Jones M., Fleming S.A. Organic Chemistry

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7.12 Additional Problems 329

(b) When menthyl chloride is treated with 80% aqueous ethyl

alcohol with no ethoxide present, once again, two products

are formed in the ratio shown. Explain mechanistically.

Steric effects are important in solvolysis reactions. Problems

7.75 and 7.76 illustrate this point.

PROBLEM 7.75 Explain the following rates for S

1 solvolysis.

Construct an Energy versus Reaction progress diagram showing

the two reactions.

COOH, 25 ⬚C

CH OTs

Rate

(CH

)

CH(CH

)

(

32%

)(

68%

)

CH(CH

)

CH(CH

)

CH(CH

)

CH(CH

)

(25%) (75%)

–

Neomenthyl

chloride

CH(CH

)

PROBLEM 7.76 Explain the relative rates of the following

compounds in the S

1 reaction:

PROBLEM 7.77 Rationalize the following curious experimental

results. When the optically active chlorosulfite 1 is heated in

toluene, the chloride with inverted configuration is obtained.

However, when 1 is decomposed in 1,4-dioxane (2), retention

of configuration in the resulting chloride is observed. Hint:

See Figure 7.43, and also think about the differences between

toluene and 1,4-dioxane. A good Lewis structure for

1,4-dioxane may also be helpful.

For Problems 7.78 and 7.79 you need to know that treatment of

amines, , with nitrous acid, HONO, gives diazonium

ions, .

PROBLEM 7.78 Provide arrow formalism mechanisms for the

following reactions:

HONO

(a)

(b)

toluene

Retention

Inverted

CHCl

Relative Rate

(CH

)

(CH

)

540

125

330 CHAPTER 7 Substitution and Elimination Reactions

PROBLEM 7.79 Rationalize the different products formed on treat-

ment of the stereoisomeric aminocyclohexanols 1–4 with nitrous

acid.

PROBLEM 7.80 You are given a supply of 2-bromo-1-methoxybi-

cyclo[2.2.2]octane. Your task is to prepare 1-methoxybicyclo-

[2.2.2]oct-2-ene (A) from this starting material. In your first

attempt, you treat the bromide with sodium ethoxide in ethyl alco-

hol. Although a small amount of the desired alkene is formed, most

of the product is 2-ethoxy-1-methoxybicyclo[2.2.2]octane (B).

(minor)

(major)

HOCH

–

OCH

COH

OCH

HONO

C(CH

)

C(CH

)

HONO

C(CH

)

C(CH

)

HONO

C(CH

)

CHO

C(CH

)

C(CH

)

Suggest a simple experimental modification that should

result in an increase in the amount of the desired alkene,

and a decrease in the amount of the unwanted substitution

product, B.

PROBLEM 7.81 A common procedure to increase the rate of

an S

2 displacement is to add a catalytic (very small) amount of

sodium iodide. Notice that even though the rate increases, the

structure of the product is not the iodide. Explain the function

of the iodide. How does it act to increase the rate, and why is

only a catalytic amount necessary?

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 7.82 Select the animation titled “Halide formation”

in the panel of reactions labeled Semester 2A. Write out the

reaction (see Fig. 7.45). How many intermediates are included

in the animation? According to the energy diagram shown for

the reaction, which step is the fastest?

PROBLEM 7.83 The “Halide formation” reaction has several

intermediates formed that aren’t shown in the animation.

Predict what they are. It might help to know that one of the

final products in the reaction is P(OH)

PROBLEM 7.84 After you watch the “Halide formation”

reaction, you should be able to describe the mechanism

of the second step. What kind of reaction would you call it?

Do you suppose a tertiary alcohol goes through the same

mechanism?

PROBLEM 7.85 Select the “Bimolecular elimination: E2” reac-

tion. You know about the stereochemistry of the reaction from

Section 7.9c. Select the LUMO track for the reaction and

observe the starting alkyl halide. Which hydrogens have

LUMO character? Why?

PROBLEM 7.86 If you were to add the methoxide base to

propane, would propane have a hydrogen acidic enough to react

with the base? In other words, would you predict a reaction?

See Table 6.4 (p. 235) if you need pK

information.

–

(small amount)

acetone

(solvent)

–

Equilibria

331

8.1 Preview

8.2 Equilibrium

8.3 Entropy in Organic Reactions

8.4 Rates of Chemical Reactions

8.5 Rate Constant

8.6 Energy Barriers in Chemical

Reactions: The Transition

State and Activation Energy

8.7 Reaction Mechanism

8.8 The Hammond Postulate:

Thermodynamics versus

Kinetics

8.9 Special Topic: Enzymes and

Reaction Rates

8.10 Summary

8.11 Additional Problems

ENERGY BARRIERS Reactions have energy barriers that must be crossed to go from

reactants to products. A reasonable analogy is the mountain range that must be

crossed to go from your starting point to the other side of the mountain.There may

be intermediates in the reaction path, just like the valleys that separate the peaks.

–

Transition state

Energy

Reaction progress

–

FIGURE 8.1 The S

2 reaction

is an equilibrium in which two

nucleophiles (the Lewis bases Nu

−

and L

−

) compete for a Lewis acid.

“I hope,” the Golux said, “that this is true. I make things up, you know.”

—JAMES THURBER,

THE 13 CLOCKS

8.1 Preview

The study of how reactions occur depends on an analysis of the structures of the

starting materials, the products, and the transition states separating them. We have

spent much time on structure in the last several chapters. However, there are other,

nonstructural factors that are important. To understand chemical reactivity, and to

be able to use our knowledge in a predictive way, we also need to know something

of the energetics involved in the reaction. Which are more stable, starting materi-

als or products? Of course those designations—starting materials and products—

are arbitrary. Nature doesn’t care about such synthetic labels! How high is the energy

of the transition state separating “starting materials and products”? If the transition-

state energy is too high, no reaction will occur. If there are several available low-

energy pathways—several accessible transition states—there will be a mixture of

products, and the synthetic chemist’s longed-for specificity will be impossible. So

we need to know about energy.

In Chapter 7, we saw some examples of chemical equilibria. For instance, the

2 reaction involves an equilibrium in which two nucleophiles (Nu



and L



)

compete for the R group (Fig. 8.1).

James Grove Thurber (1894–1961) was an American writer and cartoonist, much of whose work first

appeared in The New Yorker magazine. The 13 Clocks is a fairy tale of surpassing charm and wit written in

Bermuda.

332

CHAPTER 8 Equilibria

8.1 Preview 333

In practice, thermodynamics may dictate that one side of an equilibrium be

overwhelmingly favored over the other. Nevertheless, we can still apply techniques

for studying equilibria to the reaction, deriving information about the position

of the equilibrium (thermodynamics) and the rate of reaction (kinetics). The

concepts we will develop in this short chapter are extremely general and very

important.

All chemical reactions can be regarded as equilibria. Multistep reactions are

simply series of equilibria in which species occupying energy minima are sepa-

rated by barriers, the tops of which are transition states. The S

1 solvolysis

reaction of tert-butyl iodide in water makes a good example. Here we have three

equilibria.The first is between tert-butyl iodide and the tert-butyl cation and iodide

ion. This equilibrium is highly unfavorable, because the cation–anion pair is far

less stable than the covalent iodide. The second equilibration is the exothermic

capture of the cation by water to give the protonated alcohol, an oxonium ion.

Finally, the oxonium ion is deprotonated to give the ultimate product, tert-butyl

alcohol (Fig. 8.2).

Energy

Reaction progress

(CH

)

(CH

)

(CH

)

(CH

)

(CH

)

Equilibrium 1

Equilibrium 2

Equilibrium 3

(CH

)

(CH

)

(CH

)

(CH

)

(CH

)

2 H

–

2 H

–

2 H

FIGURE 8.2 S

1 reactions—and all

multistep reactions—are series of

equilibria.

334 CHAPTER 8 Equilibria

8.2 Equilibrium

For a general chemical reaction at equilibrium,

(8.1)

the concentrations of the species on the two sides of the arrows are related to each

other by an equilibrium constant, K:

(8.2)

or, more familiarly:

(8.3)

The small letter superscripts are the coefficients in the balanced equation for

the reaction [Eq. (8.1) in this case], and the capital letters enclosed in brackets rep-

resent the molar concentrations of the products and reactants at equilibrium. The

concepts of starting materials and products are arbitrary ones, arising from practical

considerations—we generally are interested in making the reaction run one way or

the other, and by convention the desired reaction is written left to right. But it need

not be; it is just as valid to look at the reaction in the other direction.

A reaction will be a successful one if the compounds we seek, the products, are

strongly favored at equilibrium. So we are very interested in what determines

the magnitude of the equilibrium constant (K ): the difference in the free energies

(ΔG°  ΔH°  T ΔS°; see below for a fuller discussion) of the compounds involved

in the reaction.The exact relationship between K and the relative free energies of

the starting materials and products is shown in Eq. (8.4):

(8.4)¢G

ⴰ

=-RT ln K

K =

[C]

[D]

[A]

[B]

[C]

[D]

= K [A]

[B]

aA + bB

cC + dD

ESSENTIAL SKILLS AND DETAILS

1. Thermodynamics. Understanding the importance of the relative energies of starting

materials and products, as well as the heights of the barriers separating these species,

is critical to understanding reactivity. The relative energies of starting materials and

products determine the ratio of products under conditions that allow all energy barriers

to be passed.

2. Kinetics. The heights of the barriers separating starting materials from products

determine the rates of formation of products. If there is not sufficient energy to return

to starting material from products, it is the heights of energy barriers—topped by

transition states—that will determine the ratio of products.

3. A little bit of energy goes a long way at equilibrium.The outcome of a reaction at

equilibrium is determined by the energy difference between the starting materials

and products: ΔG RT ln K is the operative equation.

4. Rates.The rate of a reaction has nothing to do directly with the energy difference

between starting materials and products. Rates are determined solely by the energy

difference between the starting molecules and the transition states separating those

molecules from products.

8.2 Equilibrium 335

Energy

Reaction progress

ΔG⬚ or ΔH⬚

Starting

material

Product

Exothermic (ΔH⬚) left-to-right

Exergonic (ΔG⬚) left-to-right

Transition state

FIGURE 8.3 An exothermic (exergonic) reaction.

Energy

Reaction progress

ΔG⬚ or ΔH⬚

Starting

material

Product

Endothermic (ΔH⬚) left-to-right

Endergonic (ΔG⬚) left-to-right

Transition state

FIGURE 8.4 An endothermic (endergonic) reaction.

or, if we want to use base 10 rather than natural logarithms, in Eq. (8.5):

(8.5)

R is the gas constant, 1.986 × 10

3

kcal/deg

mol, T is the absolute temperature, and

ΔG°, the Gibbs free energy change, is the difference in energy between starting

materials and products in their standard states (Figs. 8.3 and 8.4).

¢G

ⴰ

=-2.3 RT log K

Up to now, we have more or less equated energy with H°, the enthalpy. The

enthalpy change ΔH°, or the heat of reaction,is a measure of the bond energy changes

in a reaction.The Gibbs free energy (ΔG°) is more complicated than this, as we will

see in Section 8.3, and includes not only ΔH°, but entropy change (⌬S°) as well.

The entropy of a reaction is a measure of the change in amount of order in starting

materials and products.The higher the degree of ordering of reactants necessary, the

greater are the energy requirements for the reaction. Conversely, an increase in

disorder will require less energy for the reaction.

A number of times in Chapter 7 we drew diagrams that described the course

of a reaction. We can now label those diagrams a bit more completely. The dia-

grams are plots of energy versus something called,most often,“reaction progress.”

Figures 8.3 and 8.4 show an exothermic reaction (products more stable than

starting materials) and an endothermic reaction (starting materials more stable

than products).

The terms “exothermic” and “endothermic” really refer to the enthalpy change,

ΔH°,which may or may not be in the same direction as the free energy change,ΔG°,

which involves entropy, ΔS°, as well as enthalpy, ΔH°. If we really mean free ener-

gy, ΔG°, the proper terms are exergonic and endergonic.

For an exothermic reaction, the difference in enthalpy between products and

starting materials is negative and the equilibrium constant (K ) is greater than 1. For

an endothermic reaction (read the first reaction backward to produce an endother-

mic process), the enthalpy difference is positive and K is less than 1.

336 CHAPTER 8 Equilibria

More stable products are favored at equilibrium. By how much? Note that the

relationship between the difference in free energy between starting material and

products (ΔG°) and the equilibrium constant K is logarithmic. We can rewrite

Eqs. (8.4) and (8.5) to emphasize that point:

(8.4)

or, in base 10:

(8.5)

These equations show that a small difference in energy between starting materials and

products results in a large excess of one over the other at equilibrium. Let’s do a few

calculations to show this. First,let’s take a case in which products are more stable than

starting materials by the seemingly tiny amount of 1 kcal/mol, ΔG° 1 kcal/mol,

so K  10

1/(1.364)

 10

0.7331

 5.4.

An equilibrium constant of 5.4 translates into 84.4% product at equilibrium! One

would very often be quite happy to find a reaction that gives about 85% product, and

a mere 1 kcal/mol suffices to ensure this. Conversely, if we are “fighting”a 1-kcal/mol

endothermic process, it will be a futile fight indeed because equilibrium will settle out

at 84.4% starting material. A little bit of energy goes a long way at equilibrium.Table 8.1

summarizes a number of similar calculations.

¢G

ⴰ

=-2.3RT log K

K = 10

-¢G

ⴰ

>2.3RT

¢G

ⴰ

=-RT ln K

K = e

-¢G

ⴰ

>RT

TABLE 8.1 The Relationship between ⌬G° and K at 25 °C

G° (kcal/mol) K Equilibrium Ratio

0.1 1.2 54.5/45.5

0.5 2.3 69.7/30.3

1 5.4 84.4/15.6

2 29.3 96.7/3.3

5 4631 99.98/0.02

10 2.1  10

99.999996/0.000004

At 25 °C, 2.3RT  1.364 kcal/mol. This number is useful to remember because with it one can do calcula-

tions easily at odd moments without having to look up the value of R and multiply it by T.

PROBLEM 8.1 Calculate the energy difference between starting material and prod-

ucts at 25 °C given equilibrium constants of 7 and 14.

The principle is named for Henri Louis Le Châtelier (1850–1936). The mathematics of this principle had

been worked out earlier by the American Josiah Willard Gibbs (1839–1903), who donated the G in Gibbs to

ΔG. Le Châtelier knew of Gibbs’s contributions and was a great popularizer of Gibbs’s work in France.

In practice, there often is something we can do to drive a reaction in the direc-

tion we want.Le Châtelier’s principle

states that a system at equilibrium responds

to stress in such a way as to relieve that stress. This statement means that if we

increase the concentration of one of the starting materials, an equilibrium process

will be driven toward products. In Eqs. (8.1–8.3), if we increase [A]

, the quantity

[C]

[D]

/[B]

must also increase to maintain K, and we will have more product.

Similarly, if we can remove one product as it is formed, the equilibrium will be driv-

en toward products. Perhaps either C or D is a solid that crystallizes out of solution

or is a gas that can be driven off. As the concentration of this compound drops, the

equilibrium will be disturbed and more C or D will be formed.

8.3 Entropy in Organic Reactions 337

In a multistep reaction,it often does not matter that an early

step involves an unfavorable equilibrium, as long as the overall

reaction is exergonic.As long as all barriers can be traversed,the

ratio of C and A will be governed by the ΔG° between C and

A. So, A B is endergonic, but, overall, A C is exergonic

(ΔG°  0). Figure 8.5 shows such a process for the conversion

of A into C via an intermediate B.

As long as some B is present in equilibrium with starting

material A, the reaction will be successful. As the equilibrium

is established, B will be converted into the much more

stable final product C. As the small amount of B present in

equilibrium with A is depleted, the equilibrium will be

reestablished and more B will be formed from A. In the

long run, A will be largely converted into C, and as long as

there is enough energy to reach the intermediate B, the final

mixture of A and C will depend on the ΔG° between these

two compounds.

8.3 Entropy in Organic Reactions

The parameter ΔG° is composed of enthalpy (ΔH°, the change in bond strengths in

the reaction) and entropy (ΔS°, the change in freedom of motion in the reaction).

The exact relationship is

(8.6)

where T is the absolute temperature.We can make rather accurate estimates of the ΔH°

term by knowing the bond dissociation energies for a variety of bonds. We know a

few of these values already from our discussion of alkanes.Table 8.2 gives several more.

¢G

ⴰ

=¢H

ⴰ

- T ¢S

ⴰ

TABLE 8.2 Some Homolytic Bond Dissociation Energies (BDE)

Bond BDE Bond BDE

(kcal/mol) (kcal/mol)

36.1 101.1

46.1 98.6

59.0 96.5

38.0 57.6

51 72.1

38 83.7

104.2 115

71.3 92.1

87.5 83.2

103.2 85.2

136.3 90.1

118.8 89.0

104.6 66

107.6 174.1

105.0 230.7

178.8H

O (total)

CH (total)H

(total)H

(π bond only)H

OCH

(CH

)

OC(CH

)

C(CH

)

CH(CH

)

Energy

Reaction progress

ΔG⬚

Starting

material A

Product C

Intermediate B

FIGURE 8.5 As long as equilibrium is reached between

A and B, it does not matter that formation of B from A

is endergonic, compound C will be the major product.

338 CHAPTER 8 Equilibria

PROBLEM 8.2 Estimate the ΔH° for the following reactions:

(a)

(b)

(c)

(d)

PROBLEM 8.3 The bond dissociation energies for the carbon–hydrogen bonds in

Table 8.2 show a steady decline as the breaking carbon–hydrogen bond becomes

more substituted. What can you infer from these data about the stability of the

neutral species, , called “free radicals”?R

HCl + H

ClCH

+ H

BrCH

+ H

OH + H

OCH

The entropy term (ΔS°) is related to freedom of movement. The more restricted

or “ordered”the product(s) of a reaction, the more negative is the entropy change in

the reaction. Since it is the negative of ΔS° that is related to ΔG° [Eq. (8.6)], the

more negative the entropy change, the higher is the free energy change (ΔG°) of

the reaction. Note also that the entropy term is temperature-dependent. Entropy

becomes more important at high temperature. Imagine the reaction in Figure 8.6,

in which strong carbon–carbon σ bonds are broken and weaker carbon–carbon π

bonds are made (Table 8.2). The ΔH° term will surely favor the left-hand side. At

the same time, one molecule is made into two other, smaller molecules, and this

change will be favored by the entropy change, ΔS°. This reaction, in which entropy

change favors the right-hand side, can be driven to the right by using a high tem-

perature,at which the T ΔS° term is large.An unfavorable reaction can even become

favorable at high temperature where the entropy term becomes more important in

determining the value of ΔG°.

WORKED PROBLEM 8.4 Use ΔH° values from Table 8.2, as well as the value for

the BDE of of 80 kcal/mol,

to estimate ΔH° for the reaction in Figure 8.6.

ANSWER Compare the bonds that are breaking with those that are being made.

Two carbon–carbon single bonds are broken. As the problem says, each is worth

80 kcal/mol. In addition, a π bond (66 kcal/mol) is broken in the reaction. So, the

bonds broken are worth a total of 226 kcal/mol. Three π bonds are made, each

worth 66 kcal/mol, for a total of 198 kcal/mol. The left-to-right reaction is

endothermic by about 28 kcal/mol, because the bonds made are weaker than

those broken.

¢H

ⴰ

=+28 kcal/mol

3 * C

C (π) =-198 kcal/mol (bonds made)

C (π) =+66 kcal/mol (bond broken)

2 * C

C =+160 kcal/mol (bonds broken)

FIGURE 8.6 This reaction is a process

in which one direction is favored by

enthalpy and the other by entropy.

Although ΔH° favors cyclohexene, at

very high temperature the formation

of two compounds from one can

drive this reaction to the right.