Jones M., Fleming S.A. Organic Chemistry

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8.4 Rates of Chemical Reactions 339

Summary

Now we know something of the factors that affect equilibrium.Enthalpy change,

ΔH, is important. Basically, the side of the equilibrium with the stronger bonds

will be favored by enthalpy. Also important is the temperature-dependent entropy

change, ΔS, term. The two terms may act in concert or oppose each other.

We have said nothing so far of the factors influencing the rate of establishing

equilibrium.

8.4 Rates of Chemical Reactions

Just because a reaction is exothermic does not mean that it will be a rapid reaction

or even that it will proceed at all under normal conditions. First of all, an exother-

mic reaction can be endergonic if entropy plays an especially important role. Entropy

does not raise its ugly head in this way often, but there is another, much more fre-

quently encountered reason why an exergonic reaction might not be a fast reaction.

As we have seen, there are energy barriers to chemical reactions. In principle, a very

stable product can be separated from a much less stable starting material by a bar-

rier high enough to stop all reaction (Fig. 8.7).

Energy

Reaction pro

ress

ΔG⬚

Starting

material A

Barrier

Product B

Transition state

This barrier must be

surmounted in order to

produce product, even

though the reaction

A B is exergonic

FIGURE 8.7 Rates of reactions are

determined not by the ΔG° between

starting material and product, but by

the height of the energy barrier

separating starting material from

product. Some very unstable

materials are protected by high

barriers and thus can be isolated.

PROBLEM SOLVING

Whenever you see the word “rate” in a problem, or when you see words such as

“faster” that talk about rates, you are very likely to have to answer the question by

drawing the transition state for the reaction. Remember that the rate of a

reaction is determined by ΔG

‡

, the energy difference between starting material

and the transition state, and not by the energy difference between starting

material and product. You need a pull-down menu that says, “Think about the

transition state” when the word “rate”appears in a problem.

340 CHAPTER 8 Equilibria

Sometimes a reaction we expect to be second order appears not to be. The S

solvolysis reaction of methyl iodide with solvent (therefore in great excess) ammonia

is an example.This reaction is first order in substrate, methyl iodide,because the con-

centration of one of the reactants, ammonia, does not change appreciably during the

reaction. Such reactions are called pseudo-first-order reactions (Fig. 8.10).

Number of molecules

Kinetic energy

X = energy necessary

to surmount the

activation barrier

FIGURE 8.8 Boltzmann distributions

of molecules at two temperatures.

The blue line is the distribution at

higher temperature, and the red line

is the distribution at lower

temperature.

OH OH

Rate = k [(CH

)

Rate = k [CH

][

–

OH]

–

FIGURE 8.9 Rate laws for the S

1 and S

2 reactions.

+++++

Solvent (present

in great excess)

Rate = k [CH

]

I CH

–

FIGURE 8.10 A pseudo-first-order

reaction.

As we already saw in our discussion of the S

2 and S

1 reactions,rates of chem-

ical reactions also depend on the concentrations of reactants. The rates of most

organic reactions depend either only on the concentration of a single reactant (S

for example) and thus are first-order reactions or on the concentrations of two species

2) and are second-order reactions (Fig. 8.9).There are examples of higher-order

reactions, but they are relatively rare.

At a given temperature, collections of molecules exist in ranges of energies called

Boltzmann distributions. In a Boltzmann distribution, the energies of most

molecules cluster about an average value. As energy is provided to the molecules,

usually through the application of heat, the average energy increases.Although there

is always a small number of highly energetic molecules present (the high-energy tail

of the Boltzmann distribution), in practice there may not be enough molecules of

sufficient energy to make the reaction proceed at an observable rate if the barrier

is high enough. The application of energy in the form of heat produces more

molecules with sufficient energy to traverse the barrier (Fig. 8.8). A reaction rate

approximately doubles for every increase in temperature of 10 °C.

8.5 Rate Constant 341

8.5 Rate Constant

In the rate laws mentioned so far, there has always been a proportionality constant,

the rate constant (k), which is a fundamental property of any given chemical reac-

tion. A rate must have the units of concentration per unit time,often moles per liter

per second. For a first-order process such as the S

1 reaction, the rate constant k

must have the units of reciprocal seconds (s

1

) (if we use the second as our time

unit, Fig. 8.11).

Rate = k [A]

concentration/time = (rate constant) (concentration)

 (s)

–1



= k

Rate =

in units,

mol

 (s)

–1



= k

rearranging we have:

mol

(s)

–1

= k

so, or

FIGURE 8.11 For a first-order

reaction, the rate constant (k) has the

units of time

1

 s

1

For a second-order reaction, k must have units of reciprocal moles per liter per

second [(mol/L)

1

] (Fig. 8.12).

Rate = concentration/time = (rate constant) (concentration) (concentration)

 (s)

–1



= k

in units,

mol



mol

 (s)

–1



= k

= k = kor

rearranging we have:

mol

 (s)

–1

mol

1mol  s2

mol

–1



mol

so,

Rate = k[A][B]

FIGURE 8.12 For a second-order

reaction, the rate constant (k) has the

units of (mol/L)

1

We can determine the value of k by measuring how the reaction rate varies with

the concentration(s) of the reactants.

It is easy to confuse the rate of a reaction, which is simply how fast the concen-

trations of the reactants are changing, with the rate constant, which tells us how the

rate will change as a function of reactant concentration. The rate of a reaction will

vary with concentration.This variation is easy to see by imagining the limiting case

in which one reactant is used up. When its concentration goes to zero, the reaction

rate must also be zero. On the other hand, the rate constant is an intrinsic property

of the reaction. It will vary with temperature, pressure, and solvent, but does not

depend on the concentrations of reactants.

Only when the concentrations of all reactants are 1 M are the rate and the rate

constant the same. The rate constant is thus a measure of the rate under standard

conditions (Fig. 8.13).

Rate = k [A][B][C][D] … [X]

if all [ ] = 1 M

Rate = k [1][1][1][1] … [1]

and therefore, rate = k

FIGURE 8.13 If the concentrations

of the reactants are all 1 M, the rate

constant and the rate are the same.

342 CHAPTER 8 Equilibria

CONVENTION ALERT

PROBLEM 8.5 Draw an arrow to show the activation energy for the reverse of the

reaction in Figure 8.14, the formation of methane and oxygen from carbon diox-

ide and water.

8.6 Energy Barriers in Chemical Reactions:

The Transition State and Activation Energy

We have already mentioned that even exothermic reactions may be very slow,

indeed immeasurably slow. And it’s a good thing. Combustion is a very exothermic

chemical reaction, and yet we and our surroundings exist more or less tranquilly in

air without bursting into flame. Somehow we are protected by an energy barrier.

For example, methane and oxygen do not react unless energy, called activation

energy, is supplied. If we are considering free energy, the activation energy is

ΔG

‡

; if we are speaking of enthalpy, the activation energy is ΔH

‡

. The double

dagger (

‡

) is always used when we are referring to activation parameters involv-

ing the energy of a transition state. A match will suffice to provide the energy in

our methane plus oxygen example. This highly exothermic process (p. 68) then

provides its own energy as very stable molecules are produced from less stable

ones (Fig. 8.14).

Energy

Reaction progress

ΔG⬚

+ 2 H

+ 2 O

Very negative for this highly

exergonic reaction; the

energy given off can be

used to supply ΔG

This amount of energy

must be supplied to

start the reaction

ΔG

†

FIGURE 8.14 An exergonic reaction

can provide the energy necessary to

pass over an activation barrier. Once

such a reaction is started, it continues

until one reactant is used up.

We are protected from spontaneous combustion in the same way we are

protected from hungry bears and tigers during a visit to the zoo. At the zoo the

barrier is tangible and operates in a clear way to ensure our survival. Activation

barriers are less visible but no less necessary for the continuance of our lives.

In this important section, we will discuss these energy barriers to chemical

reactions.

In a process such as the S

1 reaction, a bond must be broken. Consider the ion-

ization of tert-butyl iodide.The first step in this reaction is the ionization of the sub-

strate, and surely this step will involve an energy cost in breaking the bond from

8.6 Energy Barriers in Chemical Reactions 343

This energy cost is partially offset by an increase in entropy as we form two particles

out of one. But we have to be careful when we estimate entropy changes because there

are invisible participants in most reactions,including this one.These invisible participants

are the solvent molecules.They will have to be disordered and reordered in any ionic reac-

tion in solution, and the net change in entropy may not be at all easy to estimate.

(CH

)

C (CH

)

–

WEB 3D WEB 3D

FIGURE 8.15 The enthalpy change

ΔH° (and ΔG°) is positive for this

endothermic reaction.

WORKED PROBLEM 8.6 (a) Construct an Energy versus Reaction progress dia-

gram for the reaction of tert-butyl iodide and water (see Fig. 8.2), showing

clearly the activation energies and transition states for all steps. A discussion of

this reaction is coming, but try to anticipate it here. (b) Show the activation ener-

gies for the reverse reactions involved in the formation of tert-butyl iodide from

tert-butyl alcohol and hydrogen iodide (HI).

ANSWER

PROBLEM 8.7 Show that at equilibrium the equilibrium constant (K ) is given by

the ratio of the rate constants (k) for the forward and reverse reactions. Hint:At

equilibrium the rates of the forward and reverse reactions are equal.

Energy

Reaction progress

Transition state 1 Transition state 2 Transition state 3

(CH

)

(CH

)

(CH

)

COH

(CH

)

Activation energies

for the forward

reactions

Activation energies

for the reverse

reactions

–

The situation is a little different for the S

2 reaction. Here we must also break

the bond from carbon to the leaving group, but this energy cost is partially offset

by the formation of the bond from carbon to the entering nucleophile. We might

imagine that there must be a substantial entropy cost in this reaction as we must

carbon to the leaving group.The enthalpy change ΔH° (and ΔG°) is positive for this

reaction (Fig. 8.15).

344 CHAPTER 8 Equilibria

order the reactants in a special way for the reaction to proceed. The S

2 reaction

can occur only if the nucleophile attacks from the rear of the departing leaving group,

and this requirement demands a specific arrangement of the molecules and ions

involved in the reaction (Fig. 8.16).

Energy

Reaction progress

ΔG⬚

This amount of activation

energy must be supplied

to start the reaction

Very negative for this

highly exergonic reaction

(read left to right)

ΔG

–

The transition state

for this S

2 reaction

–

†

FIGURE 8.17

An energy diagram for

the S

2 displacement of iodide by

cyanide (and its reverse reaction, the

much less favorable displacement of

the strong nucleophile cyanide by

iodide).The activation energy for

the exergonic displacement of iodide

is shown.

2 with cyanide

WEB 3D WEB 3D

–

FIGURE 8.16 The transition state for

the S

2 reaction has severe ordering

requirements. Attack of the

nucleophile must be from the rear for

the reaction to succeed.

We can follow the course of both these reactions in diagrams that plot Energy

versus Reaction progress. We usually don’t specify reaction progress in detail because

there are many possible measures of this quantity, and for this kind of qualitative

discussion it doesn’t much matter what we pick. In the S

2 reaction, for example,

we might choose to plot energy versus the length of the bond between carbon and

the leaving group (which increases in the reaction), or the length of the bond between

carbon and the nucleophile, which decreases through much of the process.The over-

all picture is the same.

If we have enough energy for molecules to reach the transition state, to cross

the barrier in reasonable numbers, the reaction will run at a reasonable rate,

and the final mixture of starting materials and products will depend on the

ΔG° between starting materials and products. Figure 8.17 shows a diagram for

8.6 Energy Barriers in Chemical Reactions 345

Of course, this separation of the two reactions is absolutely artificial—we are

really describing a single equilibrium between methyl iodide and cyanide ion on one

side and methyl cyanide and iodide ion on the other (Fig. 8.19).

the exothermic formation of methyl cyanide (CH

CN) and potassium iodide

(KI) from methyl iodide (CH

I) and potassium cyanide (KCN). The activa-

tion energy is the height of the transition state above the starting material.

As there is enough energy to pass over the transition state in both direc-

tions, the ratio of product and starting material will depend on ΔG°, the

difference in free energy between starting material and product. The reaction

shown in Figure 8.17 has a negative ΔG°, and will be successful in producing

product.

In Figure 8.18, this same reaction is considered from the reverse point of

view: the endothermic formation of methyl iodide from methyl cyanide and

iodide ion. The activation energy ΔG

‡

is much higher than that for the

forward reaction, and ΔG° is positive for the endergonic formation of methyl

iodide. The transition state is exactly the same as that for the forward reaction.

This reaction will not be very successful. It not only requires overcoming a

high activation energy but is “fighting” an unfavorable equilibrium because ΔG°

is positive.

Energy

Reaction progress

ΔG⬚

ΔG

–

The transition state

for this S

2 reaction

–

This amount of activation

energy must be supplied

to start the reaction

Very positive for this

highly endergonic reaction

(read right to left)

†

FIGURE 8.18 This energy diagram

for the same reaction shown in

Figure 8.17 emphasizes the reversal.

Here the activation energy for

displacement of cyanide by iodide

is shown.

–

FIGURE 8.19 The competition

between cyanide and iodide for a

Lewis acidic carbon.

The exergonic formation of methyl cyanide and the endergonic formation of

methyl iodide are one and the same reaction. The reactions share exactly the same

transition state; only the ease of reaching it is different (ΔG

‡

forward

 ΔG

‡

reverse

).The

energy difference between the starting material and the product (ΔG°) is the differ-

ence between the two activation energies.

The rates of the forward and reverse reactions are determined by the magni-

tudes of the activation energies for the two halves of the reaction. They are inti-

mately related to the equilibrium constant for the reaction K, determined by ΔG°

[Eq. (8.7)].

ΔG

‡

reverse

 ΔG

‡

forward

 ΔG° RT ln K (8.7)

The principle of microscopic reversibility says that if we know the mechanism

of a forward reaction, we know the mechanism of the reverse reaction. One diagram

suffices to tell all about both forward and reverse reactions. Figures 8.17 and 8.18

are in reality a single diagram describing both reactions, in this case one “forward”

and exergonic, the other “reverse” and endergonic. There is no lower-energy route

from methyl cyanide and iodide ion back to methyl iodide and cyanide ion.The low-

est-energy path back is exactly the reverse of the path forward.

The classic analogy for this idea pictures the molecules as crossing a high moun-

tain pass.The best path (lowest energy, or lowest altitude) from village A to village B

is the best path (lowest energy, lowest altitude) back from village B to village A. If

there were a lower-energy route back from B to A, it would have been the better route

from A to B as well (Fig. 8.20).

346 CHAPTER 8 Equilibria

Village A

Village B

Had there been a lower-energy path (green)

it would still be the best path in both directions

Best = lowest-energy path from A to B. This

is also the best path back from B to A

Village A

Village B

FIGURE 8.20 Molecules, unlike

explorers, take the best (lowest-

energy) path. You might take the red

path from A to B and the orange

path back, but a molecule wouldn’t.

The Energy versus Reaction progress diagram for the S

1 reaction is more

complicated than that for the S

2 reaction, because there are more steps than in

the S

2 reaction. We dealt with the description of the S

1 reaction as a sequence

of equilibria earlier (Fig. 8.2). Let’s again take the reaction of tert-butyl iodide with

8.6 Energy Barriers in Chemical Reactions 347

solvent water as an example.The first step is the formation of an intermediate car-

bocation. We expect this first reaction to be quite endothermic as a carbon–leav-

ing group bond must be broken and ions must be formed. The high-energy point

in this reaction, transition state 1, occurs as the carbon–iodine bond stretches

(Fig. 8.21).

Energy

Reaction progress

(CH

)

Transition state 1

–

C H

(CH

)

–

FIGURE 8.21 The first step in the

1 reaction of tert-butyl iodide in

water. The formation of the cation is

highly endothermic.

Energy

Reaction progress

(CH

)

(CH

)

–

Transition state 1

(CH

)

–

Transition state 2

FIGURE 8.22 By contrast, the second

step in this S

1 reaction, the capture

of the cation by water, is highly

exothermic.Transition state 2 is

lower than transition state 1;

therefore, capture of the carbocation

by water is favored.

The second step involves the reaction of the nucleophilic solvent water with the

carbocation, and is highly exothermic because a bond is made and none is broken.

There is a second barrier, and therefore another transition state (2) for this second

step of the reaction (Fig. 8.22).

348 CHAPTER 8 Equilibria

As you saw in Worked Problem 8.6,there are six activation energies in the com-

pleted diagram for this S

1 reaction, three for the forward reactions and three for

the reverse reactions (Fig. 8.24).

Energy

Reaction progress

Transition state 3Transition state 1 Transition state 2

(CH

)

(CH

)

(CH

)

(CH

)

–

FIGURE 8.23 Transition state for the

final step in this S

1 reaction.

PROBLEM 8.8 Draw an Energy versus Reaction progress diagram for the S

1 reac-

tion of tert-butyl iodide in H

O/KOH containing an additional nucleophile,

Note that the major product is not necessarily tert-butyl alcohol (Fig. 7.60, p. 295).

Activation energies for

the reverse reactions

Activation energies for

the forward reactions

Energy

Reaction progress

(CH

)

–

(CH

)

(CH

)

–

(CH

)

–

Transition state 1

Transition state 2

Transition state 3

FIGURE 8.24 The three black double-

headed arrows represent activation

energies for the forward reactions and

the three red double-headed arrows

represent activation energies for the

reverse reactions.

The third and final step in this particular S

1 reaction is removal of a proton

from the oxonium ion by a base to give the ultimate product, tert-butyl alcohol. So

there is a third barrier and a third transition state (Fig. 8.23).