Hammett Plots2

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


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


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


km-Me

RDS

MeO

O

OEt

OH

MeO

O

OEt

OHd-

d-

d-

MeO

O

OEt

OH


km-OMe

RDS

O

OEt

OH

O

OEt

OHd-

d-

d- O

OEt

OH


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


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


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


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


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


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.)


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


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.


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


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


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


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


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

?

?

?

?


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


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

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