Richard F. Daley and Sally J. Daley
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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
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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.
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Chapter 21
Radical Reactions
Chapter Outline
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 “antiMarkovnikov” product
Understand the autooxidation processes
Be able to use radical reductions in synthesis
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Whoever in discussion adduces authority uses not
intellect but rather memory.
—Leonardo da Vinci
A
Radical polymerization
is discussed in several
sections in Chapter 22.
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 Ph3COOCPh3, or with iodine to form
Ph3CI. 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
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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 sp2 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 sp3 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
Less repulsion
•
••
•
C
X
C
••
X
Radical stability is also similar to carbocation stability. Thus,
the order of stability for radicals is 3o > 2o > 1o > 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.
C
•
C
•
C
C
C
C
Allylic radical
CH2•
CH2
CH2
•
•
CH2
•
Benzylic radical
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An inhibitor is some
chemical species, either
a molecule or radical,
which is particularly
reactive with a radical.
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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.
Cl2
CH4
CH3Cl
h
+
CH2Cl2
+
CHCl3
+
CCl4
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
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 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.
Cl
The single barbed
mechanism arrows in
this reaction indicate
the movement of single
electrons.
Cl
h or
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• +
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The resulting methyl radical, which also is very electrophilic, then
removes a chlorine atom from a chlorine molecule.
Cl
Each step in a chain
reaction produces a
chemical species that
initiates another step
in the reaction.
Initiation forms the
initial radicals to begin
a chain reaction.
Propagation continues
the chain reaction.
Termination stops the
chain reaction.
Cl
CH3•
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.
A chain reaction mechanism consists of three categories of
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
Cl2
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.
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When 2-methylbutane reacts at 300oC with one mole of
chlorine, the result is a mixture of four monochlorinated products in
the following relative amounts.
CH3
CH3CHCH2CH3
CH3
Cl2
300oC
CH3
ClCH2CHCH2CH3 + CH3CCH2CH3 +
Cl
22%
2-Methylbutane
33.3%
CH3
CH3
CH3CHCHCH3 + CH3CHCH2CH2Cl
Cl
28%
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
CH3
CH3CHCH2CH3
22% of the product
28% of the product
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.
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Secondary 28/2
Primary = 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.
CH3
CH3CHCH2CH3
CH3
Br2
h , 127oC
CH3
CH3CCH2CH3 + CH3CHCH2CH2Br
Br
90.3%
+
0.2%
CH3
CH3CHCHCH3
Br
9.1%
CH3
+
BrCH2CHCH2CH3
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
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The regioselectivity of chlorine is dependent on the temperature of the
reaction. The relative rates for chlorination of 2-methylbutane at
600oC are 1 : 2.1 : 2.5 rather than the 1 : 2.5 : 4 at 300oC. 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
Roy Plunkett is the
inventor of Teflon. See
Section 0.3, page 000.
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 mid1970s, 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.
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h
O2
2 O•
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.
O2 + O •
h
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.
h
O3
O2 + O •
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, CFCl3 and CF2Cl2,
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.
h
CFCl3
CF2Cl2
h
CFCl2 • + Cl •
CF2Cl • + Cl •
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.
Net:
Cl • + O3
ClO + O2
ClO + O •
Cl • + O2
O3 + O •
2 O2
These reactions take place more readily than do reactions involving
just oxygen and ozone. Moreover, the reactions with chlorine take
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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 CCl4, in the presence of light and heat:
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O
O
N
Br
h or
N • + Br •
CCl4
O
O
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 •
HBr
H
•
Allyl radical
The HBr then reacts with another molecule of NBS to form Br2 and
succinimide. Succinimide is a by-product of the reaction.
••
••
O••
O
H
•
•
••
H
••
N
O ••
••
Br
•
Br
•
••
•
•
••
N
Br
••
O
H
••
•
•
Br
••
N ••
O••
O••
••
••
NBS
Tautomerize
••
O••
••
N
H
O••
••
Succinimide
At this point in the reaction, the reaction mixture contains a low
concentration of bromine molecules. These bromine molecules react
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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
•
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.
+
Br•
Because the allylic radical reacts with Br2 instead of NBS to
form the final product of the reaction, you may be wondering why Br2
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 CCl4 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.
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O
O
O
O•
O
O
Benzoyl peroxide
N
CN
N
+
CN
N2
CN
Azobisisobutyronitrile
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
NBS
CCl4,
•
CH3CHCH
CHCH3
CH3CHCH
•
CH3CH CHCHCH3
CHCH3
Br
Exercise 21.3
When 3-phenyl-1-propene is heated with NBS in CCl4, 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?
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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.
CH3
CH2Br
NBS
CCl4,
(88%)
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 •
•
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.
NBS
CCl4,
CH2
CH
Br
(84%)
CH2Br
CH3
NBS
CCl4,
(90%)
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Synthesis of 1-Bromo-1-phenylethane
Br
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. 94oC/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
See Section 14.3, page
000, for more on
hydrohalogenation
reactions.
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.
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CH3CH
CH2
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CH3CH2CH2Br
Anti-Markovnikov
product
+
CH3CHCH3
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.
HBr
CH3CH
CH2
CH3CHCH3
Br
(91%)
Markovnikov
product
HBr
O2
CH3CH2CH2Br Anti-Markovnikov
product
(78%)
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 antiMarkovnikov product in comparison to the yield without the initiator.
A mixture of propene, HBr, and benzoyl peroxide at –78oC rapidly
reacts to produce 1-bromopropane in a 97% yield. The yield without
the radical initiator is 78%.
CH3CH
CH2
HBr
O
CH3CH2CH2Br
(97%)
(PhCO)2
o
–78 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.
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O
Daley & Daley
O
O
O•
2
Benzoyl peroxide
Benzoyl radical
H
Br
O
OH
+
Br •
Benzoic acid
The propagation step follows the sequence shown below:
CH3CH
CH2
Br•
•
CH3CHCH2Br
H
Br
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 > 3o > 2o > 1o > 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 antiMarkovnikov 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
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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.
CH3CH2SH
SCH2CH3
1-Ethylthio-2-methylcyclohexane
(91%)
Br
CH3CH2C
CH2
CH3
Cl3CBr
O
CH3CH2CCH2CCl3
CH3
(CH3CO)2
h
3-Bromo-1,1,1-trichloro-3-methylpentane
(85%)
CH3(CH2)4CH
CH2
CH3SiCl2H
O
(CH3CO)2
CH3(CH2)5CH2SiCl2CH3
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.
O
(PhCO)2
n
Polystyrene
Solved Exercise 21.2
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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.
Br
HBr
O
(PhCO)2
The first step in the mechanism forms a bromine atom. This step initiates the
radical chain reaction.
O
O
O
PhCO
OCPh
PhCO •
H
Br
O
PhCOH + Br •
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.
•
•
Br
Br
H
Br
H
Br
+ Br •
Exercise 21.6
Predict the major products of each of the following reactions.
a)
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(CH3)3CCH
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Cl3CBr
O
CH2
(PhCO)2
h
b)
CH3SH
h
c)
HBr
H2O2, warm
d)
(CH3)3CCH
Cl3SiH
O
CH2
(CH3CO)2
e)
CH3SH
h
C(CH3)3
Sample Solution
b)
CH3SH
h
SCH3
21.6 Radical Oxidations
In the process of
autooxidation, a
molecule spontaneously
reacts with oxygen.
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
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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.
CH3CH2OCH2CH3
O2
CH3CH2OCHCH3
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
•
CH3CH2OCHCH3
••
••
••
••
O
O•
••
CH3CH2••
O
•
CHCH3
••
HO
••
••
O•
••
OOH
CH3CH2OCHCH3
Fatty acids are long
chain carboxylic acids.
Generally, fatty acids
contain at least 12
carbon atoms.
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.
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H
H H
C
C
CH3(CH2)4
O
H
C
(CH2)7COH
H
H
••
O•
••
Daley & Daley
O
C
C
H
••
•
••
1114
H H
C
Linoleic acid
C
CH3(CH2)4
H
C
•
O
C
(CH2)7COH
C
H
H
H
C
C
R
R
From another
linoleic acid
H
H H
C
Polymer
C
H
Repeat reaction
CH3(CH2)4
H
with double bond
•
C
H
C
(CH2)7COH
C
C
R
O
C
H
R
Antioxidants are
compounds that react
more readily with
molecular oxygen than
the molecules in food or
other sensitive products
react with oxygen.
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
CH3
C
OH
CH3
CH3
CH3
CH3
CH3
C
OCH3
2-tert-Butyl-4methoxyphenol
3-tert-Butyl-4methoxyphenol
C
CH3
C
CH3
CH3
CH3
CH3
CH3
OCH3
OH
"Butylated Hydroxy Anisole"
BHA
CH3
2,6-di-tert-Butyl-4-methylphenol
"Butylated Hydroxy Toluene"
BHT
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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.
O
H
R
O
R
•
O
••
•
•
••
O•
O
•
R
R
R
•
R
R
R
O
O
R
R
R
H
••
••
••
••
O
R
••
O
R
O•
R
R
•
•
HOO
R
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.
CH3CH2CH2CH3
O2
initiator
O
CH3COH
Exercise 21.7
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
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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
Radical reductions are
reduction reactions
that proceed via a
radical mechanism.
See Section 14.7, page
000, for the trans
addition of hydrogen to
an alkyne.
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
In a dissolving metal
reaction, the reaction
proceeds by using a
metal as an electron
source to effect the
reaction.
See Section 7.7, page
000, and Section 8.5,
page 000, for more
about metal hydrides.
C
C
Na
R'
NH3, -33oC
H
R'
C
R
C
H
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.
The substrates for a large number of dissolving metal reactions
are carbonyl groups.
••
• •
•
•
O
Na •
O•• Na
••
•
••
H
OCH(CH3)2
OH
••
•
Na •
•
•
••
OH
•
•
(CH3)2CHO
H
••
OH
•
•
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
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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.
O
OH
Na, EtOH
H
Cycloheptanol
(83%)
O
CH3(CH2)5COCH2CH3
Na, EtOH
CH3(CH2)5CH2OH + CH3CH2OH
1-Heptanol
(77%)
Ethanol
CH3
CH3
Li, EtOH
NH3,
H3O
-33oC
O
O
H
trans-6-Methylbicyclo[4.4.0]decane-3-one
(95%)
Exercise 21.9
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
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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
•
C
CH3
CH3 CH3
CH3
••
•
•
O
••
•
•
Mg2
••
O
••
•
CH3
C
CH3
C
•
•
C
•
•
O
••
•
•
CH3
•
•
O
••
Mg2
H2O
••
H
H
OH2
••
CH3 CH3
CH3
C
•
•
OH
••
C
•
•
CH3
OH
••
2,3-Dimethylbutane-2,3-diol
(Pinacol)
(55%)
The pinacol reaction is
a dimerization of a
ketone using a
magnesium amalgam.
An acyloin
condensation is the
dimerization of an
ester using sodium.
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.
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O
O
O
CH3COMe
Na
O
CH3COMe
CH3COMe
CH3C
OMe
O
CCH3
OMe
Na
O
CH3C
O
O
CCH3
O
CH3C
CCH3
Na
O
CH3C
O
CCH3
H2O
H
H
O
O
CH3C
CCH3
Tautomerize
O
CH3CHCCH3
OH
Acyloin
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.
O
COOMe
Na
COOMe
Xylene,
OH
2-Hydroxytetradecanone
(48%)
The Birch reduction is
the reaction of an
aromatic ring with
sodium metal forming
a cyclohexadiene.
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.
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Na, EtOH
NH3, -35oC
1,4-Cyclohexadiene
(88%)
The mechanism for the Birch reduction reaction is as follows:
H
H
H
H
••
Na •
H
H
H
OEt
•
•
H
Na •
H
H
H
EtO
H
H
••
H
H
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.
CH3
Na, EtOH
CH3
NH3, -33oC
1-Methyl-1,4-cyclohexadiene
(84%)
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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 sp3 carbons in the product.
COOH
COOH
Na, EtOH
NH3, -33oC
1,4-Cyclohexadiene-3-carboxylic acid
(96%)
Synthesis of 1-Methoxy-1,4-cyclohexadiene
OCH3
OCH3
Li, NH3
(CH3)3COH
Anisole
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. 30-40oC).
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. 40oC/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
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Predict the major products of each of the following reactions.
a)
O
Na, EtOH
b)
O
1) Mg(Hg), C6H6
2) H3O
c)
OCH3
Li, EtOH
NH3, –33oC
d)
O
O
CH3OC(CH2)8COCH3
Li, EtOH
e)
O
Li, EtOH
NH3, –33oC
CH2CH3
Sample Solution
a)
O
OH
Na, EtOH
[Special Topic]
Electron Spin Resonance Spectroscopy
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Electron spin
resonance looks at
unpaired electrons in
ways similar to how
NMR looks at nuclei.
1123
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
Hyperfine splitting is
analogous to the spinspin splitting in NMR.
Daley & Daley
First derivative curve
Any nuclei possessing a magnetic moment—1H 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.
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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.
3.7 Gauss
ESR spectrum of
Exercise 21.11
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.
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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 sp2 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 (sp3) 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.
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❑
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 sp2 carbon of the product. An
electron-withdrawing substituent stabilizes the intermediate,
so the substituent ends up on the sp3 carbon.
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5 July 2005