Organic Chemistry

Section 3.7 Thermodynamics and Kinetics 135

Mechanistic Tutorial:

Addition of HBr to an alkene

Progress of the reaction

ab

Free energy Free energy

Progress of the reaction

+∆G°

∆G‡

∆G‡

−∆G°

reactants

transition state transition state

product

carbocation carbocation

>Figure 3.5

Reaction coordinate diagrams for

the two steps in the addition of HBr

to 2-butene: (a) the first step;

(b) the second step.

PROBLEM 23

a. Which reaction has the greater equilibrium constant, one with a rate constant of

for the forward reaction and a rate constant of for the reverse reac-

tion or one with a rate constant of for the forward reaction and a rate constant

of for the reverse reaction?

b. If both reactions start with a reactant concentration of 1 M, which reaction will form the

most product?

Reaction Coordinate Diagram for the Addition

of HBr to 2-Butene

We have seen that the addition of HBr to 2-butene is a two-step reaction (Section 3.6).

The structure of the transition state for each of the steps is shown below in brackets.

Notice that the bonds that break and the bonds that form during the course of the reac-

tion are partially broken and partially formed in the transition state—indicated by

dashed lines. Similarly, atoms that either become charged or lose their charge during

the course of the reaction are partially charged in the transition state. Transition states

are shown in brackets with a double-dagger superscript.

A reaction coordinate diagram can be drawn for each of the steps in the reaction

(Figure 3.5). In the first step of the reaction, the alkene is converted into a carbocation

that is less stable than the reactants. The first step, therefore, is endergonic is

positive). In the second step of the reaction, the carbocation reacts with a nucleophile

to form a product that is more stable than the carbocation reactant. This step, therefore,

is exergonic (¢G°is negative).

(¢G°

CH 3 CH ++CH 3 CH CHCH (^3) +
‡
CH 3 CHCH 2 CH 3
H
Br
transition state
CHCH 3 HBr
δ+
δ−
Br−
‡
transition state
δ+
δ−Br Br
CH 3 CHCH+ 2 CH 3 + Br− CH 3 CHCH 2 CH 3 CH 3 CHCH 2 CH 3
1 10 -^3
1 10 -^2
1 10 -^31 10 -^5
Because the product of the first step is the reactant in the second step, we can hook
the two reaction coordinate diagrams together to obtain the reaction coordinate
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