Jones M., Fleming S.A. Organic Chemistry

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11.2 Formation and Simple Reactions of Radicals 469

Hydrogen

Methane

Ethane

1s 1s

(sp

)

(sp

)

(

)

1ssp

FIGURE 11.1 The formation of

hydrogen, methane, and ethane

through reactions of the radicals, H

and CH

Can we reverse the bond-forming process to produce radicals from these mol-

ecules? Energy will surely have to be added in the form of heat, because bond form-

ing is highly exothermic and bond breaking is highly endothermic. Figure 11.2a

Energy

Antibonding

molecular orbital σ*

Bonding

molecular orbital σ

(sp

)

(sp

)

Stabilization

~90 kcal/mol

Reaction progress

(sp

)

(sp

)

90 kcal/mol

(bond dissociation

energy)

(a) (b)

FIGURE 11.2 (a) A schematic picture of the formation of a covalent bond through overlap of two half-filled sp

orbitals.

(b) To reverse the process, energy must be supplied. For example, methyl radicals can be formed by supplying enough

energy (90 kcal/mol) to cause homolytic cleavage of the carbon–carbon bond in ethane.

11.2 Formation and Simple Reactions of Radicals

We saw radicals in Chapters 1 and 2 when we considered the formation of molecules

such as hydrogen, methane, and ethane from their constituent parts: the hydrogen

atom, which is the smallest radical, and the methyl radical (see Sections 1.5, 2.4

and 2.5; Fig. 11.1).

470 CHAPTER 11 Radical Reactions

–

Methyl radicals

Methyl

cation

Homolytic

cleavage

Heterolytic

cleavage

Methyl

anion

CCH

FIGURE 11.3 Two possible bond

breakings in ethane. Homolytic

cleavage requires less energy because

no charges are generated.

WORKED PROBLEM 11.3 In addition to bond vibration, what hard-to-see process

is going on in ethane as we heat it?

ANSWER Don’t forget rotation about the carbon–carbon bond! No new product

is formed, but this process is still occurring.

Although adding heat is effective in producing methyl radicals from ethane, it

is not a generally useful method of making radicals.The carbon–carbon sigma bond

in hydrocarbons is strong, and therefore harsh conditions are required to break it,

even in a homolytic fashion. Other, more delicate functional groups will not survive

such treatment. In practice, this means that if we want to generate radicals in the

presence of other functional groups we should not attempt to do so by heating

simple hydrocarbons.

PROBLEM 11.4 When ethane is heated, why does the carbon–carbon bond break

rather than one of the carbon–hydrogen bonds?

shows schematically the stabilization gained when two singly occupied sp

orbitals

overlap to form a bond. In order to reverse the bond-making process and form two

radicals, energy must be provided up to the amount of the stabilization achieved

in the bond making. For example, we can apply heat to ethane in a process called

pyrolysis or thermolysis, eventually providing 90 kcal/mol and homolytically

breaking the carbon–carbon bond (Fig. 11.2b). Carbon–carbon bond cleavage is the

lowest energy process easily detectable in ethane.

Recall that the energy required to break the sigma bond homolytically is called

the bond dissociation energy (BDE, p. 337). Note that the bond breaks to give two

neutral species, two radicals, and not to give two polar species, a cation and an anion

(Fig. 11.3).

The pyrolysis of alkanes can be effective in decreasing the overall chain length

of the molecule and in introducing unsaturation. Let’s consider the pyrolysis of

butane. First, there are two possible homolytic cleavages of carbon–carbon bonds

11.2 Formation and Simple Reactions of Radicals 471

Ethyl radicals

Propyl

radical

Methyl

radical

FIGURE 11.4 There are two possible

homolytic cleavages of

carbon–carbon bonds in butane.

Abstraction from the

secondary position gives

the sec-butyl radical

sec-Butyl radical

Abstraction from the primary

position gives the butyl radical

Abstracting radical

Products from

radical abstraction

CH CH

+RH

Butyl radical

FIGURE 11.5 Hydrogen abstraction

from butane by a radical ultimately

leads to the formation of smaller

hydrocarbons and two possible butyl

radicals.

Now we need to evaluate the pathways available for these reactive intermedi-

ates. One reaction possible for these radicals is hydrogen abstraction. Abstraction

is used to describe the process of a radical species taking a hydrogen atom (not a

proton) from another molecule.In this case, one of the new radicals (methyl, ethyl,

and propyl) can abstract hydrogen from butane to make the smaller hydrocarbons

methane, ethane, and propane, along with the butyl radical. There are two posi-

tions from which hydrogen abstraction can take place in butane, the primary

of the methyl group or the secondary of the methylene group, to

give the primary radical or the secondary radical (Fig. 11.5). We will consider the

question of selectivity, of choosing between the two different positions,a little later

in this chapter.

PROBLEM 11.5 Draw the transition states for the two possible abstractions of

hydrogen from butane by a methyl radical.

in butane. The methyl, ethyl, and propyl radicals can all be formed from breaking

carbon–carbon bonds (Fig. 11.4).

It is also possible for one radical to abstract a hydrogen from the vicinal (adjacent

or β) carbon of another radical in a process called disproportionation,which produces

an alkane and an alkene. In the propyl radical, for example, a carbon–hydrogen bond

472 CHAPTER 11 Radical Reactions

could be broken by a methyl radical to give propene and methane (Fig. 11.6). In the

butane pyrolysis example, any of the radicals can do the abstracting so an is used

in Figure 11.6.

This carbon is in

the

α position

Hydrogen to be

removed is in

the

β position

available R

radicals

Reference point

is this carbon

(the radical itself)

Propyl radical Propene Alkane

CHCH

FIGURE 11.6 The disproportionation

reaction produces an alkane and

an alkene from a pair of radicals.

A hydrogen atom of one radical is

abstracted by another radical.

CH CH

CCH

βα

FIGURE 11.7 β Cleavage of the sec-

butyl radical.

dimerization

Two propyl radicals Hexane

New

products

FIGURE 11.8 Two radicals can form a bond in a process called dimerization.

PROBLEM 11.6 Draw the transition state for hydrogen abstraction in a dispropor-

tionation reaction.

PROBLEM 11.7 Why will β cleavage be favored at high temperature?

To make the issue even more complex, two radicals can occasionally react with

each other (dimerize) in a very exothermic formation of new hydrocarbons (dimers)

that can reenter the reaction sequence and provide new species. For example, two

propyl radicals might combine to form hexane (Fig. 11.8). Further pyrolysis of

hexane would lead to many other products.

Another reaction possible for many radicals is ␤ cleavage (Fig. 11.7). In the

terminology of radical chemistry, the point of reference is the radical itself; the adja-

cent atom is α and the next atom is β. If an α–β carbon–carbon bond is present, it

can break to give a new radical and an alkene.This cleavage is simply the reverse of

the addition of a radical to an alkene (discussed later in this chapter).In the sec-butyl

radical, for example, cleavage between the methylene and methyl groups produces

a methyl radical and propene. The β cleavage reaction is favored at high tempera-

ture and can be quite important in high-temperature reactions of alkanes.

11.2 Formation and Simple Reactions of Radicals 473

Summary

The reactions available to an alkyl radical include hydrogen abstraction to make a

saturated compound and a new radical, disproportionation (hydrogen abstraction

from another radical to give an alkene and an alkane), β cleavage (the fragmenta-

tion of a radical into a new radical and an alkene), and dimerization. (Fig. 11.9).

2 CH

Hydrogen abstraction

Disproportionation

β Cleavage

Dimerization

FIGURE 11.9 Simple reactions of radicals.

Clearly, little specificity is possible in hydrocarbon pyrolysis, and one might

at first expect that there would be little utility in such a series of reactions.That’s

not quite right. Using heat to form smaller hydrocarbons from larger ones is called

hydrocarbon cracking, and the petroleum industry relies on this process to con-

vert the high molecular weight hydrocarbons making up much of crude oil into

the much more valuable lower molecular weight gasoline fractions. It is also

possible to do the pyrolysis in the presence of hydrogen to help produce smaller

alkanes from the radicals that are produced initially.

....

Disproportionation gives the alkanes already mentioned, in addition to

ethylene, propene, 1-butene, and 1-pentene:

....

PROBLEM 11.8 What function does the hydrogen perform in hydrocarbon cracking?

WORKED PROBLEM 11.9 What products do you expect from the thermal cracking

of hexane?

ANSWER The point of this problem is to show you how incredibly complicated

this “simple” process becomes even in a relatively small alkane such as hexane.

First of all, carbon–carbon bond breakings give the methyl,ethyl, propyl, butyl,

and pentyl radicals,which can abstract hydrogen to give methane, ethane,propane,

butane, and pentane:

(continued)

474 CHAPTER 11 Radical Reactions

β Cleavage of these radicals gives more ethylene and the methyl, ethyl, and

propyl radicals:

β cleavage

Of course, the alkanes and alkenes formed in these reactions can react further.

Propane, butane, and pentane, as well as the alkenes formed in disproportionation

and β cleavage, can participate in similar radical reactions. Moreover, dimerization

reactions have the potential of producing larger hydrocarbons, only two of which

are shown here. These dimers can go on to produce other molecules as they, too,

participate in the radical reactions. It’s an incredible mess.

Decane

Nonane

Many more products

Now comes a big question. How can an alkane be modified in order to make

the carbon–carbon bond weaker and thus more easily broken?

Energy

Transition state

for diradical

formation

65 kcal/mol

89 kcal/mol

Propane

Cyclopropane

Reaction progress

Strain

energy

Transition state

for formation of

radical pair

FIGURE 11.10 Strain can raise the energy of a hydrocarbon, making it easier

to break a carbon–carbon bond.

PROBLEM 11.10 Draw a molecule with a carbon–carbon bond weaker than the one

in a typical alkane.

There are many possible answers to

Problem 11.10. For example, we could

induce some strain in the molecule,raising

its energy and making bond breaking

easier. Cyclopropane is a nice example.

This small ring compound contains both

severe angle strain and substantial tor-

sional strain (Section 5.3).The strain ener-

gy of cyclopropane results in a low bond

dissociation energy of only 65 kcal/mol

(Fig. 11.10).

Breaking a carbon–carbon bond in

cyclopropane doesn’t yield two new separat-

ed radicals, but a 1,3-diradical, in which

two nonbonding electrons are present in

one molecule. We haven’t really done exactly

what we wanted—produce two new radi-

cals from a single molecule—so let’s try

another way to modify an alkane to make

it easier to cleave. We just tried making the

starting material less stable by introducing

strain. How about taking the opposite

approach by making the products of the

reaction, the two radicals, more stable?

11.2 Formation and Simple Reactions of Radicals 475

The question now becomes: How can we make a radical more stable than the

methyl radical? We know that delocalization of electrons is stabilizing, so one

approach would be to use a starting material that will yield resonance-stabilized rad-

icals.That change should lower the bond strength of the crucial carbon–carbon bond.

For example, introduction of a single vinyl group (as in 1-butene) lowers the bond

dissociation energy by about 13 kcal/mol to a value of 76 kcal/mol. As shown in

Figure 11.11, propane breaks a carbon–carbon bond to give a methyl radical and an

ethyl radical and 1-butene breaks a carbon–carbon bond to give a methyl radical and

an allyl radical.The difference in the energy required for the two bond breakings is

reasonably attributed mostly to the stabilization of the allyl radical, and, of course,

of the transition state leading to it. Introduction of a second vinyl group, as in 1,5-

hexadiene,reduces the BDE by approximately an additional 13 kcal/mol (Fig. 11.11).

The key to lowering the bond dissociation energy is the formation of allyl radicals

with their stabilizing delocalization of the nonbonding electrons.

Bond Dissociation

Energy (kcal/mol)

Propane

CCH

An allyl radical

1-Butene

CCH

CH CH

1,5-Hexadiene

Two all

l radicals

CHCH

C CH

CH CH

FIGURE 11.11 Stabilization of the products of bond breaking through delocalization can make the bond

cleavage process easier.

PROBLEM 11.11 Construct an Energy versus Reaction progress diagram that

illustrates this second general approach of making the product of the reaction

more stable.

1,1,6,6-Tetradeuterio-

1,5-hexadiene

3,3,4,4-Tetradeuterio-

1,5-hexadiene

PROBLEM 11.12 There is an even lower energy process occurring in 1,5-hexadiene,

although no new molecule appears! Can you draw an arrow formalism for the

conversion of one molecule of 1,5-hexadiene into another?

Hint: 1,1,6,6-Tetradeuterio-1,5-hexadiene equilibrates on heating with 3,3,4,4-

tetradeuterio-1,5-hexadiene:

476 CHAPTER 11 Radical Reactions

WEB 3D

2 RO

Two carboxyl radicals

An alkyl peroxide

An acyl peroxide

FIGURE 11.12 In alkyl peroxides, the

weak oxygen–oxygen bond is easily

cleaved. In acyl peroxides, the product

radicals are stabilized and as a result

the oxygen–oxygen bond is even

weaker.

NRRRNN

An acyl peroxide

An azo compound

2 R 2 CO

(g)

FIGURE 11.13 Some molecules give

free radicals irreversibly through the

formation of gases.

(CH

)

C C(CH

)

2 (CH

)

(g)

ΔH

= 42.2 kcal/mol

(CH

)

C C(CH

)

2 (CH

)

(g)

AIBN

ΔH

= 31.9 kcal/mol

WEB 3D

(a)

(b)

FIGURE 11.14 Azobisisobutyronitrile

is an azo compound that is especially

easy to cleave.

Another technique for producing radicals is to start with molecules containing

especially weak, and therefore easily broken bonds. Examples are peroxides and acyl

peroxides (Fig. 11.12).The oxygen–oxygen bond in simple peroxides is already quite

weak (~38 kcal/mol; Table 8.2, p. 337). The addition of the carbon–oxygen double

bond of the acyl group ( ), helps delocalize the nonbonding electron in

the carboxyl radical and lowers the BDE to about 29 kcal/mol.

Some molecules give stable gases (often nitrogen or carbon dioxide) on heating

and thus produce radicals irreversibly. Azo compounds, which have the structure

, and acyl peroxides are good examples (Fig. 11.13).R

PROBLEM 11.13 Why does AIBN decompose much more easily than a simple azo

compound such as the molecule shown in Figure 11.14b?

A particularly useful source of radicals is azobisisobutyronitrile (AIBN), a mol-

ecule that decomposes much more easily than simpler azo compounds (Fig.11.14a).

11.3 Structure and Stability of Radicals 477

11.3 Structure and Stability of Radicals

We have already encountered both carbocations and car-

banions, and spent some time discussing their structures

(Section 2.4, p. 62).The methyl cation, and carbocations in

general,are flat,sp

-hybridized molecules with an empty 2p

orbital extending above and below the plane of the three

substituents on carbon (Fig.11.15).By contrast,the methyl

anion is pyramidal. One might well guess that the methyl

radical, with a single nonbonding electron, would have a

structure intermediate between that of the two ions.

Both theory and experiment agree that the structure of simple radicals is hard

to determine! However, it is clear that these molecules are not very far from planar.

If the molecules are not planar, the pyramid is very shallow and the two possible

forms are rapidly inverting (Fig. 11.16).

Planar... or... ver

shallow p

ramid

Cor

FIGURE 11.16 Alkyl radicals are

either flat or very shallow, rapidly

interconverting pyramids.

PROBLEM 11.14 Draw the transi-

tion state for the inversion

process in Figure 11.16.

As with carbocations, more sub-

stituted radicals are more stable than

less substituted ones. The first four

entries of Table 11.1 show BDEs of

four hydrocarbons that lead to a

methyl, primary, secondary, and ter-

tiary radical. The more substituted

the radical, the easier it is to form,

and the easier it is to break the bond

that produces it.The data presented

in the table are slightly suspect,

because we are forming the radicals

from different starting materials.

Might not the different energies of

the different starting materials influ-

ence the BDEs? Yes they might, and

do. But this effect isn’t large, and the

data in the table can be used to show

that more substituted radicals are

more stable than less substituted

radicals.

Carbocation,

planar

Carbanion,

pyramidal

Radical—somewhere

in between the cation

and anion

C C

–

120⬚

99.8⬚

FIGURE 11.15 Radicals are intermediate in structure between

the planar carbocations and the pyramidal carbanions.

...

105.0

101.1

98.6

96.5

89.8

88.8

TABLE 11.1 Bond Dissociation Energies for Some Hydrocarbons

Hydrocarbon Radical BDE (kcal/mol)

CHH

478 CHAPTER 11 Radical Reactions

Why should more substituted radicals be more stable than less substituted

radicals? Let’s start with an analogy. In explaining the order of stability for

carbocations (tertiary > secondary > primary > methyl), we drew resonance forms

in which a filled carbon–hydrogen bond overlapped with the empty 2p orbital in

a process called hyperconjugation (Fig. 11.17).

Energy

CH σ

Antibonding

molecular orbital

Bonding

molecular orbital

(empty)

Empty 2p

2p/σ bond

overlap

Filled C

σ bond

C CC

FIGURE 11.17 Resonance stabilization of a carbocation through hyperconjugation, shown both by an interaction energy

diagram and by an orbital overlap picture.

Half-filled 2p

2p/σ bond

overlap

Filled C

σ bond

CC CC

FIGURE 11.18 Resonance

stabilization of an alkyl free radical.

We can draw similar resonance forms for radicals. As before, the more

alkyl groups attached to the carbon bearing the nonbonding electron, the

more resonance forms are possible. But there is an important difference:

Stabilization of the carbocation involves overlap of a filled carbon–hydrogen

bond with an empty 2p orbital (Fig. 11.17), whereas in the radical, overlap

is between a filled carbon–hydrogen bond and a half-filled 2p orbital

(Fig. 11.18).

Why should this orbital mixing be stabilizing? The evident problem is that

this is a three-electron system. One electron must occupy the antibonding orbital,

and this is certainly destabilizing. Nevertheless, the overall system is still stabi-

lized because two electrons are able to occupy the low-energy bonding orbital.