Jones M., Fleming S.A. Organic Chemistry

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11.8 Allylic Halogenation: Synthetically Useful Reactions 499

Further reaction with bromine leads to the product of allylic bromination and a new

chain-carrying bromine atom (Fig. 11.54).

One might hope that the photohalogenation reaction would be general, that

photochlorination would be as useful as photobromination. Alas it is not, and Table

11.5 shows why. A chlorine atom is far less selective than a bromine atom. Utility

in synthesis comes from selectivity—from doing one reaction, not several.

Photobromination is useful because the bromine atom picks and chooses among

the various possible carbon–hydrogen bonds with fairly high discrimination. A

chlorine atom does not. For example, the allylic hydrogens in cyclohexene (H

)

are only slightly more reactive toward a chlorine atom than the H

methylene

hydrogens (Fig. 11.55).

3-Bromocyclohexene

abstract H

hν

WEB 3D

FIGURE 11.54 Only the allylic

bromide is formed.

(37.5%)

abstracted

(62.5%)

abstracted

hν

Relative

reactivity = 0.6

Relative

reactivity = 1.0

FIGURE 11.55 Chlorine is much less

selective than bromine. The allylic

chloride is not the only product

formed in this reaction.

In photohalogenation reactions, keeping the concentration of X

low is essen-

tial to success. If there is too much halogen, polar addition to give vicinal dihalides

competes with formation of the allylic halide, and the desired selectivity is lost.

500 CHAPTER 11 Radical Reactions

HBr

Br2

hν

FIGURE 11.57 The first two steps

in bromination of cyclohexene using

NBS.

FIGURE 11.58 The bromine radical

abstracts a hydrogen to form the

cyclohexenyl radical. Independently,

a molecule of bromine is formed

from the NBS. The Br

then reacts

with the cyclohexenyl radical to give

the product and a new bromine atom.

Methods have been developed to ensure that this low concentration is maintained

and to lower the temperature necessary for the initiation step. A common and very

effective method of allylic bromination involves the molecule N-bromosuccinimide,

generally known as NBS,in the solvent carbon tetrachloride (CCl

).The overall reac-

tion is shown in Figure 11.56.

NBr

trace HBr

NBSCyclohexene 3-Bromocyclohexene

CCl

NHS

WEB 3D

FIGURE 11.56 Allylic bromination

with NBS.

The initiation steps in this reaction are formation of succinimide (NHS),

followed by formation of bromine atoms (Fig.11.57).The bromine radical abstracts

the allylic hydrogen to produce the cyclohexenyl radical,which in turn reacts with

the Br

to give the allylic bromide and a new, chain-carrying bromine atom. The

NBS is only slightly soluble in carbon tetrachloride, thus ensuring the low

concentration of bromine necessary for the success of this reaction (Fig. 11.58).

11.9 Special Topic: Rearrangements (and Nonrearrangements) of Radicals 501

PROBLEM SOLVING

11.9 Special Topic: Rearrangements (and

Nonrearrangements) of Radicals

We saw in Chapters 9 and 10 that polar additions to alkenes are bedeviled by

hydride or alkyl shifts when there is a possibility of forming a more stable cation

from a less stable one. A typical example, the polar addition of hydrogen chloride

to 3,3-dimethyl-1-butene, is shown again in Figure 11.59.Yet free radical additions

PROBLEM 11.28 Suggest a mechanism for the formation of bromine from NBS

and hydrogen bromide. Hint: The oxygen of a carbon–oxygen double bond is

a base.

PROBLEM 11.29 Selectivity is not possible in a photochemical reaction of Br

with

a disubstituted alkene such as trans-2-pentene. Explain.

NBS is a reagent that comes with its own pull-down menu that says, “watch out

for allylic bromination.” When you see NBS in a problem, allylic bromination is

very likely to be involved in the answer.

methyl

shift

More stable

tertiary carbocation

Less stable

secondary carbocation

Major product

Minor product

addition

–

addition

FIGURE 11.59 Rearrangements occur

in the polar addition of hydrogen

chloride to 3,3-dimethyl-1-butene.

502 CHAPTER 11 Radical Reactions

We might have expected rearrangement through the shift of a methyl group to

give the more stable tertiary radical from the less stable secondary one, yet this reac-

tion doesn’t happen. In fact, the situation is quite dramatic: 1,2-Hydride shifts

(migration of ) are among the fastest known organic reactions but 1,2-shifts of

a hydrogen atom or an alkyl group are unknown (Fig. 11.61).

hydride shift

very common!

hydrogen atom

shift unknown

CC CC

+ +

FIGURE 11.61 The simple 1,2-shift

of a hydrogen atom ( ) is unknown.H

This situation seems mysterious, but a look at the molecular orbitals involved in

the transition state for the reaction shows the reason. There is more than one way

to approach this problem, but we think an analysis of the transition state for the

rearrangement is the easiest. In the transition state for any symmetrical 1,2-shift

of hydrogen, the hydrogen atom is half-transferred from one carbon to the other

(Fig. 11.62).

The transition state

is exactly at the

halfway point in this

symmetrical reaction

means:

+, –, or

FIGURE 11.62 In the transition

state for any symmetrical 1,2-shift,

the hydrogen is halfway between

the two carbons. Here the symbol

, , or .

are usually uncomplicated by similar rearrangements. For example, the peroxide-

initiated addition of hydrogen bromide to 3,3-dimethyl-1-butene gives only the

unrearranged product (Fig. 11.60).

methyl shift

does not occur

More stable

tertiar

radical

Less stable

secondary radical

Not formed

FIGURE 11.60 The related radical

addition gives no rearranged product.

11.9 Special Topic: Rearrangements (and Nonrearrangements) of Radicals 503

What do the molecular orbitals of the transition state look like? To answer this ques-

tion,we need only construct the molecular orbitals for the system of a hydrogen 1s orbital

interacting with the π and π* orbitals of an alkene. The combination of 1s, π, and π*

will yield three new molecular orbitals A,B,and C.They are constructed in Figure 11.63.

The 1s orbital interacts with the π orbital to form two new molecular orbitals

(A and B), but there is no interaction between 1s and π* in this geometry.

Accordingly, π* becomes C, the third molecular orbital for the transition state. To

order the energy of A, B, C, we need only count nodes. Molecular orbital A has no

new nodes, but B and C each have one additional node (Fig. 11.64).

–

π*

–

C C

Antibonding

Bonding

There is no interaction

between

π* and 1s

AB C

–

FIGURE 11.63 The three molecular orbitals for the transition state in this reaction can be easily

constructed from H (1s), π, and π*.

–

Nodes

FIGURE 11.64 The new molecular

orbitals (A–C) can be ordered in

energy by counting nodes.

The ordering of the new molecular orbitals is shown in Figure 11.65. Now, how

many electrons do we need to put into the orbital system? For the cationic reaction

of Figure 11.61, there are only two and they will go nicely into the new stabilized

Energy

CCBB

Nonbonding

Cation, two electrons

Radical, three electrons,

one in an antibonding orbital;

bad news

Nonbonding

FIGURE 11.65 The transition state

will have a single bonding molecular

orbital (A) and two, equi-energetic

antibonding molecular orbitals

(B and C). For the cationic

rearrangement, the two electrons

involved nicely fill the bonding

molecular orbital (A). In the radical

case, there is another electron and it

must occupy an antibonding

molecular orbital (B or C) in the

transition state. Occupancy of this

antibonding orbital is unfavorable.

504 CHAPTER 11 Radical Reactions

ANSWER A vinyl group, or any other group with p orbitals, is not restricted to

using the σ bond in the migration. In this case, the radical on C(2) could add to

the vinyl group to give the symmetrical intermediate A. Opening of intermediate

A can occur in two directions (a and a′) either to regenerate the starting material

(a′) or complete the vinyl migration (a).

PROBLEM 11.30 Analyze the shift of hydrogen in the related anionic system.

WORKED PROBLEM 11.31 1,2-Shifts of vinyl groups are known in radicals.

Describe how this reaction might differ from the 1,2-shift of a hydrogen atom or

a methyl radical, both of which are unknown.

bonding molecular orbital A. However, for the radical reaction of Figure 11.61 there

is another electron, and it must be placed in one of the antibonding molecular

orbitals, B or C.This high-energy electron is evidently destabilizing enough to com-

pletely shut off the reaction (Fig. 11.65).

WORKED PROBLEM 11.32 We have already seen a reaction much like the opening

of the cyclic intermediate shown in the answer to Problem 11.31. What is the

name of this process?

ANSWER It is just a β cleavage (p. 472).

11.10 Special Topic: Radicals in Our Bodies;

Do Free Radicals Age Us?

Free radicals are strongly implicated in the aging process. Cell membranes, which

are made up of long-chain hydrocarbon carboxylic acids called fatty acids, are par-

ticularly vulnerable to oxidation through free radical attack. Once a cell’s membrane

has been oxidized, the cell is damaged. In time, the immune system comes to

11.11 Summary 505

WEB 3D

Vitamin E

FIGURE 11.66 Vitamin E.

recognize damaged cells as “foreign” and attacks them. This leads to cell death and

ages the organ that the cell is in. If radicals are involved in aging, then radical trap-

ping agents might be effective in arresting the aging process. One such potential

trapping agent is Vitamin E (Fig. 11.66):

VITAMIN E

In rats, vitamin E is essential for proper liver function as well

as nerve and muscle condition. At least one observer has

pointed out that because rats and humans are so alike, vita-

min E might be vital for humans as well. Despite the obvi-

ous truth of the observation, there is no proven need for

vitamin E in humans. Still, that lack is not sufficient to stop

speculation, and it has been suggested that the ability of

vitamin E to act as a free radical interceptor, an antioxidant,

might make it effective in delaying aging. Vitamin E traps

radicals through the hydroxyl group highlighted in red on

the six-membered ring (Fig. 11.66). A radical ( ) plucks

a hydrogen atom from this hydroxyl to make and

destroys itself.

The long hydrocarbon chain of vitamin E also has its

function. It aids in the delivery of vitamin E to the cell

membrane by making the molecule soluble in fatty acids,

which also contain long greasy hydrocarbon chains.

Remember: “Like dissolves like” (p. 239).

ROH

11.11 Summary

New Concepts

One of the most important concepts of this chapter is that

of selectivity. It is not really new because you have already

seen many selective reactions. A simple example is the

protonation of isobutene to give the more stable tertiary

carbocation rather than the less stable primary carbocation

(Fig. 11.23).

In this chapter, you also see selectivity when a radical

abstracts one of several possible hydrogens from an alkane.

The ease of hydrogen abstraction is naturally determined by

the energies of the transition states leading to the products

(Figs. 11.49 and 50). A tertiary radical is more likely to be

formed than a primary radical. In the transition state for

radical formation, the factors that make a tertiary radical

more stable than a primary radical will operate to stabilize

the partially formed tertiary radical over the partially

formed primary radical, as noted in Figures 11.49 and

11.50.

What is more subtle, and at least somewhat new, is the

idea that the degree of selectivity in a reaction will be influ-

enced by how product-like the transition state is. The more the

506 CHAPTER 11 Radical Reactions

Key Terms

abstraction (p. 471)

acyl group ( ) (p. 476)

allylic halogenation (p. 497)

azo compounds (p. 476)

chain reaction (p. 468)

β cleavage (p. 472)

disproportionation (p. 471)

hydrocarbon cracking (p. 473)

inhibitors (p. 484)

initiation (p. 483)

propagation (p. 483)

pyrolysis (p. 470)

termination (p. 483)

thermolysis (p. 470)

transition state is like the product (the radical intermediate),

the more important the relative stabilities of two possible

products will be. Abstraction of a hydrogen by chlorine is

endothermic by about 2 kcal/mol. The same abstraction reac-

tion by bromine is endothermic by about 17 kcal/mol. The

transition state for the much more endothermic abstraction

reaction by bromine will be more product-like than that for

abstraction by chlorine. So we expect bromine to be more

selective than chlorine, because the carbon–hydrogen bond

will be more broken in the transition state. The stability of the

product radical will be more important in the transition state

for hydrogen abstraction by a bromine than it will in the tran-

sition state for abstraction by a chlorine.

There are a few other points to keep in mind. Radical sta-

bility parallels carbocation stability. Tertiary radicals are more

stable than secondary, secondary more stable than primary,

and primary more stable than methyl. The differences aren’t as

great for radicals as for cations. Delocalization helps stabilize

radicals by spreading the odd electron over two or more

atoms, so allylic and benzylic radicals are more stable than ter-

tiary radicals.

Radical reactions are often cyclic: Radical chain reactions

produce a chain-carrying radical in their last, product-making

step, and the new radical can begin another series of

propagation steps leading to another molecule of product.

Chain reactions only stop when starting material is exhausted

or a chain is terminated by bond formation by a pair of

radicals.

Chains can be started by small amounts of radicals

released by radical initiators. In this way, a small number of

chain-initiating radicals can have an impact far out of proportion

to their numbers.

The reactions in this chapter involve neutral free radicals, in

contrast to the ionic species of earlier chapters. Chain reactions

are characteristic of radical chemistry and consist of three steps:

(1) Initiation—a small number of radicals serves to start the

reaction; (2) Propagation—radical reactions are carried out that

generate a product molecule and a new, chain-carrying radical,

which recycles to the beginning of the propagating steps

and starts a new series of product-forming steps; and

(3) Termination—radical recombinations destroy chain-carrying

species and end the chain reaction. The success of a chain

reaction depends on the relative success of the chain-carrying

and termination steps. Figure 11.26 gives an example of a

typical chain reaction.

Important reactions of radicals include abstraction, addi-

tions to alkenes and alkynes, β cleavage, and disproportionation

(Fig. 11.9), but not simple 1,2-rearrangements.

Reactions, Mechanisms, and Tools

Syntheses

The new synthetic reactions of this chapter are shown below.

1. Alkanes

2. Alkenes

Disproportionation gives both alkanes and alkenes;

this synthesis is not efficient

Disproportionation gives both alkanes and alkenes;

this synthesis is not efficient

3. Alkyl Bromides

peroxides

HBr

Also works for Cl

; photobromination is more

selective than photochlorination

A chain reaction that works only for H

Br, not

other H

X molecules; note the anti-Markovnikov

regiochemistry

11.12 Additional Problems 507

4. Allylic Halides

hν/CCl

NBS

Bromination is specifically at the allylic position

5. Radicals

6. Vicinal Dihalides

7. Vinyl Bromides

Pyrolysis of hydrocarbons is an unselective

method for making radicals; many different

radicals are typically produced

Pyrolysis of azo compounds

2 R

peroxides

BrHC

HBr

inyl halides (see 7) can add a second molecule of

HBr to

ive vicinal

(

not

eminal

)

dihalides

peroxides

BrHC

HBr

Radical-induced addition to alkynes gives

anti-Markovnikov addition; cis and trans

products are formed

HC C

A very common mistake is to confuse the propagation step in a

chain reaction with the termination step. In the photobromina-

tion of an alkane such as methane, the product is formed in a

chain reaction in which bromine and methyl radicals carry the

chain (Fig. 11.45). It is all too tempting to imagine that the

product, methyl bromide, is formed by the combination of one

methyl radical and one bromine radical! In fact, although this

step does produce a molecule of product, it destroys two chain-

carrying radicals. This reaction would terminate the chain

process and so we know it does not contribute significantly to

product formation.

Common Errors

PROBLEM 11.33 Let’s build right away on Problem 11.30 and

Section 11.9. Suppose you are walking down the street, and a

stranger comes up to you and asks, Which is more stable, linear

or triangular H



? We maintain that you should be able to give

him or her a quick answer to this seemingly bizarre question.

Obvious hint: Look at Section 11.9.

Here’s an outline of how to do this problem: First, build the

molecular orbitals for linear and cyclic H

. You did the linear

molecule long ago in Chapter 1 (Problem 1.60, p. 48). Next,

put the appropriate number of electrons in. Finally, look at the

orbital containing those electrons in each species. Which is

lower in energy?

PROBLEM 11.34 Show the reactions you would use to synthe-

size propene if your only source of carbon is propane.

PROBLEM 11.35 Show how you would make 2-butene if your

only source of carbon is butane.

PROBLEM 11.36 Show how you would make trans-1,2-dibro-

mocyclohexane starting with cyclohexane.

PROBLEM 11.37 Predict the major product(s) for the radical

hydrohalogenation reaction (HBr, peroxides, heat) with each

of the following alkenes:

(a) 1-pentene

(b) 2-pentene

(d) cis-3-hexene

(e) trans-3-hexene

PROBLEM 11.38 Draw the monochlorinated products formed

by radical chlorination of the following compounds:

(a) cyclobutane

(b) cyclopentane

(d) pentane

PROBLEM 11.39 Draw four dibromo products that could be

obtained by radical bromination of propane. Show the reaction

pathway for the compound you think would be the major product.

PROBLEM 11.40 Which of the products in the previous prob-

lem are chiral and which are achiral?

11.12 Additional Problems

(a)

(b)

HBr

ROOR

HBr

CCH

508 CHAPTER 11 Radical Reactions

PROBLEM 11.41 Give the major product(s) expected for

each of the following reactions. Pay attention to regiochemistry

and stereochemistry where appropriate. Mechanisms are

not required, although they may be useful in finding the

answers.

PROBLEM 11.42 Predict the products of hydrogen bromide

addition to 1-butyne. Consider both ionic (a) and radical-

initiated (b) reactions. Rationalize your predictions with arrow

formalism descriptions. Remember that a second molecule

of hydrogen bromide can add to the initial product(s) in

both cases.

PROBLEM 11.43 The relative rates of recombination of methyl

and isopropyl radicals are quite different. Give two possible

reasons for this difference.

(a)

HBr

ROOR

(b)

HBr

(c)

(d)

(e)

(f)

(g)

HCl

ROOR

HBr

ROOR

HBr

hν

CCl

NBS, CCl

COOCH

CCH

2 H

Relative

Rate

CH CH

CH2

PROBLEM 11.44 Decomposition of the following azo

compounds at 300 °C occurs at quite different rates. Give two

possible explanations for these differences.

300 ⬚C

, CH

, H

, H, H

H, H, H

13,000

111

Relative Rate

PROBLEM 11.45 Predict the stereochemical results of the

photochlorination of the optically active compound 1.The (

)

indicates a single enantiomer, here (R). It is the tertiary

carbon–hydrogen bond that is chlorinated.

No stereochemistry implied

in this drawing

hν

PROBLEM 11.46 In Problem 11.25, we examine the pho-

tochlorination of ethane to give ethyl chloride. As was the

case in the photochlorination of methane (p. 490), products

containing more than one chlorine are also isolated in the

photochlorination of ethane. What is the structure of the

major dichloroethane produced in this reaction? Rationalize

your prediction with an analysis of the mechanism of the

reaction.