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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
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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.
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Organic Chemistry - Ch 21 1093 Daley & Daley
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
9 Understand the structure of a radical 9 Know the distribution
of the halogens in a radical halogenation of
an alkane
9 Recognize that radicals at the allylic and benzylic positions
are more stable than alkyl radicals
9 Know why a radical addition to an alkene leads to an
anti-Markovnikov product
9 Understand the autooxidation processes 9 Be able to use
radical reductions in synthesis
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Organic Chemistry - Ch 21 1094 Daley & Daley
Whoever in discussion adduces authority uses not intellect but
rather memory.
Leonardo da Vinci
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.
Radical polymerization is discussed in several sections in
Chapter 22.
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.
AgPh3CCPh3Ph3CCl
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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Organic Chemistry - Ch 21 1095 Daley & Daley
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.
CC
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.
C X
Less repulsion
CX
Repulsion
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 CC C C Allylic radical
CH2CH2CH2CH2
Benzylic radical
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Organic Chemistry - Ch 21 1096 Daley & Daley
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.
An inhibitor is some chemical species, either a molecule or
radical, which is particularly reactive with a radical.
21.2 Halogenation of Alkanes Alkanes react with chlorine in the
presence of ultraviolet light (represented as h) 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.
CH4 CH3Cl CH2Cl2 CHCl3 CCl4+ + +Cl2h
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. 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 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 Cl 2 Clh or
H = 58 kcal/mol
The single barbed mechanism arrows in this reaction indicate the
movement of single electrons.
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.
+ HClCH3CH3 HCl
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Organic Chemistry - Ch 21 1097 Daley & Daley
The resulting methyl radical, which also is very electrophilic,
then removes a chlorine atom from a chlorine molecule.
CH3ClClCl Cl +CH3
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.
Each step in a chain reaction produces a chemical species that
initiates another step in the 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:
Initiation forms the initial radicals to begin a chain reaction.
Propagation continues the chain reaction. Termination stops the
chain reaction.
CH3ClClCH3
CH3CH3CH3CH3
Cl2ClCl
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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Organic Chemistry - Ch 21 1098 Daley & Daley
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.
Cl2300oC
CH3CHCH2CH3
CH3
ClCH2CHCH2CH3
CH3
CH3CCH2CH3
CH3
Cl
CH3CHCHCH3
CH3
Cl
CH3CHCH2CH2Cl
CH3
+
+
+
33.3% 22%
28% 16.7%
2-Methylbutane
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%.
CH3CHCH2CH3
CH3
50% of the product
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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Organic Chemistry - Ch 21 1099 Daley & Daley
SecondaryPrimary =
28/250/9 = 2.5
TertiaryPrimary =
22/150/9 = 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.
9.1% 0.4%
BrCH2CHCH2CH3
CH3+CH3CHCHCH3
CH3
Br
+CH3CHCH2CH3
CH3
CH3CCH2CH3
CH3
Br
+ CH3CHCH2CH2Br
CH3
90.3% 0.2%
Br2h , 127oC
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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Organic Chemistry - Ch 21 1100 Daley & Daley
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 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.
Roy Plunkett is the inventor of Teflon. See Section 0.3, page
000.
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.
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Organic Chemistry - Ch 21 1101 Daley & Daley
h 2 OO2
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.
h O2 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.
+O3h
OO2 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 CCl bond homolytically cleaves to form a
chlorine atom.
ClCFCl2CFCl3 +h
ClCF2ClCF2Cl2 +h
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: + OO3 2 O2
Cl + O2O+ClO
O2+ClOO3Cl +
These reactions take place more readily than do reactions
involving just oxygen and ozone. Moreover, the reactions with
chlorine take
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Organic Chemistry - Ch 21 1102 Daley & Daley
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 HOH 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 N-bromosuccinimide (NBS). Simply
dissolve NBS in a nonpolar substance, such as CCl4, in the presence
of light and heat:
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Organic Chemistry - Ch 21 1103 Daley & Daley
CCl4h or
+ BrN
O
O
N Br
O
O After the bromine atom forms, it abstracts a hydrogen atom
from the allylic position of an alkene. This abstraction produces a
resonance-stabilized allyl radical and HBr.
HBr
+ HBr
Allyl radical The HBr then reacts with another molecule of NBS
to form Br2 and succinimide. Succinimide is a by-product of the
reaction.
Tautomerize
Br
N H
O
O
H BrN
O
O
H
N Br
O
O
H
N Br
O
O
NBS
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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Organic Chemistry - Ch 21 1104 Daley & Daley
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.
+ BrBrBr 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 wasnt 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 of bromine to a double bond was discussed in
Section 14.6, page 000.
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.
The bromonium ion is introduced in Section 14.2, page 000.
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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Organic Chemistry - Ch 21 1105 Daley & Daley
Azobisisobutyronitrile
Benzoyl peroxide
O
OO
O
+ N2
O
O
CNCN
N N
CN
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.
CCl4, NBS
CH3CHCH CHCH3
Br
CHCHCH3CH3CH
CH3CHCH CHCH3CH3CH2CH CHCH3
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-1-propene and 3-bromo-3-phenyl-1-propene. Which of
the two products forms in the higher yield? Why?
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Organic Chemistry - Ch 21 1106 Daley & Daley
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 CH2BrNBS
(88%)CCl4,
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.
CH2CH2CH2CH2
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.
CH2 CCl4, NBS CH
Br(84%)
CCl4, NBS
(90%)
CH3 CH2Br
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Organic Chemistry - Ch 21 1107 Daley & Daley
Synthesis of 1-Bromo-1-phenylethane
(82%)
Br
(PhCO)2O
NBS, CCl4
Ethylbenzene 1-Bromo-1-phenylethane
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 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.
See Section 14.3, page 000, for more on hydrohalogenation
reactions.
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Organic Chemistry - Ch 21 1108 Daley & Daley
Markovnikov product
Anti-Markovnikov product
CH3CH2CH2Br + CH3CHCH3
Br
HBrCH3CH CH2
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.
(91%)
(78%)
O2HBr
CH3CH2CH2Br
Markovnikov product
Anti-Markovnikov product
CH3CHCH3
BrHBr
CH3CH CH2
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 78oC
rapidly reacts to produce 1-bromopropane in a 97% yield. The yield
without the radical initiator is 78%.
CH3CH CH2 CH3CH2CH2BrHBr
(97%)(PhCO)2
O
78oC 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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Organic Chemistry - Ch 21 1109 Daley & Daley
O
O
2
O
O H Br
O
OH+ Br
Benzoyl peroxide Benzoyl radical
Benzoic acid The propagation step follows the sequence shown
below:
BrCH3CH CH2 CH3CHCH2Br
CH3CH2CH2BrBrH
+ Br
The bromine atom reacts with the double bond of propene to form
a 1-bromo-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 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
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Organic Chemistry - Ch 21 1110 Daley & Daley
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.
(91%)
SCH2CH3
CH3CH2SH
1-Ethylthio-2-methylcyclohexane
h (CH3CO)2
O CH3CH2CCH2CCl3
CH3
Br
CH3CH2C CH2
CH3
Cl3CBr
(85%)3-Bromo-1,1,1-trichloro-3-methylpentane
(98%)(CH3CO)2
O
CH3SiCl2HCH3(CH2)4CH CH2 CH3(CH2)5CH2SiCl2CH3
Dichloroheptylmethyl silane
An important industrial process is the radical formation of long
chains of carboncarbon bonds. These long chains of carboncarbon
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.
(PhCO)2
O
n
Polystyrene Solved Exercise 21.2
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Organic Chemistry - Ch 21 1111 Daley & Daley
Predict the major product of the following reaction and write a
mechanism to explain its formation.
HBr
(PhCO)2
O
Solution The product of this reaction is
1-bromo-2-phenylethane.
BrHBr
(PhCO)2
O
The first step in the mechanism forms a bromine atom. This step
initiates the radical chain reaction.
PhCO
O
OCPh
O
PhCO
O
H Br
PhCOH
O
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.
BrBr
BrH
Br
H
+ Br
Exercise 21.6 Predict the major products of each of the
following reactions. a)
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Organic Chemistry - Ch 21 1112 Daley & Daley
(CH3)3CCH CH2Cl3CBr
(PhCO)2
O
h b)
CH3SHh
c)
H2O2, warm
HBr
d)
(CH3)3CCH CH2
(CH3CO)2
O
Cl3SiH
e)
C(CH3)3
CH3SHh
Sample Solution b)
CH3SHh
SCH3
21.6 Radical Oxidations When an organic compound oxidizes, a new
carbonoxygen bond forms, or the oxidizing agent removes hydrogen
from two adjacent carbons to form a new 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
In the process of autooxidation, a molecule spontaneously reacts
with oxygen.
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Organic Chemistry - Ch 21 1113 Daley & Daley
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.
O2 CH3CH2OCHCH3
OOH
CH3CH2OCH2CH3
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.
CH3CH2OCHCH3
OOH
CH3CH2O CHCH3CH3CH2OCHCH3
H
HO O
O O
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.
Fatty acids are long chain carboxylic acids. Generally, fatty
acids contain at least 12 carbon atoms.
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Organic Chemistry - Ch 21 1114 Daley & Daley
Polymerwith double bond
Repeat reaction
C CC
C C(CH2)7COHCH3(CH2)4
OH H HH
HCC
HH
R
R
From anotherlinoleic acid
Linoleic acid
C CR
HH
R
C CC
C C(CH2)7COHCH3(CH2)4
OH H HH
H
C CC
C C(CH2)7COHCH3(CH2)4
OH H HH
HH
O O
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 3-tert-butyl-4-methoxyphenol. BHT is an acronym for Butylated
Hydroxy Toluene and is 2,6-di-tert-butyl-4-methylphenol.
Antioxidants are compounds that react more readily with
molecular oxygen than the molecules in food or other sensitive
products react with oxygen.
"Butylated Hydroxy Anisole" "Butylated Hydroxy Toluene"
2,6-di-tert-Butyl-4-methylphenol3-tert-Butyl-4-methoxyphenol
2-tert-Butyl-4-methoxyphenol
BHTBHA
OH
OCH3
CCH3
CH3
CH3
OH
OCH3
C
CH3
CH3CH3
OH
CH3
CC
CH3
CH3
CH3CH3
CH3
CH3
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Organic Chemistry - Ch 21 1115 Daley & Daley
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 OH
O O
O
RR
R
H O
RR
R
O
RR
R
O
RR
R
O
RR
R
O
RR
HOO 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.
CH3CH2CH2CH3O2
initiatorCH3COH
O
Exercise 21.7 The commercial synthesis of BHA involves
p-methoxyphenol and 2-methylpropene. 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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Organic Chemistry - Ch 21 1116 Daley & Daley
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 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.
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.
C CR
R'
H
H
NH3, -33oCNa
R C C R'
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.
In a dissolving metal reaction, the reaction proceeds by using a
metal as an electron source to effect the reaction. The substrates
for a large number of dissolving metal reactions
are carbonyl groups. See Section 7.7, page 000, and Section 8.5,
page 000, for more about metal hydrides.
ONa
O Na
H OCH(CH3)2OH
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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Organic Chemistry - Ch 21 1117 Daley & Daley
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.
OH
H
Na, EtOH
(83%)
O
Cycloheptanol
+ CH3CH2OH
(77%)
Na, EtOH
CH3(CH2)5COCH2CH3
O
CH3(CH2)5CH2OH1-Heptanol Ethanol
H3O
(95%)
O
CH3
H
Li, EtOHNH3, -33oC
O
CH3
trans-6-Methylbicyclo[4.4.0]decane-3-one
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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Organic Chemistry - Ch 21 1118 Daley & Daley
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:
(Pinacol)2,3-Dimethylbutane-2,3-diol
C6H6,
(55%)
C C CH3CH3
CH3 CH3
OH OH
HH2O H OH2
Mg2
CC
O O
CH3
CH3
CH3
CH3 C
CH3
CH3
OMg(Hg)CH3CCH3
O
C
CH3
CH3O
Mg2
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.
The pinacol reaction is a dimerization of a ketone using a
magnesium amalgam.
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 -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.
An acyloin condensation is the dimerization of an ester using
sodium.
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Organic Chemistry - Ch 21 1119 Daley & Daley
Acyloin
TautomerizeCH3CHCCH3
OH
O
CH3C CCH3
O O
HH
H2O
CH3C CCH3
O O
CH3C CCH3
O O
NaNa
CH3C CCH3
O O
Na CH3C CCH3
O O
OMeOMe
CH3COMe
O
CH3COMe
O
CH3COMe
O
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
COOMe Na
Xylene,
O
OH2-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.
The Birch reduction is the reaction of an aromatic ring with
sodium metal forming a cyclohexadiene.
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Organic Chemistry - Ch 21 1120 Daley & Daley
NH3, -35oC
Na, EtOH
(88%)1,4-Cyclohexadiene
The mechanism for the Birch reduction reaction is as
follows:
OEtH
H
H
Na
H
H
H
HH
HH
H
HH
H H
Na
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.
CH3 CH3
(84%)
Na, EtOH
NH3, -33oC
1-Methyl-1,4-cyclohexadiene
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Organic Chemistry - Ch 21 1121 Daley & Daley
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
(96%)
Na, EtOHNH3, -33oC
1,4-Cyclohexadiene-3-carboxylic acid
Synthesis of 1-Methoxy-1,4-cyclohexadiene
1-Methoxy-1,4-cyclohexadieneAnisole
OCH3 OCH3Li, NH3
(CH3)3COH
(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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Organic Chemistry - Ch 21 1122 Daley & Daley
Predict the major products of each of the following reactions.
a)
Na, EtOH
O
b)
2) H3O
1) Mg(Hg), C6H6O
c)
Li, EtOH
NH3, 33oC
OCH3
d)
Li, EtOHCH3OC(CH2)8COCH3
O O
e)
Li, EtOH
O
CH2CH3
NH3, 33oC
Sample Solution a)
Na, EtOH
O OH
[Special Topic]
Electron Spin Resonance Spectroscopy
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Organic Chemistry - Ch 21 1123 Daley & Daley
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.
Electron spin resonance looks at unpaired electrons in ways
similar to how NMR looks at nuclei.
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 moment1H is the most commonthat 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.
Hyperfine splitting is analogous to the spin-spin splitting in
NMR.
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Organic Chemistry - Ch 21 1124 Daley & Daley
ESR spectrum of CH3
23 Gauss
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
2,2-dimethylpropyl radical.
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Organic Chemistry - Ch 21 1125 Daley & Daley
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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Organic Chemistry - Ch 21 1126 Daley & Daley
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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
1,4-cyclohexadiene.
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.
Radical ReactionsCopyright 1996-2005 by Richard F. Daley &
Sally J. Daley