Organic Chemistry

(Dana P.) #1
Section 4.3 The Structure of the Transition State 145

George Simms Hammondwas born
in Maine in 1921. He received a B.S.
from Bates College in 1943 and a
Ph.D. from Harvard University in


  1. He was a professor of
    chemistry at Iowa State University
    and at the California Institute of
    Technology and was a scientist at
    Allied Chemical Co.


Free energy

Progress of the reaction

I II III

an endergonic
reaction

an exergonic
reaction

>Figure 4.3
Reaction coordinate diagrams for
reactions with (I) an early transition
state, (II) a midway transition state,
and (III) a late transition state.

PROBLEM 1

List the carbocations in order of decreasing stability.

a.

b.

PROBLEM 2

a. How many C H bond orbitals are available for overlap with the vacant porbital in the
methyl cation?
b. Which is more stable, a methyl cation or an ethyl cation?

4.3 The Structure of the Transition State


Knowing something about the structure of a transition state is important when you
are trying to predict the products of a reaction. In Section 3.7, you saw that the struc-
ture of the transition state lies between the structure of the reactants and the struc-
ture of the products. But what do we mean by “between”? Does the structure of the
transition state lie exactlyhalfway between the structures of the reactants and prod-
ucts (as in II in the following diagram), or does it resemble the reactants more close-
ly than it resembles the products (as in I), or is it more like the products than the
reactants (as in III)?


According to the Hammond postulate,the transition state is more similar in struc-
ture to the species to which it is more similar in energy.In the case of an exergonic re-
action, the transition state (I) is more similar in energy to the reactant than to the
product (Figure 4.3, curve I). Therefore, the structure of the transition state will more


reactants products
transition state

C (II)
(III)

(I) AB
AB+ ABCA + BC

C

ABC

symbolizes the
‡ transition state


¬

CH 3 CHCH 2 CH 2

F

+
CH 3 CHCH 2 CH 2

CH 3

CH 3 CHCH 2 CH (^2) +
Cl



  • CH 3 CH 2 CH 2 CH 2


  • CH 3 CH 2 CCH 3 CH 3 CH 2 CHCH+ 3
    CH 3





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