Organic Chemistry
Second Edition
David Klein
Chapter 11
Radical Reactions
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Klein, Organic Chemistry 2e
11.1 Free Radicals
• Free radicals form when bonds break homolytically
• Note the single-barbed or fishhook arrow used to show
the electron movement
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11-2
Klein, Organic Chemistry 2e
11.1 Free Radicals
• Recall the orbital hybridization in carbocations and
carbanions
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11-3
Klein, Organic Chemistry 2e
11.1 Free Radicals
• Free radicals can be thought of as sp2 hybridized or
quickly interconverting sp3 hybridized
sp2 hybridized
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11-4
Klein, Organic Chemistry 2e
11.1 Free Radical Stability
• Free radicals do not have a formal charge but are
unstable because of an incomplete octet
• Groups that can push (donate) electrons toward the free
radical will help to stabilize it. WHY? HOW? Consider
hyperconjugation
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11-5
Klein, Organic Chemistry 2e
11.1 Free Radical Stability
• Use arguments that involve hyperconjugation and an
energy diagram to explain the differences in bond
dissociation energy below
• Practice with conceptual checkpoint 11.1
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11-6
Klein, Organic Chemistry 2e
11.1 Free Radical Resonance
• Drawing resonance for free radicals using fishhook
arrows to show electron movement
• Remember, for resonance, the arrows don’t ACTUALLY
show electron movement. WHY?
• Draw the resonance hybrid for an allyl radical
• HOW and WHY does resonance affect the stability of the
free radical?
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11-7
Klein, Organic Chemistry 2e
11.1 Free Radical Resonance
• The benzylic radical is a hybrid that consists of 4
contributors
• Draw the remaining contributors
• Draw the hybrid
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11-8
Klein, Organic Chemistry 2e
11.1 Resonance Stabilization
• How does resonance affect the stability of a radical?
WHY?
• What is a bigger factor, hyperconjugation or resonance?
• Practice with SkillBuilder 11.1
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11-9
Klein, Organic Chemistry 2e
11.1 Resonance Stabilization
• Vinylic free radicals are especially unstable
• No resonance
stabilization
• What type of orbital is the vinylic free radical located in,
and how does that affect stability?
• Practice with SkillBuilder 11.2
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11-10
Klein, Organic Chemistry 2e
11.2 Radical Electron Movement
• Free radical electron movement is quite different from
electron movement in ionic reactions
• For example, free radicals don’t undergo rearrangement
• There are SIX key arrow-pushing patterns that we will
discuss
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11-11
Klein, Organic Chemistry 2e
11.2 Radical Electron Movement
1. Homolytic Cleavage: initiated by light or heat
2. Addition to a pi bond
3. Hydrogen abstraction: NOT the same as proton transfer
4. Halogen abstraction
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11-12
Klein, Organic Chemistry 2e
11.2 Radical Electron Movement
5. Elimination: the radical from the α carbon is pushed
toward the β carbon to eliminate a group on the β
carbon (reverse of addition to a pi bond)
–
The –X group is NOT a leaving group. WHY?
6. Coupling: the reverse of homolytic cleavage
•
Note that radical electron movement generally involves
2 or 3 fishhook arrows
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11-13
Klein, Organic Chemistry 2e
11.2 Radical Electron Movement
•
Note the reversibility of radical processes
•
Practice with SkillBuilder 11.3
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11-14
Klein, Organic Chemistry 2e
11.2 Radical Electron Movement
•
Radical electron movement is generally classified as
either initiation, termination, or propagation
Initiation
Termination
• Termination
occurs when
radicals are
destroyed
• Initiation
occurs
when
radicals are
created
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11-15
Klein, Organic Chemistry 2e
11.2 Radical Electron Movement
• Propagation occurs when radicals are moved from one
location to another
Propagation
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11-16
Klein, Organic Chemistry 2e
11.3 Chlorination of Methane
• Let’s apply our electron-pushing skills to a reaction
• We must consider each pattern for any free radical that
forms during the reaction
• Is homolytic cleavage also likely for CH4?
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11-17
Klein, Organic Chemistry 2e
11.3 Chlorination of Methane
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11-18
Klein, Organic Chemistry 2e
11.3 Chlorination of Methane
• Why do
termination
reactions
happen less
often than
propagation
reactions?
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11-19
Klein, Organic Chemistry 2e
11.3 Chlorination of Methane
• The propagation steps give the net reaction
1. Initiation produces a small amount Cl• radical
2. H abstraction consumes the Cl• radical
3. Cl abstraction generates a Cl• radical, which can go on
to start another H abstraction
• Propagation steps are self-sustaining
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11-20
Klein, Organic Chemistry 2e
11.3 Chlorination of Methane
• Reactions that have self-sustaining propagation steps
are called chain reactions
• Chain reaction: the products from one step are
reactants for a different step in the mechanism
• Polychlorination is difficult to prevent, especially when
an excess of Cl2 is present. WHY?
• Practice with SkillBuilder 11.4
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11-21
Klein, Organic Chemistry 2e
11.3 Chlorination of Methane
• Draw a reasonable mechanism that shows how each of
the products might form in the following reaction
• Which of the products above are major and which are
minor? Are other products also possible?
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11-22
Klein, Organic Chemistry 2e
11.3 Radical Initiators
• An initiator starts a free radical chain reaction
• Which initiator above initiates reactions most readily?
WHY?
• The acyl peroxide will be effective at 80 °C
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11-23
Klein, Organic Chemistry 2e
11.3 Radical Inhibitors
• Inhibitors act in a reaction to scavenge free radicals to
stop chain reaction processes
• Oxygen molecules can exist in the form of a diradical,
which reacts readily with other radicals. Use arrows to
show the process
• How can reaction conditions be modified to stop oxygen
from inhibiting a desired chain reaction?
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11-24
Klein, Organic Chemistry 2e
11.3 Radical Inhibitors
• Hydroquinone is also often used as a radical inhibitor
• Draw in necessary arrows in the reactions above
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11-25
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
• If we want to determine whether a process is product
favored, we must determine the sign (+/-) for ΔG
• In a halogenation reaction, will ΔH or -TΔS have a bigger
effect on the sign of ΔG? WHY?
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11-26
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
• Because the -TΔS term should be close to zero, we can
simplify the free energy equation
• Considering the bonds that break and form in the
reaction, estimate ΔHhalogenation for each of the halogens
in the table below
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11-27
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
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11-28
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
• Fluorination is so exothermic that it is impractical. WHY
does that make it impractical?
• Iodination is nonspontaneous, so it is not product
favored
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11-29
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
• Consider chlorination and bromination in more detail
• Chlorination is more product
favored than bromination
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11-30
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
• Which step in the mechanism is the slow step? Which
reaction has a faster rate? Which is more product
favored?
Both steps are exothermic
First step is endothermic
Second step is
exothermic
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11-31
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• With substrates more complex than ethane, multiple
monohalogenation products are possible
• If the halogen were indiscriminant, predict the product
ratio?
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11-32
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• For the CHLORINATION process, the actual product
distribution favors 2-chloropropane over 1chloropropane
• Which step in the mechanism determines the
regioselectivity?
• Show the arrow pushing for that key mechanistic step
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11-33
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• In one reaction, a 1° free radical forms, and in the other
a 2° radical forms
• Is the chlorination
process
thermodynamically
or kinetically
controlled?
• WHY?
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11-34
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• For the BROMINATION process, the product distribution
vastly favors 2-bromopropane over 1-bromopropane
• Which step in the mechanism determines
regioselectivity?
• Show the arrow pushing for that key mechanistic step
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11-35
Klein, Organic Chemistry 2e
11.4 Halogenation Thermodynamics
• Is the key step in the bromination mechanism kinetically
or thermodynamically controlled? WHY?
Both steps are exothermic
First step is endothermic
Second step is
exothermic
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11-36
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• Focus on the H abstraction step, and consider the
Hammond postulate: species on the energy diagram
that are similar in energy are similar in structure
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11-37
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
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11-38
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• When the bromine radical abstracts
the hydrogen, the carbon must be
able to stabilize a large partial
radical
• When the chlorine radical abstracts
the hydrogen, the carbon does not
carry as much of a partial radical
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11-39
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• Which process is more regioselective? WHY?
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11-40
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• Bromination at the 3° position happens 1600 times
more often than at the 1° position
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11-41
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• Which process is least regioselective? WHY?
• What is the general relationship between reactivity and
selectivity? WHY?
• Practice with SkillBuilder 11.5
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11-42
Klein, Organic Chemistry 2e
11.5 Halogenation Regioselectivity
• Ignoring possible addition products for now, draw the
structure for EVERY possible monobromination product
for the reaction below
• Rank the products in order from most major to most
minor
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11-43
Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry
• The halogenation of butane or more complex alkanes
forms a new chirality center
• 2-chlorobutane will form as a racemic mixture
• Which step in the mechanism is responsible for the
stereochemical outcome?
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11-44
Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry
• Whether the free radical carbon is sp2 or a rapidly
interconverting sp3, the halogen abstraction will occur
on either side of the plane with equal probability
sp2 hybridized
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11-45
Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry
• Three monosubstituted products form in the
halogenation of butane
• Draw all of the monosubstituted products that would
form in the halogenation of 2-methylbutane including all
stereoisomers
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11-46
Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry
• In the halogenation of (S)-3-methylhexane, the chirality
center is the most reactive carbon in the molecule.
WHY?
• Name the product and predict the stereochemical
outcome
• Practice with SkillBuilder 11.6
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11-47
Klein, Organic Chemistry 2e
11.6 Halogenation Stereochemistry
• Draw all of the monosubstituted products that would
form in the halogenation below including all
stereoisomers
• Classify any stereoisomer pairs as either enantiomers or
diasteriomers
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11-48
Klein, Organic Chemistry 2e
11.7 Allylic Halogenation
• When an C=C double bond is present it affects the
regioselectivity of the halogenation reaction
• Given the bond dissociation energies below, which
position of cyclohexene will be most reactive toward
halogenation?
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11-49
Klein, Organic Chemistry 2e
11.7 Allylic Halogenation
• When an allylic hydrogen is abstracted, it leaves behind
an allylic free radical that is stabilized by resonance
• Based on the high selectivity of bromination that we
discussed, you might expect bromination to occur as
shown below
• What other set of side-products is likely to form in this
reaction? Hint: addition reaction
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11-50
Klein, Organic Chemistry 2e
11.7 Allylic Halogenation with NBS
• To avoid the competing halogenation addition reaction,
NBS can be used to supply Br• radicals
• Show how resonance stabilizes the succinimide radical
• Heat or light initiates the process
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11-51
Klein, Organic Chemistry 2e
11.7 Allylic Halogenation with NBS
• Propagation produces new Br• radicals to continue the
chain reaction
• Where does the Br-Br above come from? The amount of
Br-Br in solution is minimal, so the competing addition
reaction is minimized
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11-52
Klein, Organic Chemistry 2e
11.7 Allylic Halogenation with NBS
• The succinimide radical that is produced in the initiation
step can also undergo propagation when it collides with
an H-Br molecule. Show a mechanism
+ H-Br
• Give some examples of some termination steps that
might occur
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11-53
Klein, Organic Chemistry 2e
11.7 Allylic Halogenation with NBS
• Give a mechanism that explains the following product
distribution. Hint: resonance
• Practice with SkillBuilder 11.7
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11-54
Klein, Organic Chemistry 2e
11.8 Atmospheric Chemistry and O3
• Ozone is both created and destroyed in the upper
atmosphere
• O3 molecules absorb harmful UV radiation
• O3 molecules are recycled as
heat energy is released
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11-55
Klein, Organic Chemistry 2e
11.8 Atmospheric Chemistry and O3
• For this process to be spontaneous, the entropy of the
universe must increase
• Heat is a more disordered form of energy
than light. WHY?
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11-56
Klein, Organic Chemistry 2e
11.8 Atmospheric Chemistry and O3
• O3 depletion (about 6% each year) remains a serious
health and environmental issue
• Compounds that are most destructive to the ozone layer
– Are stable enough to reach the upper atmosphere
– Form free radicals that interfere with the O3 recycling process
• Chlorofluorocarbons (CFCs) fit both criteria
• Atmosphere O3 is vital for protection, but what effect
does O3 have at the earth’s surface?
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11-57
Klein, Organic Chemistry 2e
11.8 Atmospheric Chemistry and O3
• CFC substitutes that generally decompose before
reaching the O3 layer include hydrochlorofluorocarbons
• Hydrofluorocarbons don’t even form the chlorine
radicals that interfere with the O3 cycle
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11-58
Klein, Organic Chemistry 2e
11.8 Combustion and Firefighting
• Like most reactions, combustion involves breaking
bonds and forming new bonds
– A fuel is heated with the necessary Eact to break bonds (C-C, CH, and O=O) homolytically
– The resulting free radicals join together to form new O-H and
C=O bonds
• Why does the process release energy overall?
• What types of chemicals might be used in fire
extinguishers to inhibit the process?
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11-59
Klein, Organic Chemistry 2e
11.8 Combustion and Firefighting
• Water deprives the fire of the Eact needed by absorbing
the energy
• CO2 and Argon gas deprive the fire of needed oxygen
• Halons are very effective fire-suppression agents
• Halons suppress combustion in three main ways
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11-60
Klein, Organic Chemistry 2e
11.8 Combustion and Firefighting
• Halons suppress combustion in three main ways
1. As a gas, they can smother the fire and deprive it of O2
2. They absorb some of the Eact for the fire by undergoing
homolytic cleavage
3. The free radicals produced can combine with C• and H• free
radicals to terminate the combustion chain reaction
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11-61
Klein, Organic Chemistry 2e
11.8 Combustion and Firefighting
• How do you think the use of halons to fight fires affects
the ozone layer?
• FM-200 is an alternative firefighting agent
• Practice with conceptual checkpoint 11.18
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11-62
Klein, Organic Chemistry 2e
11.9 Autooxidation
• Autooxidation is the process by which compounds react
with molecular oxygen
• The process is generally very slow
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11-63
Klein, Organic Chemistry 2e
11.9 Autooxidation Mechanism
• The
mechanism
illustrates
the need for
a more
refined
definition of
initiation and
propagation.
HOW?
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11-64
Klein, Organic Chemistry 2e
11.9 Autooxidation Mechanism
• Propagation can be more precisely defined as the steps
that add together to give the net chemical equation
• Steps in the mechanism that are not part of the net
equation must be either initiation or termination
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11-65
Klein, Organic Chemistry 2e
11.9 Autooxidation
• Some compounds such as ethers are particularly
susceptible to autooxidation
• Because hydroperoxides can be explosive, ethers like
diethyl ether must not be stored for long periods of time
• They should be dated and used in a timely fashion
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11-66
Klein, Organic Chemistry 2e
11.9 Autooxidation
• Light accelerates the autooxidation process
• Dark containers are often used to store many
chemicals such as vitamins
• In the absence of light, autooxidation is usually a
slow process
• Compounds that can form a relatively stable C•
radical upon H abstraction are especially susceptible
to autooxidation. WHY?
• Consider the autooxidation of compounds with
allylic or benzylic hydrogen atoms
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11-67
Klein, Organic Chemistry 2e
11.9 Antioxidants
• Triglycerides are important to a healthy diet
• Autooxidation can occur at the allylic positions causing
the food to become rancid and toxic
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11-68
Klein, Organic Chemistry 2e
11.9 Antioxidants
• Foods with unsaturated fatty acids have a short shelf life
unless preservatives are used
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11-69
Klein, Organic Chemistry 2e
11.9 Antioxidants
• Preservatives can undergo H abstraction to quench the
C• radicals that form in the first step of autooxidation
• One molecule of BHT can prevent thousands of
autooxidation reactions by stopping the chain reaction
• How does BHT’s structure make it good at taking on a
free radical? Consider resonance and sterics
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11-70
Klein, Organic Chemistry 2e
11.9 Natural Antioxidants
• Vitamins C is hydrophilic
• Vitamin E is hydrophobic
• What parts of the body do these
vitamins protect?
• For each vitamin, show its oxidation mechanism, and
explain how that protects the body from autooxidation
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11-71
Klein, Organic Chemistry 2e
11.10 Anti-Markovnikov Addition
• We learned in chapter 9 that H-X will add across a C=C
double bond with anti-Markovnikov regioselectivity
when peroxides are present
• Now that we have discussed free radicals, we can
explain the mechanism for the anti-Markovnikov
addition
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11-72
Klein, Organic Chemistry 2e
11.10 Anti-Markovnikov Addition
• The O-O single bond can break homolytically with a
relatively low Eact
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11-73
Klein, Organic Chemistry 2e
11.10 Anti-Markovnikov Addition
• The Br• radical reacts to give the more stable C• radical
• A 3° radical forms through a lower TS than
a 2° radical
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11-74
Klein, Organic Chemistry 2e
11.10 Anti-Markovnikov Addition
• Most of the radicals in solution at any given moment will
be Br• radical. WHY?
• Give some examples of other termination steps that
might occur but that would be less common
• Go back through each step in the mechanism to explain
the ΔH value for each step
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11-75
Klein, Organic Chemistry 2e
11.10 Addition Thermodynamics
• Anti-Markovnikov radical addition of HBr is generally
spontaneous (product favored)
• Yet, radical addition of HCl and HI are generally
nonspontaneous (reactant favored)
• On the next few slides, we will examine the
thermodynamics of each propagation step
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11-76
Klein, Organic Chemistry 2e
11.10 Addition Thermodynamics
• Explain the sign (+/-) for the entropy term in each
process
• Will the entropy change favor products or reactants?
WHY?
• The enthalpy change is affected most by the relative
stability of the two radical species
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11-77
Klein, Organic Chemistry 2e
11.10 Addition Thermodynamics
• The HI reaction will never favor products. WHY?
• How can the temperature be adjusted to favor products
for the HCl and HBr reaction?
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11-78
Klein, Organic Chemistry 2e
11.10 Addition Thermodynamics
• For the second propagation step, why is the entropy
term approximately zero?
• What small differences in entropy between products
and reactants might account for entropy changes being
slightly + or -?
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11-79
Klein, Organic Chemistry 2e
11.10 Addition Thermodynamics
• For HCl , the second propagation step will not be
product favored. WHY?
• HBr is the only reactant that favors product formation
for both propagation steps
• Might the overall propagation involving HCl be product
favored if the –ΔG of step 1 outweighs the
+ΔG of step 2?
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11-80
Klein, Organic Chemistry 2e
11.10 Addition Stereochemistry
• Addition reactions often form a new chirality center
• Recall that at least 1 reagent in the reaction must be
chiral for the reaction to be stereoselective
• Predicts the product distribution for the reaction below
and explain the stereochemical outcome
• Practice with SkillBuilder 11.8
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11-81
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• In chapter 9, we learned how some ionic
polymerizations occur
• Free radical conditions are also frequently used to form
polymers
• Recall that a polymerization process joins together many
small units called monomers in a long chain
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11-82
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• Radical polymerizations generally proceed through a
chain reaction mechanism
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11-83
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• Radical polymerizations generally proceed through a
chain reaction mechanism
• Note how the sum of the propagation steps yields the
overall reaction
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11-84
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• Radical polymerizations generally proceed through a
chain reaction mechanism
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11-85
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• Radical polymerizations generally produce chains of
monomers with a wide distribution of lengths
• How might experimental conditions be optimized to
control the average length of the chains?
• Because the mechanism we just learned proceeds
through 1° carbon free radical intermediates, it is usually
not facile
• In chapter 27, we will discuss specialized catalysts that
can be used to control such polymerization reactions
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11-86
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• Branching is common in some radical polymerizations
• Branching makes polymer materials
more flexible such as a polyethylene
squeeze bottle
• When catalysts are used to
minimize branching, more rigid
materials are produced such as the
squeeze bottle cap
• Why does branching affect rigidity?
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11-87
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
• Many derivatives of ethylene are also polymerized
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11-88
Klein, Organic Chemistry 2e
11.11 Radical Polymerization
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11-89
Klein, Organic Chemistry 2e
11.12 Radical Petroleum Processes
• Recall the cracking and reforming processes from
chapters 4 and 8
• Crude oil is cracked to produce smaller alkanes and
alkenes
• Reforming increases branching. Propose a mechanism
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11-90
Klein, Organic Chemistry 2e
11.13 Synthetic Utility of Halogenation
• Radical chlorination and bromination are both useful
processes
• Recall that bromination is more selective. WHY?
• Temperature can be used to help avoid polysubstitution.
HOW?
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11-91
Klein, Organic Chemistry 2e
11.13 Synthetic Utility of Halogenation
• Chlorination can be useful with highly symmetrical
substrates
• It is difficult to avoid polysubstitution. WHY?
• The synthetic utility of halogenation is limited
– Chlorination is difficult to control
– Bromination requires a substrate with 1 site that is
significantly more reactive than all others
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Klein, Organic Chemistry 2e
11.13 Synthetic Utility of Halogenation
• Synthesizing a target molecule from an alkane is
challenging because of its limited reactivity
• Often halogenation is the best option
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Klein, Organic Chemistry 2e
Additional Practice Problems
• Explain the difference between hyperconjugation and
induction.
Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
11-94
Klein, Organic Chemistry 2e
Additional Practice Problems
• Show all significant resonance contributors and a
resonance hybrid for the following molecule.
Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
11-95
Klein, Organic Chemistry 2e
Additional Practice Problems
• Explain what is
incorrect about
each of the
mechanistic
steps shown
below.
Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
11-96
Klein, Organic Chemistry 2e
Additional Practice Problems
• Give the major product(s) for the reaction below.
Carefully consider regiochemistry and stereochemistry.
Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
11-97
Klein, Organic Chemistry 2e
Additional Practice Problems
• Predict the major product for the reaction below and
explain why NBS is preferred over Br2.
Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
11-98
Klein, Organic Chemistry 2e
Additional Practice Problems
• Draw the monomer necessary to synthesize the given
polymer using a free radical mechanism.
Copyright © 2015 John Wiley & Sons, Inc. All rights reserved.
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Klein, Organic Chemistry 2e