hammett plots2
TRANSCRIPT
The Hammett Equation
Simon MeekLT1, 5pm 4/4/2005
And
The Curtin-Hammett Principle
S. J. Meek Louis P. Hammett (1894-1987); David Y. Curtin (1920- )
ß The Hammett Equation
ß Curtin-Hammett Principle
log k = sr k0
log K = sr K0
PA A BkA
kBPB
FastSlow Slow
k1 k2
k1, k2 << kA, kB
S. J. Meek Linear Free Energy Relationships
ßEquilibrium constants, K, and rate constants, k, are each related to the freeenergy changes in the relevant reactions in the following way:
-DG
2.303 RTlog K =
k' = Boltzmann's constanth = Planck's constant
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
-5 -4.8 -4.6 -4.4 -4.2 -4 -3.8 -3.6 -3.4 -3.2 -3
log K RCO2H
log
k R
CO
2Et
ßSuch relationships were first studied on a thoroughbasis by Hammett in the1930’s.
ßPlotting log k for the reaction of esters against log Kfor ionisation of acids resulted in a reasonablestraight line.
RCO2Etk
RCO2OH EtOH
RCO2H H2OK
RCO2 H2O
-DG
2.303 RTlog k = k' T
h+ log
S. J. Meek Linear Free Energy Relationships
-DG
2.303 RTlog K =
k' = Boltzmann's constanth = Planck's constant
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
-5 -4.8 -4.6 -4.4 -4.2 -4 -3.8 -3.6 -3.4 -3.2 -3
log K ARCO2H
log
k A
rCO
2E
t
ß The straight line relationship between log k for thereaction of esters and log K for ionisation in water ofthe corresponding carboxylic acids, implies there is arelationship between the free energy ofactivation for the ester reaction and the standardfree energy change of ionisation in water of the acids.
ß The straight line correlation between the freeenergy terms for two different reaction series arereferred to as linear free energy relationships.
RCO2Etk
RCO2OH EtOH
RCO2H H2OK
RCO2 H2O
DG
DG
-DG
2.303 RTlog k = k' T
h+ log
S. J. Meek Hammett Equation
ß Probably the most common linear free energy relationship.
ß Hammett reported the Linear correlations between the logarithms of rateconstants for reactions of meta- and para-substituted phenyl derivatives (logkx)and pKa values of the corresponding substituted benzoic acids.
log k = r log K + c
XOH
O
H2OX
O
O
H3O
K
XOEt
O
H2OX
O
O
EtOH
NaOHEtOH
k
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
-5 -4.8 -4.6 -4.4 -4.2 -4 -3.8 -3.6 -3.4 -3.2 -3
log K ARCO2H
log
k A
rCO
2E
t
S. J. Meek Hammett Equation
ß In 1937 Hammett presented the equation:
log k = sr k0
log K = sr K0
k = rate constant (X≠H) k0 = rate constant (X=H)K = equilibrium constant (X≠H)
or
Kinetics equation: effect of substituents ontransition state vs initial state
Equilibria equation: effect of substituents onfinal state vs initial state
Hammett, JACS, 1937, 59, 96.
K0 = equilibrium constant (X=H)s = Hammett substituent constantr = Hammett reaction constant
S. J. Meek Derivation of the Hammett Equation
Eq 1. log kx = r log Kx + c
Considering the unsubstituted carboxylic acid R=Has the standard reaction:
Eq 2. log kH = r log KH + c
Subtract 2 from 1:logkx - log kH = r(log Kx - log KH)
Which can be written in the form:
sx :
The ionisation of m- and p-benzoic acids in water (25 oC) is defined as the standardreference reaction with the reaction constant r = 1.
y = mx + c
sx = log Kx
KHor
sx = log Kx - log KH
= [(-pKa(x)) -(-pKa(H))] = pKa(H) - pKa(x)
log kx = r log Kx = rsx
kH KH
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
-5 -4.8 -4.6 -4.4 -4.2 -4 -3.8 -3.6 -3.4 -3.2 -3
log K ARCO2H
log
k A
rCO
2E
t
S. J. Meek Hammett Equation
ßThe Hammett relationship only applies tosystems where the substituents are attachedto the reaction centre via aromatic rings andare situated meta- or para-.
ßOrtho- substituted aromatic rings do no fallon the line owing to steric and through-spaceeffects. ie.increased crowding in thetetrahedral intermediate
ß Aliphatic acids do not fall on the line, due toa small Increase in steric crowding in thetransition state and their increased flexibilitymay decrese correlation between thetransition state structure and equilibriumposition.
H
p-NO2
m-NO2
p-Cl
p-Br
p-I
p-Me
p-OMe
p-NH2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
log K/Ko
log
k/
ko
OEtOOH
= o-substituent
OEtOOH
= aliphatic
S. J. Meek Hammett Substituent Constant s
ß The substituent constant (s) for meta- and para-substituents in benzenederivatives as defined, on the basis of the ionization constant of substitutedbenzoic acid in water at 25 oC. ie. X=H
ßs is a collective measure of the total electronic effects of a substitutent[resonance and inductive effects]. ie. how electron withdrawing or donating asubstituent is in its ability to supply/withdraw electrons to/from the reaction site.
ß s is independent of the nature of the reaction site.
ß s is defined separately:ßsm for meta-substituentsßsp for para-substituents
ßInductive effect: caused by polarizationof the s-bonds. sm > sp for a given substituent due toits closer proximity to the reaction site.
ßResonance effect: conjugation isstronger for p-substituents than meta-.sp > sm for a given substituent
para-: strongconjugation intocarbonyl
meta-: conjugationinto ring not carbonylbalances weak effectof X
OH
O
X
OH
O
X
S. J. Meek Physical Meaning of s
ßMagnitude of s: larger +/-values= greater inductive and/or conjugative effect of substituent.ß s is -ve = electron-donatingß s is +ve = electron-withdrawings = 0, substituent has no effect (ie. electronically the same as X=H)
Recall: sx = pKa(H) - pKa(x)
sm-NO2 = +0.71
sm-OMe = +0.12
sm-Me = -0.07
sp-OMe = -0.27
km-NO2 = 63.5KH
km-OMe > kH
kH > km-OMe
km-Me = 0.66KH
Electron-withdrawing m-NO2increases stability oftetrahedral intermediate cf.to electron-donating m-Me
OMe can be electron-withdrawing in the meta-position due to inductiveeffects or electron-donatingin the para-position due toa conjugative effect
O2N
O
OEt
OH
O2N
O
OEt
OHd-
d-
d-
O2N
O
OEt
OH
transition state tetrahedral intermediate
km-NO2
RDS
Me
O
OEt
OH
Me
O
OEt
OHd-
d-
d-
Me
O
OEt
OH
transition state tetrahedral intermediate
km-Me
RDS
MeO
O
OEt
OH
MeO
O
OEt
OHd-
d-
d-
MeO
O
OEt
OH
transition state tetrahedral intermediate
km-OMe
RDS
O
OEt
OH
O
OEt
OHd-
d-
d- O
OEt
OH
transition state tetrahedral intermediate
kp-OMe
RDS
MeOMeOMeO
S. J. Meek Hammett Reaction Constant r
ßThe constant of proportionality between log k (or K) and s values
ßThe magnitude of r reflects how sensitive a particular reaction is to theelectronic effects of the substituents.
ßSign of r (the slope) shows whether a reaction is accelerated by EW-/ED-substituents:
-positive r: A result of the build up of negative charge at the reactioncentre in the transition state of the rate determining step (rds).
\ rate will be accelerated by electron-withdrawing substituents.
-negative r: A result of the build up of positive charge at the reactioncentre in the transition state of the rate determining step (rds).
\rate will be accelerated by electron-donating substituents.
NB. rate will be retarded by the opposing EW-/ED-substituent .
S. J. Meek Standard Hammett Equations
ßIonisation of para-substituted benzoic acids:
ßIonisation of meta-substituted benzoic acids:
CO2H CO2H
K
X X
sp = logK(pXC6H4CO2H)
, at 25 oC in aqueous solution
r = 1
K(C6H4CO2H)
CO2H CO2H
KX X
sm = logK(mXC6H4CO2H)
, at 25 oC in aqueous solution
r = 1
K(C6H4CO2H)
S. J. Meek Through-conjugation
In many cases a breakdown in linearity between log K’s and s’s is observed forstrongly EW-/ED-substituents as predicted by the Hammett relationship.
Example: ionization of phenols and anilines
p-CN, p-NO2 substituted phenols are stronger acidsthan would have predicted by the Hammett correlation.ie. values lie above the line
OH O
NO
O
H
NO
O
NH2 N
NO
O
H
NO
O
H
OHH2O X
OH3OX
log Kx
KH
log Kx
KH(XC6H4CO2H)
(XC6H4OH)
..
..
.
..
p-CN, p-NO2
Reason: resonance stabilization can be extendedthrough to the reaction centre by ‘through-conjugation’ fi more stabilized species.
S. J. Meek Through-conjugation: sx- and sx
+
ßThrough-conjugation effects can be separated from the inductive effects toproduce new s-values which account for through conjugation non-linearity.
ßA Hammett correlation with r based on m-substituents only can be developed(NB. m-substituents do not exhibit a resonance effect directly to the reactioncentre). The amount by which certain substituents deviate from the line can beadded or subtracted from their s-values to produce sx
- and sx+ modified substituent
constants.
log Kx
KH
log Kx
KH
(XC6H4CO2H)
(XC6H4OH)
.. .
..
p-CN
p-NO2
ß sx- strongly electron-withdrawing groups
Substituent(in XC6H4OH)
CO2EtCOMe
CNCHONO2
sp
0.450.500.660.430.78
sp-
0.680.840.881.031.27
S. J. Meek
Substituent
C6H5Me
MeONH2
NMe2
sp
-0.01-0.17-0.27-0.66-0.83
sp+
-0.18-0.31-0.78-1.30-1.70
Through-conjugation: sx- and sx
+
p-MeO
p-Me
sx
log kx
kH
ß sx+ strongly electron-donating groups
Caveat: Choosing between s, s- and s+ presupposes the mechanism of the reaction.SNAr reactions correlate well with s-; SEAr reactions correlate well with s+.
S. J. Meek Through-conjugation: sx- and sx
+
ß sx+ reference reaction:
•solvolysis of cumyl chloridesX
Cl
XCl
sx+ = log
k(XC6H4C(CH3)2Cl
k(C6H4C(CH3)2Cl, at 25 oC in 90% aqueous acetone
r = 1
ßsx- reference reaction:
•ionisation of substituted phenols X
OHX
OH
sx- = log
Ka(XC6H4OH)
Ka(C6H5OH), at 25 oC in aqueous solution
r = 1
S. J. Meek Typical Hammett Plot & Summary of r-values
Log k
ss = -veEDG: MeO, Me, NH2,..
s = +veEWG: Cl, CO2Et, CN, NO2,..
r= +ve:electrons flow towards thearomatic ring in the RDS
r= -ve:electrons flow away from the
aromatic ring in the RDS
X
s = 0.0X=H
-5-6 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6
Large -ve:+ve charge on
ring ordelocalized
round benzenering
Moderate -ve:e-’s flow out of TS+ve charge in near
ringloss of conjugation
Small r:1.Ar too far away2.No e- change3. Two r-values
cancel each otherout
Moderate +ve:e-’s flow into TS
-ve charge in near ring loss of conjugation
Large +ve:-ve charge on
ring ordelocalized
round benzenering
r :
S. J. Meek
ßCalculation of k or K for a specific reaction of a specific compound (This is limitedto substituted arenes).
•s-values for substituents are known
•Provided r is known for the specific reaction, the rate (or equilibrium)constant for any substituent relative to the unsubstituted compound can becalculated.
ßUse r to determine information about reaction pathways:
•Magnitude and sign of r fi development of charge at reaction centre
•If sx- or sx
+ give a better correlation than s through conjugation is important
•Deviations from linearity: Hammett plots are most informative at the point atwhich they deviate from linearity with the greatest information being found inwhether they deviate concave ‘upwards’ or ‘downwards’.
Uses of the Hammett Equation
log kx = rsx
kH
S. J. Meek Deviations From Linearity
ßConcave upwards:
•Compare the plots for the hydrolysis of ArCO2Me and ArCO2Et carried out in99.9% H2SO4.
log kx
sx
r = -3.25
r = -3.25 r ≈ +2.0
ArCO2Me
ArCO2Et
ßMe shows expected correlation r= -3.25Mechanism:
-positive charge develops in RDS andreaction is accelerated by ED-substituents.
O
OMeAr
H O
OMeArH
O
Ar
-MeOH
RDSH2O
O
OHArH
H2O
O
OHAr
ßEt esters show an expected correlation r=-3.25 switching to r=+2.0.Change in mechanism:
-Et esters can form a more stable carbocation,+CH2CH3 (cf. Me esters) n.b.essentially underanhydrous conditions- EW-substituents decrease the positive chargeat the reaction centre in the rate RDS andaccelerate the reaction
O
OEtAr
H
O
OArH
Et
RDS O
OHArCH2CH3
S. J. Meek Deviations From Linearity
ßConcave Upwards:
•Hydrolysis of acid chlorides.
log kx
sx
r = -4.4 r = +2.5
ArCO2Cl
O
ClAr
O
OHAr
H2O
acetone
O
ClAr
H2O
RDS r = +2.5
Mechanism 2: rate isaccelerated by EW-
substituents
r = -4.4
O
ClAr
O
ArRDSH2O
O
OHArH OH2
O
OHAr
-Cl
fast
fast
Mechanism 1(SN1): rate isaccelerated by ED-substituents
ß Concave upwards deviation: Indicates a change in reaction mechanism
-any new pathway occurring must be faster than the original for it to becomedominant-curving upwards of the correlation means the new mechanism is faster
S. J. Meek Deviations From Linearity
ßConcave Downwards:
•Intramolecular Friedel-Crafts alkylation
•+ve r for ED-substituents•-ve r for EW-substituents
log kx
sxs=+0.5
r=+2.67 r=-2.51
OHAr
PhAr Ph
H2SO4
H2O, HOAc
OHAr
Ph
H
FastOH2Ar
Ph ArPh Ar Ph
HAr Ph
-H
Fast
E1 SEAr
Which step is RDS?- E1: a positive charge is increasing at the reaction centre (-ve r) \ E1 is RDS for EW-substituents (ie. RDS for right-hand of Hammett correlation )
- SEAr: positive charge at the reaction centre is decreasing (+ve r) \ SEAr is RDS for ED-substituents (ie. RDS for left-hand of Hammett correlation )
RDS? RDS?
EWG +sEWG -s
S. J. Meek Deviations From Linearity
ßDeviation Concave Upwards:
- Indicates a change in the reaction mechanism.
ßDeviation Concave Downwards:
- Indicates the same mechanism but a change in the rate determining step.
S. J. Meek Other Variants of the Hammett Equation
ßYukawa-Tsuno Equation: introduces a further parameter into the Hammett equation toquantify the graded response to through-conjugation for para-substituents.
r = 1.0 for the solvolysis of tertiary halides eg. cumyl chlorideIf no through-conjugation occurs r=0 and the equation simplifies to the original Hammettequation
ßBase-catalysed hydrolysis of p-phenoxytriethylsilanes gives r=0.50The extent of through conjugation by a group such as p-NO2 gives r=+3.52 suggestingthe build up of substantial negative charge in the transition state.
logkx
kH= r[sx +r(sX
+-sx)]
O
N
CMe2
OMe
SiEt3
OHd-
d-
O O
r =+3.52r = 0.50
Cl
d++
d--
r = -4.52r = 1.0 (by definition)
OSiEt3
X
OHO
X
+ Et3SiOH
s+ for electron-donating p-substituentss- for electron-withdrawing substituentsr = a measure of through conjugation operating in a particularreaction
S. J. Meek Other Variants of the Hammett Equation
ßTaft Equation: used to explain the reactivities of aliphatic esters.
logkR
kMe= r*sR* + dEs
sR* = the polar substituent constantEs = steric substituent parameterr* = measures the reactions susecptibility towards polar effectsd = a measure of a particular reaction’s susecptibility towardssteric effects (ie d is the steric parallel to r*)
ß d = 1.00 for acid catalysed ethylester hydrolysisß Es = 0 for R = Methyl
logkRCO2Et
kMeCO2Et= Es
acid
S. J. Meek Curtin-Hammett
How does the conformation of a molecule affect its reactivity?
Consider:
Do the two different conformers react at the same rate, or different rates?What factor determines product distribution?
The Situation:
Consider the two interconverting conformers A and B, each of which can undergoa reaction resulting in two different products, PA and PB.
Two limiting cases:
1. The rate of reaction is faster than the rate of conformational interconversion
2. The rate of reaction is slower than the rate of conformational interconversion
NMe
NMe13MeI 13MeI
PA A BkA
kBPB
k1 k2
major minor
S. J. Meek Curtin-Hammett
Case 1: (Kinetic Quench) The rate of reaction is faster than the rate of conformational interconversion k1,k2 >> kA,kB.
If the rates of reaction are faster than the rate of conformational interconversion,A and B cannot equilibrate during the course of the reaction, and the productdistribution (PA/PB) will reflect the initial composition.
PA
DG1‡ DG2
‡
DGo PB
DGAB‡
A
B
Ene
rgy
Æ
Rxn. Coordinate
[PB]
[PA]
[B]o
[A]o=
PA A BkA
kBPB
k1 k2
In this case, the productdistribution depends solely on theinitial ratio of the two conformers
S. J. Meek Curtin-Hammett
Case 1: (Kinetic Quench) The rate of reaction is faster than the rate of conformational interconversion k1,k2 >> kA,kB.
Example:
Padwa, JACS, 1997, 4565
In the case above while enolate conformers can be equilibrated at higher temperatures,the products of alkylation at -78 oC always reflect the initial ratio of enolate isomers.
NO
Me
Me
H
MeBr
NMe
Me
HO
MeBr
more stableless stable
N
Me
Me
O Me
N
Me
Me
O Me
major productminor product
-78 oC
DG = -3.0 kcal/mol(by by ab initio calc.)
S. J. Meek Curtin-Hammett
Case 2: (Curtin-Hammett conditions) The rate of reaction is slower than the rateof conformational interconversion kA,kB >> k1,k2.
If the rates of reaction are much slower than the rate of conformationalinterconversion, (DGAB
‡ is small relative to DG1‡ and DG2
‡), then theratio of A to B is constant throughout the course of the reaction.
PA A BkA
kBPB
FastSlow Slow
k1 k2
PA
DGoPB
DGAB‡
A
Ene
rgy
Æ
Rxn. Coordinate
DG2‡
B
DG1‡
major minor
S. J. Meek Curtin-Hammett
The Derivation: (Case 2)
Using the rate equation and we can write:
Since A and B are in equilibrium, we can substitute
d[PA]
dt= k1[A]
d[PB]
dt= k2[B]
d[PA]
d[PB]= k2[B]
k1[A]d[PA]d[PB] =
k2[B]
k1[A]or
=Keq[B]
[A]
d[PB] =k2
k1Keq∫ ∫ d[PA] Integration gives Keq[PA]
= k2
k1
[PB]
When A and B are in rapid equilibrium, we must consider the rates ofreaction of the conformers as well as the equilibrium constant whenanalyzing the product ratio.
S. J. Meek Curtin-Hammett
To relate this quantity to DG values, recall that DGo = -RT ln Keq orKeq = e-DGo/RT, k1 = e-DG1‡/RT, and k2 = e-DG2‡/RT. Substituting this into theprevious equations:
Combining terms:
Curtin-Hammett Principle: The product composition is not solelydependent on relative proportions of the conformational isomers in thesubstrate; it is controlled by the standard Gibbs energies of the respectivetransition states.
=Keq[PA]ek2
k1
[PB] -DG2/RT
-DG1/RTe
=e-DGo/RT
e-DG2/RT
e-DGo/RT -DG1/RT
e=
R = 8.314 J/Kmol = 2 cal/Kmol[PA]
[PB]e
-(DG2 + DGo - DG1 )/RT
= or[PA]
[PB]= e
-DDG /RT
S. J. Meek Curtin-Hammett
ß Three Senarios:
• If both conformers react at the same rate, the product distribution willbe the same as the ratio of conformers at equilibrium.
•If the major conformer is also the faster reacting conformer, the productfrom the major conformer should prevail, and will not reflect theequilibrium distribution.
•If the minor conformer is the faster reacting conformer, the product ratiowill depend on all three variables, and the observed product distributionwill not reflect the equilibrium distribution.
\ You can potentially isolate a product which is derived from aconformer you cannot observe in the ground state.
S. J. Meek Curtin-Hammett: Examples
NMe
NMe13MeI 13MeI
less stable more stable
NMe 13Me
Slower
minorproduct
NMe
13Me
Faster
majorproduct
ß The minor conformer reacts much faster than the more stable conformer
N
Me
t-Bu Nt-Bu Me
less stable more stableKeq=10.5
H2O2H2O2 k2k1slower faster
N
Me
t-Bu ONt-Bu Me
Omajor
productminor
product
ratio= 5 : 95
JOC 1974 319
Tet. 1972 573Tet. 1977 915
S. J. Meek Curtin-Hammett: Example
ß Stereoselective Hydrogenation:O
OMe
O H2
(R)-BINAP-Ru
OH
OMe
O99:1
92 % ee100% yield
O
HO OMe
H
Ru
X
P
P
- Enantioselectivities are the same , regardless of whether the starting material is alreadyenantiomerically enriched in the opposite enantiomer.
- Explanation: Enantiomerisation of the starting material through an achiral ester enolate(by tuning Lewis acidity of the solvent) is faster than the rate of hydrogenation.
- This is a case of Dynamic Kinetic Resolution: Two enantiomeric keto-esters areequilibrating during the course of a reaction with H2.
Why is the yield >50% ?A diastereospecific rection
should not exceed 50%
•Configuration:- C-3 is governed by thehandedness of the BINAP.- C-2 is dependent onsubstrate structures
Noyori JACS 1989 111 9134
O
OMe
O
O
OMe
O
H2
(R)-BINAP-Ru
H2
(R)-BINAP-Ru
OH
OMe
O
OH
OMe
O
syn
antifast
slowO
OMe
O
achiral esterenolate
S. J. Meek Curtin-Hammett: Example
ß Asymmetric Hydrogenation of prochiral olefins, catalysed by Rhodium
chelating diphosphine ligands give very high enantioselectivities (e.g. S,S-CHIRAPHOS)
NHAcMeO2C
Ph
[L2*Rh]+ MeO2C NHAc
Ph
> 95% ee
ß Two diastereomeric Rh-substrate complexes are formed when the chiral ligand(S,S-CHIRAPHOS), rhodium and substrate are mixed.
Observations:
ßComplex B is the only observed diastereoisomer for the catalyst-substrate complex(1H-NMR, X-ray crystallography)ß The observed enantiomer of the product is derived exclusively from the minor complex Aß The enantioselectivities are dependent on the pressure of H2
RhP
P
O
Me
MeO2CNH
Ph*
RhP
P
OMe
CO2MeHN
Ph *A
minor complexB
major complex
S. J. Meek Curtin-Hammett
ß Explaination: The Curtin-Hammett Principle
RhP
P
O
Me
MeO2CNH
Ph*
RhP
P
OMe
CO2MeHN
Ph *
Aminor complex
Bmajor complex
H2
Addition is fast
H2
Addition is slow
RhP
P
O
Me
MeO2CNH
Ph
*H
H
migration
+ SRhP
P
S
*H
O
Me
NH
CH2Ph
CO2Me reductive elimination
MeO2C NHAc
Ph
(R)
-L2RhS2
RhP
P SH
S*
Ph
MeO2C NHAc
(S)-Enantiomer
KAeq
KBeq
KAeq , K
Beq > k2[H2], k1[H2]
k2[H]
k1[H]
k2[H2] >> k1[H2]
The minor diastereoisomercomplex is 580-fold morereactive than the majordiastereoisomer complexfi 96% ee
96% ee
Halpern JACS 1987 109 1746
S. J. Meek Curtin-Hammett
Curtin-Hammett Principle: The product composition is not solely dependent onrelative proportions of the conformational isomers in the substrate; it is controlled
by the difference in the standard Gibbs energies of the respective transitionstates.
References
Hammett Equation:
1. A guide book to Mechanism in Organic Chemistry, Peter Sykes pg358-395(sixth edition)2. Advanced Organic chemistry: Part A, Carey, and Sundberg pg204-215 (fourth edition)3. Free energy relationships in organic and bio-oragnic chemistry, A. Willliams, pg17-544. Organic Chemistry, Clayden, Greeves, Warren and Wothers. pg 1090.4. A review of Hammett substituent constants Chem. Rev. 1991, 91, 165.
The Curtin-Hammett Principle:
1. Advanced Organic chemistry: Part A, Carey, and Sundberg pg215-261 (fourth edition)2. Seeman, Chem. Rev. 1983 83 833. Seeman, J. Med. Ed. 1986 63 424. Prof. D. A. Evans Chem 206 lecture 7
S. J. Meek Questions
1. The pKa of p-methoxybenzoic acid is 4.49; that of benzoic acid is 4.19, caculate for sfor p-MeO?
2.The base-catalysed hydrolysis of ethyl m-nitrobenzoate is 63.5 times faster than theunsubstituted ester under the same conditions; what will be the comparable rate ofhydrolysis of ethyl p-methoxybenzoate? [From tables sm-NO2=0.71 and sp-MeO=-0.27]
3.
The theoretical transition state energy DDG‡ = -1.43 kcal/mol, and experimentally theproduct ratio = 91:09. Determine whether this reaction obeys the Curtin-Hammett principleby calculating the expected ratio of products based on the theoretical DDG‡ value?
4. In the following B and C are diastereoisomers:
Assuming the diastereomeric excess is 99% de (B:C) calculate the difference in transitionstate energies (kcal/mol) leading to the products?
Areagents
80 oCB + C
Draw the major/minor productsexpected from this reaction?
HO
Me
Ti(OiPr)4
tBuOOH, rt
Product
S. J. Meek Questions
5. Use the proportion of enolates shown for the two cases (R=t-Bu and R=Et) to determinethe difference in the free energies of activation at -78 oC, assuming that enolate formationis kinetically controlled?
6. Fill in the following table and draw energy diagrams to illustrate each situation?
7. For the Asymmetric Hydrogenation of prochiral olefins, catalysed by Rhodium (see slide35) calculate the difference in the free energies of activation at 298K of the twodiastereomeric rhodium-substrate complexes?
R
OLDA, THF-78 oC
R
OLi
R
OLi
R = t-Bu 98% 2%R = Et 30% 70%
PA A BKeq
PB
k1 k2
Keq PB/PA k2/k1 DDG
10-1
10+1
5:1
5:1
?
?
?
?
S. J. Meek Questions
8. In a recent paper the following kobs were recorded for the 2-arylpyrrolidinone-catalysedaqueous aldol reaction shown below. From these values calculate r? What does r tell youabout the RDS and how the various substituents affect catalysis? (note that thesubstituents in the table are on the Ar-group on the catalyst).
AnswersS. J. Meek
1. The pKa of p-methoxybenzoic acid is 4.49; that of benzoic acid is 4.19, caculate for sfor p-MeO?
sx = pKa(H) - pKa(x)\ sx = 4.19 - 4.49\ sx = -0.3
2.The base-catalysed hydrolysis of ethyl m-nitrobenzoate is 63.5 times faster than theunsubstituted ester under the same conditions; what will be the comparable rate ofhydrolysis of ethyl p-methoxybenzoate? [From tables sm-NO2=0.71 and sp-MeO=-0.27]
For m-nitrobenzoate,
Using log k = sr, where k = 63.5 ko ko
\ log(63.5) = r x 0.71\ r = 2.5391
For p-methoxybenzoate
\ log k = -0.27 x 2.5391 k0\ k = 0.206
ko
S. J. Meek Answers
3.
The theoretical transition state energy DDG‡ = -1.43 kcal/mol, and experimentally theproduct ratio = 91:09. Determine whether this reaction obeys the Curtin-Hammett principleby calculating the expected ratio of products based on the theoretical DDG‡ value?
Using DDG‡ = -RTln[Pb] [Pa]
[Pb] = 11 \ transition state ratio = 92:08 and therefore,[Pa] agrees with the Curtin-Hammett principle.
4. In the following B and C are diastereoisomers:
Assuming the diastereomeric excess is 99% de (B:C) calculate the difference in transitionstate energies (kcal/mol) leading to the products?
B : C ratio, 99.5 : 0.5 T = 273 + 80 = 353 K[Pb] = 99.5/0.5 = 199 DDG‡ = -RTln[Pb] DDG‡ = -(2x10-3)(353) ln199[Pa] [Pa] = -3.7 Kcal/mol
Areagents
80 oCB + C
Draw the major/minor productsexpected from this reaction?
HO
Me
Ti(OiPr)4
tBuOOH, rt
Product
H
HOMe
Me
HHOO O
Major Minor
S. J. Meek Questions
5. Use the proportion of enolates shown for the two cases (R=t-Bu and R=Et) to determinethe difference in the free energies of activation at -78 oC, assuming that enolate formationis kinetically controlled?
[Pb]/[Pa] = 49 \ DDG‡= -(195x1.98) ln49 ≈ -1.5[Pb]/[Pa] = 0.43 \ DDG‡= -(195x1.98) ln0.43 ≈ 0.33
6. Fill in the following table and draw energy diagrams to illustrate each situation?
7. For the Asymmetric Hydrogenation of prochiral olefins, catalysed by Rhodium (see slide35) calculate the difference in the free energies of activation at 298K of the twodiastereomeric rhodium-substrate complexes?96 % ee firatio 98 : 2 T = 273 + 25 = 298 K[Pb] = 98/2 = 49 DDG‡ = -RTln[Pb] DDG‡ = -(2x10-3)(298) ln 49[Pa] [Pa] ≈ -2.3 Kcal/mol
8. r = 1.14 (JOC 2005 70 3705)
R
OLDA, THF-78 oC
R
OLi
R
OLi
R = t-Bu 98% 2%R = Et 30% 70%
PA A BKeq
PB
k1 k2
Keq PB/PA k2/k1 DDG
10-1
10+1
5:1
5:1
50
0.5
-RTln5
-RTln5