Jones M., Fleming S.A. Organic Chemistry

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WEB 3D

Styrene

(

vin

lbenzene

)

11.6 Radical-Initiated Addition of HBr to Alkynes 489

PROBLEM 11.21 Write a mechanism for the formation of polystyrene, the product

of the polymerization of styrene (vinylbenzene).

PROBLEM 11.20 Styrene (vinylbenzene) is an example of a molecule in which the

addition of a radical is especially favorable. Explain why.

cis- and trans-1-Bromopropene 2-Bromopropene

initiation

–70 ⬚C

not

FIGURE 11.34 The radical-induced addition of HBr to propyne.

Propagation steps

cis- and trans-1-

Bromopropene

2-Bromopropene

(not formed)

More stable

vinyl radical

Less stable vinyl

radical (not formed)

HCC

Initiation steps

In2 In

hν

InBr

FIGURE 11.35 Addition of a bromine atom to 1-propyne gives the more stable vinyl radical, which then abstracts the

hydrogen atom from hydrogen bromide to give the anti-Markovnikov product.

The free-radical chain reaction for addition to alkynes is quite analogous to that

operating in additions to alkenes.The crucial step is the formation of the more sta-

ble vinyl radical rather than the less stable vinyl radical (Fig. 11.35).

11.6 Radical-Initiated Addition of HBr to Alkynes

Just as with alkenes, hydrogen bromide adds to alkynes in the anti-Markovnikov

sense when free radicals initiate the reaction.The light-initiated reaction of

with propyne gives cis- and trans-1-bromopropene,not 2-bromopropene (Fig. 11.34).

490 CHAPTER 11 Radical Reactions

Vicinal dibromide

(>97%)

Geminal dibromide

(<3%)

–80 ⬚C

initiation

HBr

FIGURE 11.36 A second radical

addition of hydrogen bromide to a

vinyl bromide generates the vicinal

dibromo compound as the major

product, not the geminal dibromo

compound. Problem 11.22 asks you

to explain why.

In principle, a second radical addition of hydrogen bromide could form either a

geminal or vicinal dihalide. In contrast to ionic reactions, it is the vicinal dibromide

that is preferred (Fig. 11.36).

Cl CH

CHCl

CCl

Methyl

chloride

Methylene

chloride

Chloroform Carbon

tetrachlorid

hν

or Δ

FIGURE 11.37 Photolysis or pyrolysis

of chlorine in methane gives a

mixture of chlorinated methanes.

PROBLEM 11.22 Explain in resonance and molecular orbital terms why the vicinal

dihalide is formed in Figure 11.36. Be careful! This question is more subtle than

it looks at first. Consider the stabilization of an adjacent radical (a half-filled

orbital) by an adjacent bromine atom. Why is this a stabilizing interaction?

Hint: Recall the molecule He



from p. 41, and see p. 479 in this chapter.

PROBLEM 11.23 Now predict whether will abstract hydrogen from the

methylene or methyl position of diethyl ether (CH

OCH

). Explain

your answer.

The composition of the product mixture depends on the amount of chlorine

available and the length of time the reaction is allowed to run. If we use a suffi-

cient amount of chlorine and allow the reaction to run long enough, all the methane

can be converted into carbon tetrachloride. However, measuring the product

11.7 Photohalogenation

So far, we have seen a variety of reactions of radicals with alkenes. Alkanes are much

less reactive than alkenes, and we have not seen ways to induce these notoriously

unreactive species into reaction in a specific way. We have seen pyrolysis, the ther-

mal cracking of saturated hydrocarbons (p. 470), but this method is anything but

useful in making specific molecules. Here we find a way to use alkanes in synthe-

sis. There still will be little specificity here, and these reactions will not find a

prominent place in your catalog of synthetically important processes. Yet they are

occasionally useful, and provide a nice framework on which to base a discussion of

selectivity.

11.7a Halogenation of Methane A mixture of chlorine and methane is sta-

ble indefinitely at room temperature in the dark,but reacts vigorously to give a mix-

ture of methyl chloride, methylene chloride, chloroform (trichloromethane), and

carbon tetrachloride when the mixture is either irradiated or heated (Fig. 11.37).

11.7 Photohalogenation 491

composition as a function of time shows that the initial product of the reaction is

methyl chloride. Methylene chloride and chloroform are formed on the way to

carbon tetrachloride (Fig. 11.38).

Cl CH

CHCl

CCl

Methyl

chloride

(first formed)

Methylene

chloride

(second formed)

Chloroform

(third formed)

Carbon

tetrachloride

(the ultimate

product)

hν

FIGURE 11.38 The chlorinated

products are formed by sequential

chlorinations of methane.

105 kcal/mol 59 kcal/mol

187 kcal/molBonds made

164 kcal/mol

23 kcal/mol exothermic

Bonds broken

84 kcal/mol 103 kcal/mol

CCl Cl

ClCl

FIGURE 11.39 The formation of

methyl chloride is exothermic.

Initiation

Termination

Propagation

CCH

FIGURE 11.40 The mechanism of

methyl chloride formation.

Let’s first examine the formation of the initial product,methyl chloride.The reac-

tion is exothermic by about 23 kcal/mol (Fig. 11.39).

The first step is the thermal or photochemical breaking of the chlorine–chlorine

bond. This initiation reaction is followed by the first propagation step (Fig. 11.40),

the abstraction of a hydrogen atom from methane by a chlorine atom to produce a

methyl radical and hydrogen chloride. In the second propagation step, the methyl

radical abstracts a chlorine atom from a chlorine molecule to give methyl chloride

and another chlorine atom that can carry the chain reaction forward.There are many

possible termination reactions; Figure 11.40, which shows the overall mechanism,

includes only one.

Now let’s look at the thermochemistry of the propagation steps. The second

propagation step is exothermic by 25 kcal/mol and thus is very favorable (Fig.11.41).

First propagation step

This reaction is endothermic by 2 kcal/mol

105 kcal/mol 103 kcal/mol

Second propagation step

This reaction is exothermic

by 84 – 59 = 25 kcal/mol

59 kcal/mol 84 kcal/mol

FIGURE 11.41 The thermochemistry of the two propagation steps for methyl chloride formation.

492 CHAPTER 11 Radical Reactions

The first propagation step is endothermic by about 2 kcal/mol. Will this endothermic

step be enough to stop the reaction? No.Although an energy difference of 2 kcal/mol

produces an unfavorable equilibrium mixture, the second highly exothermic step

results in the overall reaction being favored. The few methyl radicals formed react

rapidly to give methyl chloride,thus disrupting the slightly unfavorable equilibrium

Reestablishment of this equilibrium generates more methyl radicals, which go on

to make more methyl chloride and chlorine radicals (Fig. 11.42), which continue

the chain reaction. Note that a chlorine radical is just a chlorine atom.

+ CH

Z——

H +

H H

Energy

Reaction pro

ress

Cl ClH

+2 kcal/mol

(slightly endothermic)

–25 kcal/mol

(very exothermic)

FIGURE 11.42 In the formation of

methyl chloride from methane and a

chlorine radical, the high exothermicity

of the second step overcomes the slight

endothermicity of the first step.

Methylene

chloride

Methyl chloride

FIGURE 11.43 The formation of

methylene chloride (H

CCl

) from

methyl chloride and a chlorine atom.

Moreover, the carbon–hydrogen bond in methyl chloride is weaker than the one

in methane, and thus easier to break.So methyl chloride competes favorably with

methane and cannot build up in the reaction unless there is a vast excess of

methane.

As methyl chloride builds up in the reaction chamber, it, too,begins to react with

chlorine atoms in another chain reaction (Fig. 11.43).

11.7 Photohalogenation 493

Methylene chloride is more reactive than either methyl chloride or methane,and

so it reacts to form chloroform as it builds up. Chloroform is more reactive still,

and rapidly produces carbon tetrachloride through abstraction of its single hydro-

gen (Fig. 11.44).

Chloroform

Carbon

tetrachloride

FIGURE 11.44 The formation of

chloroform from methylene chloride

and of carbon tetrachloride from

chloroform.

Methane can also be photohalogenated with bromine, but the overall reaction

is much less exothermic than is chlorination. Moreover, the first propagation step is

endothermic by the substantial amount of 17 kcal/mol (Fig. 11.45). As we will see,

this strongly endothermic step will have important consequences for the selectivity

of the reaction when a choice of carbon–hydrogen bonds exists.

Initiation

105 kcal/mol

This first propagation step is

105 – 88 = 17 kcal/mol endothermic!

88 kcal/mol

Termination

Propagation

CCH

FIGURE 11.45 The formation of

methyl bromide from bromine and

methane.

PROBLEM 11.24 Explain why the carbon–hydrogen bond in methyl chloride is

easier to break than the carbon–hydrogen bond in methane itself.

494 CHAPTER 11 Radical Reactions

carbons of butane are more reactive than the hydrogens on the primary carbons.

Abstraction by a chlorine atom occurs more easily at the secondary position than at

the primary position. The percentages of products shown in Figure 11.46 indicate

that the preference is 72.2:27.8, or a factor of 2.6. However, this simple analysis

ignores the fact that in butane there are more primary hydrogens (6) than second-

ary hydrogens (4). In order to correct this value for the statistical advantage of the

primary bonds, we must multiply it by 6/4 to produce the true factor of 3.9. On a

per hydrogen basis, secondary hydrogens are abstracted more easily than primary

hydrogens by a factor of 3.9.

sec-Butyl chloride

(72.2%)

Butyl chloride

(27.8%)

ClCH

CHCH

hν, 35 ⬚C

FIGURE 11.46 On a per hydrogen

basis, abstraction of a secondary

hydrogen is favored over abstraction

of a primary hydrogen by a factor of

(72.2/27.8)  6/4.

hν, 35 ⬚C

1-Chloro-2-methylpropane

(64%)

2-Methylpropane

2-Chloro-2-methylpropane

(36%)

FIGURE 11.47 For a chlorine radical,

abstraction of a tertiary hydrogen is

favored over abstraction of a primary

hydrogen by a factor of about 5 on a

per hydrogen basis.

However, there are nine primary hydrogens and only a single tertiary hydrogen

in 2-methylpropane, so on a per hydrogen basis the primary:tertiary reactivity

ratio is 1.78  1/9  0.198. Put another way, the tertiary hydrogen is favored by

1:0.198  5.1.

Bromine is much more selective in the photohalogenation reaction than is chlo-

rine. That is, bromine is able to choose among the various reaction possibilities

with more discrimination and pick out the most favorable abstraction pathway by

a wider margin than chlorine. For example, a bromine atom prefers secondary to

11.7b Halogenation of Other Alkanes If we irradiate a mixture of chlorine

and a hydrocarbon more complicated than methane or ethane, there is usually more

than one possible initial product.For example, butane can give either 1-chlorobutane

or 2-chlorobutane. Figure 11.46 shows that the hydrogens on the secondary

PROBLEM 11.25 Write a reaction mechanism for the photochlorination of ethane.

PROBLEM 11.26 Analyze the thermochemistry of the two propagation steps for

the photochlorination of ethane. The bond dissociation energies of the

chlorine–chlorine and carbon–chlorine bonds of ethyl chloride and the

carbon–hydrogen bond of ethane are 59, 84.8, and 101.1 kcal/mol, respectively.

Selectivity is increased when the possibilities include the formation of a tertiary

radical. Photochlorination of 2-methylpropane, for example, gives a 64:36 mixture of

1-chloro-2-methylpropane (isobutyl chloride) and 2-chloro-2-methylpropane

(tert-butyl chloride). The primary:tertiary ratio is 64:36  1.78 (Fig. 11.47).

Alkane halogenation

11.7 Photohalogenation 495

PROBLEM 11.27 Use the bond dissociation energies of Table 8.2 (p.337) to estimate

the thermochemistry of the propagation steps for photochemical fluorination and

iodination of methane.

(

~100%

)(

~0%

)

hν, 127 ⬚C

(98.2%) (1.8%)

CHCH

FIGURE 11.48 Photobromination

is far more selective than

photochlorination.

TABLE 11.5 Selectivities of Halogen Atoms in Carbon–Hydrogen Bond Abstraction

Halogen Primary Secondary Tertiary

F 1 1.2 1.4

Cl 1 3.9 5.1

Br 1 82 1600

Energy

Reaction progress

Transition state—much

closer to product

17 kcal/mol

Energy

Reaction

ress

Transition state—nearly

at the midpoint for this

reaction

2 kcal/mol

The C

H bond is much more

fully broken than in the case

of chlorine; the H

Br bond

is nearly fully made, and the

free electron is nearly fully

developed on carbon

Cl CH

H and H

bonds approximately

equally formed

HCl

Br CH

FIGURE 11.49 For the nearly thermoneutral abstraction of hydrogen by a chlorine atom, the transition state is almost

halfway between starting material and products. For the highly endothermic abstraction of hydrogen by the bromine

atom, the transition state will be quite product-like.

Neither fluorine nor iodine undergoes a useful photohalogenation reaction.The

reaction of fluorine with carbon–hydrogen bonds is overwhelmingly exothermic and

can even occur explosively. Almost no selectivity is possible (Table 11.5). In direct

contrast to the reaction of fluorine atoms, reaction of iodine atoms with hydrocar-

bons is highly endothermic. It is so slow as to be of no synthetic utility.

What is the reason for bromine being more selective than chlorine? Recall that

the abstraction reaction is close to thermoneutral for chlorine (Fig. 11.49), but very

primary hydrogens by a factor of 82, and tertiary hydrogens to primary hydrogens

by a factor of about 1600 (Fig. 11.48; Table 11.5).

496 CHAPTER 11 Radical Reactions

endothermic for bromine (p. 493, Fig. 11.45).

The Hammond postulate (the

transition state for an endothermic process resembles the product) tells us that the

transition state for the abstraction of hydrogen by chlorine is nearly halfway between

starting material and product (the radical intermediate) (Fig. 11.50a), but the

transition state is very product-like for bromine (Fig. 11.50b). Because the reaction

is more endothermic, the transition state for abstraction of hydrogen by bromine is

product-like. There is a strongly developed hydrogen–bromine bond and a large

degree of radical character on carbon in the transition state (Fig. 11.50b).

In these transition states the

carbon–hydrogen bonds are

only about half-broken, so transition

states for abstraction of different

hydrogens don’t differ much

HCl

In these transition states, the carbon–hydrogen

bonds are largely broken and the radicals are

very well developed on the carbons; differences

in radical stability will be important in selection

of the hydrogen to be abstracted

HBr

(a)

(

)

FIGURE 11.50 (a) Because the

transition state for abstraction by

chlorine lies roughly half way

between starting material and

product there can be little selectivity.

(b) The product-like transition state

for abstraction by bromine means

that the radical on carbon will be well

developed and the substitution

pattern of the carbon atom (tertiary,

secondary, primary, or methyl) will be

relatively important in determining

the ultimate major product.

Watch out! One of the things chemistry professors laugh about is the number of people who call “thermoneu-

tral” reactions “thermonuclear” reactions. Believe it or not, this happens a lot!

11.8 Allylic Halogenation: Synthetically Useful Reactions 497

The transition states for abstraction of hydrogen from an alkane by chlorine and

bromine radicals are very different from each other. In the transition states for

abstractions by bromine, the radical on carbon is far more developed than it is in

the abstractions by chlorine, which means that differences in the stabilities of dif-

ferent radicals will be far more important in determining the major product of the

bromine abstractions than in the chlorine abstractions (Fig. 11.50).

Summary

When questions of selectivity arise, it is always important to consider what the

transition state looks like. Selective radicals will have more product-like transi-

tion states in any abstraction reaction, and therefore the factors influencing the

stability of those products will be important in the transition state as well.

11.8 Allylic Halogenation: Synthetically

Useful Reactions

One of the goals of synthetic organic chemists is to increase selectivity—to be able

to do chemical transformations at specific sites in a molecule without affecting the

rest of the molecule.Selectivity is useful—even necessary—when one is dealing with

a complicated molecule. We must have ways of inducing change at one place in a

molecule without affecting others if we are to be able to manipulate molecules ration-

ally. It’s quite clear that the photochlorination reaction of alkanes is not a good way

of doing this because this reaction is far too unselective. Chlorination takes place all

too easily at the various positions in the hydrocarbon.

Photobromination more closely approaches the goal of perfect selectivity because

the bromine atom abstracts tertiary hydrogens much more readily than secondary

or primary hydrogens. But even this reaction is not without its problems.It is a slow

process and the side products of di- and polybromination are formed.

There are some instances where the chlorination or bromination reaction can

be used to good effect, however. Alkenes that have allylic hydrogens can some-

times be halogenated specifically at an allylic position in a process called allylic

halogenation, another free radical chain reaction. For example, when low con-

centrations of bromine are photolyzed in the presence of the complicated cyclo-

hexene shown in Figure 11.51, the product is exclusively brominated in one allylic

position.

(70%)

Cyclohexene in red

Brominated only in

this allylic position

Cholesterol benzoate

COO

hν

WEB 3D

FIGURE 11.51 When low concentrations of bromine are photolyzed in the presence of the

substituted cyclohexene, the product is exclusively the allylic bromide shown.

498 CHAPTER 11 Radical Reactions

The mechanism is a typical radical chain reaction in which a low concentration

of bromine atoms is first produced by breaking of the weak bromine–bromine bond

(initiation step). Abstraction of hydrogen from the cyclohexene is the first propa-

gation step, but there are three possible kinds of hydrogen that could be removed:

(vinylic), H

(allylic), and H

(methylene) (Fig. 11.52).

Br Initiation step

Three possible propagation steps

abstract H

HBr

FIGURE 11.52 In principle, a bromine

atom could abstract a vinylic (H

allylic (H

) or methylene (H

)

hydrogen.

FIGURE 11.53 In a propagation step,

any of the carbon radicals could

react with bromine to give a

bromocyclohexene and a chain-

carrying bromine atom.

As shown in Figure 11.53, each of these radicals could lead to a bromocyclo-

hexene through reaction with bromine in the second propagation step.The prop-

agation is then carried forward by the new bromine atom formed in this step.The

formation of only the allylic bromide tells us that only the allylic radical is pro-

duced originally.

Because there is only one product formed, the allylic bromide, the allylic radical

must be formed with high selectivity. Why? It is only this radical that is resonance sta-

bilized.There are two equivalent resonance forms for the allylic radical and this delo-

calization makes it far more stable than the other two possibilities (Table 11.1).