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MODULE TITLE LINEAR FREE ENERGY RELATIONSHIP (HAMMETT-TAFT

EQUATION)

MODULE NUMBER 1

COURSE USED CHM3103 –CHEMICAL KINETICS

MODULE CODE UPM/FS/CHM3103-C1

VERSION VERSION 4.0

LEARNING RESOURSES

1. Laidler, K. J. (1997). Chemical Kinetics. 3rd edition. New

York: Addison-Wesley Pub Co.

2. Mortimer, M. & Taylor, P. G. (2002). Chemical Kinetics and

Mechanism. Milton Keynes: The Open University.

3. Espenson, J. H. (2002). Chemical Kinetis and Reaction

Mechanisms. 2nd edition. New York: McGraw Hill.

4. Hammett, L.P. Some Relations between Reaction Rates and

Equilibrium Constants. Chem. Rev. 1935. 17(1): 125–136.

5. Hansch, c., Leo, A., Taft, R.W. A Survey of Hammett

Substituent Constants and Resonance and Field Parameters.

Chem. Rev. 199l. 97:165-195.

6. http://en.wikipedia.org/wiki/Hammett_equation#cite_note-3

7. Any organic chemistry books.

BACKGROUND KNOWLEDGE

Reaction rates; Chemical equilibrium; Role of substituent in the

chemistry of aromatic compounds

TERMS TO KNOW (TERMINOLOGY)

Gibbs free energy; Equilibrium constant; m- & p-substitution;

Inductive effect

LEARNING OUTCOMES

On completion of this module student should be able to:

1. Explain The Hammett-Taft Equation Using The Linear Free

Energy Concept.

2. Relate the electronic effects of substituent on the strengths of

m- and p-substituted benzoic acids as an example & apply that.

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COURSE CONTENTS

The term linear free energy relationship (LFER) applies to a

variety of relationships between kinetic and thermodynamic

quantities that are very important in both organic and inorganic

chemical reactions. The Hammett equation which was developed

in 1937 relates the dissociation constants of substituted benzoic

acid to benzoic acid. This equation shows the linear relationship

between the change in free energy and the product of ρσ.

Equation 1

Equation 2

The Hammett equation was originally used to explain the

electronic effects of substituent on the strengths of m- and p-

substituted benzoic acids. For example, if the group X is electron

withdrawing, the acidity of the -COOH group is increased and σ is

positive. Conversely, the σ values are negative for electron

releasing group.

Note that this equation is satisfactory if the reactive site is

sufficiently removed from the substituent so that steric factors do

not enter into the rate-determining step. If the reaction involves a

series of substituents that greatly alter the way in which either the

reactant or the transition state is solvated, the relationship may be

less satisfactory.

While the original Hammett LFER was applied to aromatic

compounds but it can be extended to aliphatic compounds as well.

The approach was taken by Taft is similar to that of Hammet;

Equation 3

Where σ* is a constant related to polar substituents effects and ρ*

is a reactant constant like the case before, and δEs is a steric energy

term.

For a given series of reactants, δEs is frequently considered to be

zero because for any pair of similar species subtraction of two

equations having the above equation form. So, Taft equation is

written as;

3

Equation 4

The Taft and Hammett equations are the same, but Taft used

constants that are also appropriate to aliphatic and restricted

aromatic materials.

ASSIGNMENTS

1. Explain the two constants of "sigma" and "rho" in the

Hammett-Taft equation.

2. The Hammett-Taft equation is used only for substituent at para

and meta positions not for ortho. Explain why.

3. The values of σ for m-NH2 & p-NO2 have been reported as -

0.71 & +0.78. The values are so small compared to pKa value

of benzoic acid. Explain why.

4. The base-catalysed hydrolysis of ethyl m-nitrobenzoate is 63.5

times faster than the unsubstituted ester under the same

conditions; what is the comparable rate of hydrolysis of ethyl

p-methoxybenzoate? [ m-NO2= +0.71 and p-MeO= -0.27]

5. Using equation 3 shows that which of the following reactions is

faster?

i. CCl3CO2Et + H2O CCl3CO2H + EtOH

(Es= -0.07; *= -0.10)

ii. C2H5CO2Et + H2O C2H5CO2H + EtOH

(Es= -2.06; *= +2.65)

6. The linear free energy relationships are very useful in organic

and inorganic chemistry. Explain why.

EVALUATION

In a group manner (refer to the grouping list), you must submit the

module by 23/10/2013 by group discussion and doing the

assignment.

COPYRIGHT

Dr. ROGHAYEH ABEDI KARJIBAN

roghayeh@upm.edu.my

03-89467487

Date Created: 31 March 2013

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Grouping – CHM3103 – Semester 2 – 2013/2014

Group 1 1 1 5 6 3 8 5 S I T I N O R H A S M A H B I N T I A H M A D J A N I2 1 5 7 1 6 9 S I T I N U R ' I Z Z A T I B I N T I M O H D S O F I3 1 5 7 5 3 2 N U R H A S A N A H B I N T I H A M Z A H4 1 5 7 5 3 7 N U R S H A F I K A B I N T I R A H I M5 1 5 7 6 8 1 T E E X I N R U I6 1 5 8 0 1 7 N U R H A Y A T I B I N T I A B D U L R A H M A N

Group 2 7 1 5 8 4 7 4 T A N M E I W E I8 1 5 8 9 0 4 N U R U L H A Z W A N I B I N T I S A L L E H9 1 6 1 1 7 4 A I N A Z I R A B I N T I A Z L A N1 0 1 6 1 2 5 0 S I T I A I S A H B I N T I I S M A I L1 1 1 6 1 5 9 7 S H A H I B U L B A R I A H B I N T I M A T G H A N I1 2 1 6 1 9 2 6 N U R K H A L I S A B I N T I R A M L I

Group 3 1 3 1 6 1 9 8 1 N U R S H A F I Q A B I N T I S H A K I M I1 4 1 6 2 0 5 2 P A R H A W A H I D A B I N T I S A I R I1 5 1 6 2 0 8 9 Z A R I N A B I N T I A H M A D1 6 1 6 2 1 3 5 R A Z I L A B I N T I A B D U L L A H1 7 1 6 2 2 5 0 M A T H I V A N I A / P M U R U G E E S A N1 8 1 6 2 2 9 3 I N T A N I ' D Z A R N I B I N T I A B D U L M A N A N

Group 4 1 9 1 6 2 3 0 1 N U R U L A S Y I Q Q I N B I N T I A H M A D S H A H2 0 1 6 2 3 1 2 N U R R A S Y I D A H B I N T I A W A N G A H M A D2 1 1 6 2 4 1 1 H A D H I N A H A N A T I B I N T I Y A N D A S R I L2 2 1 6 2 4 1 5 H A M R A A S S Y A I M A B I N T I A B D U L B A S H I D2 3 1 6 2 4 3 4 L E E X I N J I E2 4 1 6 2 5 1 3 N A Z A T U L A K M A B I N T I S O B R I

Group 5 2 5 1 6 2 6 8 9 N A J W A B I N T I N O R B I2 6 1 6 2 7 2 6 M U H A M M A D ' A M I R B I N Y A H A Y A2 7 1 6 2 7 2 8 L U Q M A N H A K I M B I N M O H D S A I D2 8 1 6 2 7 3 7 S I T I A L W A N I B I N T I R O S L I2 9 1 6 2 7 4 5 M O H D H A Z I Q B I N R E D H U A N3 0 1 6 2 7 8 2 N U R N A D Z I E R A B I N T I N A Z R I

5

Group 6 3 1 1 6 2 8 2 0 S I T I S A R A H Y A S M I N B I N T I A H M A D3 2 1 6 2 9 9 2 S I T I N U R A I S H A H B I N T I I S M A I L3 3 1 6 3 1 0 7 M U H A M A D N O R E M Y B I N P A I M I N3 4 1 6 3 2 2 6 Z H A R I F B I N C H E Z A I N U D D I N3 5 1 6 3 2 9 5 N U R F A Z R E E N A T I L A B I N T I H A S H I M3 6 1 6 3 3 2 5 C H E E W E I J I A N

Group 7 3 7 1 6 3 4 2 3 N U R U L H U S N A B I N T I R O S D I3 8 1 6 3 4 7 0 N O R A I N A B I N T I A H M A D K A M A L H A S N I3 9 1 6 3 8 1 3 N U R F A Z L I A N A B I N T I A Z M I4 0 1 6 3 8 4 9 Z A E T O N B I N T I M U H A M M A D4 1 1 6 3 8 5 4 U N G W A I K E O N G4 2 1 6 4 1 6 2 N U R U L N A B I L A B I N T I Z E K E R I A

Group 8 4 3 1 6 4 1 6 7 E F A R I Z A N B I N T I S U R A D I4 4 1 6 4 2 6 9 M U H A M M A D N O R H A F F I S B I N M U S T A F A4 5 1 6 4 2 8 9 N U R K H A I R U N I S A B I N T I R A Z A L I4 6 1 6 4 3 5 2 S I T I N O R F A D I L A H B I N T I A B U H A N I F A H4 7 1 6 4 7 3 3 A L V I N L I M T E I K Z H E N G4 8 1 6 5 1 5 0 N U R U L N A S H A S Y A F I Q A H B I N T I C H E K H A M S A H