Jones M., Fleming S.A. Organic Chemistry

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11.3 Structure and Stability of Radicals 479

Energy

Antibonding

molecular orbital

Bonding

molecular orbital

Antibonding

molecular orbital

Bonding

molecular orbital

Antibonding

molecular orbital

Bonding

molecular orbital

Cationic, two-electron

system

Radical, neutral,

three-electron system

FIGURE 11.19 A comparison of the stabilizations of carbocations, radicals, and He



. Notice the electron in the

antibonding orbital of He



and the radical.

Recall He



in which a similar orbital system was encountered (p.41). Figure 11.19

compares hyperconjugation of a carbocation, hyperconjugation of a radical, and

the He



molecule. As we saw in Chapter 1, He



is held together (bound) by

over 60 kcal/mol.

2-Propyl radical 1-Propyl radical

Both sp

/sp

Weaker than the others

FIGURE 11.20 The 2-propyl radical

contains two relatively strong sp

/sp

bonds, whereas the 1-propyl radical

has one strong sp

/sp

bond and one

less strong sp

/sp

bond.

Consider the bond strengths in the carbon radicals, as we did earlier in discussing

alkenes and carbocations (p. 378, Problem 9.7). In the 2-propyl radical we have two

/sp

carbon–carbon bonds, whereas in the less stable 1-propyl radical there is one

/sp

bond and one less strong sp

/sp

carbon–carbon bond (Fig. 11.20).

Of course, we should analyze the carbon–hydrogen bonds as well. If we do this,

we find that the differences in the strengths of the bonds should favor the

1-propyl radical. As with carbocations, it is the bonds that dominate and

determine the issue. As a practical matter, it appears that we can “get away with” an

analysis of the bonds alone.C

PROBLEM 11.15 Analyze the difference in bond strength between the

bonds in the 2-propyl and 1-propyl radical.

Radicals substituted with electron-delocalizing groups, such as the vinyl or

phenyl groups, are even more stable. As Table 11.1 shows, breaking a carbon-

hydrogen bond to give an allyl radical or benzyl radical is relatively easy.

The allyl and benzyl radicals are even more stable than a tertiary radical

(Fig. 11.21).

480 CHAPTER 11 Radical Reactions

CH HCH

CH CH

C H

FIGURE 11.21 Resonance stabilization in the allyl and benzyl radicals.

Dimer

Triphenylmethyl radical

FIGURE 11.22 The triphenylmethyl

radical exists in equilibrium with its

dimer.

PROBLEM 11.16 As you can see from Figure 11.22, the dimer is not the “obvious”

one, hexaphenylethane, but a much more complicated molecule. Can you see why

hexaphenylethane is a most unstable molecule, and draw an arrow formalism

mechanism for the formation of the real dimer?

PROBLEM SOLVING

Although the order of radical stability is the same as that for cation stability, the

absence of a charge makes a big difference. Primary and methyl radicals are not

“mechanistic stop signs,” as are the related carbocations. Their energies are not so

high as to make them unknown.

If they are sufficiently stabilized, radicals can be made and studied

spectroscopically at low temperature. A few highly delocalized radicals are even

stable under normal, room temperature conditions. For example, the triphenyl-

methyl radical happily exists in solution as an equilibrium mixture with its dimer

(Fig. 11.22).

11.4 Radical Addition to Alkenes 481

Less stable,

primary carbocation

(

not formed

)

Much more stable

tertiary carbocation

is formed

–

FIGURE 11.23 The addition of hydrogen bromide to an unsymmetrical alkene follows the Markovnikov

rule because protonation of the alkene must give the more substituted (more stable) carbocation.

11.4 Radical Addition to Alkenes

In Section 9.3 (p. 366), the addition of hydrogen chloride to alkenes was presented

with the reaction always leading to Markovnikov addition. Chapter 9 stressed the

initial formation of the more stable carbocation followed by addition of the resid-

ual halide ion.The product of Markovnikov addition must be produced in this reac-

tion (Fig. 11.23). Of course, HBr reacts the same way as HCl.

The historical reality was far from that simple. The regiochemistry of the

reaction appeared strangely dependent on the source of the starting materials and

even on the place in which the reaction was carried out. In some worker’s hands, the

reaction did follow the Markovnikov path; in other places, with other investigators,

anti-Markovnikov addition was dominant. Confusion and argument were the rule

of the day in the late 1800s. Oddly enough, the problem was restricted to hydrogen

bromide addition. Both hydrogen chloride and hydrogen iodide addition remained

securely Markovnikov in everyone’s hands (Fig. 11.24).

Sometimes Markovnikov, sometimes not

Always Markovnikov

and/or

Always Markovnikov

FIGURE 11.24 Both hydrogen

chloride and hydrogen iodide

addition always follow the

Markovnikov rule, but hydrogen

bromide addition is strangely

capricious.

Summary

Radical stability parallels carbocation stability: Benzylic and allylic are more

stable than tertiary, tertiary more stable than secondary, secondary more stable

than primary, and primary more stable than methyl.The determining factors are

resonance and hyperconjugative stabilization.

The product of

anti-Markovnikov

addition of H

RO OR

(a peroxide)

FIGURE 11.25 Addition of hydrogen

bromide to alkenes in the presence of

peroxide is anti-Markovnikov.

In the 1930s, largely through the efforts of M. S. Kharasch,

it was discovered

that the purity of the compounds used was vital.Pure materials led to Markovnikov

addition, and we can remain content with the mechanism for the polar reaction.

Impure reagents led to predominately anti-Markovnikov addition. A special culprit

was found to be peroxides (Fig. 11.25). Breaking the weak oxygen–oxygen bond in

peroxides provides a relatively easy route to free radicals. The mechanism of addi-

tion in the presence of peroxides is different from the familiar polar one and can be

triggered by these small amounts of radicals.

482 CHAPTER 11 Radical Reactions

or h

OStep 1

.. ..

O O

OStep 2

Step 3

Step 4

Recycle to Step 3 and

start all over again

ORR

RRRR

FIGURE 11.26 The mechanism for the radical addition of HBr to an alkene.

11.4a The Mechanism of Radical HBr Addition As noted earlier (p. 476),

peroxides contain a very weak oxygen–oxygen bond and are easily cleaved by heat ( )

or light (

h ) to give two radicals (Fig. 11.26, Step 1). The radical can

react with by hydrogen abstraction to form and a new radical, a

bromine atom (Step 2). Addition of the bromine atom to an alkene gives still

another radical (Step 3), which can react with to produce an alkyl bromide

and another bromine atom (Step 4). The bromine atom can recycle to Step 3 and

start the reaction over again with another alkene molecule.The overall result is the

addition of hydrogen bromide to the alkene.

OHH

Kharasch, 1895–1957, is reported by one of his students, F. R. Mayo (b. 1908), to have suggested that the

direction of addition may have been affected by the phases of the moon.

11.4 Radical Addition to Alkenes 483

RRRR

FIGURE 11.27 Whenever two

radicals come together they can react

to form a bond. The bond formation

removes two radicals and terminates

this chain reaction.

Other products can be formed in this reaction.When two radicals meet they can

react in a very exothermic fashion to form a new bond and end the sequence, called

a chain reaction. There are many possibilities for this kind of termination reaction

using the radicals in Figure 11.26. A few ways are shown in Figure 11.27.

We have just outlined the radical-induced addition of hydrogen bromide to an

alkene. The steps of this chain reaction naturally divide into three kinds:

Initiation. The reaction cannot start without the presence of a radical. A radical

must be formed to start the process,in a reaction known as an initiation step.Peroxides

are good generators of radicals because of the weakness of their oxygen–oxygen bond

(Fig. 11.26, Step 1). Once a radical is created in the presence of hydrogen bromide, a

second initiation step, abstraction of hydrogen (Fig. 11.26,Step 2),takes place to form

a bromine atom, which is the radical that propagates the chain reaction.

Propagation. Once a bromine atom has been formed in the initiation steps, the

chain reaction can begin.The reactions making up the chain reaction are called prop-

agation steps. In this example, the two propagation steps are the addition of the

bromine atom to the alkene to give a new, carbon-centered alkyl radical and the sub-

sequent abstraction of hydrogen from hydrogen bromide by the alkyl radical

(Fig. 11.26, Steps 3 and 4).

The two radicals involved in the propagation steps, the bromine atom and the

alkyl radical, are called chain-carrying radicals.The chain reaction continues as the

newly formed bromine atom adds to another alkene to repeat the process over and

over again.

Termination.Any time one of the chain-carrying radicals (in this case a bromine

atom or an alkyl radical) is annihilated by combination with another radical, the chain

reaction is stopped (Fig. 11.27). If such a step does not happen, a single radical could

initiate a chain reaction that would continue until starting material was completely

used up. In practice, a competition is set up between the propagation steps carrying

the chain reaction and the steps ending it.Such reactions are called termination steps

and there are many possibilities.

For a chain reaction to produce large amounts of products effectively, the prop-

agation steps must be much more successful than the termination steps. Such is often

the case, because in order for a termination step to occur, two radicals must wander

through the solution until they find each other. Typically, the propagation steps

require reaction with a reagent present in relatively high concentration. So, even

though termination steps often can take place very easily once the reactants find each

other, propagation steps can compete successfully, and reaction cycles of many thou-

sands are common.Under such circumstances,the small amounts of product formed

by the occasional termination step are inconsequential.

PROBLEM 11.17 Why is abstraction of H from preferred to abstraction of Br?

Hint: The bond dissociation energy of is 56 kcal/mol.

11.4b Inhibitors Small amounts of substances that react with the chain-carrying

radicals can stop chain reactions by removing the chain-carrying radicals from the

process. These substances are called inhibitors. They act as artificial terminators of

the propagation steps (Fig. 11.28).

484 CHAPTER 11 Radical Reactions

The first propagation step in a typical

radical chain addition reaction

The chain reaction can be stopped if the

chain-carrying radical reacts with an inhibitor, X

CC X

FIGURE 11.28 An inhibitor of radical chain reactions can act by intercepting one of the

chain-carrying radicals.

WEB 3D

Hydroquinone BHT BHA

(CH

)

C(CH

)

C(CH

)

OCH

(CH

)

OCH

FIGURE 11.29 Some radical inhibitors.

substituted

radical

Less

substituted

radical

FIGURE 11.30 When a radical adds

to an unsymmetrical alkene, there are

two possible radical products.

A few molecules can have an impact out of proportion to their numbers, because

stopping one chain reaction can mean stopping the formation of thousands of prod-

uct molecules. Oxygen (O

), itself a diradical, can act in this way, and many radical

reactions will not proceed until the last oxygen molecules are scavenged from the

system, thereby allowing the radical chains to carry on. Radical inhibitors are often

added to foods to retard radical reactions causing spoilage.Figure 11.29 shows a few

typical radical inhibitors. Both butylated hydroxy toluene (BHT) and butylated

hydroxy anisole (BHA) are widely used in the food industry. It is not important that

you memorize these structures. What is worth remembering is the general structur-

al features. You should also think a bit about how these molecules might function

in intercepting radicals.

11.4c Regiospecificity of Radical Additions: Anti-Markovnikov Addition

When a radical adds to an unsymmetrical alkene, there are two possible out-

comes: formation of a more substituted or less substituted alkyl radical (Fig. 11.30).

11.4 Radical Addition to Alkenes 485

So, the regiospecificity of radical addition to an unsymmetrical alkene such as

2-methylpropene (isobutene) is determined by the direction of radical attack on

the alkene, and thus by the stability of the alkyl radical formed. In the case of

alkene radical hydrohalogenation, the bromine atom adds to give the more sub-

stituted, more stable radical.The reaction continues when the newly formed alkyl

radical abstracts a hydrogen atom from hydrogen bromide to give, ultimately,

the compound in which the bromine is attached to the less substituted end of the

original alkene, the anti-Markovnikov product (Fig. 11.31a). This reaction is in

contrast to polar hydrohalogenation, in which the first step is protonation of

the alkene to give the more substituted carbocation, which is then attacked by

bromide ion to give the more substituted bromide, the Markovnikov product

(Fig. 11.31b).

(a)

(b)

–

This less stable primary

radical is not formed

—

The more stable,

tertiary radical

is formed

This anti-Markovnikov

product is determined by

the initial addition of a

bromine atom to give the

more stable radical

Radical addition

Contrast radical addition with polar addition of H

This less stable primary

carbocation is not formed

The more stable,

tertiary carbocation

is formed

This Markovnikov product

is determined by the initial

protonation to give the

more stable carbocation

FIGURE 11.31 (a) The first step in a radical addition to an alkene gives the more

substituted, and therefore more stable radical. (b) The first step in the polar addition gives

the more stable intermediate, which leads to the Markovnikov product. Notice, however,

that in the polar reaction it is the hydrogen atom that adds first, and in the radical reaction

it is the bromine atom that adds first.

The historical difficulty in determining the regiospecificity of hydrogen bromide

addition had to do only with recognizing that radical initiators such as peroxides had

to be kept out of the reaction. If even a little peroxide were present, the chain nature

of the radical addition of hydrogen bromide could overwhelm the nonchain polar reac-

tion and produce large amounts of anti-Markovnikov product.This understanding of

the causes of the regiochemical outcome of these reactions presents an opportunity

for synthetic control.We now have ways to do Markovnikov addition of hydrogen chlo-

ride and hydrogen iodide, and either Markovnikov or anti-Markovnikov addition of

hydrogen bromide. Of course, the halides formed in these reactions can be used for

further synthetic work.

Radical alkene hydrohalogenation

486 CHAPTER 11 Radical Reactions

TABLE 11.2 The Thermochemistry of the First Propagation Step:

Addition of Halogen Radicals to Ethylene

Bond Strength Bond Strength

X(π bond) ( in ) (kcal/mol)

Cl 66 85 19

Br 66 72 6

I66 57 9

≤H

ⴙ H

PROBLEM 11.18 Show how to carry out the following conversions. You need not

write mechanisms in such problems unless they help you to see the correct

synthetic pathways.

Devise two syntheses

for each

Neither hydrogen chloride nor hydrogen iodide will add in anti-Markovnikov

fashion, even if radical initiators are present. Now we have to see why.

11.4d Thermochemical Analysis of Radical HX Additions to Alkenes

Why does the chain reaction of Figure 11.26 succeed with but not for other

hydrogen halides? We can write identical radical chain mechanisms for and

, but these must fail for some reason.

We can understand the difference if we examine the thermochemistry of the two

propagation steps for the different reactions.

The energetics of the first propagation step are shown in Table 11.2. These val-

ues are estimated by noting that in this first reaction a π bond is broken and a

bond is made.The overall exothermicity or endothermicity of the reaction is a com-

bination of these two bond strengths.A π bond is worth 66 kcal/mol,and we approx-

imate the bond strength in the product by taking the values for ethyl halides.X

HCl

HBr

We are making all sorts of assumptions, but the errors involved turn out to be rel-

atively small. For both bromine and chlorine this first step is exothermic—the

product is more stable than starting material and is favored at equilibrium. For

iodine, the reaction is substantially endothermic.The product is strongly disfavored

and the reaction proceeds very slowly. So now we know why the addition of

fails; the first propagation step of the reaction is highly endothermic and does not

compete with the termination steps. But both the bromine case, which we know

succeeds, and the chlorine case, which we know fails, have exothermic first steps.

11.5 Other Radical Addition Reactions 487

TABLE 11.3 The Thermochemistry of the Second Propagation Step:

Abstraction of a Hydrogen Atom from

Bond Strength Bond Strength

X ( ) ( ) (kcal/mol)

Cl 103 98 5

Br 87 98 11

I71 98 27

C in H

≤H

X ⴙ

X ⴙ H

The second step is exothermic for Br and I, but endothermic for Cl.So,we can

expect this reaction to succeed for Br and I, but fail for Cl. If we sum up the analy-

sis for the two propagation steps we see that it is only for addition that

both steps are exothermic. For and , one of the two propagation steps is

endothermic,and the radical addition reaction cannot successfully compete with the

termination steps. The polar addition of wins out.HX

HIHCl

HBr

ROORStep 1 RO OR+

CCl

ROStep 2

CCl

Step 3 CH

CCl

Step 4

CCl

ClCH

CCl

ROCl +

Propagation

Initiation

There are man

possible termination stepsStep 5

(recycle

to Step 3)

FIGURE 11.32 The mechanism for

the radical-induced addition of

carbon tetrachloride to an alkene.

11.5 Other Radical Addition Reactions

Hydrogen bromide addition to alkenes is only one example of a vast number of

radical chain reactions. Another example involves carbon tetrahalides, such as CCl

and CBr

, which add to alkenes in the presence of initiators. The steps in the

peroxide-initiated reaction are shown in Figure 11.32.

Let’s look at the second propagation step in Table 11.3.

Summary

The polar addition of HBr to alkenes leads to the more substituted bromide,

whereas the radical-initiated addition of HBr gives the less substituted bromide.

Thus, you have the ability to make either regioisomer when the alkene is unsym-

metrically substituted. The addition of HCl and HI always gives the more sub-

stituted halide.

PROBLEM 11.19 Write as many termination steps as you can imagine for the

reaction of Figure 11.32.

Under some conditions, alkenes can be polymerized by free radicals. In the

examples we have discussed, radical addition to an alkene produced an alkyl radi-

cal, which then abstracted hydrogen (or halogen in the case of CX

) in a second

propagation step to give the product of overall addition to the alkene. What if

Step 4 is not easy? If the concentration of alkene is very high, for example, or if

the addition step (Step 3) is especially favorable, the abstraction reaction (Step 4)

might not compete effectively with it. Under such conditions the alkyl radical can

undergo addition to the alkene to form a new alkyl radical (the dimer).The dimer

radical can react with alkene to form a trimer radical and so forth, until either a

termination step intercedes or the supply of alkene is exhausted (Fig. 11.33).

488 CHAPTER 11 Radical Reactions

IN IN 2 IN

Products of hydrogen

abstraction

Dimer radical Trimer radical

repeat many times

until a termination

step or hydrogen

abstraction stops

the chain reaction

C C H

C C

FIGURE 11.33 In the radical-induced polymerization of alkenes, repeated

additions grow a long polymer chain. A high concentration of alkene favors this

process ( radical).IN

= initiator

TABLE 11.4 Some Monomers and Polymers

Monomer Polymer Formula NameMonomer Polymer Formula Name

Styrene

Ethylene

CCH

(CH

)

Polyethylene,

polymethylene

Polystyrene

(CF

)

COOCH

Polyacrylate

Poly(tetrafluoro-

ethylene), Teflon

Poly(vinyl chloride)

(PVC)

Tetrafluoroethylene

CCF

ClCH

Vinyl chloride

OOCCH

Methyl acrylate

WEB 3D

Such polymerization reactions are common and the products often have very long

chain lengths. Very high molecular weight products can be formed. Table 11.4

shows a number of common polymers and the monomers from which they are

built up.