Jones M., Fleming S.A. Organic Chemistry

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5.10 Summary 219

produces “diamantane,” two caps gives us “triamantane,” three caps produces

“tetramantane,” and so on. The ultimate result of the capping process is a

molecule composed of a network of adamantanes, and this is the structure of

diamond.

Adamantane chemistry remains full of astonishing surprises. For years it has

been known that synthetic diamonds could be made from another polymeric form

of carbon, graphite, by treatment at high temperature ( 2300 K) and very high

pressure (7  10

kg/cm

). However, in the 1980s in the Soviet Union and Japan,

it was discovered that diamond films could be grown at low temperature and pres-

sure by passing a stream of hydrogen gas containing a few percent methane through

an electric discharge (Fig. 5.69).The mechanism of this reaction remains obscure.

Any ideas?

electrical

discharge

Diamond

FIGURE 5.69 A simple (and

astonishing) synthesis of diamond.

PROBLEM 5.30 What is the empirical formula of diamond? For a large polymeric

molecule such as diamond, the edges of the molecule are insignificant in figuring

out the formula.

PROBLEM 5.31 Can you draw other isomers of tetramantane?

Adamantane remains a favorite of chemists because of its intrinsic symmetry

and beauty, and probably because of the difficulty of working out the mechanism

of its formation. It does not seem of much practical interest. However, some of its

simple derivatives have remarkable properties and some are quite active biologi-

cally. For example, 1-aminoadamantane, a compound easily made from adaman-

tane itself, is one of the few antiviral agents known. This remarkable property

was discovered during routine empirical screening at duPont in the 1960s, and

1-aminoadamantane has since been marketed, mostly as an agent against influenza

A and C. It apparently works by migrating through the cell membrane to attack

the virus within. It is the adamantane cage, acting as a molecular ball of grease,

that helps in the membrane penetration. It has been speculated that other, large,

symmetrical hydrocarbon “blobs” might act in the same way. Unfortunately, they

are not so easy to make.

This chapter deals exclusively with the structural properties of

ring compounds. Two kinds of destabilizing effects on rings are

discussed. Torsional strain, the destabilizing effect of eclipsed car-

bon–hydrogen bonds, was mentioned in Chapter 2 when the

acyclic hydrocarbons were discussed.The planar forms of ring

compounds are particularly subject to torsional strain. In addition,

rings contain varying amounts of angle strain, depending on the

ring size. Any deviation from the ideal tetrahedral angle of 109.5°

will introduce angle strain. Rings adopt nonplanar forms in order

to minimize the combination of torsional and angle strain.

5.10 Summary

New Concepts

The diamond in this impressive bling

is composed of a multi-ring carbon

framework of adamantanes.

220 CHAPTER 5 Rings

angle strain (p. 187)

axial hydrogens (p. 191)

bridged (p. 211)

equatorial hydrogens (p. 191)

fused (p. 211)

spiro (p. 211)

torsional strain (p. 187)

van der Waals strain (p. 198)

Key Terms

The few reactions encountered here are not really new. There is,

however, a somewhat more detailed treatment of the formation

of carbon dioxide and water from hydrocarbons and oxygen

(combustion) than we saw in Chapter 2 (p. 68). There is also a

discussion of the effects of strain on the bond energy of the car-

bon–carbon bond.

Reactions, Mechanisms, and Tools

You are presented in this chapter with the most difficult chal-

lenge so far in visualizing molecules. Moreover, the difficulty in

translating from two dimensions to three is compounded by the

mobile equilibria present in many ring compounds. In analyzing

complicated phenomena, such as the equilibrations of disubsti-

tuted cyclohexanes, be sure to use models, at least at first.

Two small points seem to give students trouble. First, one

hears constantly, I’m no artist, I can’t draw a good chair cyclo-

hexane! Nonsense. It does not take an artist, merely an artisan.

Go slowly, do not scribble, and follow the simple procedure

outlined in this chapter (p. 191ff ). Drawing a perfect chair

cyclohexane is one thing that everyone can do.

Determining which groups are cis and trans in ring com-

pounds causes problems. Remember: cis means “on the same side,”

and trans, “on opposite sides.” Bonds to cis groups do not have to

be parallel, merely both up or both down. Figure 5.70 gives two

common examples, one straightforward, the other not so obvious.

Common Errors

Down

FIGURE 5.70 In both these compounds the two hydrogens shown

are cis, as are the two X groups.This relationship is easy to see for

the flat cyclopropane, but harder in the nonplanar cyclohexane.

PROBLEM 5.32 By now you have encountered most structural

types and have some experience with rings. Write and name

all the isomers of the formula C

. Stay alert for isomerism

of the cis/trans (Z/E) and (R/S) types. Indicate chiral molecules

with an asterisk. This problem is very hard at this point.

There are all sorts of odd structural types among these iso-

mers. It should not be hard to get most of them, but getting

them all is really tough. Think in three dimensions and

organize. Hint: The total number of isomers, not including

enantiomers, is 28.

PROBLEM 5.33 Write out the pairs of enantiomers for the

chiral isomers in Problem 5.32.

PROBLEM 5.34 Assign (R) and (S) configurations for the ver-

sions of the chiral cyclic molecules shown on the left in the

answers to Problem 5.33 given in the Study Guide.

PROBLEM 5.35 Draw the most stable chair conformation of

tert-butylcyclohexane. Put a circle around each of the axial

hydrogens on the cyclohexane. Put a square around each of the

equatorial hydrogens attached to the ring.

5.11 Additional Problems

Indeed, this problem constituted the entire first hour examination in MJ’s first course in organic chemistry.

One nonplanar ring deserves special mention.

Cyclohexane avoids torsional and angle strain by adopting a

chair conformation in which a nearly ideal tetrahedral angle

is achieved and all carbon–hydrogen bonds are perfectly

staggered. In cyclohexane, there is a set of six axial hydrogens

that is converted into the set of six equatorial hydrogens

through rotations about carbon–carbon bonds called a

ring flip.

There are many possible quantitative evaluations of strain.

This chapter uses related heats of formation ( ) and heats

of combustion ( ) to arrive at values for the strain energies

of various rings. As rings increase in size, strain initially

decreases, reaching a minimum at the strain-free cyclohexane.

Strain then increases until large ring sizes are reached.

By combining rings in spiro, fused, and bridged fashion,

very complex polycyclic molecules can be constructed.

¢H

¢H °

5.11 Additional Problems 221

PROBLEM 5.36

Draw the two possible chair forms of cis- and

trans-1,4-dimethylcyclohexane. Are the two forms identical,

enantiomeric, or diastereomeric? In each case, indicate which

chair form will be more stable and explain why. Is either of

these molecules chiral?

PROBLEM 5.37 Draw the two possible chair forms of cis- and

trans-1-isopropyl-4-methylcyclohexane. Are the two forms

identical, enantiomeric, or diastereomeric? In each case, indicate

which chair form will be more stable and explain why. Is either

of these molecules chiral?

PROBLEM 5.38 Draw the two possible chair forms of cis- and

trans-1,3-dimethylcyclohexane. Are the two forms identical,

enantiomeric, or diastereomeric? In each case, indicate which

chair form will be more stable and explain why. Is either of

these molecules chiral?

PROBLEM 5.39 Draw the possible chair forms of cis- and

trans-1-isopropyl-3-methylcyclohexane. Are the two forms

identical, enantiomeric, or diastereomeric? In each case, indicate

which chair form will be more stable and explain why. Is either

of these molecules chiral?

PROBLEM 5.40 Draw the double Newman projection as

shown in Figure 5.12 for the following compounds:

(a) 1,1-dimethylcyclohexane looking down the

and bonds.

(b) cis-1,2-dimethylcyclohexane looking down the

and bonds.

hexane looking down the and

bonds.

(d) trans-1,3-dimethylcyclohexane looking down the

and bonds.

PROBLEM 5.41 In Section 5.6b (p. 205), you saw that for pur-

poses of stereochemical analysis you could treat the decidedly

nonplanar cis- and trans-1,2-dimethylcyclohexanes as if they

were planar. The planar forms represent the average positions of

ring atoms in the rapid chair–chair interconversions. Use planar

representations of the following cyclohexanes to determine

which molecules are chiral.

(a) cis-1,3-dimethylcyclohexane

(b) trans-1,3-dimethylcyclohexane

(d) trans-1-isopropyl-3-methylcyclohexane

(e) cis-1,4-dimethylcyclohexane

(f) trans-1,4-dimethylcyclohexane

(g) cis-1-isopropyl-4-methylcyclohexane

(h) trans-1-isopropyl-4-methylcyclohexane

PROBLEM 5.42 Draw the planar structure for each of the fol-

lowing compounds. Next, draw the most stable conformation

for each in three dimensions.

(a) trans-1,2-dibromocyclohexane

C(5)

C(4)C(1)

C(2)

C(5)

C(4)C(1)

C(2)

C(5)

C(4)

C(1)

C(2)

C(4)

C(5)

C(2)

C(1)

(b) cis-1,3-dichlorocyclohexane

PROBLEM 5.43 Which of the molecules in Problem 5.42 are

chiral and which are achiral?

PROBLEM 5.44 Draw a chair conformation of cis-1-bromo-4-

fluorocyclohexane. Draw the ring-flipped structure. Which

structure do you expect to be favored? Why?

PROBLEM 5.45 Draw a chair conformation of cis-1-bromo-3-

chlorocyclohexane. Draw the ring-flipped structure. Which is

favored? Why? Is the molecule chiral?

PROBLEM 5.46 Use the data in Table 5.3 (p. 202) to calculate

the energy difference between the possible isomers of cis- and

trans-1,4-dimethylcyclohexane.

PROBLEM 5.47 Use the data in Table 5.3 (p. 202) to calculate

the energy difference between the possible isomers of cis- and

trans-1-isopropyl-4-methylcyclohexane.

PROBLEM 5.48 Cholestanol (shown below) is a natural prod-

uct found in gallstones and eggs. How many stereogenic atoms

are there in this molecule? Draw the compound with the rings

in their most stable conformation. Hint: The molecule is rela-

tively flat when viewed in three dimensions.

Cholestanol

PROBLEM 5.49 How many signals will appear in the

NMR spectrum of 1,1-dimethylcyclohexane?

(a) At low temperature.

(b) At high temperature.

Hint: Start by figuring out what effect the increase in tempera-

ture will have.

PROBLEM 5.50 How many signals will appear for cis-1,3-

dimethylcyclohexane in its

C NMR spectrum? How many

signals will appear for trans-1,3-dimethylcyclohexane in its

NMR spectrum?

(a) At low temperature.

(b) At high temperature.

PROBLEM 5.51 How many signals will appear in the

NMR spectrum of adamantane (p. 218)?

222 CHAPTER 5 Rings

Use Organic Reaction Animations (ORA) to answer the fol-

lowing questions. In these problems you will encounter new

reactions. Don’t be psyched out or intimidated—follow the ani-

mation carefully and you should be able to get these problems.

PROBLEM 5.59 Observe the reaction titled “Bimolecular

nucleophilic substitution.” Notice that a bromide is being dis-

placed from a carbon atom, in this case C(2) of 2-bromo-

propane. We can imagine a closely related reaction in which

displacement takes place in bromocyclopropane. All we have

(mentally) done is to connect the two end carbons of 2-bromo-

propane with a carbon–carbon bond. Would the reaction with

bromocyclopropane be faster or slower (it can’t be the same!) as

the reaction with 2-bromopropane? Hint: Think about the tran-

sition state for this reaction—what is the hybridization of car-

bon in the transition state?

PROBLEM 5.60 Now consider the same substitution reaction

with axial bromocyclohexane. What would be the conformation

of the product? Would this reaction be faster or slower than the

reaction with 2-bromopropane?

PROBLEM 5.61 Would the rate of reaction with equatorial

bromocyclohexane be faster, slower, or the same as that with

axial bromocyclohexane? Would the reaction with equatorial

bromocyclohexane be faster or slower than the reaction with

2-bromopropane?

PROBLEM 5.62 Observe the reaction titled “Bimolecular elim-

ination.”This reaction also starts with 2-bromopropane. Look

at the orientation of the breaking carbon–hydrogen and carbon–

bromine bonds in this reaction. Now imagine the same

bimolecular elimination occurring in axial and equatorial bro-

mocyclohexane. Which stereoisomer would undergo this reac-

tion faster, the axial or equatorial bromide?

PROBLEM 5.53 Name the following compounds:

CH(CH

)

(a)

CH(CH

)

(b)

CH(CH

)

(e)

CH(CH

)

(f)

CH(CH

)

(c)

CH(CH

)

(d)

PROBLEM 5.52 Draw both chair forms for the following cyclo-

hexanes. Indicate the more stable chair form in each case. Not all

examples will be obvious; some choices may be too close to call.

PROBLEM 5.54 Write structures for the following

compounds:

(a) 2,2-dibromo-3,3-dichloro-5,5-difluoro-6,6-diiodobicy-

clo[2.2.1]heptane

(b) bicyclo[1.1.1]pentan-1,3-diol

(b)

(c)(a)

HOH

PROBLEM 5.56 One of the isomers of methylbicyclo[2.2.1]hep-

tane can exist in two diastereomeric forms (we are not count-

ing mirror images, which increases the total number of

stereoisomers to four). Write the possible isomers of methylbi-

cyclo[2.2.1]heptane and explain how one of them can exist in

two forms.

PROBLEM 5.57 (a) Can one of the isomers of methylbicy-

clo[2.2.2]octane also exist in two diastereomeric forms? That is,

does it behave just like the methylbicyclo[2.2.1]heptane men-

tioned in Problem 5.56? (b) Can one of the isomers of methyl-

bicyclo[2.2.2]octane exist in two enantiomeric forms?

PROBLEM 5.58 Draw the favored envelope conformation of

the cyclopentane shown below.

PROBLEM 5.55 The sugar glucopyranose is a six-membered

ring containing an oxygen atom. There are groups attached at

each of the five ring carbons as shown. The red “squiggly” bond

merely means that the hydroxyl group at the 1-position may be

either up or down; that is, there are two structures for gluco-

pyranose. Draw these two molecules in three dimensions.

Alkyl Halides, Alcohols,

Amines, Ethers, and Their

Sulfur-Containing Relatives

223

6.1 Preview

6.2 Alkyl Halides: Nomenclature

and Structure

6.3 Alkyl Halides as Sources of

Organometallic Reagents:

A Synthesis of Hydrocarbons

6.4 Alcohols

6.5 Solvents in Organic Chemistry

6.6 Diols (Glycols)

6.7 Amines

6.8 Ethers

6.9 Special Topic: Thiols

(Mercaptans) and Thioethers

(Sulfides)

6.10 Special Topic: Crown Ethers

6.11 Special Topic: Complex

Nitrogen-Containing

Biomolecules—Alkaloids

6.12 Summary

6.13 Additional Problems

CANOLA Canola oil is obtained from the rapeseed plant. Rapeseed also

pumps a significant amount of methyl bromide into the atmosphere.

So, if Sunday you’re free,

Why don’t you come with me,

And we’ll poison the pigeons in the park.

My pulse will be quickenin’,

With each drop of strych-i-nin’,

We feed to a pigeon,

(It just takes a smidgin),

To poison a pigeon in the park.

—TOM LEHRER

6.1 Preview

In this chapter, we will see some old friends, alkyl halides and alcohols, as well as

some new molecules rather closely related to them: amines, ethers, thiols, and

thioethers. Both alkyl halides and alcohols can be made through the addition

reactions encountered in Chapter 3 (p. 131). Amines somewhat resemble alcohols,

and understanding the chemistry of amines is a necessary prelude to working

through Chapter 23, which discusses the chemistry of some of the many biolog-

ically important molecules containing nitrogen. A nitrogen atom provides one of

the crucial linking agents in the formation of polyamino acids, called peptides or

proteins (Fig. 6.1). In this chapter, we examine the qualities that make these mol-

ecules similar and prepare for Chapter 7, where we will take a long, detailed look

at four prototypal new reactions in which alkyl halides and alcohols are especially

important.

224 CHAPTER 6 Alkyl Halides, Alcohols, Amines, Ethers, and Their Sulfur-Containing Relatives

several

reactions

several

reactions

new amino acid

Growing peptide (or protein)

Amino acids

FIGURE 6.1 A nitrogen-to-carbon link is vital in the polymeric molecules known as proteins or peptides.

Thomas Lehrer (b. 1928) is an American mathematician and songwriter, most of whose work was done in

the 1950s and 1960s. Despite the unfortunate fact that there has been little new material for many years, his

permission.)

6.2 Alkyl Halides: Nomenclature and Structure 225

ESSENTIAL SKILLS AND DETAILS

1. So much of chemistry, including organic chemistry, involves acids and bases. In this

chapter, we review what we know already about this subject, and focus on a measure of

acidity, pK

. It is important to have a reasonable idea of the pK

values of the general classes

of compounds.There is a table of pK

values on the inside of the back cover of this book.

2. A low pK

means that the molecule in question is a strong acid, a good proton donor.

Conversely, a high pK

means that the compound is a poor acid.

3. We also encounter solvation in this chapter. Being able to choose an appropriate

solvent for a reaction, and to understand how solvents operate, is an important skill

both in the practical world and in this course.

4. The synthetic sequence “alkene halide Grignard reagent” is one you will use

many times in solving synthetic problems.

6.2 Alkyl Halides: Nomenclature and Structure

Alkyl iodides, bromides,chlorides, and fluorides are collected under the name of alkyl

halides. Alkyl halides have the formula C

2n  1

X, where X  F, Cl, Br, or I.

Quite appropriately, alkyl halides are named as fluorides, chlorides, bromides, and

iodides. Both common and systematic nomenclature names are used, although as

the complexity of the molecule increases, the system naturally takes over.The impor-

tant rules to remember for naming saturated halides are (1) minimize the number

given to the substituent, (2) name multiple substituents in alphabetical order, and

(3) base the name on the longest straight chain that contains the halide. If a

carbon–carbon double bond is present, the double bond takes precedence over the

halogen when positional numbers are assigned.Table 6.1 gives some examples with

both common and systematic names given. There are some important common

names that are used very often. For example, haloforms are CHX

, vinyl halides are

, and allyl halides are .H

CHX

TABLE 6.1 Some Alkyl Halides, Names, and Known Properties

Compound Common Name Systematic Name mp (°C)

Methyl fluoride Fluoromethane

Bromoethane

–141.8

–118.6Ethyl bromide

Isopropyl iodide 2-Iodopropane –90.1

tert-Butyl chloride 2-Chloro-2-methylpropane –25.4

Methylene bromide Dibromomethane –52.7

CHCl

Chloroform Trichloromethane –63.5

Vinyl bromide Bromoethene –139.5

Allyl chloride 3-Chloropropene –134.5

2-Bromo-1-chlorobutane

3-Chloro-1-pentene

Br Bromocyclopentane

bp (°C)

–78.4

38.4

89.4

51.0

61.5

15.8

147

138

(CH

)

CH I

C CHBr

CCH

ClCH

(CH

)

CCl

WEB 3D

226 CHAPTER 6 Alkyl Halides, Alcohols, Amines, Ethers, and Their Sulfur-Containing Relatives

CHH

CH I

(CH

)

CHI

Iodomethane

methyl iodide (a methyl halide)

(CH

)

C I

Iodoethane

ethyl iodide (a primary halide)

2-Iodopropane

isopropyl iodide (a secondary halide)

Iodoethene

vinyl iodide

2-Iodo-2-methylpropane

tert-butyl iodide (a tertiary halide)

FIGURE 6.2 Some alkyl and vinyl

halides at various levels of

abstraction.

means

δδ

–

FIGURE 6.3 A substantial dipole moment exists in the polar

carbon–halogen bond, X  F, Cl, Br, or I.

Halides are classified as methyl, primary, secondary, tertiary, and vinyl (Fig. 6.2).

Chapter 2 (p. 76) introduced this nomenclature. Remember: A primary carbon is

attached to only one other carbon, a secondary carbon to two others, and a tertiary

carbon to three others. In a primary halide, the halogen atom is attached to a pri-

mary carbon, and so on. In vinyl halides, there is always a carbon–carbon double bond

to which the halide is directly attached.

TABLE 6.2

Dipole Moments of Some

Simple Alkyl Halides

Alkyl Dipole

Halide Moment (D)

Methyl fluoride 1.85

Ethyl fluoride 1.94

Methyl chloride 1.87

Ethyl chloride 2.05

Methyl bromide 1.81

Ethyl bromide 2.03

Methyl iodide 1.62

Ethyl iodide 1.91

Substituted alkanes are polar molecules.The halogens are relatively electroneg-

ative atoms and strongly attract the electrons in the carbon–halogen bond. Alkyl

halides have dipole moments of approximately 2 debye, abbreviated D (Fig. 6.3;

Table 6.2).

Simple alkyl halides are nearly tetrahedral, and therefore the carbon to which

the halogen is attached is approximately sp

hybridized. The bond to halogen

involves an approximately sp

carbon orbital overlapping with an orbital on the

halogen for which the principal quantum number varies from n  2 for fluorine

to n  5 for iodine. The bond lengths increase and the bond strengths

decrease as we read down the periodic table. Some typical values for methyl

halides are shown in Figure 6.4, which gives both bond lengths (in

angstrom units) and bond strengths (in kilocalories per mole). Remember:

These values refer to homolytic bond breaking (formation of two neutral radi-

cals, Fig. 1.44).

6.3 Alkyl Halides as Sources of Organometallic Reagents: A Synthesis of Hydrocarbons 227

Bond lengths (A

⬚

)

Overlap of two

singly occupied

orbitals

1.78 1.93 2.14

1.39

115 84 72 58

Bond strengths (kcal/mol)

FIGURE 6.4 The bond of an

alkyl halide is formed by the overlap

of a singly occupied carbon sp

orbital

with a singly occupied halogen

orbital.

(a) (b)

(d)

(c)

(f)

(g)(e)

CHI

PROBLEM 6.1 Name the following compounds:

PROBLEM 6.2 Draw structures for the following compounds:

(a) 3,3-dichlorohexane (d) meso-3,4-dichlorohexane (three dimensions!)

(b) 1-bromocyclohexene (e) (1R,2R)-dichlorocyclopropane

6.3 Alkyl Halides as Sources of Organometallic

Reagents: A Synthesis of Hydrocarbons

Halides have all sorts of uses in the “real” practical world. They appear as anes-

thetics, insecticides, herbicides, the building blocks of polymers such as polyvinyl

chloride (PVC), and myriad other products. As you read this paragraph you are

probably thinking, “Sure, but haven’t there been serious problems with insecticides

such as DDT and herbicides like Agent Orange, and aren’t chlorofluorocarbons

eating up the ozone layer?”The answer is a resounding, Yes! There seems to be no

“free lunch” as unintended consequences often appear when we try to solve prob-

lems with technology. The issues are complex, but it is clear that the chemists of

today and the future need to supply the technical expertise to avoid these problems

while retaining the benefits.

Halides are important in synthetic chemistry because they allow for the forma-

tion of carbon–hydrogen and carbon–carbon bonds. Central to these uses of alkyl

and other halides is their easy conversion into organometallic reagents, molecules

228 CHAPTER 6 Alkyl Halides, Alcohols, Amines, Ethers, and Their Sulfur-Containing Relatives

RX R X

Diethyl ether

MgX

FIGURE 6.6 The Grignard reagent

has a complex structure.

An alkyl radical

X = Cl, Br, or I

RRX X

Mg Mg

FIGURE 6.7 Formation of a Grignard

reagent.

containing carbon–metal bonds. We look at only one limited use of these synthet-

ically important compounds in this chapter, but many other uses will appear later.

When an alkyl halide (fluorides are generally exceptions) is added to a cold

mixture of magnesium or lithium metal and an ether solvent, the metal begins to

disappear in an exothermic reaction. In general, the solvent used for this reaction,

an ether, has the structure . Diethyl ether or tetrahydrofuran (THF) is

most often used. The end result of the reaction is either a Grignard reagent or an

organolithium reagent (Fig. 6.5).

R = alkyl or alkenyl

X = Cl, Br, or I

Grignard reagent synthesis

RMgX

R MgX

–

Organolithium reagent synthesis

RLi

R Li

–

FIGURE 6.5 Formation of

organometallic reagents.

Grignard reagents were discovered by P. A. Barbier (1848–1922), and their

chemistry was worked out extensively by his student Victor Grignard (1871–1935),

who received the 1912 Nobel Prize in Chemistry for his work in this area. Neither

the formation of the Grignard reagent nor its structure is an easy subject.The sim-

plest formulation is “RMgX” and should be amplified by the addition of the ionic

resonance form shown in Figure 6.5. The Grignard reagent is also in equilibrium

with a mixture of the magnesium halide and dialkylmagnesium compound.

Moreover, the ether solvent is essential for its formation. Grignard reagents incor-

porate two molecules of solvent, conveniently left out in the traditional formulation

as RMgX (Fig. 6.6). Whatever the detailed structure, the result is a highly polar

reagent that is a very strong Lewis base.

The mechanism of formation of the Grignard reagent involves radical-transfer

reactions in which a transient alkyl radical is formed in the presence of a magnesium-

centered radical (Fig. 6.7).

We met the methyl radical in Chapter 2 (p. 62). Radicals, sometimes called free

radicals, are neutral species that have a single nonbonding electron and can under-

go many reactions, as we will see in Chapter 11. Typical and important reactions

of radicals include combining with other radicals to form bonds.The combination

of two carbon-based radicals is especially important because it results in the for-

mation of a carbon–carbon bond. The arrow formalism for radical reactions uses

single-barbed arrows.

CONVENTION ALERT