Jones M., Fleming S.A. Organic Chemistry

Подождите немного. Документ загружается.

8.7 Reaction Mechanism 349

Let us now concentrate on these energy barriers separating the various intermedi-

ates.The barriers are made up of the same quantities, enthalpy and entropy, that define

the energy of any chemical species. Recall that ΔG°  ΔH°  TΔS° [Eq. (8.6)].

Transition state theory assumes that we can treat a transition state as if it were a real

molecule occupying a potential energy minimum. It is important to emphasize that it

is not, however. It occupies an energy maximum and can never be bottled and exam-

ined as real molecules can.Nonetheless it has an energy,as indeed does every other point

in the energy diagrams on the preceding pages.The energy of the transition state, and

thus the height of the energy barrier in the reaction, can be calculated using Eq. (8.8).

ΔG

‡

 ΔH

‡

 T ΔS

‡

(8.8)

Remember: The double daggers (

‡

) simply indicate that we are referring to a transi-

tion state,not to an energy minimum.The energy barrier, ΔG

‡

,is the Gibbs free ener-

gy of activation,ΔH

‡

is the enthalpy of activation,and ΔS

‡

is the entropy of activation.

Clearly, the rate constant for any reaction will depend on the height of the bar-

rier,and thus on ΔG

‡

.The theoretical study of reaction rates gives us the exact equa-

tion relating the rate constant k and ΔG

‡

[Eq. (8.9)].

k  ν

‡

or k  ν

‡

(8.9)

It is worth dissecting this equation a bit further. The quantity ν

‡

is a frequency—

its units are reciprocal seconds (s

1

) in this case. In qualitative terms, ν is the fre-

quency with which molecules with enough energy to cross the barrier actually do

so. So,the rate constant k is equal to e

ΔG

‡

/RT

(the fraction of molecules with enough

energy to cross the barrier) times ν

‡

(the frequency with which such molecules actu-

ally do cross the barrier).

¢S

‡

-¢H

‡

/RT

-¢G

‡

/RT

Summary

Reaction rates are determined not by the relative energies of the starting material

and product, but by the height of the transition state separating them. Even

though the transition state is an energy maximum, a transient species that can-

not be isolated, its energy is determined by enthalpy and entropy factors, just as

is the energy of any stable compound.

8.7 Reaction Mechanism

Throughout this book we will frequently be concerned with the determination of

reaction mechanisms.A mechanism is nothing less than the description of the struc-

tures and energies of the starting materials and products of a reaction, as well as of

any reaction intermediates. In addition, all of the transition states (energy maxima)

separating the stable molecules lying in energy minima must be described. It is rel-

atively easy to deal with energy minima—we can often isolate the molecules them-

selves, take their spectra, and measure their properties. As we can by definition never

isolate a transition state, it is exceptionally difficult to make a direct measurement

on it. Yet if we are to have a good picture of a reaction, we must arrive at good

descriptions of the transition states involved.

We can usually measure the rate law, which tells us the number and kinds of

molecules involved in the transition state. The rate law does not tell us anything

about the orientation of the molecules in the transition state, however. Usually,

350 CHAPTER 8 Equilibria

stereochemical experiments do that. Recall the S

2 reaction (p.268), in which inver-

sion of configuration is observed, thus specifying the direction of approach of the

entering nucleophile in the transition state (Fig. 8.25).

frontside

backside

–

FIGURE 8.25 In the S

2 reaction,

kinetics cannot tell frontside from

backside displacement. It takes a

stereochemical experiment to do that.

This observation of inversion in the S

2 reaction tells us that the incoming nucleo-

phile must approach from the rear of the departing leaving group. The rate law,

,does not tell us that. It shows only that both the nucleophile

and the substrate are involved in the transition state. As we work through the impor-

tant reactions of organic chemistry, we will see many similar experiments designed

to determine the stereochemistry of the reaction. The energy of the transition state

can be measured if good quantitative kinetic experiments can be performed. If the

rate constant for the reaction is determined at a variety of temperatures, the height

of the transition state can be calculated. But we can do this only for the step that

has the highest-energy transition state. In a multistep reaction, we can only meas-

ure the rate for the slowest step in the sequence.

This slowest step is the rate-determining step, sometimes called the rate-limiting

step. It is always the step in the reaction with the highest-energy transition state. We

cannot measure the rates of faster steps in the reaction.In the S

1 reaction,for exam-

ple, the slowest step is the ionization of the substrate to a carbocation. This ion may

then be captured by a host of nucleophiles in following faster steps. There will be a

different product for every capturing nucleophile (p. 295; Fig. 8.26).

rate = k[Nu

][R

Energy

Reaction progress

Transition state 1 Transition state 2

Transition state 3

Transition state 4

(CH

)

(CH

)

(CH

)

(CH

)

(CH

)

Activation

energy

Slow step

ionization

Fast steps

capture by nucleophiles

–

FIGURE 8.26 In this S

1 reaction,

the rate-determining step is the

ionization of the starting iodide to

give the tert-butyl cation.The faster,

product-determining steps follow, but

their rates cannot be measured

kinetically (unless the intermediate

carbocation can be isolated and

investigated independently).

8.8 The Hammond Postulate: Thermodynamics versus Kinetics 351

If we determine the rate of this process by measuring the disappearance of

tert-butyl iodide over time, we can ultimately find the height of transition state 1.

The rates of the faster captures of the carbocation do not affect the rate of ioniza-

tion of tert-butyl iodide. The carbocation is gobbled up by the nucleophiles at a

rate faster than that at which it is formed. In the limit, each molecule of the tert-

butyl cation is captured before another is produced from tert-butyl iodide. The

rate of formation of product depends on how fast the tert-butyl cation is pro-

duced in the slow, rate-determining step but not on the rate of the fast capture

by a nucleophile.

Imagine a cage of especially ravenous rats into which we are slowly dropping

incredibly delicious food pellets. At first, the pellets are gobbled up by the hun-

gry rats at a rate far faster than they are delivered to the cage.The slowest step in

this reaction, the step that determines the rate of disappearance of the food, is the

rate at which the pellets are delivered to the cage. This process is analogous to a

reaction in which the first step (food dropping into the cage) is slower than the

second step (the rats eating the food). Later, however, a different situation obtains.

Even these rats can become satiated, and the food pellets accumulate in the cage

as the no-longer-starving rodents largely ignore them, only taking a nibble now

and then. The second step is now slower than the first step. Now the step that

determines the rate of disappearance of the food is the rate at which the rats eat

the pellets. This second scenario has a different rate-determining step than the

first scenario.

Energy

Reaction progress

Product X

Product Y

Product X

+ Product Y

Starting

material

Starting material

ΔG⬚, the energy

difference between

X and Y

FIGURE 8.27 The energy difference

between X and Y does not necessarily

affect their rates of formation.

8.8 The Hammond Postulate:

Thermodynamics versus Kinetics

We are used to saying, “Product Y is more stable than product X and therefore

is formed faster.” This statement has a comfortable sound to it and intuitively

makes sense. But it need not be true! The first half of the statement describes

thermodynamics, the relationship of the energies of product X and product Y.

The second half speaks of the rates of formation of X and Y, which is a

statement about kinetics. Despite the good intuitive feel of the original state-

ment, there need be no direct connection between thermodynamic stabilities and

kinetics.

Figure 8.27 shows the relative energies of two products, X and Y, being formed

from a starting material. One might expect that it is the ΔG° between the two

352 CHAPTER 8 Equilibria

products X and Y that determines the amount of the two compounds formed, but

this notion is true only if X and Y are in equilibrium via the starting material

(Fig. 8.28).

Starting materialProduct X

Product

FIGURE 8.28 Only if X and Y are in

equilibrium does ΔG° determine their

ratio.

It is conceivable that the products are not in equilibrium with each other. For

instance, there might not be enough energy for the products to go back to the start-

ing material. Under these conditions, the amounts of X and Y are determined not

by their relative energies but only by the heights of the transition states leading to

them. Figure 8.29a shows a conventional picture in which the transition state lead-

ing to the more stable product, Y, is lower in energy than the transition state lead-

ing to the higher-energy product, X. Under such conditions, the major product will

be the more stable compound,Y.We have seen many examples of reactions like this.

The scenario of Figure 8.29a is not the only possibility. It could be the case that the

transition state leading to the more stable compound, Y, is actually higher in energy

than the transition state leading to the less stable product, X (Fig. 8.29b).

(a)

Energy

Reaction progress

The energy

difference

between

the products

Starting

material

Product X

Product Y

ΔΔG

, the energy

difference between the

two transition states

ΔG⬚

†

(b)

Energy

Reaction progress

The energy

difference

between

the products

Starting

material

Product X

Product Y

ΔG⬚

ΔΔG

, the energy difference

between the two transition

states; the transition state

for formation of Y is higher

in energy

†

FIGURE 8.29 (a) The usual picture is this: The higher transition state leads to the less stable product. The lower transition state

leads to the more stable product. (b) The situation could be otherwise; the higher transition state could lead to the lower-energy

product, and the lower-energy transition state to the higher-energy product.This second scenario is rare, but it does happen.

PROBLEM 8.9 In Figure 8.29a what is the activation energy for the reaction of X

going to starting material? What is the activation energy for the reaction of Y

going to starting material?

8.8 The Hammond Postulate: Thermodynamics versus Kinetics 353

This inversion will not matter if X and Y equilibrate via the starting material.

In such a case, the greater stability of Y will win out eventually, and the ratio of X

and Y depends only on the ΔG° between them. Such a reaction is said to be under

thermodynamic control. But what if the two products do not equilibrate? What if there

is enough energy available so that the forward reactions are possible but the backward

reactions from X or Y to starting material are not? In the exothermic processes shown,

the forward reaction is easier—has a lower activation barrier—than the backward

reaction (Fig. 8.30).

Under these circumstances, the relative amounts of X and Y depend only on the

relative heights of the transition states leading to the two products.The product for

which the transition state is lower is the product that is formed in greater amount.

Such a reaction is said to be under kinetic control.

There is no necessary connection between thermodynamics and kinetics. In our

original statement, “Product Y is more stable than product X and therefore is

formed faster,” we are equating thermodynamics and kinetics. We tacitly assume

that it will always be the case that the product of greater stability (lower energy)

will be formed at a faster rate via the lower-energy (more stable) transition state

(Fig. 8.29a).

In principle, it could be that the transition state for the reaction leading to the

more stable product is actually higher in energy than the transition state leading to

the less stable product. Such a situation can lead to the preferred formation of the

less stable product, if the reaction is under kinetic control (Fig. 8.29b).

Although there are a few reactions for which the scenario of Figure 8.29b applies,

fortunately they are rare because there are factors that operate to preserve a paral-

lelism between the energies of products and the transition states leading to them.

The rest of this section will try to show you why the parallelism between product

energy and transition state energy exists for many reactions. Consider, for a start,

Worked Problem 8.10.

Energy

Reaction progress

ΔG

, activation energy for

Y starting material

Product X

Product Y

ΔG⬚

ΔΔG

, the energy difference between

the two transition states; this quantity

will determine the relative amounts

of Y and X if these products cannot

go back to starting material

The energy difference

between the products;

this makes no difference

if the products cannot go

back to starting material

ΔG

, activation energy for

X starting material

Starting

material

†

FIGURE 8.30 If there is enough

energy so that all barriers can be

crossed, the ratio of products will be

determined by ΔG°, and the reaction

is said to be under thermodynamic

control. If, however, there is not

enough energy to cross the high

barriers for the reverse reactions

leading from X and Y to starting

material, the ratio of products is

determined by the relative energies

of the two transition states, and the

reaction is said to be under kinetic

control.

354 CHAPTER 8 Equilibria

Before we accept the comfortable reasoning laid out in the answer to Worked

Problem 8.10, we had better think it through. Look first at the Energy versus

Reaction progress diagram (Fig. 8.31).

Energy

Reaction progress

The energy difference

between the two cation

products

Transition state for

formation of the

secondary carbocation B

⌬G⬚

Transition state for

formation of the

tertiary carbocation A

Starting material

FIGURE 8.31 The more substituted

cation is at lower energy than the less

substituted cation. Note also that the

transition state for formation of the

tertiary cation is lower than the

transition state for formation of the

secondary cation. Why?

WORKED PROBLEM 8.10 As we saw in Chapter 3, alkenes can be protonated to

give carbocations. Unsymmetrical alkenes, such as 2-methyl-2-butene, could be

protonated at either end of the double bond to give two different carbocations.

Predict the direction of protonation of 2-methyl-2-butene and explain your

answer.

ANSWER The choice is between the formation of secondary carbocation B and

tertiary carbocation A. It is surely tempting to predict that the more stable ter-

tiary carbocation will be formed faster than the less stable secondary carbocation.

However, this comfortable reasoning equates thermodynamics (more stable)

with kinetics (will be formed faster), and this equation is not necessarily valid.

Read on.

Carbocation A Carbocation B2-Methyl-2-butene

The tertiary cation is more substituted than the secondary cation and thus is more

stable by virtue of hyperconjugative overlap of a sigma bond with the empty 2p

orbital (p.292). Substitution stabilizes the positive charge on carbon in these species.

8.8 The Hammond Postulate: Thermodynamics versus Kinetics 355

In the transition states for formation of the carbocations, the new bond from

carbon to hydrogen is partially formed and the carbons have developed partial

positive charges (Fig. 8.32).

C H

Transition state contains

a partial positive charge

on a tertiary carbon

Transition state contains

a partial positive charge

on a secondary carbon

FIGURE 8.32 In the two transition

states, partial positive charge has built

up on carbon. A partial positive

charge on a tertiary carbon is more

stable than a partial positive charge

on a secondary carbon.

Energy

Reaction pro

ress

Starting material

Product

Endothermic

Energy

Reaction pro

ress

Starting material

Transition state

Product

Exothermic

Transition state

FIGURE 8.33 Equivalent energy

diagrams for endothermic and

exothermic reactions.

The factors that operate to stabilize a full positive charge also operate to stabi-

lize a partial positive charge. We therefore expect the transition state for the forma-

tion of a tertiary carbocation to be lower in energy than the transition state for

formation of a secondary carbocation, although the difference will not be as great

as that for the fully developed cations (Fig. 8.31).

This specific situation has been generalized in what is called the Hammond pos-

tulate after its originator, G. S.Hammond (1921–2005): “For an endothermic reac-

tion, the transition state will resemble the product.” Note immediately that this

statement can be turned around. “For an exothermic reaction, the transition state

will resemble the starting material.” Figure 8.33 shows an endothermic reaction,

but we know that all we have to do to see an exothermic reaction is look at the

356 CHAPTER 8 Equilibria

endothermic reaction in the other direction! So these two versions of the Hammond

postulate are really the same.

In the endothermic protonation of 2-methyl-2-butene, we rationalized the

faster formation of the more stable tertiary cation rather than the less stable

secondary carbocation by pointing out that in the transition state for carbocation

formation partial positive charge builds up on a tertiary, rather than secondary

carbon. The tertiary carbocation intermediate and the transition state leading to it

are more stable than their secondary counterparts. Because the transition state for

tertiary carbocation formation is lower energy than the transition state for

secondary carbocation formation, the tertiary carbocation will be formed faster.

The key here is that the transition states resemble the final products, just as the

Hammond postulate would have it.

We can also do a thought experiment to see that the more endothermic a reac-

tion is, the more the transition state will look like the product. Imagine an

endothermic reaction with the energy profile shown in Figure 8.34. Now apply a

pair of molecule-grabbing chemical tweezers and raise the energy of the product

without changing the energy of anything else. As we magically increase the ener-

gy of the product, it becomes less and less stable. Eventually, it becomes the

transition state, as it reaches a point where it is so unstable that it no longer lies

in an energy minimum. Now think about the moment just before this happens.

At this point (ε in Fig. 8.34), the product and the transition state are so close to

each other that the slightest increase in energy in the product makes them the

same (Fig. 8.34).

Energy

Reaction

ress

Starting material

Product

Transition

state

FIGURE 8.34 A thought experiment

in which we raise the energy of the

product without changing the energy

of the starting material.

By reflecting on this limiting situation,we can see how the transition state and the

product approach each other as the energy of the product increases and the reaction

becomes more endothermic.

8.9 Special Topic: Enzymes and Reaction Rates 357

It is a simple matter to turn this all around and examine the other side of the

Hammond postulate, “The more exothermic a reaction, the more the transition

state resembles starting material.” Figure 8.34 will do it for you if you simply read

it backward, from right to left. See what happens to the transition state as the

right-to-left reaction becomes less and less exothermic.

The take-home lesson of all this discussion is that although one cannot

take the correspondence between thermodynamics and kinetics as a given—there

will be counterexamples—in general, the statement “Y is more stable and thus

formed faster” will be true. There is more to it than it appears at first. This

statement is tricky indeed, and it is worth stopping and examining it every time

we make it.

8.9 Special Topic: Enzymes and Reaction Rates

In this chapter, we have discussed the interconversion of compounds by passing

from starting material to product over a high-energy point called the transition

state. The higher the transition state, the slower is the reaction. In the laboratory,

we affect the rate of reaction by adding acid or base catalysts,by changing solvents,

or sometimes simply by supplying lots of energy. Our bodies run on the same chem-

istry found in the laboratory. For example, the substitution and elimination reac-

tions described in Chapter 7 are all reasonably common in the transformations of

biomolecules occurring in each of us.

Nature can’t overcome activation barriers to reactions important to us by adding

catalysts such as sulfuric acid,and turning up the heat in our bodies to provide ener-

gy becomes uncomfortable very quickly indeed. So,Nature does it another way, and

it is most successful. Nature uses enzymes, which are polyamino acid chains with

catalytic activity, to lower the barriers to chemical transformations. Aconitase, an

enzyme of molecular weight 89,000 (!), catalyzes the conversion of citrate into cis-

aconitate, an elimination reaction. Aconitate is then hydrated to give isocitrate. So,

this enzyme makes possible one of the key reactions of Chapter 7, elimination, and

then reverses the process, but putting the OH on the other alkene carbon, to give

the isocitrate (Fig. 8.35).

Citrate cis-Aconitate Isocitrate

–

OOC

COO

–

COO

–

OOC

–

OOC

COO

–

COO

–

COO

–

COO

–

FIGURE 8.35 The interconversions catalyzed by the enzyme aconitase.

The vast molecular architecture of the enormous molecule aconitase serves

to bind the substrates precisely, holding them in perfect position for the side

chains of the amino acids to act as catalysts, speeding the elimination and addi-

tion reactions. Spectacular rate accelerations can be achieved by enzymes, allow-

ing all sorts of reactions to take place at body temperature, 37 °C, that could not

occur otherwise.

358 CHAPTER 8 Equilibria

3-Hydroxy-3-methylglutaryl

(or HMG)

Mevalonic acid

Cholesterol

CoA

HMG-CoA

reductase

rate-limiting

step

many

steps

–

O OH

O O

–

H H

–

CHOLESTEROL FORMATION

Sometimes, our well-being depends on interfering with

an enzyme-mediated rate acceleration. Cholesterol (p. 121)

is formed in the body by a lengthy process in which the

thioester A is reduced to mevalonic acid, which is then

converted in a series of reactions into cholesterol.

Controlling cholesterol levels in the body is an important

part of healthy living. Recent advances in medicinal and

organic chemistry have allowed the development of the

“statin” drugs (i.e., atorvastatin, fluvastatin, lovastatin, pravas-

tatin, simvastatin, and rosuvastatin), which act to reduce the

level of “unhealthy” low-density lipoprotein (LDL) choles-

terol. Examples of a healthy artery and a partially clogged

artery, which can result from high levels of LDL cholesterol,

are shown below. The benefit of taking a statin drug is a sig-

nificant reduction in the risk of heart attacks and strokes.

The statins function by inhibiting the enzyme HMG-

CoA reductase, which is involved in the rate-limiting or

slowest step in the formation of mevalonic acid, and hence

cholesterol. Without the enzyme, the reduction to mevalonic

acid in the body is much too slow. We have learned about

hydride reagents (p. 315), and we do have biological hydride

sources. You will meet one, NADH, in Chapter 16. But simply

mixing this “natural” hydride reagent with the HMG-CoA

thioester A results in no reaction. However, HMG-CoA

reductase can bring the hydride source and the thioester A

together in a fashion that allows the reaction to occur. The

energy barrier for the reaction is lowered, and mevalonic acid

is formed and goes on to make cholesterol. The statins inter-

fere with the reduction by inhibiting the enzyme necessary—

no reduction, no cholesterol!

This chapter is devoted entirely to concepts. You won’t find a

new mechanism or new synthetic route in it. Yet, it is extremely

important because what we say here about equilibrium and rates

is relevant to all the reactions that will appear in subsequent

chapters.

All reactions are equilibria or, in the case of multistep

reactions, sequences of equilibria between a starting material

and a product. The equilibrium constant K is related to the

difference in energy between starting materials and products

in a logarithmic fashion (ΔG° RT ln K or K  e

ΔG°/RT

8.10 Summary

New Concepts