Free Radical Reactions

Richard F. Daley and Sally J. Daley 

www.ochem4free.com 

Organic 

Chemistry 

Chapter 21 

Radical Reactions 

21.1 Radical Structure and Stability 1093 

21.2 Halogenation of Alkanes 1095 

Sidebar - Atmospheric Ozone Depletion 1099 

21.3 Allylic Bromination 1102 

21.4 Benzylic Bromination 1105 

Synthesis of 1-Bromo-1-phenylethane 1106 

21.5 Radical Addition to Alkenes 1107 

21.6 Radical Oxidations 1112 

21.7 Radical Reductions 1115 

Synthesis of 1-Methoxy-1,4-cyclohexadiene 1121 

Special Topic - Electron Spin Resonance Spectroscopy 1122 

Key Ideas from Chapter 21 1125

Organic Chemistry - Ch 21 1092 Daley & Daley 

Copyright 1996-2005 by Richard F. Daley & Sally J. Daley 

All Rights Reserved. 

No part of this publication may be reproduced, stored in a retrieval system, or 

transmitted in any form or by any means, electronic, mechanical, photocopying, 

recording, or otherwise, without the prior written permission of the copyright 

holder. 

www.ochem4free.com 5 July 2005


Chapter Outline 

Chapter 21 

Radical Reactions 

21.1 Radical Structure and Stability 

A chemical species with an unpaired 

electron in the valence shell 

21.2 Halogenation of Alkanes 

The reaction of alkanes and halogens with 

energy provided by light or heat to form 

alkyl halides 

21.3 Allylic Bromination 

The reaction of bromine radicals with 

alkenes in the allylic position 

21.4 Benzylic Bromination 

The reaction of bromine radicals with alkyl 

benzenes in the benzylic position 

21.5 Radical Addition to Alkenes 

Anti-Markovnikov additions to double bonds 

21.6 Radical Oxidations 

A brief survey of autooxidation processes in 

organic chemistry 

21.7 Radical Reductions 

A brief survey of radical reduction reactions 

Objectives 

Understand the structure of a radical 

Know the distribution of the halogens in a radical halogenation of 

an alkane 

Recognize that radicals at the allylic and benzylic positions are 

more stable than alkyl radicals 

Know why a radical addition to an alkene leads to an “anti- 

Markovnikov” product 

Understand the autooxidation processes 

Be able to use radical reductions in synthesis 



Whoever in discussion adduces authority uses not 

intellect but rather memory. 

—Leonardo da Vinci 

Radical polymerization 

is discussed in several 

sections in Chapter 22. 

A ny atom, or group of atoms, that bears an unpaired 

electron is a radical. Although a radical may be 

charged or uncharged, most organic radicals are uncharged. This 

chapter covers only the uncharged species. Because electrons tend to 

exist in pairs and because radicals have an unpaired electron, radicals 

are usually highly reactive. Unlike the reactions discussed to this 

point, radical reactions involve the movements of single electrons 

instead of pairs of electrons. 

This chapter is an introduction to some of the many laboratory, 

industrial, and biological processes that involve radicals. For example, 

many polymers of commercial importance are synthesized via radical 

reaction processes. Additionally, the oxygen carrying capability of 

hemoglobin depends on the diradical nature of oxygen. Biochemical 

degradation processes often involve radicals, too. 

21.1 Radical Structure and Stability 

During the latter part of the nineteenth century, most chemists 

thought that radicals were sufficiently unstable to preclude their 

observation. Many also thought that radicals were so unstable that 

they could not even exist. However, in 1900, Moses Gomberg at the 

University of Michigan generated the first laboratory example of a 

radical, although it was another 30 years before anyone realized what 

it was that he had made. Gomberg had successfully synthesized 

tetraphenylmethane in 1897 and wished to synthesize 

hexaphenylethane to study its properties. Gomberg's plan was to 

produce hexaphenylethane by reacting triphenylmethyl chloride with 

silver ion. 

Ph3CCl 

Ag 

Ph3CCPh3 

When Gomberg ran the reaction, he obtained a yellow solution 

that contained a very reactive material. This material reacted rapidly 

with oxygen from the air to form Ph 3 COOCPh 3 , or with iodine to form 

Ph 3 CI. When Gomberg reported this reaction, he suggested that the 

intermediate was a trivalent carbon. However, he proposed that it was 

a carbocation instead of a radical. Although chemists have come to 



understand the part radicals had in Gomberg's experiment, no one has 

yet accomplished Gomberg's goal of synthesizing hexaphenylethane. 

The structure of radicals is very similar to the structure of 

carbocations because both are sp 2 hybridized. However, carbocations 

have an empty p orbital, whereas radicals have an unpaired electron 

in the p orbital. 

C 

• 

C 

Carbocation Radical 

The structure of a radical varies somewhat depending on the 

substituents bonded to the carbon atom. When the substituents are 

hydrocarbons, the radicals have a mostly planar structure. When one 

of the substituents is a heteroatom with nonbonding electron pairs, 

however, the radical tends towards an sp 3 arrangement due to the 

repulsive influence that the nonbonding electrons exert on the single 

electron of the radical. Note that nonbonding electrons, particularly if 

close to the radical site as with an oxygen or nitrogen, can also 

stabilize the radical. 

Repulsion 

C X 

Less repulsion 

C X 

Radical stability is also similar to carbocation stability. Thus, 

the order of stability for radicals is 3 o > 2 o > 1 o > methyl. A vinyl or 

phenyl group bonded adjacent to the site of the radical makes the 

radical more stable than a tertiary radical. This is because allylic and 

benzylic radicals are resonance stabilized. 

CH2 

C C C 

Allylic radical 

CH2 

Benzylic radical 

C C C 

CH2 

CH2 


An inhibitor is some 

chemical species, either 

a molecule or radical, 

which is particularly 

reactive with a radical. 

When a bond breaks in 

a homolytic bond 

dissociation, each atom 

takes one electron. 

When a bond breaks in 

a heterolytic bond 

dissociation, one atom 

takes both electrons. 

The single barbed 

mechanism arrows in 

this reaction indicate 

the movement of single 

electrons. 


With uncharged radicals, the polarity of the solvent does not 

usually affect the rate of radical reaction. However, the presence of an 

inhibitor does affect the rate. Oxygen is a common inhibitor. It 

normally exists as a diradical with two unpaired electrons in two 

different degenerate orbitals. 

21.2 Halogenation of Alkanes 

Alkanes react with chlorine in the presence of ultraviolet light 

(represented as hν) or heat (usually 200-300oC) to produce alkyl 

chlorides. Generally, the reaction gives a mixture of products, as does 

the reaction of methane with chlorine. 

Cl 2 

CH4 CH3Cl + CH2Cl2 + CHCl3 + CCl4 h 

The composition of this mixture of alkyl chlorides varies with the 

concentrations of the chlorine and the alkane. However, even if you 

use a large excess of the alkane, the reaction still forms a mixture. 

Radicals and Atoms 

The reaction of chlorine with an alkane is a radical reaction. Chemists refer to the 

species that forms when the chlorine molecule dissociates as chlorine atoms. They are 

called chlorine atoms because chlorine has seven valence electrons, giving the chlorine 

atom an unpaired electron. The reaction is a radical reaction because the chlorine 

atom reacts with an alkane forming an alkyl radical. 

The bond dissociation energy of the chlorine molecule is only 

58 kcal/mol, so chlorine readily undergoes a homolytic bond 

dissociation. All the reactions that you have studied in the previous 

chapters underwent heterolytic bond dissociations. 

h or 

Cl Cl 2 Cl 

H = 58 kcal/mol 

The chlorine atoms that form in a homolytic bond dissociation reaction 

are very reactive because each has an unpaired electron. They are 

electrophilic, thus each seeks an electron to complete its unfilled shell 

of electrons. In a reaction with methane, a chlorine atom readily 

removes a hydrogen from the methane. 

CH3 H 

Cl 

CH3 

+ HCl 


Each step in a chain 

reaction produces a 

chemical species that 

initiates another step 


The resulting methyl radical, which also is very electrophilic, then 

removes a chlorine atom from a chlorine molecule. 

CH3 

Cl Cl Cl + 

CH3Cl 

Notice that the last step in the mechanism produced another 

chlorine atom. This chlorine atom can then remove a hydrogen atom 

from another methane molecule to produce another methyl radical. 

The methyl radical can then react with another chlorine molecule to 

produce another chlorine atom to start the cycle again. This type of 

reaction is known as a chain reaction. 

in the reaction. A chain reaction mechanism consists of three categories of 

Initiation forms the 

initial radicals to begin 

a chain reaction. 

Propagation continues 

the chain reaction. 

Termination stops the 

chain reaction. 

steps: 1) the initiation step, 2) the propagation steps, and 3) the 

termination steps. The initiation step produces the reactive species, 

or radicals. In the radical chlorination reaction above, the initiation 

step is the formation of chlorine atoms. The propagation steps produce 

the major portion of the reaction product and are repeated many 

times. With each propagation series a new reactive species forms, 

keeping the reaction going. The next two steps of the radical 

chlorination above, consuming a chlorine atom then producing 

another, are the propagation steps. The termination steps are the 

steps that stop the chain reaction. For the radical chlorination, the 

possible termination steps are as follows: 

Cl 

CH3 

Cl 

CH3 

CH3 

Cl 

CH3Cl 

CH3CH3 

The initiation step is generally the slowest step in the radical 

halogenation reaction because it requires 58 kcal/mol to produce the 

reactive halogen atom. The propagation steps carry the reaction 

forward. The propagation steps in an alkane halogenation reaction 

produce one molecule of the product and a new halogen atom. For 

radical halogenation, about 10,000 propagation steps occur for each 

initiation step. Moreover, termination happens infrequently because 

the concentrations of the radicals are low compared to the 

concentrations of the other reagents. 

Cl2 



When 2-methylbutane reacts at 300 o C with one mole of 

chlorine, the result is a mixture of four monochlorinated products in 

the following relative amounts. 

CH3 

CH3CHCH2CH3 

2-Methylbutane 

Cl2 

300oC 

CH3 

ClCH2CHCH2CH3 

33.3% 

CH3 

CH3CHCHCH3 

Cl 

28% 

+ 

CH3 

CH3CCH2CH3 

Cl 

22% 

+ 

CH3 

+ 

CH3CHCH2CH2Cl 

16.7% 

Using the above percentages of the reaction's products, you can 

determine the relative reactivity of each of the hydrogens in the 

substrate, 2-methylbutane. Nine of the 12 hydrogens are primary 

hydrogens. Reactions involving these nine hydrogens form only 50% 

(the 33.3% and 16.7% products) of the total amount of product. In 

comparison, the two secondary hydrogens forms 28% of the product 

and the single tertiary hydrogen forms 22%. 

50% of the product 

CH 3 

CH 3CHCH 2CH 3 



Based on statistical predictions, if these three classes of 

hydrogens all had the same reaction rate, you would expect 75% (9/12) 

of the product to form from the primary hydrogens, 16.7% (2/12) from 

the secondary hydrogens and 8.3% (1/12) from the tertiary hydrogen. 

However, the primary hydrogens have less than the statistical amount 

of product and the secondary and tertiary hydrogens have more, so 

there is a difference in their reactivity. 

To calculate the relative rates for the reaction that occurs at 

each of the hydrogens, assume that the rate of reaction for primary 

hydrogens is 1. Then perform the following calculations. 



Secondary 

Primary 

28/2 

= 

50/9 

= 2.5 

Tertiary 22/1 

Primary 

= 

50/9 

= 4 

These calculations show you that the secondary hydrogens react 2.5 

times faster than the primary hydrogens, and the tertiary hydrogens 

react four times faster than the primary hydrogens. 

This difference in reactivity in the various types of hydrogens is 

the result of how readily the various radicals form. The tertiary 

radical is the most stable and the easiest to form. The primary radical 

is the least stable radical and the hardest to form. 

The differences in radical reactivity are less important in 

reactions that involve fluorine radicals and alkanes than in reactions 

that involve chlorine radicals and alkanes. The fluorine atom is more 

reactive than the chlorine atom. Thus, the fluorine atom is much less 

selective than the chlorine atom. In contrast, the iodine atom is so 

unreactive that it does not even react with alkanes. 

Although bromine radicals are much more selective than 

chlorine atoms, they are sufficiently reactive to allow some reaction to 

occur. For example, the radical bromination of 2-methylbutane gives 

more than 90% 2-bromo-2-methylbutane. The reaction requires both 

heat and light to proceed. 

CH 3 

CH 3CHCH 2CH 3 

Br 2 

h , 127 oC 

CH 3 

CH 3CCH 2CH 3 

Br 

90.3% 

CH 3 

CH 3CHCHCH 3 

CH 3 

+ CH 3CHCH 2CH 2Br 

0.2% 

+ 

CH 3 

BrCH 2CHCH 2CH 3 

Br 

9.1% 0.4% 

If you perform the same calculations for bromine as with chlorine, the 

relative reactivities are 1 : 83 : 1640. Thus, radical bromination is 

much more selective for the tertiary position than is chlorination. This 

increased selectivity makes the reaction synthetically useful for the 

preparation of tertiary alkyl bromides. 

Exercise 21.1 

www.ochem4free.com 5 July 2005 

+

Roy Plunkett is the 

inventor of Teflon. See 

Section 0.3, page 000. 


The regioselectivity of chlorine is dependent on the temperature of the 

reaction. The relative rates for chlorination of 2-methylbutane at 

600 o C are 1 : 2.1 : 2.5 rather than the 1 : 2.5 : 4 at 300 o C. Explain this 

observation. 

Exercise 21.2 

Even at relatively high temperatures and in the presence of light, 

neopentane (2,2-dimethylpropane) reacts much faster with chlorine 

than it does with bromine. Explain this observation. 

[SIDEBAR] 

Atmospheric Ozone Depletion 

Seventy-five years ago, refrigerators used toxic and noxious 

gases such as ammonia and sulfur dioxide as refrigerants. If a leak 

developed in a refrigerator, dangerous amounts of these gases escaped 

into the air of the home or workplace. In the 1920s, Roy Plunkett and 

his assistant, Jack Rebok, experimented to find an odorless, tasteless, 

and nontoxic substitute for these substances. 

After a careful survey of the chemical literature, they decided 

that the best possible candidates were the organic compounds that 

contained both chlorine and fluorine. They synthesized a sample of a 

gaseous compound of chlorine and fluorine and placed some of the 

substance, along with a guinea pig, under a bell jar. The guinea pig 

was unharmed. Although this test seems crude by today's 

experimental standards, it was a standard practice then. 

Encouraged by the low toxicity demonstrated by this test, they 

synthesized a variety of these chlorofluorocarbons (CFCs). Further 

tests indicated that these compounds were indeed nontoxic to animals 

and, by inference, nontoxic to humans as well. Du Pont introduced 

these CFCs under the trade name of Freon. 

For a number of years, industry used the CFC chemicals 

widely. Not only were they used as refrigerants, but they were also 

used for such things as propellants in aerosol products and foaming 

agents for foam plastics. As a result of their extensive use, thousands 

of tons of CFCs were introduced into the atmosphere. In the mid- 

1970s, environmental chemists proposed that these otherwise inert 

materials could destroy the stratospheric ozone layer. 

To understand the problem, review the process of ozone 

formation in the upper atmosphere. Incoming ultraviolet radiation 

causes a homolytic bond dissociation in molecular oxygen. 



O 2 

h 

More incoming ultraviolet radiation provides the energy needed by 

each of these oxygen atoms to either react with another oxygen atom 

to reform a molecule of oxygen or to react with a molecule of oxygen to 

produce ozone. 

2 O 

h 

O + 2 O O3 As well as dissociating the molecular oxygen, ultraviolet radiation also 

dissociates molecules of ozone to produce an electronically excited 

oxygen atom and an oxygen molecule. 

O3 

h 

These reactions make up a chain reaction that will continue as long as 

oxygen and ultraviolet radiation are available. The net result of these 

three reactions is the absorption of most of the incoming ultraviolet 

radiation that would otherwise reach earth's surface damaging the 

plant and animal life there. 

Oxygen and ozone are not the only compounds that absorb 

ultraviolet radiation. Two widely used CFCs, CFCl 3 and CF 2 Cl 2 , 

absorb radiation at the same wavelengths as molecular oxygen and 

ozone. When these CFCs absorb ultraviolet radiation, a C—Cl bond 

homolytically cleaves to form a chlorine atom. 

CFCl3 

CF2Cl2 

h 

h 

O2 

CFCl2 

+ 

CF2Cl 

Once formed, the chlorine atom can react with ozone to produce 

ClO and molecular oxygen. The ClO, in turn, reacts with atomic 

oxygen to form a chlorine atom and a molecule of oxygen. 

Cl + 

ClO 

+ 

O3 

O 

Net: + O 

O3 

O 

+ 

+ 

ClO 

Cl 

Cl 

Cl + O2 

2 O2 

These reactions take place more readily than do reactions involving 

just oxygen and ozone. Moreover, the reactions with chlorine take 

+ 


O2


place without the presence of ultraviolet radiation. The net result is a 

catalytic cycle that destroys a molecule of ozone while regenerating 

the chlorine atom. Notice that the oxygen atom reacts with the ClO 

instead of with another oxygen atom to form an oxygen molecule, or 

with an oxygen molecule to form ozone. 

Both chain reactions take place between 25 and 40 km above 

the earth's surface. The low reactivity that makes CFCs so attractive 

for their industrial uses also gives them a long lifetime in the 

atmosphere. Environmental chemists estimate that it will take from 

40 to 150 years for the CFCs to diffuse into the upper atmosphere and 

react there. This means that even if CFCs were immediately removed 

from the marketplace, their concentration in the upper atmosphere 

would continue to increase for a number of years. 

Not all scientists accept that CFCs are responsible for the 

decline of the ozone layer. Some feel that there is not enough data to 

even conclude that there is a genuine loss of the ozone layer. From 

their viewpoint, because the baseline of data covers only a few years, 

there is insufficient data to justify the conclusion that human 

activities are damaging the ozone layer. Perhaps what is happening 

with the ozone is a part of some, as yet unknown, natural cycle. All do 

agree, however, that the loss of the ozone is a potentially serious 

problem and must be closely monitored. 

The chlorine in the CFCs is not the only potential culprit in the 

destruction of the ozone layer in the upper atmosphere. 

Environmental chemists know of other chemical substances that react 

with ozone in similar ways to the CFCs. Two of these are nitrogen 

oxides and hydroxyl radicals. The nitrogen oxides originate in 

automobile exhaust gases and other high temperature processes. The 

hydroxyl radicals form in nature as a result of the homolytic cleavage 

of an H—OH bond of water. 

If human activity is responsible for the decline of the ozone 

layer, it is urgent to understand the extent of the problem and to 

correct it. If the decline of the ozone is a natural process, measures 

must be taken to minimize the damages from the resulting increase in 

UV levels at the earth's surface. Perhaps you could be instrumental in 

solving these problems. 

21.3 Allylic Bromination 

In general, when chemists want to substitute a halogen onto an 

allylic carbon of an alkene, they use a radical halogenation reaction. 

An excellent source of bromine atoms for this reaction is Nbromosuccinimide 

(NBS). Simply dissolve NBS in a nonpolar 

substance, such as CCl 4 , in the presence of light and heat: 



O 

O 

N Br 

h or 

CCl4 

O 

O 

N 

+ Br 

After the bromine atom forms, it abstracts a hydrogen atom from the 

allylic position of an alkene. This abstraction produces a resonancestabilized 

allyl radical and HBr. 

Br 

H 

Allyl radical 

+ HBr 

The HBr then reacts with another molecule of NBS to form Br 2 and 

succinimide. Succinimide is a by-product of the reaction. 

O 

O 

NBS 

N Br 

H Br 

O 

O 

H 

N Br 

Br 

Tautomerize 

O 

O 

O 

O 

N 

H 

N H 

Succinimide 

At this point in the reaction, the reaction mixture contains a low 

concentration of bromine molecules. These bromine molecules react 


The addition of 

bromine to a double 

bond was discussed in 

Section 14.6, page 000. 

The bromonium ion is 

introduced in Section 

14.2, page 000. 


with the allylic radicals to produce the allyl bromide and a bromine 

atom. The new bromine atom can then react with the alkene to form 

another allylic radical. 

Br Br 

Br 

+ Br 

Because the allylic radical reacts with Br 2 instead of NBS to 

form the final product of the reaction, you may be wondering why Br 2 

wasn’t used to begin with instead of NBS? The problem is, the 

bromine would add to the double bond instead of substituting onto the 

allylic carbon. With NBS as the reagent, the addition reaction does not 

occur. 

The addition reaction does not occur with NBS as the reagent 

because the concentration of bromine is too low to have much 

probability of occurring. Recall from Chapter 14 that the first step in 

the addition of bromine to the double bond is the reversible formation 

of a bromonium ion. The next step is an attack of a bromide ion on this 

intermediate. If no bromide ion is nearby, the bromonium ion 

dissociates. Another reason that the addition reaction does not occur is 

that NBS competes with the bromonium ion for bromide ions. Because 

there is a far higher concentration of NBS, most bromide ions in 

solution will find a molecule of NBS before they will find a bromonium 

ion. 

Monitoring the NBS Reaction 

Chemists can easily monitor the progress of an allylic halogenation reaction being run 

in CCl 4 because both the NBS and the by-product, succinimide, are nearly insoluble 

whereas the product is soluble. Furthermore, the NBS is denser than the solvent, so it 

sinks below the solvent whereas succinimide is lighter than the solvent so it floats on 

top of the reaction mixture. The reaction is complete when the NBS on the bottom of 

the reaction mixture disappears. 

The reaction also proceeds well in the presence of a radical 

initiator. Two good radical initiators are benzoyl peroxide and 

azobisisobutyronitrile (AIBN). Both molecules readily form radicals 

that initiate the chain reaction of NBS with an alkene. 



O 

Benzoyl peroxide 

CN 

O 

O 

N N 

O 

CN 

Azobisisobutyronitrile 

CN 

O 

+ N2 

Chemists have extensively studied the mechanism for this 

reaction, but do not yet clearly understand it. For simple cases, the 

mechanism proposed in this section explains the outcome of the 

reaction; but in more complicated cases, it doesn't. Chemists are still 

working to answer the questions that arise, so they can clearly 

understand the mechanism. 

Solved Exercise 21.1 

How many isomeric bromoalkenes are formed from the reaction of 2-pentene 

with NBS? 

Solution 

There are two allylic positions in 2-pentene: one primary at C1 and one 

secondary at C4. The secondary radical is more stable than the primary 

radical, so the secondary radical forms almost exclusively. The resulting 

radical is symmetrical and only one bromoalkene is formed. 

CH3CH2CH CHCH3 

Exercise 21.3 

NBS 

CCl4, 

CH3CHCH CHCH3 

CH3CHCH CHCH3 

Br 

CH3CH 

O 

CHCHCH3 

When 3-phenyl-1-propene is heated with NBS in CCl 4 , it forms two 

products in a 5:1 ratio. The two products are 3-bromo-1-phenyl-1propene 

and 3-bromo-3-phenyl-1-propene. Which of the two products 

forms in the higher yield? Why? 



21.4 Benzylic Bromination 

The hydrogens attached to the carbon in the benzylic position 

of an alkyl benzene react similarly to the hydrogens attached to the 

carbon in the allylic position of an alkene. Both sites readily react in a 

radical halogen substitution reaction. The reaction of NBS with 

toluene produces an excellent yield of benzyl bromide. 

CH 3 

NBS 

CCl4, (88%) 

CH 2Br 

The reaction proceeds via a resonance-stabilized benzylic 

radical in which the electron deficiency spreads over four carbon 

atoms. This resonance stabilization makes the benzylic radical a 

relatively stable species. 

CH2 

CH2 

Following a mechanistic pathway similar to the allylic radical, the 

benzylic radical reacts in a radical chain mechanism resulting in a 

substitution on the benzylic carbon. 

The following additional examples of benzylic substitutions 

react similarly to toluene. Thus, the reaction mechanism is quite 

general for all benzylic substitutions and usually produces a good yield 

of the product. 

CH 2 

CH 3 

NBS 

CCl4, NBS 

CCl4, CH2 

(90%) 

CH 

Br 

(84%) 

CH 2Br 

CH2 


See Section 14.3, page 

000, for more on 

hydrohalogenation 

reactions. 


Synthesis of 1-Bromo-1-phenylethane 

NBS, CCl4 

O 

(PhCO)2 

Ethylbenzene 1-Bromo-1-phenylethane 

(82%) 

In a 25 mL round bottom flask place 5.5 mL of dry carbon tetrachloride and a 

magnetic stir bar. Add 1.17 g (1.1 mmol) of ethylbenzene, 1.78 g (1.0 mmol) of NBS, 

and 0.03 g of benzoyl peroxide. Stir to dissolve the reactants and flush the flask with 

nitrogen. Reflux the solution for 30 minutes. Cool the reaction mixture and filter out 

the insoluble succinimide. Wash the succinimide with two portions of 2 mL of carbon 

tetrachloride. Remove the carbon tetrachloride on a rotary evaporator. Distill the 

residue under reduced pressure. The yield of product is 1.52 g (82%), b.p. 94 o C/16 mm. 

Discussion Questions 

1. Why is this reaction run in a nitrogen atmosphere? What effect might the presence 

of oxygen have on the outcome of the reaction? 

Exercise 21.4 

In Section 21.1, you studied the triphenylmethyl radical. The 

triphenylmethyl radical is stable enough to be isolated and studied. 

Propose an explanation for its stability. 

21.5 Radical Addition to Alkenes 

In the hydrohalogenation reaction, which was discussed in 

Chapter 14, hydrogen adds to the least substituted carbon of a double 

bond, and a halogen adds to the most substituted carbon. This pattern 

of addition follows Markovnikov's rule. However, in the 1920s and 

1930s, as chemists studied the hydrohalogenation reaction, they saw 

that when they reacted HBr with an alkene the reaction did not 

always form a product that followed Markovnikov's rule. In fact, the 

reaction gave variable results. On one occasion, it produced mostly the 

expected Markovnikov product; on another occasion, it produced 

significant amounts of anti-Markovnikov product. 


Br


CH3CH CH2 

HBr 

CH3CH2CH2Br + CH3CHCH3 

Anti-Markovnikov 

product 

Br 

Markovnikov 

product 

Morris S. Karasch of the University of Chicago was able to 

trace the unpredictability of the reaction to the presence of oxygen in 

the reaction mixture. When he excluded oxygen from the reaction 

mixture by using carefully purified reagents, he received an excellent 

yield of the expected Markovnikov product. But when he deliberately 

added oxygen, his product was predominately the anti-Markovnikov 

product. 

CH3CH CH2 

HBr 

O2 

HBr 

CH3CHCH3 

Br 

(91%) 

CH3CH2CH2Br 

(78%) 

Markovnikov 

product 

Anti-Markovnikov 

product 

The formation of the anti-Markovnikov product in the presence 

of oxygen, a diradical, suggests that the reaction follows a radical 

mechanism. Furthermore, adding a radical initiator, such as benzoyl 

peroxide, to the reaction mixture increases the yield of the anti- 

Markovnikov product in comparison to the yield without the initiator. 

A mixture of propene, HBr, and benzoyl peroxide at –78 o C rapidly 

reacts to produce 1-bromopropane in a 97% yield. The yield without 

the radical initiator is 78%. 

HBr 

CH3CH CH2 CH 

O 

3CH2CH2Br (97%) 

(PhCO) 2 

–78 o C 

The reaction mechanism for this process begins with a 

homolytic cleavage of the benzoyl peroxide to form the benzoyl radical. 

The hydrogen from the HBr then reacts with the benzoyl radical to 

form benzoic acid and a bromine atom. This sequence makes up the 

initiation step of the reaction. 



O 

O 

2 

Benzoyl peroxide Benzoyl radical 

O 

O H Br 

O 

Benzoic acid 

The propagation step follows the sequence shown below: 

OH 

+ Br 

Br 

H Br 

CH3CH CH2 CH3CHCH2Br CH3CH2CH2Br + Br 

The bromine atom reacts with the double bond of propene to form a 1bromo-2-propyl 

radical. The 1-bromo-2-propyl radical reacts with a 

molecule of HBr to give 1-bromopropane and a bromine atom. The 

bromine atom is then available to propagate the chain reaction by 

reacting with the double bond of another propene molecule. Of the 

hydrogen halides, only HBr can form radicals reactive enough to 

undergo anti-Markovnikov addition to the double bond of an alkene. 

Exercise 21.5 

Write the termination steps for the radical addition of HBr to an 

alkene. 

Radical addition reactions to alkenes are regioselective due to 

the stability of the alkyl radical and steric factors. Alkyl radical 

stability follows the same sequence as carbocation stability: allyl, 

benzyl > 3 o > 2 o > 1 o > methyl. However, the difference in stability is 

smaller for radicals than for carbocations, so radical reactions are 

often less selective than reactions with carbocations. Additionally, 

incoming radicals are very sensitive to steric factors, so they attack 

the least hindered carbon of the double bond. 

Although HBr is the only hydrogen halide that forms the anti- 

Markovnikov product in a radical addition reaction to an alkene, there 

are other reagents that also do so. Examples include thiols, 

bromotrichloromethane, chlorosilanes, and even other alkenes. With 

each reagent you must adjust the reaction conditions appropriately to 



generate the radical. These reaction conditions vary from adding a 

radical initiator, such as oxygen or a peroxide, to heating the reaction 

mixture to a high temperature and using ultraviolet radiation. 

CH3CH2C CH2 

CH3 

CH3CH2SH 

Cl3CBr 

O 

SCH2CH3 

1-Ethylthio-2-methylcyclohexane 

(91%) 

Br 

CH3CH2CCH2CCl3 

CH3 

(CH3CO)2 

h 

3-Bromo-1,1,1-trichloro-3-methylpentane 

(85%) 

CH3(CH2) 4CH CH2 CH3SiCl2H O 

CH3(CH2) 5CH2SiCl2CH3 (CH3CO) 2 

Dichloroheptylmethyl silane 

(98%) 

An important industrial process is the radical formation of long 

chains of carbon—carbon bonds. These long chains of carbon—carbon 

bonds form when alkenes react in the presence of a radical initiator. 

The compound that forms is called a polymer. The plastics and fibers 

that you use in your daily life are polymers. Chapter 22 discusses 

polymers in greater depth. 

Solved Exercise 21.2 

O 

(PhCO) 2 

n 

Polystyrene 



Predict the major product of the following reaction and write a mechanism to 

explain its formation. 

HBr 

O 

(PhCO)2 

Solution 

The product of this reaction is 1-bromo-2-phenylethane. 

HBr 

O 

(PhCO)2 

The first step in the mechanism forms a bromine atom. This step initiates the 

radical chain reaction. 

O 

PhCO 

O 

OCPh 

O 

PhCO 

H Br 

O 

Br 

PhCOH 

In the propagation steps, the bromine atom reacts with the double bond to 

form a benzylic radical. The benzylic radical then reacts with HBr to form the 

product and a bromine atom ready to begin another propagation sequence. 

Exercise 21.6 

Br 

Predict the major products of each of the following reactions. 

a) 

Br 

+ 

H 

H 

Br 

Br 

Br 

+ Br 


In the process of 

autooxidation, a 

molecule spontaneously 

reacts with oxygen. 


b) 

c) 

d) 

e) 

(CH 3) 3CCH CH 2 

CH 3SH 

h 

(CH 3) 3CCH CH 2 

C(CH 3) 3 

Sample Solution 

b) 

CH 3SH 

h 

CH 3SH 

h 

Cl 3CBr 

O 

(PhCO) 2 

h 

HBr 

H2O2, warm 

Cl 3SiH 

O 

(CH 3CO) 2 

21.6 Radical Oxidations 

SCH 3 

When an organic compound oxidizes, a new carbon—oxygen 

bond forms, or the oxidizing agent removes hydrogen from two 

adjacent carbons to form a new π bond. Many organic molecules 

oxidize spontaneously in the presence of oxygen in a process called 

autooxidation. Light can also catalyze the autooxidation reaction of 

some molecules, so those organic compounds must be stored in dark 

colored bottles and cans. The more stable the radical, the more readily 

autooxidation occurs to form that radical. Compounds especially 


Fatty acids are long 

chain carboxylic acids. 

Generally, fatty acids 

contain at least 12 

carbon atoms. 


susceptible to autooxidation are benzylic and allylic compounds, as 

well as ethers, amines, and similar compounds containing 

heteroatoms. All these compounds readily form radicals. 

Autooxidation occurs so readily with ethers that ether solvents 

that are stored for a long time oxidize to form some amount of 

hydroperoxide products. Hydroperoxide products are unstable and 

decompose violently when heated. Therefore, when chemists want to 

use diethyl ether from old bottles in the lab, they must first remove 

the hydroperoxide to prevent possible explosions. 

CH 3CH 2OCH 2CH 3 

O 2 

CH 3CH 2OCHCH 3 

OOH 

Diethyl ether Hydroperoxide of diethyl ether 

The mechanism for the autooxidation of organic molecules is 

not known for sure. An oxygen molecule is believed to abstract a 

hydrogen from the carbon bearing the ether oxygen, which produces a 

radical and a hydroperoxide radical. These two radicals then react 

with each other to form the ether hydroperoxide. 

H 


O O 

CH 3CH 2O CHCH 3 

HO O 

OOH 


Autooxidation is a process that has practical value. For 

example, autooxidation accounts for the drying of many oil based 

paints. The most commonly used oil in these oil based paints is linseed 

oil, which contains a mixture of esters of various long chain carboxylic 

acids, called fatty acids. Approximately 90% of the fatty acids in 

linseed oil contain one or more double bonds. The allylic position of 

these double bonds readily forms a radical. These radicals dimerize, 

then trimerize, then tetramerize, etc., ultimately producing high 

molecular weight polymers. Linoleic acid, a fatty acid, is a major 

constituent of linseed oil. As the linoleic acid reacts in an 

autooxidation reaction and forms a polymer, the paint dries. 


Antioxidants are 

compounds that react 

more readily with 

molecular oxygen than 

the molecules in food or 

other sensitive products 

react with oxygen. 


H 

C C 

H H 

C C 

H 

O 

CH3(CH2)4 C (CH2)7COH 

O O 

H H 

H 

C C 

H H 

C C 

H 

O 

Linoleic acid 

CH3(CH2)4 C (CH2)7COH 

Polymer 

H H 

C C 

R R 

From another 

linoleic acid 

H 

C 

Repeat reaction 

CH3(CH2)4 

with double bond H 

H H 

C C 

H C 

C H 

C 

H 

O 

(CH2)7COH 

C R 

Autooxidation is a process that has practical consequences, too. 

For example, the main causes of food spoilage are microbial (mold, and 

bacteria) and autooxidation. Since autooxidation takes place in so 

many foods, food processors add antioxidants to the food or 

packaging materials. Using an antioxidant gives the food a longer 

shelf life by preserving the taste and nutrient levels. The antioxidants 

that food processors commonly use are BHA and BHT. BHA is an 

acronym for Butylated Hydroxy Anisole and is a mixture of 2- and 3tert-butyl-4-methoxyphenol. 

BHT is an acronym for Butylated 

Hydroxy Toluene and is 2,6-di-tert-butyl-4-methylphenol. 

OH 

OCH3 

2-tert-Butyl-4methoxyphenol 

CH3 

C 

CH3 

CH3 

OH 

OCH3 

C 

CH3 

3-tert-Butyl-4methoxyphenol 

CH3 

CH3 

"Butylated Hydroxy Anisole" "Butylated Hydroxy Toluene" 

BHA 

BHT 

R 

CH3 

CH3 

H 

CH3 

C 

OH 

CH3 

CH3 

C 

CH3 

CH3 

2,6-di-tert-Butyl-4-methylphenol 



Both BHA and BHT form resonance-stabilized phenoxy radicals with 

oxygen or other radicals. These phenoxy radicals then react with a 

hydroperoxide radical to form a dienone. The structures below are 

generalized for both BHA and BHT, as the mechanisms for both are 

identical. 

R 

R 

O 

O 

R 

HOO R 

H O 

R 

R 

H 

O O 

O O 

R 

R 

Oxidations with molecular oxygen are seldom used in 

laboratory syntheses, but they are used extensively in industry. For 

example, acetic acid is made industrially via the oxidation of butane 

with oxygen in the presence of an initiator. 

Exercise 21.7 

CH 3CH 2CH 2CH 3 

R 

O 

R 

R 

R 

O2 initiator CH O 

3COH 

The commercial synthesis of BHA involves p-methoxyphenol and 2methylpropene. 

Describe the laboratory method used to synthesize 

BHA. BHT is prepared in a similar fashion, but the reaction is much 

more regioselective than the synthesis of BHA. Why? 

Exercise 21.8 


R 

R 

O 

R 

O 

R 

R 

R

Radical reductions are 

reduction reactions 

that proceed via a 

radical mechanism. 


000, for the trans 

addition of hydrogen to 

an alkyne. 

In a dissolving metal 

reaction, the reaction 

proceeds by using a 

metal as an electron 

source to effect the 


Would you expect the autooxidation of ethyl ether to occur more 

readily than the autooxidation of isopropyl ether? Explain your 

answer. 

21.7 Radical Reductions 

A radical reduction reaction generally involves the addition 

of hydrogen to a π bond. The reduction reaction types that you studied 

previously were catalytic reductions, some ionic reductions, and one 

radical reduction. The radical reduction was the trans addition of 

hydrogen to an alkyne. 

R C C R' 

Na 

NH3, -33oC 

H 

R 

C C 

To proceed, a radical reduction reaction requires a source of 

electrons. Some of the older methods for reducing organic molecules 

used metals (especially alkali metals) as an electron source. To make 

the electrons from the alkali metal available, the reaction needs some 

solvent that dissolves the metal. The solvents most commonly used 

are alcohols and liquid ammonia. Because the reaction proceeds by 

dissolving the metal in the solvent, chemists call this reaction a 

dissolving metal reaction. However, since the introduction of metal 

hydrides, dissolving metal reactions are not used very often anymore. 

reaction. The substrates for a large number of dissolving metal reactions 

are carbonyl groups. 


000, and Section 8.5, 

page 000, for more 

about metal hydrides. 

O 

Na 

R' 

H 

O Na H OCH(CH3)2 

(CH3)2CHO H 

The reaction begins when an electron from the metal transfers to the 

carbonyl group forming a radical anion. A hydrogen from the alcohol 

solvent protonates the radical anion producing a neutral radical 

OH 

H 

Na 

OH 

OH 



intermediate. Then an electron from another metal atom transfers to 

the neutral radical intermediate to form a strongly basic carbanion. A 

second protonation of the carbanion produces the alcohol. 

The dissolving metal reduction reaction readily reduces 

aldehydes, ketones, and esters. The reaction does not work with 

isolated double bonds, but it does reduce triple bonds, conjugated 

double bonds, and conjugated carbonyl systems. An alkyne forms a 

trans alkene. A conjugated diene, by a radical 1,4-addition reaction, 

forms an alkene. The double bond is reduced in a conjugated carbonyl. 

CH3(CH2)5COCH2CH3 

CH 3 

Exercise 21.9 

O 

O 

O 

Na, EtOH 

Li, EtOH 

NH 3, -33 oC 

Na, EtOH 

Cycloheptanol 

(83%) 

CH3(CH2)5CH2OH 

H 3O 

OH 

H 

+ CH3CH2OH 

1-Heptanol Ethanol 

(77%) 

CH 3 

H 

trans-6-Methylbicyclo[4.4.0]decane-3-one 

(95%) 

In the last reaction above, the carbonyl group was not reduced, but in 

the first example above, the carbonyl group was reduced. Provide an 

explanation for this difference. (Hint: Temperature is not the 

important difference.) 

All the reduction reactions that you have looked at to this point 

included a proton source. If the reaction mixture provides no source of 

protons, or if the radical anion is stabilized, dimerization of the 

substrate occurs. To get the best yield of the desired dimer, chemists 

choose a metal that has two or more electrons to donate, such as 

magnesium, zinc, or aluminum. These metals react most effectively 


O

The pinacol reaction is 

a dimerization of a 

ketone using a 

magnesium amalgam. 

An acyloin 

condensation is the 

dimerization of an 

ester using sodium. 


when they are used in the form of a mercury alloy called an amalgam. 

The synthesis of pinacol is an example of a radical dimerization 

reaction: 

O 

CH3CCH3 

Mg(Hg) 

C6H6, 

CH3 

CH3 

C 

O 

Mg 2 

O 

C 

CH3 

CH3 

CH3 

Mg 

H2O H H OH2 

2 

CH3 

CH3 CH3 

C 

C 

O O 

CH3 CH3 

C C CH3 

OH OH 

CH3 

2,3-Dimethylbutane-2,3-diol 

(Pinacol) 

(55%) 

The reaction starts with acetone and forms a radical anion. The 

radical anion then dimerizes to form a vicinal diol. This reaction is 

sometimes called the pinacol reaction after the common name of the 

product, 2,3-dimethylbutane-2,3-diol. 

Esters undergo a dimerization reduction reaction that is called 

an acyloin condensation. This name comes from the common name 

of the simplest reaction product, acyloin, which is an α-hydroxy 

ketone. The initial product of the reaction is the disodium salt of an 

enediol. To form the acyloin product, the disodium salt hydrolyzes in 

aqueous acid. 


The Birch reduction is 

the reaction of an 

aromatic ring with 

sodium metal forming 

a cyclohexadiene. 


H 

O 

CH3COMe 

O O 

CH3C CCH3 

H 

O O 

H2O 

CH3C CCH3 

Na 

O 

CH3COMe 

O O 

CH3C CCH3 

O 

Tautomerize CH3CHCCH3 

OH 

Acyloin 

O 

CH3COMe 

CH3C CCH3 

Na 

O O 

OMe 

O O 

OMe 

CH3C CCH3 

The acyloin condensation is one of the best methods to use 

when synthesizing medium to large sized rings. The synthesis begins 

with the metal and solvent. Then the diester substrate is added very 

slowly. This procedure allows the two ends of the substrate to find 

each other in an intramolecular reaction, while it suppresses any 

intermolecular reactions. 

COOMe Na 

COOMe 

Xylene, 

2-Hydroxytetradecanone 

(48%) 

Using sodium or lithium metal with a benzene ring forms a 

cyclohexadiene. This reaction is called the Birch reduction. When 

running the Birch reduction in liquid ammonia with two equivalents 

of an alcohol, the reaction produces a 1,4-cyclohexadiene ring. 

O 

Na 

OH 



Na, EtOH 

NH 3, -35 oC 

1,4-Cyclohexadiene 

(88%) 

The mechanism for the Birch reduction reaction is as follows: 

H 

H 

Na 

H 

H 

H 

OEt 

H 

H H 

H 

H H 

EtO H 

The mechanism begins with an electron transfer from the alkali metal 

to the aromatic ring, which forms a radical anion. A hydrogen from the 

alcohol then protonates the radical anion, followed by another electron 

transfer to the radical from the metal. The final step in the reaction 

sequence is another protonation of the anion by another molecule of 

alcohol. 

When a Birch reduction occurs on a benzene ring with an 

electron-donating substituent, the substituent destabilizes the radical 

anion intermediate. As a result of this destabilization, the substituent 

usually ends up on a carbon of the double bond in the product. 

CH 3 

Na, EtOH 

NH 3, -33 oC 

H 

H 

CH 3 

Na 

1-Methyl-1,4-cyclohexadiene 

(84%) 


H 

H


If the substituent is an electron-withdrawing group, the substituent 

stabilizes the radical anion intermediate. In this case, the substituent 

ends up on one of the sp 3 carbons in the product. 

COOH COOH 

Na, EtOH 

NH3, -33oC 

1,4-Cyclohexadiene-3-carboxylic acid 

(96%) 

Synthesis of 1-Methoxy-1,4-cyclohexadiene 

Anisole 

OCH3 

Li, NH3 

(CH3)3COH 

OCH3 

1-Methoxy-1,4-cyclohexadiene 

(75%) 

Fit a 500 mL round bottom flask with an inlet tube, mechanical stirrer, and a dry ice 

condenser. Place 15 mL of dry tetrahydrofuran, 25 mL of tert-butyl alcohol, and 5 g 

(0.047 mol) of anisole into the flask. Fill the trap of the condenser with dry ice and 

acetone. Dry the ammonia by transferring 160 mL into a flask cooled in a dry 

ice/acetone bath. Add about 0.5 g of sodium to the ammonia and stir about 15-20 

minutes. Warm the flask and distill about 150 mL of the dried liquid ammonia into 

the round bottom flask. Cautiously, add 1.15 g (0.38 mol) of lithium with stirring. As 

the lithium dissolves, the solution will become deep blue. Reflux for 1 hour. 

Cautiously add methanol dropwise to discharge the blue color. About 10 mL of 

methanol is required. Then add 75 mL of water. Remove the dry ice condenser and let 

the reaction mixture stand in the hood overnight to evaporate the excess ammonia. If 

any lithium salts are not dissolved, add enough water to dissolve them. Extract the 

reaction mixture with three 10 mL portions of petroleum ether (b.p. 30-40 o C). 

Combine the petroleum ether extracts and wash them four times with 10 mL portions 

of water to remove the excess tert-butyl alcohol and methanol. Dry the petroleum 

ether layer over anhydrous magnesium sulfate. Fractionally distill the solution under 

reduced pressure to remove the solvent, then distill the residue. The yield of product 

is 3.9 g (75%), b.p. 40 o C/20 mm. 

Discussion Questions 

1. Commercial anisole is purified by washing with sodium hydroxide, then washing 

with water, followed by distillation. This process removes the phenol from which 

anisole is synthesized. What product is produced by the Birch reduction of phenol? 

2. Using a rotary evaporator to remove the solvent results in a considerably lower 

yield of product. Why? 

Exercise 21.10 



Predict the major products of each of the following reactions. 

a) 

b) 

c) 

d) 

e) 

OCH3 

O 

O 

O O 

2) H3O 

Na, EtOH 

1) Mg(Hg), C6H6 

Li, EtOH 

NH3, –33oC 

CH 3OC(CH 2) 8COCH 3 

O 

CH2CH3 

Sample Solution 

a) 

O 

[Special Topic] 

Li, EtOH 

NH3, –33oC 

Li, EtOH 

Na, EtOH 

Electron Spin Resonance Spectroscopy 

OH 


Electron spin 

resonance looks at 

unpaired electrons in 

ways similar to how 

NMR looks at nuclei. 

Hyperfine splitting is 

analogous to the spinspin 

splitting in NMR. 


The development of the electron spin resonance (ESR) 

technique has enabled chemists to spectroscopically detect radicals. 

ESR is similar to NMR in that ESR is also a type of magnetic 

spectroscopy. Electrons possess a magnetic moment similar to the 

magnetic moments associated with nuclei. Because paired electrons 

have opposite spins, their magnet moments cancel one another. Thus, 

ESR does not detect them. However, when they are unpaired, the 

magnetic moment takes on one of two possible alignments as specified 

by its spin. ESR detects this spin. 

The two possible alignments for the spin of unpaired electrons 

are either in a parallel or an antiparallel direction to the applied 

magnetic field. These alignments are similar to the alignments of the 

nuclei in the magnetic field of an NMR instrument. ESR generally 

requires radio frequencies in the microwave region. For a given 

magnetic field, the frequency for ESR spectroscopy is approximately 

1000 times higher than the frequency for an NMR. 

Similar to an NMR spectrometer, an ESR spectrometer records 

the spectrum with the magnetic field increasing from left to right on 

the graph. Unlike NMR, however, the ESR spectrum records the first 

derivative of the absorption signal rather than the typical absorption 

peak. Recording the first derivative of the absorption provides a 

cleaner spectrum than does an absorption spectrum. 

Absorption curve First derivative curve 

Any nuclei possessing a magnetic moment— 1 H is the most 

common—that are located adjacent to the radical site give rise to 

hyperfine splitting of the peak. If a single hydrogen atom is on the 

carbon that bears an unpaired electron, the signal for that electron 

splits into a doublet. The methyl radical contains a four line spectrum 

as a result of the interaction of the three equivalent hydrogens with 

the unpaired electron. Similar to a methyl signal in NMR, these four 

peaks have a 1:3:3:1 integration ratio. 



23 Gauss 

ESR spectrum of CH3 

The benzene radical anion has a total of seven peaks in its spectrum. 

Benzene's resonance distributes the unpaired electron over the six 

carbon atoms. 

Exercise 21.11 

3.7 Gauss 

ESR spectrum of 

The hyperfine coupling constant for a hydrogen attached to carbon 

bearing an unpaired electron is about 20-25 gauss. 

a) The hyperfine coupling constant for a hydrogen atom is 500 

gauss. Why is this value so much larger than for the methyl 

radical? 

b) Why is the hyperfine coupling constant for the benzene 

radical anion so much smaller than for the methyl radical? 

c) Sketch the appearance of the ESR spectrum for the 2,2dimethylpropyl 

radical. 



Key Ideas from Chapter 21 

❑ A radical is a chemical species that has a single unpaired 

electron in one of its orbitals. 

❑ Structurally, radicals and carbocations are similar because 

both have a planar trigonal sp 2 geometry. However, radicals 

contain a single electron in the unhybridized p orbital, whereas 

the unhybridized p orbital of carbocations is empty. 

❑ A nonbonding pair of electrons on an atom adjacent to the 

radical site causes some electron repulsion and tends to give 

the radical a more tetrahedral (sp 3 ) geometry. 

❑ An alkane reacts with chlorine or bromine in the presence of 

either heat or light to form alkyl chlorides or alkyl bromides. 

❑ The halogenation of an alkane is a radical chain reaction. The 

initiation step is the homolytic cleavage of the halogen. In the 

chain propagation steps that follow, each time the reaction uses 

a halogen, it produces a new one. The reaction terminates 

when two radicals react together. 

❑ For chlorination, the reactivity of various sites in an alkane is 

tertiary > secondary > primary. The relative rates of reaction 

are 4 : 2.5 : 1. 

❑ A radical bromination is a slower reaction than a radical 

chlorination. Because of this slowness, a radical bromination is 

much more selective than a radical chlorination. The reactivity 

of the various sites in an alkane is tertiary > secondary > 

primary. The relative rates of reaction are 1640 : 83 : 1. 

❑ N-Bromosuccinimide is an excellent source of bromine in low 

concentrations. The low concentrations of bromine react at the 

allylic position of an alkene. 

❑ Benzylic and allylic radicals readily form because both are 

resonance-stabilized. 

❑ The anti-Markovnikov addition of HBr proceeds via a radical 

intermediate. The amount of the anti-Markovnikov product 

increases as the amount of radical initiator such (e.g. peroxide) 

increases. 



❑ Autooxidation is the reaction of some substrate with oxygen. 

Autooxidation occurs when a substrate can form some 

particularly stable radical. Light usually accelerates the 

reaction. 

❑ A dissolving metal reduction uses a reactive metal as a source 

of electrons. The most common metals are sodium, lithium, 

magnesium, and zinc. 

❑ Dissolving metal reductions reduce aldehydes, ketones, and 

esters to alcohols. 

❑ When no source of protons is available, a dissolving metal 

reduction causes dimerization. Examples of this dimerization 

are the pinacol reaction and the acyloin condensation. 

❑ The Birch reduction adds two hydrogens to positions 1 and 4 on 

the benzene ring. The product of a Birch reduction is a 1,4cyclohexadiene. 

❑ In the Birch reduction, electron-donating substituents 

destabilize the radical anion intermediate. This destabilization 

directs the substituent to the sp 2 carbon of the product. An 

electron-withdrawing substituent stabilizes the intermediate, 

so the substituent ends up on the sp 3 carbon.

Free Radical Reactions

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