any of the four typical examples results in typical amides that have
the usual amide resonance with the carbonyl. Since when the H is lost, the resulting
lone pair can establish resonance with the carbonyl, as shown for one example.
O
O
CH3
+H
N
N
CH3
-H
O
+
N
CH3
+
-
However, with compound A, when the H is removed, the lone pair cannot do typical
amide resonance, because the lone pair of the N cannot align properly with the
orbitals of the carbonyl group. So, if the free lone pair does not participate in
resonance with the carbonyl, the electrons have little else to do, so they are less
inclined to lose the proton. Less acidic means a higher pKa value.
N+
H
-H
O
+
X
N
O
+
N
O-
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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If nomenclature is based on reactive character, compound A should really be named
an amino ketone rather than an amide.
7. If the anions of any of these deprotonated compounds are to be stabilized by way
of resonance with the three rings, they all have to lay (lie?) in one plane. Because of
the ortho hydrogens of adjacent rings bumping into one another (i.e., steric
crowding), this full three-ring stabilization does not occur completely.
crowding H's
H
C
HH
+
-H
4
HH
C
H
H
Only one ring is allowed full resonance, while the other two are twisted out of the
plane to make room for the ortho H’s.
H
H
C
crowding H's
causes the other
rings to twist out
of the plane
So when you add one nitro, you get a strong and complete -R and -I stabilizing effect
which lowers the pKa substantially. The addition of more NO2 groups to the other
rings does not add any more -R stabilization, since the twisting of the other two rings
prevents transmission of the resonance effects. The addition of those other two nitro
groups adds only a -I effect and the increment is of the size that we are used to
seeing.
This is an example of steric inhibition of resonance.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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8.
O
O
-
+
CO2 Na
HO
9, pKa = 5.59
HO
SO3-Na+
10, pKa = 4.05
As discussed in class, the –CO2-M+ group is an electron donating one and you can
corroborate this by looking at its σ values. Hence this EDG hinders ionization of the
other carboxyl group.
The pKa of 4.05, being less than 4.74, means that the –SO3-M+ must be an EWG,
despite carrying a – charge. It was important to mention this in your solution.
The reasons could be a combination of many:
1. As many said, the extra oxygen offers a third resonance structure to help
stabilize the – charge. The sulfur is also electronegative and adds some
inductive withdrawal.
2. This one was not in anybody’s solution: having three oxygens assuming some
of the – charge means there are three sites for water to H-bond to, in order to
assist in the stabilization or distribution of charge.
3. As some people did, you could inspect the σ values for –SO3-M+ (and maybe
calculate its σI value) and learn that it is a EWG inductively.
9 a)
b) 16 is the pKa of methanol, which could be viewed as the starting point onto which all
the groups and atoms of 1 are introduced. Methanol is the simplest alcohol, so 16 is the
‘base’ pKa of the simplest alcohol.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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10 a)
b) Adding an EWG to a +-charged
species will destabilize it. Thus it
has less inclination to maintain its
+-charge and therefore will be more
acidic. The pKa of the conjugate
acid will decrease.
c)
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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c) Reaction Constants, ρ
Some examples of reaction constants are shown in the table. Further examples are given
in O.. Exner, In Advances in Linear Free Energy Relationships, Edited by N.B. Chapman
and J. Shorter, Plenum Press, 1972, Chapt. 1, and in the references cited at the end of the
substituents constants table.
ρ
Reaction
ArCO2 H
ArCO 2 + H , w ater
-
+
ArCO2 H
ArCO 2
+
ArOH
-
ArCH
2 CH 2CO 2 H
-
+
0.56
2 CH 2 CO 2
-
+
+ H , water
+
+
3.19
ArNH2 + H , w ater
+
+
ArCH2 NH 2 + H , w ater
1.05
-
2.61
-
ArCO2 + EtOH
ArCO2Et + OH
-
ArCH2 CO 2 Et + OH
ArCH2 CO 2
+ EtOH
ArCH 2 OH + HCl
ArCH 2 Cl + H 2 O
ArC(Me) 2 Cl + H 2 O
ArNH2 + PhC(O)Cl
-
0.24
2.26
ArO + H , w ater
+
ArNH3
ArCH 2NH 3
1.57
+ H , EtOH
ArCH2CO 2 + H , w ater
ArCH2 CO 2 H
ArCH
1.00
ArC(O)Ph + Me Li
+
ArC(Me) 2 OH + HCl
ArNHC(O)Ph + HC
-
ArC(O )(Me)Ph
1.00
-1.31
-4.48
-3.21
0.94
Data taken from P.R. Wells, Linear Free Energy Relationships, Academic Press, New York, 1968, pp 12-13,
except for last entry which comes from J. Org. Chem. 2002, 67, 4370.
The sign and size of ρ for a reaction rate is one kind of experimental probe of transition
state structure. In particular, it can provide some indication of the difference in electron
density at the reaction site between the reactant and the transition state. In order to be
able to interpret ρ values in this way, it is necessary to understand some of the factors that
affect ρ.
i) Transmission of Electrical Effects to the Reaction Site
Since σ is a measure of the electrical effect of the substituent, ρ must depend in part on
how much of this effect is transmitted to the reaction site. This factor is illustrated by ρ
values for several acid-base equilibria.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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ρ, Ka (H2O, 25 °C)
Acid
CO2H
ρ, Ka (H2O, 25 °C)
Acid
H
1.00
COOH
C C
H
0.42
X
X
CH2CO2H
SCH2CO2H
0.56
X
0.30
X
CH2CH2CO2H
+
NH3
0.24
X
3.2
X
OH
2.3
X
ii) Solvent Effect on ρ
The magnitude of ρ increases with decreasing solvent polarity.
Some examples:
solvent
ρ
H2O
1.00
50:50 EtOH/H2O
1.57
EtOH
1.96
CO 2H
Ka, 25 °C
X
CO2H
ρ
solvent
MeOH, 30 °C
0.844
toluene, 25 °C
2.22
N2
+
X
O
C
Ph
Ph
X
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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O
CHPh2
B FAILURE OF THE SIMPLE ρσ RELATIONSHIP
The failure of correlations involving reactions of ortho-substituted benzene derivatives and
aliphatic compounds has been mentioned. Such failures are reasonably attributed to steric
effects. Three other kinds of failures will now be discussed.
a) Ka's of Phenols and Anilinium Ions
-
OH
+
H 2O
O
Ka
X
X
NH3
+
+
H 2O
ρ = 2.26
NH2
Ka
X
+
H 3O
+
+
X
H 3O
+
ρ = 3.19
If we examine the ρσ plots for these equilibria we see that the meta substituents give a
reasonably good linear correlation but that certain kinds of substituents deviate badly and
in a systematic way from the best fit line for the para substituents. For example, in the plot
of log (Ka/Ka°)(phenols) vs. log (Ka/Ka°)(benzoic acids) (i.e., σ) we see that the para -R
group points all lie above the line. For such groups the substituted phenol is more acidic
than predicted. For example, p-NO2 and p-Ac-phenol are ca. 1 log unit (10X) more acidic
than predicted.
Reason: -R groups have a special effectiveness when conjugated with a +R group (i.e., -Oor -NH2).
Called “through resonance” or: “direct resonance”
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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-
-
O
+
N
NH2
O
O
O
-
O
+
N
-
O
+
NH2
C
C
N
N-
No similar direct resonance interaction is possible in the p-NO2 or p-CN substituted
benzoate ions, consequently the σ values for these groups measure mostly a polar (-I)
effect (refer to the σIσR dissection of σ values, page 1-19). Stated another way, ionization
of benzoic acids is a poor model for ionization of phenols when we have strong -R
substituents.
We need two sets of σρ constants for strong -R groups: one where through resonance (in
reactant, product, or TS) is possible and one where no through resonance is possible. The
special σp values for use with reactions of the -OH group in phenols, of the NH2 group in
anilines (and other reactions where direct resonance interaction with an unshared electron
pair is possible) are called σ - constants.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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b) Ionization of Phenols and Anilinium Ions
log
Ka,
phenols
H2O, 25 °C
σ
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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log Ka,
anilinium ions,
H2O, 25 °C
σ
Table of corrections
group
σp
σp-
(σp- - σp)
p-C(O)CH3
0.50
0.82
0.32
p-C(O)OEt
p-CN
p-NO2
0.45
0.66
0.78
0.74
0.99
1.23
0.29
0.33
0.45
-
σ =
1
2.11
log (K/Ko) for phenols or
log Ka
phenols
σ scale
σ(p-CN)
1
2.77
log (K/Ko) for anilinium
ions
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−
σ (p-CN)
-
σ values for para substituents
σ
0.36
0.98
0.82
0.73
0.74
0.74
0.62
0.99
0.56
0.69
substituent
C≡CPh
CHO
Ac
CO2H
CO2ME
CO2Et
C(O)NH2
CN
CF3
N≡NPh
σ
3.20
1.23
0.17
0.73
1.05
1.36
0.94
1.32
0.70
1.16
substituent
N2+
NO2
SiMe3
S(O)Me
SO2Me
SO2CF3
SO2NH2
SO2F
SF6
SMe2+
c) Solvolysis of 2-Aryl-2-chloropropanes (t-cumyl chlorides)
H.C. Brown and Y. Okamoto, J. Am. Chem. Soc. 80, 4979 (1958).
Cl
C CH
3
CH3
X
OH
C CH
3
CH3
90% aq. acetone
25 °C
+
HCl
X
A plot of log k vs σ gives a reasonable straight line for all meta substituents and for -R para
substituents, with ρ = -4.54. The points for p-F, p-Cl, p-Br, p-I, p-CH3 and p-OCH3 all lie
above this line (more reactive than predicted by factors of ca 10 and 100 (1 and 2 log units)
respectively (see plot nearby). The transition state for this reaction resembles the 2-aryl-2propyl cation. It is clear that we should not expect Ka's of benzoic acids to provide a good
measure of how well a group can supply electrons by resonance on extreme electron
demand. Although a through-resonance +R effect of p-CH3O- and other +R groups is
measured to some extent in their σ values, the electron demand by the carboxyl group in
benzoic acids is much less than that of the carbocation-like group in the solvolysis
transition state.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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σ+ values for
para substituents
substituent
Me
Et
i-Pr
t-Bu
Ph
NH2
NMe2
NHAc
OH
OMe
OPh
OF
SMe
Cl
Br
I
σ+
-0.32
-0.31
-0.29
-0.27
-0.21
-1.47
-1.67
-0.58
-0.91
-0.79
-0.50
-2.30
-0.08
-0.62
0.11
0.14
0.13
log k/kH
solvolysis
of t-cumyl
chlorides
90% aq.
acetone
25 °C
Using the best fit straight line defined by the meta substituents H.C. Brown obtained a set
of σ+ values for use in reactions where through-resonance between a +R group and a
carbocation-like reaction site is possible.
+
σ = -
1
4.54
log k
log (k/ko)
for t-cumyl chloride solvolysis,
90% aq. acetone, 25 °C
σ scale
+
σ (p-OMe)
σ(p-OMe)
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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PRACTICE PROBLEMS 2
1. The ρ for benzoic acid dissociation at 25 °C in DMSO (dimethyl sulfoxide) is 2.60.
Why is this value different from the same equilibrium in water?
2. How many different σ values have you been introduced to? Explain exactly what
each of them is.
3. The reaction shown below gives a very strange and non-linear Hammett plot, also
shown below. This occurrence usually means that the reaction is actually proceeding
by two different mechanisms, depending on the identity and more specifically, the
electronic character of X. Please provide two different mechanisms and identify
where you would invoke these mechanisms.
X
X
CH2Cl
+
N
benzene
N+
100 oC
CH2
log (k X/kH)
σ+
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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4. The rates of chlorination of acetylenes are sensitive to substituents:
X
p-OMe
p-Me
p-F
H
p-Cl
p-Br
m-NO2
p-NO2
19, 500
190
14.9
10.6
4.15
2.81
0.0165
0.00325
Cl
C
C
H
X
C
+ Cl2
X
Cl
C
H
Do two Hammett analyses on the data in order to learn if the data better correlated
with σ or with σ+? Based on your answer, sketch a structure for the transition state of
the rate determining step? Hint, Cl2 reacts similarly to Br2.
5. In the accompanying scheme, when X = Cl, compounds 13 are known as O-isopropvl
benzohydroximoyl chlorides. These compound undergo methoxy substitution when
treated with methoxide ion (48 °C).
X
N
Y
13
OiPr
STEP 1
MeO
MeO X
NaOMe
10% MeOH/
90% DMSO
NY
OiPr
N
-
-Cl
STEP 2
OiPr
Y
a) Given a ρ value of 2.20, using regular σ values for this reaction, please indicate,
with an explanation, which of the two steps is rate-determining. The explanation
should address both steps of the mechanism.
b) Which reaction presented in the list on page 1-22 of the notes closely resembles
the reaction at hand in both general mechanism and approximate ρ value.
c) The overall transformation also occurs when compounds 13 (X = Cl) are exposed
to MeOH, but the mechanism is different. That mechanism involves initial ratedetermining loss of chloride and the ρ value based on σ+ values was found to be
-2.40. Provide a mechanism for this particular reaction and identify important
resonance structures involved, including those consistent with the use of σ+
values.
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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d) Reverting back to the scheme shown, if X = OEt, the reaction does not occur,
and the starting material is recovered, partially isomerized. Provide an
explanation for the isomerization of the C=N bond and for the lack of reaction.
6. Which set of σ's would be needed to handle the substituent effects on the
ionization of para-X-2,6-dimethyl-N,N-dimethylanilinium ions?
7. In the solvolysis of trifluoroacetates and chlorides, the methoxy oximino group has
been demonstrated to be a rate accelerating functionality when placed at the proper
position.
Examples:
N OMe
Me2C C
Cl H
14
Me2C CH3
Cl
15
Rate ratio 14/15 = 10 in CF3CH2OH
N OMe
16
Rate ratio 16/17 = 7 in CF3CH2OH
CH3
H
OC(O)CF3
17
OC(O)CF3
a) Explain with structures how the oximino group accelerates the rate of the
solvolysis reaction in these cases.
b) Explain why the oximino group in the
examples below severely retards the rate of
the solvolysis reaction.
Rate ratio 18/19 = 6.1 X 10-5 in CF3CH2OH.
N OMe
H
OMs
19
18
8. Predict the sign of the Hammett equation in each of the following reactions.
a)
-
O + CH3CH2 I
98% EtOH in H2O
X
70 °C
OCH2CH3 + IX
b)
+
+ -OCH
3
25 °C
N(CH3)2 + CH3OCH3
X
X
CH3CN
Is this a circumstance where one could invoke through-resonance?
N(CH3)3
OMs
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9. In a J. Org. Chem. (JOC 1993, 58, 5434) article, there is a mechanistic palladium
catalyzed cross-coupling reaction.
tBu
OSO2CF3
+ X
SnBu3
Pd2(dba)3
AsPh3, 333 K
NMP (solvent)
tBu
X
The following rate data on the following page were obtained.
Now it is recognized that a 4th yr.
undergraduate is not particularly familiar with
this reaction, but you now have been
introduced to a tool to provide some light on
the mechanism. The paper presents a plot of
these rate data vs. σ, which gave a
somewhat linear plot with ρ = -.89 and r2 =
.954.
Para substituent
log (kx/ko)
CF3
-0.22
Cl
-0.18
H
0.0
OMe
0.42
NMe2
0.90
a) Prepare a graph using these data and confirm the slope and regression data
previously obtained. What role are the electron donating substituents playing in the
transition state of the rate determining step of the reaction.
b) Have these authors done a good job in this particular aspect of their study? Is there
any experiment that should be done to gain more detail regarding the role of the
substituents? Is so, perform that experiment.
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SOLUTIONS TO PRACTICE PROBLEMS 2
1. The ρ for benzoic acid dissociation at 25 oC in H2O is defined as 1.00. The
stabilization of the dissociated form of the molecule arises from substituents and from
interaction with water, much of which is H-bonding. The ρ in DMSO (dimethyl
sulfoxide) is 2.60, which implies that substituents are involved to a much larger
extent. The reason is that the solvent is much less involved. The dipolar character of
DMSO has no effective means of stabilizing the negative charge of the carboxylate
and so, there is greater need for participation by the substituents.
2. σ -general usage when there in no opportunity for “through resonance”, but there
is still some simple +R and -R interactions of the substituents with the aromatic ring
σI -used when there is absolutely no chance for resonance of any kind. Typically the
situation is when the atoms separating reaction site and substituents are saturated
σR -not really used at all, since use would imply no induction going on and it is
difficult to envision a situation where induction can be eliminated
σ+ -should be employed when +R substituents have a chance to stabilize starting
material, transition state, intermediate or product
σ- -should be employed when -R substituents have a chance to stabilize starting
material, transition state, intermediate or product
3. The two mechanisms are SN1 and SN2.
CH2-Cl
X
N
r.d.s.
+
N+
X
CH2
+
CH2-Cl
N
X
CH2
r.d.s.
The data can be interpreted as follows.
SN1 region. Since the rate plot is flat, substituents have no influence on the reaction,
and it follows that since the pyridine possesses the substituents, that the pyridine has
no bearing on the reaction. That is, the pyridine is not part of the rate equation. An
ORGANIC REACTIVITY - CHEM*4720 - COURSE NOTES W2014
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SN1 reaction is consistent with this data and is invoked herein. The flat section is on
the more right region of the plot and includes mildly electron donating and all types of
electron withdrawing substituents. These pyridines are not a strong enough
nucleophile to attack the benzylic position before the Cl- leaves by itself.
SN2 region. With the powerful electron donating substituents, the pyridine is now a
stronger nucleophile and can perform its chemistry BEFORE the Cl- leaves by itself.
The left side of the graph has the strong donating substituents. Now, since the
pyridine is part of the rate equation, we would expect to see a non-zero slope, since
substituents have an opportunity to exert their influence. The slope is negative,
consistent with the fact that the strongest electron donators can activate the pyridine
the most and prompt it to react as a nucleophile. σ+ values were used since through
resonance can be invoked on the pyridines bearing strong +R substituents.
SN2 region,pyridine is most nucleophilic
log (KX/KH)
SN1 region, loss of Cl is r.d.s.
σ+
4. The data table is expanded to the following:
X
σ
σ+
p-OMe
p-Me
p-F
H
p-Cl
p-Br
m-NO2
p-NO2
-.27
-.17
.06
0
.23
.23
.71
,78
-.79
-.32
-.08
0
.11
.14
.71
.78
log (k/ko)
19, 500
190
14.9
10.6
4.15
2.81
0.0165
0.00325
3.26
1.25
0.148
0.00
-0.407
-0.576
-2.81
-3.51
C C H
X
For an analysis of the data vs. σ
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+ Cl2
Y = M0 + M1*X
M0
0.73677
M1
-5.4389 = ρ
r
0.96817
For an analysis of the data vs. σ+
Y = M0 + M1*X
M0
-0.044166
M1
-4.1667 = ρ
r
0.99814
The two Hammett analyses indicate that the data is better correlated with σ+ The
following mechanism is consistent with the result.
C C
H
Cl Cl
+
r.d.s.
Cl
C C
X
X
or
Cl
-
H
T.S. to this intermediate
can be stabilized by
through resonance
Cl
C C
X
Cl
H
5.
a)
The transition state (T.S.) for STEP 1 involves the addition of a methoxide, with its
additional electron density to the carbon α to the aryl ring. In the reaction intermediate
that follows the T.S., that charge ends up on the atom twice removed from the aryl ring.
In the T.S., the few atoms around the aryl ring have more electron density than the
starting neutral material and ρ should be positive. Its magnitude is about right for a
nucleophilic attack on a carbon so close to the ring. (see part b) of this question). I
cannot see any room for through resonance, so simple σ values are suitable.
In STEP 2, the substrate goes from an electron rich negatively charged compound to a
neutral. I would expect a negative ρ value here.
Based on the given ρ value of 2.20, STEP 1 is rate determining.
b)
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The following entry is very similar to the reaction under analysis. It is the 8th entry on
page 1-22. Be sure that you appreciate the mechanistic analogies between the
systems.
-
-
2.61
ArCO2 + EtOH
ArCO2Et + OH
c)
In this case, the p value means we are making a cation from a neutral and the
correlation with σ+ values means we should invoke through resonance in the transition
state for chloride loss. The reaction is SN1 and Cl- loss is rate determining. Eventual
MeOH addition provides the product.
Drawing the intermediate after complete Cl- loss will serve as an indication of how
through resonance stabilizes the T.S.
Cl
C
N
OiPr
OiPr
-ClEDG
EDG
C
+
EDG
+ N
C
N
OiPr
C
OiPr
N
+
EDG
MeOH
-H
MeO
N
OiPr
Y
d)
An easy, but wrong explanation is the following. The first step of the reaction occurs,
but the last step does not. This is because alkoxide is a poor leaving group compared to
halide. We know the first step occurs since it is the only method of isomerizing the C=N
bond. Another way of saying all this is to say STEP 2 is now rate determining and it
determines that any further reaction will not even occur.
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EtO
N
OiPr
13
Y
MeO
MeO OEt
NaOMe
10% MeOH/
90% DMSO
N-
OiPr
N
X
OiPr
Y
Y
bond rotation
EtO
MeO OEt
loss of MeO-
N
OiPr
Y
NOiPr
Y
A analogy of this has already been introduced to you. Recall how the Meisenheimer
complex as isolated in Chem 3750 lecture during the aromatic addition elimination
mechanism. THIS IS WRONG because there is no reason why MeO- should leave with
a preference over EtO-. Hence if MeO- can leave and revert back to starting material,
why can’t EtO- leave to provide product? (Alkyl groups are boring)
So the better explanation must require that MeO- does not even add to the substrate.
So why is 13 (X = Et) so much less reactive that 13 (X = Cl)? I think the answer lies in
the comparison of the +R/-I capacities of EtO and of Cl. Cl is primarily a –I substituent,
permitting and promoting attack of an Nu on the C=N bond. EtO is primarily a +R
substituent which hinders attack by the resonance structure shown:
EtO+
EtO
N
Y
- OiPr
N
OiPr
13
Y
Such a structure clearly reduces the electrophilicity of the C=N feeding the nitrogen
some of the electron density it is already striving for. Think about the reactivity
differences between a carboxylic acid chloride and a carboxylic ester.
More importantly, the increased C-N single bond character permits bond rotation to
the other isomer.
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EtO+
EtO
N
- OiPr
N
OiPr
13
Y
Y
bond rotation
EtO+
EtO
N
OiPr
N
OiPr
Y
Y
This is offered as the mechanism by which the isomerization can occur.
6. An initial analysis would suggest that through resonance would be appropriate,
since in the non-protonated form, there is through resonance of the N’s lone pair onto
a para-X group when the X is a -R group. This would mean that σ- values would give
the best correlation. However, it should be noted that the 2 and 6 methyl groups
create a steric barrier to through resonance as they force the dimethylamino group to
twist out of resonance and relieve steric strain. Hence on both the left side and right
side of the equilibrium, there is no opportunity for through resonance.
CH3 CH3
N+ H
CH3
CH3
CH3
X
-H
+
N
CH3
CH3 CH3
CH3
N
CH3
X
steric problems
CH3
CH3
X
probable rotational
isomer
7. a)
The intermediates arising from compounds 14 and 16 benefit from resonance
stabilization as shown in the diagrams. The MeO is a +R group that is capable of
transferring its electrons to the carbocation via resonance.
Please note that the diagram shows stabilization of the cationic intermediate. This is
taken as an indication that there is also significant resonance stabilization in the
transition state for loss of the leaving group.
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N OMe
Me2C C
Cl H
14
Me2C CH3
Cl
15
SN1
+
N OMe
Me2C C
H
N OMe
+
Me2C C
H
SN1
+
Me2C CH3
N OMe
16
H
OC(O)CF3
SN1
+
H
CH3
17
N
N
OMe
H
OMe
+
SN1
+
CH3
OC(O)CF3
b)
In the case of 18, the methoximino group is one carbon removed from the reaction site
and is separated by a saturated, insulating carbon. Hence the resoance as shown
above is not possible. Furthermore, the bicyclic nature of the homoadamantyl carbon
backbone of 18 does not permit intramolecular backside attack by the N or O of the
methoximino group.
So, the methoximino group as a conglomeration of electronegative atoms can only act
as a -I group and retard the solvolysis as compared to compound 19.
8. Predict the sign of the Hammett equation in each of the following reactions.
a)
-
O + CH3CH2 I
X
98% EtOH in H 2O
70 °C
OCH2CH3 + IX
ρ should be negative, the oxygen is going from high electron density to lower electron
density.
b)
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+
N(CH3)3 + -OCH3
X
heat
N(CH3)2 + CH3OCH3
X
ρ should be positive since the nitrogen gains electron density as the reaction proceeds.
Through resonance is likely since the lone pair which is getting restored onto the
nitrogen can be delocalized onto the X groups when they are -R groups in the para
position.
9. The J. Org. Chem. (JOC 1993, 58, 5434) article gives a Hammett plot using σ and
the data simply does not correlate very well.
tBu
OSO2CF3
+ X
SnBu3
Pd2(dba)3
AsPh3, 333 K
NMP (solvent)
tBu
X
So, I have plotted on the same graph the log (kx/ko) vs σ+ and I get a better fit and a
reduced ρ value. The plot of both curves is on the next page.
Para substituent
σ
σ+
log (kx/ko)
CF3
.54
.54
0.90
Cl
.23
.11
0.42
H
0
0
0.0
OMe
-.27
-.79
-0.18
NMe2
-.83
-1.67
-0.22
Following the paper:
Using σ+:
ρ = -.89 and r2 = .946.
ρ = -.54 and r2 = .979.
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The new data suggest that there
is reduced substituent
involvement, but that when the
substituents do participate, the
+R groups among them have an
opportunity to utilize their through
resonance capabilities.
The authors were very neglectful
in this analysis. The weak
correlation coefficient should
have prompted them to try σ+
values in an effort to get a better
fit.
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