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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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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Organic Chemistry - Ch 21

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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HCl

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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 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• ••

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O

C

C H

•• • ••

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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.

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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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