CHAPTER 4
ALCOHOLS AND ALKYL HALIDES
O
ur first three chapters established some fundamental principles concerning the
structure of organic molecules. In this chapter we begin our discussion of organic
chemical reactions by directing attention to alcohols and alkyl halides. These
two rank among the most useful classes of organic compounds because they often serve
as starting materials for the preparation of numerous other families.
Two reactions that lead to alkyl halides will be described in this chapter. Both illus-
trate functional group transformations. In the first, the hydroxyl group of an alcohol is
replaced by halogen on treatment with a hydrogen halide.
In the second, reaction with chlorine or bromine causes one of the hydrogen substituents
of an alkane to be replaced by halogen.
Both reactions are classified as substitutions, a term that describes the relationship
between reactants and products—one functional group replaces another. In this chapter
we go beyond the relationship of reactants and products and consider the mechanism of
each reaction. A mechanism attempts to show how starting materials are converted into
products during a chemical reaction.
While developing these themes of reaction and mechanism, we will also use alco-
hols and alkyl halides as vehicles to extend the principles of IUPAC nomenclature, con-
H11001H11001R±H
Alkane
X
2
Halogen
R±X
Alkyl halide
H±X
Hydrogen halide
H11001H11001R±OH
Alcohol
H±X
Hydrogen halide
R±X
Alkyl halide
H±OH
Water
126
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
tinue to develop concepts of structure and bonding, and see how structure affects prop-
erties. A review of acids and bases constitutes an important part of this chapter in which
a qualitative approach to proton-transfer equilibria will be developed that will be used
throughout the remainder of the text.
4.1 IUPAC NOMENCLATURE OF ALKYL HALIDES
The IUPAC rules permit alkyl halides to be named in two different ways, called func-
tional class nomenclature and substitutive nomenclature. In functional class nomencla-
ture the alkyl group and the halide ( fluoride, chloride, bromide, or iodide) are desig-
nated as separate words. The alkyl group is named on the basis of its longest continuous
chain beginning at the carbon to which the halogen is attached.
Substitutive nomenclature of alkyl halides treats the halogen as a halo- ( fluoro-,
chloro-, bromo-, or iodo-) substituent on an alkane chain. The carbon chain is numbered
in the direction that gives the substituted carbon the lower locant.
When the carbon chain bears both a halogen and an alkyl substituent, the two substituents
are considered of equal rank, and the chain is numbered so as to give the lower number
to the substituent nearer the end of the chain.
PROBLEM 4.1 Write structural formulas, and give the functional class and sub-
stitutive names of all the isomeric alkyl chlorides that have the molecular formula
C
4
H
9
Cl.
Substitutive names are preferred, but functional class names are sometimes more
convenient or more familiar and are frequently encountered in organic chemistry.
4.2 IUPAC NOMENCLATURE OF ALCOHOLS
Functional class names of alcohols are derived by naming the alkyl group that bears the
hydroxyl substituent (±OH) and then adding alcohol as a separate word. The chain is
always numbered beginning at the carbon to which the hydroxyl group is attached.
Substitutive names of alcohols are developed by identifying the longest continu-
ous chain that bears the hydroxyl group and replacing the -e ending of the
5-Chloro-2-methylheptane
CH
3
CHCH
2
CH
2
CHCH
2
CH
3
CH
3
W
Cl
W
1234567
2-Chloro-5-methylheptane
CH
3
CHCH
2
CH
2
CHCH
2
CH
3
Cl
W
CH
3
W
1234567
CH
3
CH
2
CH
2
CH
2
CH
2
F
1-Fluoropentane 2-Bromopentane
CH
3
CHCH
2
CH
2
CH
3
Br
W
1234512345
3-Iodopentane
CH
3
CH
2
CHCH
2
CH
3
I
W
12345
CH
3
CH
2
CH
2
CH
2
CH
2
Cl
Pentyl chloride
CH
3
F
Methyl fluoride
CH
3
CH
2
CHCH
2
CH
2
CH
3
Br
W
1-Ethylbutyl bromide
H
I
Cyclohexyl iodide
12 3 4
4.1 IUPAC Nomenclature of Alkyl Halides 127
The IUPAC rules permit cer-
tain common alkyl group
names to be used. These in-
clude n-propyl, isopropyl,
n-butyl, sec-butyl, isobutyl,
tert-butyl, and neopentyl
(Section 2.10).
Prior to the 1993 version of
the IUPAC rules, the term
“radicofunctional” was used
instead of “functional class.”
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
corresponding alkane by the suffix -ol. The position of the hydroxyl group is indicated
by number, choosing the sequence that assigns the lower locant to the carbon that bears
the hydroxyl group.
Hydroxyl groups take precedence over (“outrank”) alkyl groups and halogen substituents
in determining the direction in which a carbon chain is numbered.
PROBLEM 4.2 Write structural formulas, and give the functional class and sub-
stitutive names of all the isomeric alcohols that have the molecular formula
C
4
H
10
O.
4.3 CLASSES OF ALCOHOLS AND ALKYL HALIDES
Alcohols and alkyl halides are classified as primary, secondary, or tertiary according to
the classification of the carbon that bears the functional group (Section 2.10). Thus, pri-
mary alcohols and primary alkyl halides are compounds of the type RCH
2
G (where G
is the functional group), secondary alcohols and secondary alkyl halides are compounds
of the type R
2
CHG, and tertiary alcohols and tertiary alkyl halides are compounds of
the type R
3
CG.
PROBLEM 4.3 Classify the isomeric C
4
H
10
O alcohols as being primary, secondary,
or tertiary.
Many of the properties of alcohols and alkyl halides are affected by whether their
functional groups are attached to primary, secondary, or tertiary carbons. We will see a
number of cases in which a functional group attached to a primary carbon is more reac-
tive than one attached to a secondary or tertiary carbon, as well as other cases in which
the reverse is true.
6-Methyl-3-heptanol
(not 2-methyl-5-heptanol)
CH
3
CHCH
2
CH
2
CHCH
2
CH
3
CH
3
W
OH
W
7654321
3-Fluoro-1-propanol
FCH
2
CH
2
CH
2
OH
321
OH
CH
3
1
5
4
3
2
trans-2-Methylcyclopentanol
CH
3
CH
2
OH
Ethyl alcohol
Ethanol
1-Methylpentyl alcohol
2-Hexanol
CH
3
CHCH
2
CH
2
CH
2
CH
3
OH
W
1,1-Dimethylbutyl alcohol
2-Methyl-2-pentanol
CH
3
CCH
2
CH
2
CH
3
OH
CH
3
W
W
Functional class
name:
Substitutive
name:
128 CHAPTER FOUR Alcohols and Alkyl Halides
CH
3
CCH
2
OH
CH
3
W
W
CH
3
2,2-Dimethyl-1-propanol
(a primary alcohol)
CH
3
CCH
2
CH
2
CH
3
CH
3
W
C
W
Cl
2-Chloro-2-methylpentane
(a tertiary alkyl halide)
CH
3
CH
2
CHCH
3
Br
W
2-Bromobutane
(a secondary alkyl halide)
CH
3
OH
1-Methylcyclohexanol
(a tertiary alcohol)
Several alcohols are com-
monplace substances, well
known by common names
that reflect their origin
(wood alcohol, grain alcohol)
or use (rubbing alcohol).
Wood alcohol is methanol
(methyl alcohol, CH
3
OH),
grain alcohol is ethanol
(ethyl alcohol, CH
3
CH
2
OH),
and rubbing alcohol is
2-propanol [isopropyl alco-
hol, (CH
3
)
2
CHOH].
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.4 BONDING IN ALCOHOLS AND ALKYL HALIDES
The carbon that bears the functional group is sp
3
-hybridized in alcohols and alkyl halides.
Figure 4.1 illustrates bonding in methanol. The bond angles at carbon are approximately
tetrahedral, as is the C±O±H angle. A similar orbital hybridization model applies to
alkyl halides, with the halogen substituent connected to sp
3
-hybridized carbon by a H9268
bond. Carbon–halogen bond distances in alkyl halides increase in the order C±F (140
pm) H11021 C±Cl (179 pm) H11021 C±Br (197 pm) H11021 C±I (216 pm).
Carbon–oxygen and carbon–halogen bonds are polar covalent bonds, and carbon
bears a partial positive charge in alcohols (
H9254H11001
C±O
H9254H11002
) and in alkyl halides (
H9254H11001
C±X
H9254H11002
).
The presence of these polar bonds makes alcohols and alkyl halides polar molecules. The
dipole moments of methanol and chloromethane are very similar to each other and to water.
PROBLEM 4.4 Bromine is less electronegative than chlorine, yet methyl bromide
and methyl chloride have very similar dipole moments. Why?
Figure 4.2 shows the distribution of electron density in methanol and
chloromethane. Both are similar in that the sites of highest electrostatic potential (red)
are near the electronegative atoms—oxygen and chlorine. The polarization of the bonds
Water
(H9262 H11005 1.8 D)
H
O
H
Chloromethane
(H9262 H11005 1.9 D)
CH
3
Cl
Methanol
(H9262 H11005 1.7 D)
O
H
3
C H
4.4 Bonding in Alcohols and Alkyl Halides 129
C
H
H
H
C
H
O
O
H
HH
H
Lone-pair orbitals
(a) (b)
bondσ
C±O±H angle H11005 108.5H11034
C±O bond distance H11005 142 pm
FIGURE 4.1 Orbital hybrid-
ization model of bonding in
methanol. (a) The orbitals
used in bonding are the 1s
orbitals of hydrogen and sp
3
-
hybridized orbitals of carbon
and oxygen. (b) The bond an-
gles at carbon and oxygen
are close to tetrahedral, and
the carbon–oxygen H9268 bond is
about 10 pm shorter than a
carbon–carbon single bond.
Methanol (CH
3
OH) Chloromethane (CH
3
Cl)
FIGURE 4.2 Electro-
static potential maps of
methanol and chloro-
methane. The most posi-
tively charged regions are
blue, the most negatively
charged ones red. The elec-
trostatic potential is most
negative near oxygen in
methanol and near chlorine
in chloromethane.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
to oxygen and chlorine, as well as their unshared electron pairs, contribute to the con-
centration of negative charge on these atoms.
Relatively simple notions of attractive forces between opposite charges are suffi-
cient to account for many of the properties of chemical substances. You will find it help-
ful to keep the polarity of carbon–oxygen and carbon–halogen bonds in mind as we
develop the properties of alcohols and alkyl halides in later sections.
4.5 PHYSICAL PROPERTIES OF ALCOHOLS AND ALKYL HALIDES:
INTERMOLECULAR FORCES
Boiling Point. When describing the effect of alkane structure on boiling point in Sec-
tion 2.14, we pointed out that the forces of attraction between neutral molecules are of
three types listed here. The first two of these involve induced dipoles and are often
referred to as dispersion forces, or London forces.
1. Induced-dipole/induced-dipole forces
2. Dipole/induced-dipole forces
3. Dipole–dipole forces
Induced-dipole/induced-dipole forces are the only intermolecular attractive forces
available to nonpolar molecules such as alkanes. In addition to these forces, polar mol-
ecules engage in dipole–dipole and dipole/induced-dipole attractions. The dipole–dipole
attractive force is easiest to visualize and is illustrated in Figure 4.3. Two molecules of
a polar substance experience a mutual attraction between the positively polarized region
of one molecule and the negatively polarized region of the other. As its name implies,
the dipole/induced-dipole force combines features of both the induced-dipole/induced-
dipole and dipole–dipole attractive forces. A polar region of one molecule alters the elec-
tron distribution in a nonpolar region of another in a direction that produces an attrac-
tive force between them.
Because so many factors contribute to the net intermolecular attractive force, it is
not always possible to predict which of two compounds will have the higher boiling
point. We can, however, use the boiling point behavior of selected molecules to inform
us of the relative importance of various intermolecular forces and the structural features
that influence them.
Consider three compounds similar in size and shape: the alkane propane, the alco-
hol ethanol, and the alkyl halide fluoroethane.
Both polar compounds, ethanol and fluoroethane, have higher boiling points than the
nonpolar propane. We attribute this to a combination of dipole/induced-dipole and
dipole–dipole attractive forces that stabilize the liquid states of ethanol and fluoroethane,
but that are absent in propane.
The most striking aspect of the data, however, is the much higher boiling point of
ethanol compared with both propane and fluoroethane. This suggests that the attractive
forces in ethanol must be unusually strong. Figure 4.4 shows that this force results from
a dipole–dipole attraction between the positively polarized proton of the H5007OH group of
one ethanol molecule and the negatively polarized oxygen of another. The term
hydrogen bonding is used to describe dipole–dipole attractive forces of this type. The
Ethanol (H9262 H11005 1.7 D)
bp: 78°C
CH
3
CH
2
OH
Fluoroethane (H9262 H11005 1.9 D)
bp: H1100232°C
CH
3
CH
2
F
Propane (H9262 H11005 0 D)
bp: H1100242°C
CH
3
CH
2
CH
3
130 CHAPTER FOUR Alcohols and Alkyl Halides
H11002H11001 H11002H11001
FIGURE 4.3 A dipole–dipole
attractive force. Two mole-
cules of a polar substance are
oriented so that the posi-
tively polarized region of
one and the negatively po-
larized region of the other
attract each other.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
proton involved must be bonded to an electronegative element, usually oxygen or nitro-
gen. Protons in C±H bonds do not participate in hydrogen bonding. Thus fluoroethane,
even though it is a polar molecule and engages in dipole–dipole attractions, does not
form hydrogen bonds and, therefore, has a lower boiling point than ethanol.
Hydrogen bonding can be expected in molecules that have ±OH or ±NH groups.
Individual hydrogen bonds are about 10–50 times weaker than typical covalent bonds,
but their effects can be significant. More than other dipole–dipole attractive forces, inter-
molecular hydrogen bonds are strong enough to impose a relatively high degree of struc-
tural order on systems in which they are possible. As will be seen in Chapter 27, the
three-dimensional structures adopted by proteins and nucleic acids, the organic mole-
cules of life, are dictated by patterns of hydrogen bonds.
PROBLEM 4.5 The constitutional isomer of ethanol, dimethyl ether (CH
3
OCH
3
),
is a gas at room temperature. Suggest an explanation for this observation.
Table 4.1 lists the boiling points of some representative alkyl halides and alcohols.
When comparing the boiling points of related compounds as a function of the alkyl
group, we find that the boiling point increases with the number of carbon atoms, as it
does with alkanes.
4.5 Physical Properties of Alcohols and Alkyl Halides: Intermolecular Forces 131
TABLE 4.1 Boiling Points of Some Alkyl Halides and Alcohols
Name of
alkyl group
Methyl
Ethyl
Propyl
Pentyl
Hexyl
Formula
CH
3
X
CH
3
CH
2
X
CH
3
CH
2
CH
2
X
CH
3
(CH
2
)
3
CH
2
X
CH
3
(CH
2
)
4
CH
2
X
Functional group X and boiling point, H11543C (1 atm)
X H11549 F
H1100278
H1100232
H110023
65
92
X H11549 Cl
H1100224
12
47
108
134
X H11549 Br
3
38
71
129
155
X H11549 I
42
72
103
157
180
X H11549 OH
65
78
97
138
157
FIGURE 4.4 Hydrogen
bonding in ethanol involves
the oxygen of one molecule
and the proton of an ±OH
group of another. Hydrogen
bonding is much stronger
than most other types of
dipole–dipole attractive
forces.
Hydrogen bonds between
±OH groups are stronger
than those between ±NH
groups, as a comparison of
the boiling points of water
(H
2
O, 100°C) and ammonia
(NH
3
, H1100233°C) demonstrates.
For a discussion concerning
the boiling point behavior of
alkyl halides, see the January
1988 issue of the Journal of
Chemical Education,
pp. 62–64.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
With respect to the halogen in a group of alkyl halides, the boiling point increases
as one descends the periodic table; alkyl fluorides have the lowest boiling points, alkyl
iodides the highest. This trend matches the order of increasing polarizability of the halo-
gens. Polarizability is the ease with which the electron distribution around an atom is
distorted by a nearby electric field and is a significant factor in determining the strength
of induced-dipole/induced-dipole and dipole/induced-dipole attractions. Forces that
depend on induced dipoles are strongest when the halogen is a highly polarizable iodine,
and weakest when the halogen is a nonpolarizable fluorine.
The boiling points of the chlorinated derivatives of methane increase with the num-
ber of chlorine atoms because of an increase in the induced-dipole/induced-dipole attrac-
tive forces.
Fluorine is unique among the halogens in that increasing the number of fluorines
does not produce higher and higher boiling points.
Thus, although the difluoride CH
3
CHF
2
boils at a higher temperature than CH
3
CH
2
F, the
trifluoride CH
3
CF
3
boils at a lower temperature than either of them. Even more striking
is the observation that the hexafluoride CF
3
CF
3
is the lowest boiling of any of the fluo-
rinated derivatives of ethane. The boiling point of CF
3
CF
3
is, in fact, only 11° higher
than that of ethane itself. The reason for this behavior has to do with the very low polar-
izability of fluorine and a decrease in induced-dipole/induced-dipole forces that accom-
panies the incorporation of fluorine substituents into a molecule. Their weak intermole-
cular attractive forces give fluorinated hydrocarbons (fluorocarbons) certain desirable
physical properties such as that found in the “no stick” Teflon coating of frying pans.
Teflon is a polymer (Section 6.21) made up of long chains of ±CF
2
CF
2
±units.
Solubility in Water. Alkyl halides and alcohols differ markedly from one another in
their solubility in water. All alkyl halides are insoluble in water, but low-molecular-
weight alcohols (methyl, ethyl, n-propyl, and isopropyl) are soluble in water in all pro-
portions. Their ability to participate in intermolecular hydrogen bonding not only affects
the boiling points of alcohols, but also enhances their water solubility. Hydrogen-bonded
networks of the type shown in Figure 4.5, in which alcohol and water molecules asso-
ciate with one another, replace the alcohol–alcohol and water–water hydrogen-bonded
networks present in the pure substances.
Higher alcohols become more “hydrocarbon-like” and less water-soluble.
1-Octanol, for example, dissolves to the extent of only 1 mL in 2000 mL of water. As
the alkyl chain gets longer, the hydrophobic effect (Section 2.14) becomes more impor-
tant, to the point that it, more than hydrogen bonding, governs the solubility of alcohols.
Density. Alkyl fluorides and chlorides are less dense, and alkyl bromides and iodides
more dense, than water.
1,1-Difluoroethane
H1100225°C
CH
3
CHF
2
1,1,1-Trifluoroethane
H1100247°C
CH
3
CF
3
Hexafluoroethane
H1100278°C
CF
3
CF
3
Fluoroethane
H1100232°C
CH
3
CH
2
F
Boiling
point:
Dichloromethane
(methylene dichloride)
40°C
CH
2
Cl
2
Trichloromethane
(chloroform)
61°C
CHCl
3
Tetrachloromethane
(carbon tetrachloride)
77°C
CCl
4
Chloromethane
(methyl chloride)
H1100224°C
CH
3
Cl
Boiling
point:
132 CHAPTER FOUR Alcohols and Alkyl Halides
These boiling points illus-
trate why we should do
away with the notion that
boiling points always in-
crease with increasing molec-
ular weight.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
Because alkyl halides are insoluble in water, a mixture of an alkyl halide and water sep-
arates into two layers. When the alkyl halide is a fluoride or chloride, it is the upper
layer and water is the lower. The situation is reversed when the alkyl halide is a bro-
mide or an iodide. In these cases the alkyl halide is the lower layer. Polyhalogenation
increases the density. The compounds CH
2
Cl
2
, CHCl
3
, and CCl
4
, for example, are all
more dense than water.
All liquid alcohols have densities of approximately 0.8 g/mL and are, therefore,
less dense than water.
4.6 ACIDS AND BASES: GENERAL PRINCIPLES
A solid understanding of acid–base chemistry is a big help in understanding chemical
reactivity. This and the next section review some principles and properties of acids and
bases and examine how these principles apply to alcohols.
According to the theory proposed by Svante Arrhenius, a Swedish chemist and
winner of the 1903 Nobel Prize in chemistry, an acid ionizes in aqueous solution to lib-
erate protons (H
H11001
, hydrogen ions), whereas bases ionize to liberate hydroxide ions
(HO
H11002
). A more general theory of acids and bases was devised independently by Johannes
Br?nsted (Denmark) and Thomas M. Lowry (England) in 1923. In the Br?nsted–Lowry
approach, an acid is a proton donor, and a base is a proton acceptor.
H11001H11001B
Base
B H
H11001
Conjugate
acid
A
H11002
Conjugate
base
Acid
H A
4.6 Acids and Bases: General Principles 133
0.89 g/mL
CH
3
(CH
2
)
6
CH
2
Cl
1.12 g/mL
CH
3
(CH
2
)
6
CH
2
Br
1.34 g/mL
CH
3
(CH
2
)
6
CH
2
I
0.80 g/mL
CH
3
(CH
2
)
6
CH
2
F
Density
(20°C):
FIGURE 4.5 Hydrogen
bonding between molecules
of ethanol and water.
Curved arrow notation is
used to show the electron
pair of the base abstracting a
proton from the acid. The
pair of electrons in the H±A
bond becomes an unshared
pair in the anion
H11002
:
A. Curved
arrows track electron move-
ment, not atomic movement.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
The Br?nsted–Lowry definitions of acids and bases are widely used in organic
chemistry. As noted in the preceding equation, the conjugate acid of a substance is
formed when it accepts a proton from a suitable donor. Conversely, the proton donor is
converted to its conjugate base. A conjugate acid–base pair always differ by a single
proton.
PROBLEM 4.6 Write an equation for the reaction of ammonia (
:
NH
3
) with hydro-
gen chloride (HCl). Use curved arrows to track electron movement, and identify
the acid, base, conjugate acid, and conjugate base.
In aqueous solution, an acid transfers a proton to water. Water acts as a Br?nsted base.
The systematic name for the conjugate acid of water (H
3
O
H11001
) is oxonium ion. Its com-
mon name is hydronium ion.
The strength of an acid is measured by its acid dissociation constant or
ionization constant K
a
.
K
a
H11005
Table 4.2 lists a number of Br?nsted acids and their acid dissociation constants.
Strong acids are characterized by K
a
values that are greater than that for hydronium ion
(H
3
O
H11001
, K
a
H11005 55). Essentially every molecule of a strong acid transfers a proton to water
in dilute aqueous solution. Weak acids have K
a
values less than that of H
3
O
H11001
; they are
incompletely ionized in dilute aqueous solution.
A convenient way to express acid strength is through the use of pK
a
, defined as
follows:
pK
a
H11005H11002log
10
K
a
Thus, water, with K
a
H11005 1.8 H11003 10
H1100216
, has a pK
a
of 15.7; ammonia, with K
a
H11015
10
H1100236
, has a pK
a
of 36. The stronger the acid, the larger the value of its K
a
and the
smaller the value of pK
a
. Water is a very weak acid, but is a far stronger acid than ammo-
nia. Table 4.2 includes pK
a
as well as K
a
values for acids. Because both systems are
widely used, you should practice converting K
a
to pK
a
and vice versa.
PROBLEM 4.7 Hydrogen cyanide (HCN) has a pK
a
of 9.1. What is its K
a
? Is HCN
a strong or a weak acid?
An important part of the Br?nsted–Lowry picture of acids and bases concerns the
relative strengths of an acid and its conjugate base. The stronger the acid, the weaker
the conjugate base, and vice versa. Ammonia (NH
3
) is the second weakest acid in Table
4.2. Its conjugate base, amide ion (H
2
N
H11002
), is therefore the second strongest base.
Hydroxide (HO
H11002
) is a moderately strong base, much stronger than the halide ions F
H11002
,
Cl
H11002
, Br
H11002
, and I
H11002
, which are very weak bases. Fluoride is the strongest base of the halides
but is 10
12
times less basic than hydroxide ion.
[H
3
O
H11001
][A
H11002
]
[HA]
H11001H11001
H
H
O
Water
(base)
H A
Acid
A
H11002
Conjugate
base
Conjugate
acid of water
H
H
HO
H11001
134 CHAPTER FOUR Alcohols and Alkyl Halides
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
PROBLEM 4.8 As noted in Problem 4.7, hydrogen cyanide (HCN) has a pK
a
of
9.1. Is cyanide ion (CN
H11002
) a stronger base or a weaker base than hydroxide ion
(HO
H11002
)?
In any proton-transfer process the position of equilibrium favors formation of the
weaker acid and the weaker base.
Table 4.2 is set up so that the strongest acid is at the top of the acid column, with the
strongest base at the bottom of the conjugate base column. An acid will transfer a pro-
ton to the conjugate base of any acid that lies below it in the table, and the equilibrium
constant for the reaction will be greater than one.
Table 4.2 contains both inorganic and organic compounds. Organic compounds are
similar to inorganic ones when the functional groups responsible for their acid–base prop-
erties are the same. Thus, alcohols (ROH) are similar to water (HOH) in both their Br?n-
sted acidity (ability to donate a proton from oxygen) and Br?nsted basicity (ability to
accept a proton on oxygen). Just as proton transfer to a water molecule gives oxonium
ion (hydronium ion, H
3
O
H11001
), proton transfer to an alcohol gives an alkyloxonium ion
(ROH
2
H11001
).
Stronger acid H11001 stronger base weaker acid H11001 weaker base
K H11022 1
4.6 Acids and Bases: General Principles 135
TABLE 4.2
Acid Dissociation Constants K
a
and pK
a
Values for Some
Br?nsted Acids*
HI
HBr
HCl
HOSO
2
OH
H±NH
3
HOH
CH
3
OH
CH
3
CH
2
OH
(CH
3
)
2
CHOH
(CH
3
)
3
COH
H
2
NH
(CH
3
)
2
NH
Formula
?
CH
3
COH
O
X
H11001
H11001
H±OH
2
HF
Acid
Hydrogen iodide
Hydrogen bromide
Hydrogen chloride
Sulfuric acid
Hydronium ion
Hydrogen fluoride
Acetic acid
Ammonium ion
Water
Methanol
Ethanol
Isopropyl alcohol
tert-Butyl alcohol
Ammonia
Dimethylamine
H11015H1100210
H11015H110029
H11015H110027
H110024.8
H110021.7
3.5
4.7
9.2
15.7
H1101516
H1101516
H1101517
H1101518
H1101536
H1101536
pK
a
CH
3
CO
H11002
O
X
I
H11002
Br
H11002
Cl
H11002
HOSO
2
O
H11002
H
2
O
F
H11002
NH
3
HO
H11002
CH
3
O
H11002
CH
3
CH
2
O
H11002
(CH
3
)
2
CHO
H11002
(CH
3
)
3
CO
H11002
H
2
N
H11002
(CH
3
)
2
N
H11002
Conjugate
base
H1101510
10
H1101510
9
H1101510
7
1.6 H11003 10
5
1.8 H11003 10
H110025
5.6 H11003 10
H1100210
1.8 H11003 10
H1100216?
H1101510
H1100216
H1101510
H1100216
H1101510
H1100217
H1101510
H1100218
H1101510
H1100236
H1101510
H1100236
Dissociation
constant, K
a
55
3.5 H11003 10
H110024
*Acid strength decreases from top to bottom of the table. Strength of conjugate base increases from top
to bottom of the table.
?
The most acidic proton—the one that is lost on ionization—is highlighted.
?
The “true” K
a
for water is 1 H11003 10
H1100214
. Dividing this value by 55.5 (the number of moles of water in 1 L of
water) gives a K
a
of 1.8 H11003 10
H1100216
and puts water on the same concentration basis as the other substances in
the table. A paper in the May 1990 issue of the Journal of Chemical Education (p. 386) outlines the
justification for this approach. For a dissenting view, see the March 1992 issue of the Journal of Chemical
Education (p. 255).
This is one of the most
important equations in
chemistry.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
We shall see that several important reactions of alcohols involve strong acids either as
reagents or as catalysts to increase the rate of reaction. In all these reactions the first step
is formation of an alkyloxonium ion by proton transfer from the acid to the oxygen of
the alcohol.
PROBLEM 4.9 Write an equation for proton transfer from hydrogen chloride to
tert-butyl alcohol. Use curved arrows to track electron movement, and identify
the acid, base, conjugate acid, and conjugate base.
PROBLEM 4.10 Is the equilibrium constant for proton transfer from hydrogen
chloride to tert-butyl alcohol greater than 1 or less than 1?
Alkyl halides are neither very acidic nor very basic and are absent from Table 4.2.
In general, compounds, including alkyl halides, in which all the protons are bonded to
carbon are exceedingly weak acids—too weak to be included in the table.
4.7 ACID–BASE REACTIONS: A MECHANISM FOR PROTON
TRANSFER
Potential energy diagrams of the type used in Chapter 3 to describe conformational
processes can also help us understand more about chemical reactions. Consider the trans-
fer of a proton from hydrogen bromide to water:
A potential energy diagram for this reaction is shown in Figure 4.6. Because the
transfer of a proton from hydrogen bromide to water is exothermic, the products are
placed lower in energy than the reactants. The diagram depicts the reaction as occurring
in a single elementary step. An elementary step is one that involves only one transition
state. A reaction can proceed by way of a single elementary step, in which case it is
described as a concerted reaction, or by a series of elementary steps. In the case of pro-
ton transfer from hydrogen bromide to water, breaking of the H±Br bond and making
of the H
2
O
H11001
±H bond occur “in concert” with each other. The species present at the
transition state is not a stable structure and cannot be isolated or examined directly. Its
structure is assumed to be one in which the proton being transferred is partially bonded
to both bromine and oxygen simultaneously, although not necessarily to the same extent.
Br H O
H
H
H9254H11002 H9254H11001
H11001Br
H11002
H
H
H
O
H11001
H H11001
H
H
Br O
H11001H11001
R
H
O
Alcohol
H A
Acid
A
H11002
Conjugate baseAlkyloxonium ion
R
H
HO
H11001
136 CHAPTER FOUR Alcohols and Alkyl Halides
Dashed lines in transition-
state structures represent
partial bonds, that is, bonds
in the process of being made
or broken.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
The molecularity of an elementary step is given by the number of species that
undergo a chemical change in that step. The elementary step
HBr H11001 H
2
O BA Br
H11002
H11001 H
3
O
H11001
is bimolecular because it involves one molecule of hydrogen bromide and one mole-
cule of water.
PROBLEM 4.11 Represent the structure of the transition state for proton trans-
fer from hydrogen chloride to tert-butyl alcohol.
Proton transfer from hydrogen bromide to water and alcohols ranks among the
most rapid chemical processes and occurs almost as fast as the molecules collide with
one another. Thus the height of the energy barrier separating reactants and products, the
activation energy for proton transfer, must be quite low.
The concerted nature of proton transfer contributes to its rapid rate. The energy
cost of breaking the H±Br bond is partially offset by the energy released in making the
H
2
O
H11001
±H bond. Thus, the activation energy is far less than it would be for a hypothet-
ical stepwise process involving an initial, unassisted ionization of the H±Br bond, fol-
lowed by a combination of the resulting H
H11001
with water.
4.8 PREPARATION OF ALKYL HALIDES FROM ALCOHOLS AND
HYDROGEN HALIDES
Much of what organic chemists do is directed toward practical goals. Chemists in the
pharmaceutical industry synthesize new compounds as potential drugs for the treatment
of disease. Agricultural chemicals designed to increase crop yields include organic com-
pounds used for weed control, insecticides, and fungicides. Among the “building block”
molecules used as starting materials to prepare new substances, alcohols and alkyl halides
are especially valuable.
The procedures to be described in the remainder of this chapter use either an alkane
or an alcohol as the starting material for preparing an alkyl halide. By knowing how to
4.8 Preparation of Alkyl Halides From Alcohols and Hydrogen Halides 137
Potential energy
Transition state
Reaction coordinate
H
2
O---H---Br
H
2
O H11001 H–Br
H
3
O
H11001
H11001 Br
H11002
E
act
H9254H11001
H9254H11002
The 1967 Nobel Prize in
chemistry was shared by
Manfred Eigen, a German
chemist who developed
novel methods for measur-
ing the rates of very fast
reactions such as proton
transfers.
FIGURE 4.6 Energy
diagram for concerted bi-
molecular proton transfer
from hydrogen bromide to
water.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
prepare alkyl halides, we can better appreciate the material in later chapters, where alkyl
halides figure prominently in key chemical transformations. The preparation of alkyl
halides also serves as a focal point to develop the principles of reaction mechanisms.
We’ll begin with the preparation of alkyl halides from alcohols by reaction with hydro-
gen halides.
The order of reactivity of the hydrogen halides parallels their acidity: HI H11022
HBr H11022 HCl H11022H11022 HF. Hydrogen iodide is used infrequently, however, and the reaction
of alcohols with hydrogen fluoride is not a useful method for the preparation of alkyl
fluorides.
Among the various classes of alcohols, tertiary alcohols are observed to be the
most reactive and primary alcohols the least reactive.
Tertiary alcohols are converted to alkyl chlorides in high yield within minutes on
reaction with hydrogen chloride at room temperature and below.
Secondary and primary alcohols do not react with hydrogen chloride at rates fast
enough to make the preparation of the corresponding alkyl chlorides a method of prac-
tical value. Therefore, the more reactive hydrogen halide HBr is used; even then, ele-
vated temperatures are required in order to increase the rate of reaction.
The same kind of transformation may be carried out by heating an alcohol with
sodium bromide and sulfuric acid.
CH
3
CH
2
CH
2
CH
2
OH
1-Butanol
(n-butyl alcohol)
CH
3
CH
2
CH
2
CH
2
Br
1-Bromobutane (70–83%)
(n-butyl bromide)
NaBr, H
2
SO
4
heat
H
2
O
Water
H11001Br
Bromocyclohexane (73%)
80–100°C
HBr
Hydrogen bromide
H11001OH
Cyclohexanol
H11001
Water
H
2
OH11001CH
3
(CH
2
)
5
CH
2
Br
1-Bromoheptane
(87–90%)
HBr
Hydrogen
bromide
CH
3
(CH
2
)
5
CH
2
OH
1-Heptanol
120°C
H11001H11001(CH
3
)
3
COH
2-Methyl-2-propanol
(tert-butyl alcohol)
HCl
Hydrogen chloride
(CH
3
)
3
CCl
2-Chloro-2-methylpropane
(tert-butyl chloride) (78–88%)
H
2
O
Water
25°C
Increasing reactivity of alcohols
toward hydrogen halides
H11021 H11021CH
3
OH
Methyl
Least reactive
RCH
2
OH
Primary
R
2
CHOH
Secondary
H11021 R
3
COH
Tertiary
Most reactive
H11001H11001R±OH
Alcohol
H±X
Hydrogen halide
R±X
Alkyl halide
H±OH
Water
138 CHAPTER FOUR Alcohols and Alkyl Halides
The efficiency of a synthetic
transformation is normally
expressed as a percent yield,
or percentage of the theo-
retical yield. Theoretical
yield is the amount of prod-
uct that could be formed if
the reaction proceeded to
completion and did not lead
to any products other than
those given in the equation.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
We’ll often find it convenient to write chemical equations in the abbreviated form shown
here, in which reagents, especially inorganic ones, are not included in the body of the
equation but instead are indicated over the arrow. Inorganic products—in this case,
water—are usually omitted. These simplifications focus our attention on the organic reac-
tant and its functional group transformation.
PROBLEM 4.12 Write chemical equations for the reaction that takes place
between each of the following pairs of reactants:
(a) 2-Butanol and hydrogen bromide
(b) 3-Ethyl-3-pentanol and hydrogen chloride
(c) 1-Tetradecanol and hydrogen bromide
SAMPLE SOLUTION (a) An alcohol and a hydrogen halide react to form an alkyl
halide and water. In this case 2-bromobutane was isolated in 73% yield.
4.9 MECHANISM OF THE REACTION OF ALCOHOLS WITH
HYDROGEN HALIDES
The reaction of an alcohol with a hydrogen halide is a substitution. A halogen, usually
chlorine or bromine, replaces a hydroxyl group as a substituent on carbon. Calling the
reaction a substitution tells us the relationship between the organic reactant and its prod-
uct but does not reveal the mechanism. In developing a mechanistic picture for a par-
ticular reaction, we combine some basic principles of chemical reactivity with experi-
mental observations to deduce the most likely sequence of elementary steps.
Consider the reaction of tert-butyl alcohol with hydrogen chloride:
The generally accepted mechanism for this reaction is presented as a series of three ele-
mentary steps in Figure 4.7. We say “generally accepted” because a reaction mechanism
can never be proved to be correct. A mechanism is our best present assessment of how
a reaction proceeds and must account for all experimental observations. If new experi-
mental data appear that conflict with the mechanism, the mechanism must be modified
to accommodate them. If the new data are consistent with the proposed mechanism, our
confidence grows that it is likely to be correct.
We already know about step 1 of the mechanism outlined in Figure 4.7; it is an
example of a Br?nsted acid–base reaction of the type discussed in Section 4.6 and formed
the basis of Problems 4.9 through 4.11.
Steps 2 and 3, however, are new to us. Step 2 involves dissociation of an alky-
loxonium ion to a molecule of water and a carbocation, a species that contains a posi-
tively charged carbon. In step 3, this carbocation reacts with chloride ion to yield tert-
butyl chloride. Both the alkyloxonium ion and the carbocation are intermediates in the
reaction. They are not isolated, but are formed in one step and consumed in another dur-
ing the passage of reactants to products. If we add the equations for steps 1 through 3
together, the equation for the overall process results. A valid reaction mechanism must
H11001H11001(CH
3
)
3
COH
tert-Butyl
alcohol
HCl
Hydrogen
chloride
(CH
3
)
3
CCl
tert-Butyl
chloride
H
2
O
Water
H11001H11001HBr
Hydrogen bromide
H
2
O
Water2-Butanol
CH
3
CHCH
2
CH
3
OH
2-Bromobutane
CH
3
CHCH
2
CH
3
Br
4.9 Mechanism of the Reaction of Alcohols With Hydrogen Halides 139
If you have not already writ-
ten out the solutions to
Problems 4.9 to 4.11, you
should do so now.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
account for the consumption of all reactants and the formation of all products, be they
organic or inorganic. So that we may better understand the chemistry expressed in steps
2 and 3, we need to examine carbocations in more detail.
4.10 STRUCTURE, BONDING, AND STABILITY OF CARBOCATIONS
Carbocations are classified as primary, secondary, or tertiary according to the number of
carbons that are directly attached to the positively charged carbon. They are named by
appending “cation” as a separate word after the IUPAC name of the appropriate alkyl
group. The chain is numbered beginning with the positively charged carbon (the posi-
tive charge is always at C-1).
Common names that have been incorporated into IUPAC nomenclature such as isopropyl,
sec-butyl, and so on, are permitted. Thus 1,1-dimethylethyl cation (CH
3
)
3
C
H11001
may be
called tert-butyl cation.
1-Methylcyclohexyl cation
(tertiary carbocation)
CH
3
H11001
1-Ethylbutyl cation
(secondary carbocation)
H
CH
2
CH
3
CH
3
CH
2
CH
2
C
H11001
Pentyl cation
(primary carbocation)
H
H
CH
3
CH
2
CH
2
CH
2
C
H11001
140 CHAPTER FOUR Alcohols and Alkyl Halides
(CH
3
)
3
C±O H11001 H±Cl BA (CH
3
)
3
C±O±H H11001 Cl
H11002
W
H
Overall Reaction:
tert-Butyl
alcohol
Hydrogen
chloride
tert-Butyl
chloride
Water
Step 1: Protonation of tert-butyl alcohol to give an oxonium ion:
Step 2: Dissociation of tert-butyloxonium ion to give a carbocation:
Step 3: Capture of tert-butyl cation by chloride ion:
(CH
3
)
3
COH H11001 HCl ±￡ (CH
3
)
3
CCl H11001 HOH
tert-Butyl
alcohol
tert-Butyl
cation
Hydrogen
chloride
tert-Butyloxonium
ion
Chloride
ion
W
H
H11001
tert-Butyl
cation
Chloride
ion
tert-Butyl
chloride
(CH
3
)
3
C±O±H BA (CH
3
)
3
C
H11001
H11001 O±H
W
H
W
H
H11001
Watertert-Butyloxonium
ion
(CH
3
)
3
C
H11001
H11001 Cl
H11002
±￡ (CH
3
)
3
C±Cl
Carbocations are sometimes
called carbonium ions or car-
benium ions. An article in
the November 1986 issue of
the Journal of Chemical Edu-
cation, pp. 930–933, traces
the historical development
of these and related terms.
An electrostatic potential
map of tert-butyl cation appears
on the first page of this chapter.
FIGURE 4.7 The mechanism
of formation of tert-butyl
chloride from tert-butyl alco-
hol and hydrogen chloride.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
The properties of carbocations are intimately related to their structure, and so let’s
think about the bonding in methyl cation, CH
3
H11001
. The positively charged carbon con-
tributes three valence electrons, and each hydrogen contributes one for a total of six elec-
trons, which are used to form three C±H H9268 bonds. As we saw in Section 1.17, carbon
is sp
2
-hybridized when it is bonded to three atoms or groups. We therefore choose the
sp
2
hybridization model for bonding shown in Figure 4.8. Carbon forms H9268 bonds to three
hydrogens by overlap of its sp
2
orbitals with hydrogen 1s orbitals. The three H9268 bonds
are coplanar. Remaining on carbon is an unhybridized 2p orbital that contains no elec-
trons. The axis of this empty p orbital is perpendicular to the plane defined by the three
H9268 bonds.
Evidence from a variety of sources convinces us that carbocations can exist, but
are relatively unstable. When carbocations are involved in chemical reactions, it is as
reactive intermediates, formed in one step and consumed rapidly thereafter.
Numerous studies have shown that the more stable a carbocation is, the faster it
is formed. These studies also demonstrate that alkyl groups directly attached to the pos-
itively charged carbon stabilize a carbocation. Thus, the observed order of carbocation
stability is
As carbocations go, CH
3
H11001
is particularly unstable, and its existence as an intermediate
in chemical reactions has never been demonstrated. Primary carbocations, although more
stable than CH
3
H11001
, are still too unstable to be involved as intermediates in chemical reac-
tions. The threshold of stability is reached with secondary carbocations. Many reactions,
including the reaction of secondary alcohols with hydrogen halides, are believed to
involve secondary carbocations. The evidence in support of tertiary carbocation inter-
mediates is stronger yet.
PROBLEM 4.13 Of the isomeric C
5
H
11
H11001
carbocations, which one is the most
stable?
Because alkyl groups stabilize carbocations, we conclude that they release elec-
trons to the positively charged carbon, dispersing the positive charge. They do this
through a combination of effects. One involves polarization of the H9268 bonds to the posi-
tively charged carbon. As illustrated for ethyl cation in Figure 4.9, the positively charged
carbon draws the electrons in its H9268 bonds toward itself and away from the atoms attached
to it. Electrons in a C±C H9268 bond are more polarizable than those in a C±H bond, so
replacing hydrogens by alkyl groups reduces the net charge on the sp
2
-hybridized car-
bon. The electron-donating or electron-withdrawing effect of a group that is transmitted
through H9268 bonds is called an inductive effect.
Increasing carbocation stability
tert-Butyl
cation
(tertiary)
Most stable
C
CH
3
CH
3
H
3
C
H11001
Isopropyl
cation
(secondary)
C
H
CH
3
H
3
C
H11001
Ethyl
cation
(primary)
C
H
HH
3
C
H11001
Methyl
cation
Least stable
C
H
HH
H11001
H11021 H11021H11021
4.10 Structure, Bonding, and Stability of Carbocations 141
FIGURE 4.8 Structure
of methyl cation CH
3
H11001
. Car-
bon is sp
2
-hybridized. Each
hydrogen is attached to car-
bon by a H9268 bond formed by
overlap of a hydrogen 1s
orbital with an sp
2
hybrid or-
bital of carbon. All four
atoms lie in the same plane.
The unhybridized 2p orbital
of carbon is unoccupied, and
its axis is perpendicular to
the plane of the atoms.
FIGURE 4.9 The
charge in ethyl cation is sta-
bilized by polarization of the
electron distribution in the H9268
bonds to the positively
charged carbon atom. Alkyl
groups release electrons bet-
ter than hydrogen.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
A second effect, called hyperconjugation, is also important. We’ll again consider
ethyl cation, but this time direct our attention to the electrons in the C±H bonds of the
methyl group. Figure 4.10 illustrates how an orbital associated with the methyl group
can overlap with the vacant p orbital of the positively charged carbon to give an extended
orbital that encompasses both CH
3
and C
H11001
. This allows the electrons of the methyl group
to be shared by both carbons (thereby increasing their delocalization) and to stabilize the
carbocation. Notice that according to hyperconjugation, electrons in the C±H bond of
a
H11001
C±C±H unit are more stabilizing than
H11001
C±H electrons. Thus, successive replace-
ment of the hydrogens attached to CH
3
H11001
by alkyl groups increases the opportunities for
hyperconjugation, which is consistent with the observed order of increasing carbocation
stability: methyl H11021 primary H11021 secondary H11021 tertiary. Finally, although we have devel-
oped this picture for hyperconjugation of a
H11001
C±C±H unit, it also applies to
H11001
C±C±C as well as many others.
The positive charge on carbon and the vacant p orbital combine to make carboca-
tions strongly electrophilic (“electron-loving,” or “electron-seeking”). Nucleophiles are
just the opposite. A nucleophile is “nucleus-seeking”; it has an unshared pair of elec-
trons that it can use to form a covalent bond. Step 3 of the mechanism of the reaction
of tert-butyl alcohol with hydrogen chloride is an example of a reaction between an elec-
trophile and a nucleophile and is depicted from a structural perspective in Figure 4.11.
142 CHAPTER FOUR Alcohols and Alkyl Halides
FIGURE 4.10 Hyper-
conjugation in ethyl cation.
Ethyl cation is stabilized by
delocalization of the elec-
trons in the C±H bonds of
the methyl group into the
vacant 2p orbital of the posi-
tively charged carbon.
FIGURE 4.11 Combi-
nation of a carbocation and
a halide anion to give an
alkyl halide.
R
R
R
R
R
R
C
––––
C
X
X
R
R
R
XC
––––
Carbocation
(electrophile)
Halide anion
(nucleophile)
Alkyl halide
H11001 H11002
P
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
The crucial electronic interaction is between an unshared electron pair of the nucleophilic
chloride anion and the vacant 2p orbital of the electrophilic carbocation.
Figure 4.12 maps the electrostatic potential in methyl cation and shows that the
region of positive charge coincides with where we expect the vacant 2p orbital to be—
centered on carbon and above and below the plane of the atoms.
A number of years ago G. N. Lewis extended our understanding of acid–base
behavior to include reactions other than proton transfers. According to Lewis, an acid is
an electron-pair acceptor and a base is an electron-pair donor. Thus, carbocations are
electron-pair acceptors and are Lewis acids. Halide anions are electron-pair donors and
are Lewis bases. It is generally true that electrophiles are Lewis acids, and nucleophiles
are Lewis bases.
4.11 POTENTIAL ENERGY DIAGRAMS FOR MULTISTEP REACTIONS:
THE S
N
1 MECHANISM
The mechanism for the reaction of tert-butyl alcohol with hydrogen chloride presented
in Figure 4.7 involves a sequence of three elementary steps. Each step has its own tran-
sition state, and the potential energy diagram in Figure 4.13 for the overall process is a
composite of the energy diagrams for the three steps.
Reading from left to right in Figure 4.13, the first maximum corresponds to the
transition state for proton transfer from hydrogen chloride to tert-butyl alcohol. This step
is bimolecular. The proton that is transferred is partially bonded both to chlorine and to
the oxygen of the alcohol at the transition state.
4.11 Potential Energy Diagrams for Multistep Reactions: The S
N
1 Mechanism 143
FIGURE 4.12 Electro-
static potential map of
methyl cation (CH
3
H11001
). The re-
gions of lowest electron den-
sity are blue, are centered on
carbon, and are located
above and below the plane
defined by the four atoms.
Potential energy
[(CH
3
)
3
C---OH
2
] Cl
H11002
(CH
3
)
3
C
H11001
H
2
O Cl
H11002
(CH
3
)
3
COH
HCl
[(CH
3
)
3
C---Cl] H
2
O
(CH
3
)
3
CCl
H
2
O
Reaction coordinate
(CH
3
)
3
COH
2
Cl
H11002
[(CH
3
)
3
CO---H---Cl]
H
E
act
H9254H11001
H11001
H9254H11002
H9254H11001
H9254H11002
H9254H11001
H9254H11001
FIGURE 4.13 Energy dia-
gram depicting the interme-
diates and transition states
involved in the reaction of
tert-butyl alcohol with hy-
drogen chloride.
H11001(CH
3
)
3
C
H
O
tert-Butyl
alcohol
Hydrogen
chloride
H Cl
Transition state
for proton transfer
(CH
3
)
3
C
H
HO Cl
H9254H11002
H9254H11001
(CH
3
)
3
CO
H
H
H11001
tert-Butyloxonium ion
Cl
H11002
Chloride
ion
H11001
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
This is a rapid process, and therefore the activation energy for the first step is relatively
low.
Once formed, the alkyloxonium ion dissociates by cleavage of its carbon–oxygen
bond, giving a carbocation.
Only one species, the alkyloxonium ion, undergoes a chemical change in this step, mak-
ing it unimolecular. Unlike the bimolecular proton transfer step that precedes it, in which
formation of a new bond accompanies the cleavage of an old one, unimolecular disso-
ciation of the alkyloxonium ion gives a carbocation without simultaneous formation of
a new bond. Thus, the activation energy for carbocation formation is relatively high.
In the third step, the carbocation intermediate is captured by a chloride ion, and
the energy barrier for this cation–anion combination is relatively low. The transition state
is characterized by partial bond formation between the nucleophile (chloride anion) and
the electrophile (tert-butyl cation).
Two species, the carbocation and the anion, react in this step, making it bimolecular.
Note that molecularity refers only to individual elementary steps in a multistep mecha-
nism, not to the overall reaction itself. Step 1 of the mechanism (proton transfer) is
bimolecular, step 2 (dissociation of the alkyloxonium ion) is unimolecular, and step 3
(cation–anion combination) is bimolecular.
Of the three steps in the mechanism, step 2 has the highest activation energy and
is the slowest step. A reaction can proceed no faster than its slowest step, which is
referred to as the rate-determining step. In the reaction of tert-butyl alcohol with hydro-
gen chloride, formation of the carbocation by dissociation of the alkyloxonium ion is the
rate-determining step.
Substitution reactions, of which the reaction of alcohols with hydrogen halides is
but one example, will be discussed in more detail in Chapter 8. There, we will make
extensive use of a shorthand notation for a mechanism originally introduced by Sir
Christopher Ingold in the 1930s. Ingold proposed the symbol, S
N
, to stand for substitu-
tion nucleophilic, to be followed by the number 1 or 2 according to whether the rate-
determining step is unimolecular or bimolecular. The reaction of tert-butyl alcohol with
hydrogen chloride, for example, is said to follow an S
N
1 mechanism because its slow
step (dissociation of tert-butyloxonium ion) is unimolecular.
H11001 C
H
3
C
H9254H11001
H9254H11002
H
3
C
CH
3
Cl
Transition state
for cation–anion
combination
C
H
3
C
H
3
C
CH
3
Cl
tert-Butyl
chloride
C
H11001
CH
3
H
3
CCH
3
tert-Butyl
cation
Cl
H11002
Chloride
anion
H11001C
H
3
C
H
3
C
H
H
CH
3
O
H11001
tert-Butyloxonium ion
C
H
3
C
O
H
H
H9254H11001 H9254H11001
H
3
C
CH
3
Transition state
for dissociation
of alkyloxonium ion
C
H11001
CH
3
H
3
CCH
3
tert-Butyl
cation
Water
H
H
O
144 CHAPTER FOUR Alcohols and Alkyl Halides
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.12 Effect of Alcohol Structure on Reaction Rate 145
4.12 EFFECT OF ALCOHOL STRUCTURE ON REACTION RATE
We saw in Section 4.8 that the reactivity of alcohols with hydrogen halides increases in
the order primary H11021 secondary H11021 tertiary. To be valid, the mechanism proposed in Fig-
ure 4.7 and represented by the energy diagram in Figure 4.13 must account for this order
of relative reactivity. When considering rate effects, we focus on the slow step of a reac-
tion mechanism and analyze how that step is influenced by changes in reactants or reac-
tion conditions.
As mentioned, the slow step in the S
N
1 mechanism is the dissociation of the alkyl-
oxonium ion to the carbocation. The rate of this step is proportional to the concentra-
tion of the alkyloxonium ion:
Rate H11005 k[alkyloxonium ion]
where k is a constant of proportionality called the rate constant. The value of k is related
to the activation energy for alkyloxonium ion dissociation and is different for different
alkyloxonium ions. A low activation energy implies a large value of k and a rapid rate
of alkyloxonium ion dissociation. Conversely, a large activation energy is characterized
by a small k for dissociation and a slow rate.
Consider what happens when the alkyloxonium ion dissociates to a carbocation
and water. The positive charge resides mainly on oxygen in the alkyloxonium ion but is
shared between oxygen and carbon at the transition state.
The transition state for carbocation formation begins to resemble the carbocation. If we
assume that structural features that stabilize carbocations also stabilize transition states
that have carbocation character, it follows that alkyloxonium ions derived from tertiary
alcohols have a lower energy of activation for dissociation and are converted to their
corresponding carbocations faster than those derived from secondary and primary alco-
hols. Figure 4.14 depicts the effect of alkyloxonium ion structure on the activation energy
for, and thus the rate of, carbocation formation. Once the carbocation is formed, it is
rapidly captured by halide ion, so that the rate of alkyl halide formation is governed by
the rate of carbocation formation.
Inferring the structure of the transition state on the basis of what is known about
the species that lead to it or may be formed by way of it is a practice with a long his-
tory in organic chemistry. A justification of this practice was advanced in 1955 by George
S. Hammond, who reasoned that if two states, such as a transition state and an inter-
mediate derived from it, are similar in energy, then they are similar in structure. This
rationale is known as Hammond’s postulate. In the formation of a carbocation from an
alkyloxonium ion, the transition state is closer in energy to the carbocation than it is to
the alkyloxonium ion, and so its structure more closely resembles the carbocation and it
responds in a similar way to the stabilizing effects of alkyl substituents.
H11001C O
H11001
H
H
Alkyloxonium
ion
CO
H
H
H9254H11001 H9254H11001
Transition state
C
H11001
Carbocation
H
H
O
Water
The rate of any chemical re-
action increases with increas-
ing temperature. Thus the
value of k for a reaction is
not constant, but increases
as the temperature increases.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
146 CHAPTER FOUR Alcohols and Alkyl Halides
4.13 REACTION OF PRIMARY ALCOHOLS WITH HYDROGEN
HALIDES. THE S
N
2 MECHANISM
Unlike tertiary and secondary carbocations, primary carbocations are too high in energy
to be intermediates in chemical reactions. Since primary alcohols are converted, albeit
rather slowly, to alkyl halides on treatment with hydrogen halides, they must follow some
other mechanism that avoids carbocation intermediates. This alternative mechanism is
believed to be one in which the carbon–halogen bond begins to form before the car-
bon–oxygen bond of the alkyloxonium ion is completely broken.
The halide nucleophile helps to “push off” a water molecule from the alkyloxonium ion.
According to this mechanism, both the halide ion and the alkyloxonium ion are involved
in the same bimolecular elementary step. In Ingold’s terminology, introduced in Section
4.11 and to be described in detail in Chapter 8, nucleophilic substitutions characterized
by a bimolecular rate-determining step are given the mechanistic symbol S
N
2.
PROBLEM 4.14 1-Butanol and 2-butanol are converted to their corresponding
bromides on being heated with hydrogen bromide. Write a suitable mechanism
for each reaction, and assign each the appropriate symbol (S
N
1 or S
N
2).
H11001X
H11002
Halide
ion
RCH
2
OH
2
H11001
Primary
alkyloxonium
ion
CH
2
R
X OH
2
H9254H11002
H9254H11001
Transition state
CH
2
RX
Primary
alkyl halide
H11001 H
2
O
Water
Energy
CH
3
--- OH
2
RCH
2
--- OH
2
R
2
CH
--- OH
2
CH
3
+
, H
2
O
RCH
2
H11001
, H
2
O
R
2
CH
H11001
, H
2
O
R
3
C
--- OH
2
R
3
C
+
, H
2
O
Methyloxonium
ion
A primary
alkyloxonium ion
RCH
2
OH
2
A secondary
alkyloxonium ion
R
2
CHOH
2
A tertiary
alkyloxonium ion
R
3
COH
2
CH
3
OH
2
E
act
E
act
E
act
E
act
H11001H11001H11001H11001
H9254H11001 H9254H11001
H9254H11001 H9254H11001
H9254H11001 H9254H11001
H9254H11001 H9254H11001
FIGURE 4.14 Diagrams com-
paring energies of activation
for formation of carboca-
tions from alkyloxonium ions
of methyl, primary, second-
ary, and tertiary alcohols.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.14 OTHER METHODS FOR CONVERTING ALCOHOLS TO ALKYL
HALIDES
Alkyl halides are such useful starting materials for preparing other functional group types
that chemists have developed several different methods for converting alcohols to alkyl
halides. Two methods, based on the inorganic reagents thionyl chloride and phosphorus
tribromide, bear special mention.
Thionyl chloride reacts with alcohols to give alkyl chlorides. The inorganic byprod-
ucts in the reaction, sulfur dioxide and hydrogen chloride, are both gases at room tem-
perature and are easily removed, making it an easy matter to isolate the alkyl chloride.
Because tertiary alcohols are so readily converted to chlorides with hydrogen chlo-
ride, thionyl chloride is used mainly to prepare primary and secondary alkyl chlorides.
Reactions with thionyl chloride are normally carried out in the presence of potassium
carbonate or the weak organic base pyridine.
Phosphorus tribromide reacts with alcohols to give alkyl bromides and phospho-
rous acid.
Phosphorous acid is water-soluble and may be removed by washing the alkyl halide with
water or with dilute aqueous base.
Thionyl chloride and phosphorus tribromide are specialized reagents used to bring
about particular functional group transformations. For this reason, we won’t present the
mechanisms by which they convert alcohols to alkyl halides, but instead will limit our-
selves to those mechanisms that have broad applicability and enhance our knowledge of
fundamental principles. In those instances you will find that a mechanistic understand-
ing is of great help in organizing the reaction types of organic chemistry.
(CH
3
)
2
CHCH
2
OH
Isobutyl alcohol
(CH
3
)
2
CHCH
2
Br
Isobutyl bromide (55–60%)
PBr
3
PBr
3
HOH
Cyclopentanol
HBr
Cyclopentyl bromide (78–84%)
H11001H110013ROH
Alcohol
PBr
3
Phosphorus
tribromide
3RBr
Alkyl
bromide
H
3
PO
3
Phosphorous
acid
CH
3
CH(CH
2
)
5
CH
3
OH
W
2-Octanol
CH
3
CH(CH
2
)
5
CH
3
Cl
W
2-Chlorooctane (81%)
SOCl
2
K
2
CO
3
(CH
3
CH
2
)
2
CHCH
2
OH
2-Ethyl-1-butanol
(CH
3
CH
2
)
2
CHCH
2
Cl
1-Chloro-2-ethylbutane (82%)
SOCl
2
pyridine
H11001H11001ROH
Alcohol
SOCl
2
Thionyl
chloride
RCl
Alkyl
chloride
SO
2
Sulfur
dioxide
H11001 HCl
Hydrogen
chloride
4.14 Other Methods for Converting Alcohols to Alkyl Halides 147
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.15 HALOGENATION OF ALKANES
The rest of this chapter describes a second method for preparing alkyl halides, one that
uses alkanes as reactants. It involves substitution of a halogen atom for one of the
alkane’s hydrogens.
The alkane is said to undergo fluorination, chlorination, bromination, or iodination
according to whether X
2
is F
2
, Cl
2
, Br
2
, or I
2
, respectively. The general term is
halogenation. Chlorination and bromination are the most widely used.
The reactivity of the halogens decreases in the order F
2
H11022 Cl
2
H11022 Br
2
H11022 I
2
. Fluo-
rine is an extremely aggressive oxidizing agent, and its reaction with alkanes is strongly
exothermic and difficult to control. Direct fluorination of alkanes requires special equip-
ment and techniques, is not a reaction of general applicability, and will not be discussed
further.
Chlorination of alkanes is less exothermic than fluorination, and bromination less
exothermic than chlorination. Iodine is unique among the halogens in that its reaction with
alkanes is endothermic and alkyl iodides are never prepared by iodination of alkanes.
4.16 CHLORINATION OF METHANE
The gas-phase chlorination of methane is a reaction of industrial importance and leads
to a mixture of chloromethane (CH
3
Cl), dichloromethane (CH
2
Cl
2
), trichloromethane
(CHCl
3
), and tetrachloromethane (CCl
4
) by sequential substitution of hydrogens.
One of the chief uses of chloromethane is as a starting material from which sili-
cone polymers are made. Dichloromethane is widely used as a paint stripper.
Trichloromethane was once used as an inhalation anesthetic, but its toxicity caused it to
be replaced by safer materials many years ago. Tetrachloromethane is the starting mate-
rial for the preparation of several chlorofluorocarbons (CFCs), at one time widely used
as refrigerant gases. In 1987, most of the world’s industrialized nations agreed to phase
out all uses of CFCs by the year 2000 because these compounds have been implicated
in atmospheric processes that degrade the earth’s ozone layer.
H11001H11001CH
4
Methane
Cl
2
Chlorine
CH
3
Cl
Chloromethane
(bp H1100224°C)
HCl
Hydrogen
chloride
400–440°C
H11001H11001CH
3
Cl
Chloromethane
Cl
2
Chlorine
CH
2
Cl
2
Dichloromethane
(bp 40°C)
HCl
Hydrogen
chloride
400–440°C
H11001H11001CH
2
Cl
2
Dichloromethane
Cl
2
Chlorine
CHCl
3
Trichloromethane
(bp 61°C)
HCl
Hydrogen
chloride
400–440°C
H11001H11001CHCl
3
Trichloromethane
Cl
2
Chlorine
CCl
4
Tetrachloromethane
(bp 77°C)
HCl
Hydrogen
chloride
400–440°C
H11001H11001R±H
Alkane
X
2
Halogen
R±X
Alkyl halide
H±X
Hydrogen halide
148 CHAPTER FOUR Alcohols and Alkyl Halides
Volume II of Organic Reac-
tions, an annual series that
reviews reactions of interest
to organic chemists, contains
the statement “Most organic
compounds burn or explode
when brought in contact
with fluorine.”
Chlorination of methane
provides approximately one-
third of the annual U.S. pro-
duction of chloromethane.
The reaction of methanol
with hydrogen chloride is
the major synthetic method
for the preparation of
chloromethane.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.17 Structure and Stability of Free Radicals 149
The chlorination of methane is carried out at rather high temperatures (400–440°C),
even though each substitution in the series is exothermic. The high temperature provides
the energy to initiate the reaction. The term “initiation step” has a specific meaning in
organic chemistry, one that is related to the mechanism of the reaction. This mechanism,
to be presented in Section 4.18, is fundamentally different from the mechanism by which
alcohols react with hydrogen halides. Alcohols are converted to alkyl halides in reac-
tions involving ionic (or “polar”) intermediates—alkyloxonium ions and carbocations.
The intermediates in the chlorination of methane and other alkanes are quite different;
they are neutral (“nonpolar”) species called free radicals.
4.17 STRUCTURE AND STABILITY OF FREE RADICALS
Free radicals are species that contain unpaired electrons. The octet rule notwithstand-
ing, not all compounds have all of their electrons paired. Oxygen (O
2
) is the most famil-
iar example of a compound with unpaired electrons; it has two of them. Compounds that
have an odd number of electrons, such as nitrogen dioxide (NO
2
), must have at least one
unpaired electron.
Nitrogen monoxide (“nitric oxide”) is another stable free radical. Although known for
hundreds of years, NO has only recently been discovered to be an extremely important
biochemical messenger and moderator of so many biological processes that it might be
better to ask “Which ones is it not involved in?”
The free radicals that we usually see in carbon chemistry are much less stable than
these. Simple alkyl radicals, for example, require special procedures for their isolation
and study. We will encounter them here only as reactive intermediates, formed in one
step of a reaction mechanism and consumed in the next. Alkyl radicals are classified as
primary, secondary, or tertiary according to the number of carbon atoms directly attached
to the carbon that bears the unpaired electron.
An alkyl radical is neutral and has one more electron than the corresponding car-
bocation. Thus, bonding in methyl radical may be approximated by simply adding an
electron to the vacant 2p orbital of sp
2
-hybridized carbon in methyl cation (Figure 4.15a).
Alternatively, we could assume that carbon is sp
3
-hybridized and place the unpaired elec-
tron in an sp
3
orbital (Figure 4.15b).
Of the two extremes, experimental studies indicate that the planar sp
2
model
describes the bonding in alkyl radicals better than the pyramidal sp
3
model. Methyl rad-
ical is planar, and more highly substituted radicals such as tert-butyl radical are flattened
pyramids closer in shape to that expected for sp
2
-hybridized carbon than for sp
3
.
Free radicals, like carbocations, have an unfilled 2p orbital and are stabilized by
substituents, such as alkyl groups, that release electrons. Consequently, the order of free-
radical stability parallels that of carbocations.
HC
H
H
Methyl
radical
R
H
H
C
Primary
radical
R
H
R
C
Secondary
radical
R
R
R
C
Tertiary
radical
O±O
Oxygen
O?N±O
Nitrogen dioxide
N?O
Nitrogen monoxide
The journal Science selected
nitric oxide as its “Molecule
of the Year” for 1992.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
150 CHAPTER FOUR Alcohols and Alkyl Halides
PROBLEM 4.15 Write a structural formula for the most stable of the free radi-
cals that have the formula C
5
H
11
.
Some of the evidence indicating that alkyl substituents stabilize free radicals comes
from bond energies. The strength of a bond is measured by the energy required to break
it. A covalent bond can be broken in two ways. In a homolytic cleavage a bond between
two atoms is broken so that each of them retains one of the electrons in the bond.
In contrast, in a heterolytic cleavage one fragment retains both electrons.
We assess the relative stability of alkyl radicals by measuring the enthalpy change
(H9004H°) for the homolytic cleavage of a C±H bond in an alkane:
The more stable the radical, the lower the energy required to generate it by C±H bond
homolysis.
RRHH11001H
X
H11001
H11001 Y
H11002
X Y
Heterolytic bond cleavage
XXH11001Y Y
Homolytic bond cleavage
HC
H
H
Methyl
radical
(least stable) (most stable)
R
H
H
C
Primary
radical
R
H
R
C
Secondary
radical
R
R
R
C
Tertiary
radical
Increasing free radical stability
A curved arrow shown as a
single-barbed fishhook
signifies the movement of
one electron. “Normal”
curved arrows track the
movement of a pair of elec-
trons.
120H11034
Half-filled
2p orbital
(a)
109.5H11034
Half-filled
sp
3
orbital
(b)
FIGURE 4.15 Orbital hy-
bridization models of bond-
ing in methyl radical. (a) If
the structure of the CH
3
radi-
cal is planar, then carbon is
sp
2
-hybridized with an un-
paired electron in a 2p or-
bital. (b) If CH
3
is pyramidal,
carbon is sp
3
-hybridized with
an electron in an sp
3
orbital.
Model (a) is more consistent
with experimental observa-
tions.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.17 Structure and Stability of Free Radicals 151
The energy required for homolytic bond cleavage is called the bond dissociation
energy (BDE). A list of some bond dissociation energies is given in Table 4.3.
As the table indicates, C±H bond dissociation energies in alkanes are approxi-
mately 375 to 435 kJ/mol (90–105 kcal/mol). Homolysis of the H±CH
3
bond in
methane gives methyl radical and requires 435 kJ/mol (104 kcal/mol). The dissociation
energy of the H±CH
2
CH
3
bond in ethane, which gives a primary radical, is somewhat
less (410 kJ/mol, or 98 kcal/mol) and is consistent with the notion that ethyl radical (pri-
mary) is more stable than methyl.
The dissociation energy of the terminal C±H bond in propane is exactly the same
as that of ethane. The resulting free radical is primary (RC
H11554
H
2
) in both cases.
H9004H° H11005 H11001410 kJ
(98 kcal)
H11001
Propane
CH
3
CH
2
CH
2
H
n-Propyl
radical
(primary)
CH
3
CH
2
CH
2
Hydrogen
atom
H
TABLE 4.3
Bond Dissociation Energies of Some Representative
Compounds*
Bond dissociation
energy
Bond dissociation
energy
(kcal/mol)
(104)
(38)
(58)
(46)
(36)
(104)
(98)
(98)
(95)
(98)
(91)
(108)
(83.5)
(70)
(56)
(81)
(82)
(119)
(102)
(91)
kJ/mol
435
159
242
192
150
435
410
410
397
410
380
451
349
293
234
338
343
497
426
380
kJ/mol
568
431
366
297
368
355
351
334
439
339
284
330
263
380
385
380
Bond
H±H
F±F
Cl±Cl
Br±Br
I±I
CH
3
±H
CH
3
CH
2
±H
CH
3
CH
2
CH
2
±H
(CH
3
)
2
CH±H
(CH
3
)
2
CHCH
2
±H
(CH
3
)
3
C±H
Diatomic molecules
Alkanes
CH
3
±F
CH
3
±Cl
CH
3
±Br
CH
3
±I
CH
3
CH
2
±Cl
CH
3
CH
2
CH
2
±Cl
Alkyl halides
HO±H
CH
3
O±H
CH
3
±OH
Water and alcohols
Bond
H±F
H±Cl
H±Br
H±I
CH
3
±CH
3
CH
3
CH
2
±CH
3
(CH
3
)
2
CH±CH
3
(CH
3
)
3
C±CH
3
(CH
3
)
2
CH±F
(CH
3
)
2
CH±Cl
(CH
3
)
2
CH±Br
(CH
3
)
3
C±Cl
(CH
3
)
3
C±Br
CH
3
CH
2
±OH
(CH
3
)
2
CH±OH
(CH
3
)
3
C±OH
(kcal/mol)
(136)
(103)
(87.5)
(71)
(88)
(85)
(84)
(80)
(105)
(81)
(68)
(79)
(63)
(91)
(92)
(91)
*Bond dissociation energies refer to bond indicated in structural formula for each substance.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
152 CHAPTER FOUR Alcohols and Alkyl Halides
Note, however, that Table 4.3 includes two entries for propane. The second entry corre-
sponds to the cleavage of a bond to one of the hydrogens of the methylene (CH
2
) group.
It requires slightly less energy to break a C±H bond in the methylene group than in the
methyl group.
Since the starting material (propane) and one of the products (HH11554) are the same in both
processes, the difference in bond dissociation energies is equal to the energy difference
between an n-propyl radical (primary) and an isopropyl radical (secondary). As depicted
in Figure 4.16, the secondary radical is 13 kJ/mol (3 kcal/mol) more stable than the pri-
mary radical.
Similarly, by comparing the bond dissociation energies of the two different types
of C±H bonds in 2-methylpropane, we see that a tertiary radical is 30 kJ/mol (7
kcal/mol) more stable than a primary radical.
H9004H° H11005 H11001410 kJ
(98 kcal)
H11001
Isobutyl
radical
(primary)
CH
3
CHCH
2
CH
3
Hydrogen
atom
H
2-Methylpropane
CH
3
CHCH
2
CH
3
H
H9004H° H11005 H11001380 kJ
(91 kcal)
H11001
tert-Butyl
radical
(tertiary)
CH
3
CCH
3
CH
3
Hydrogen
atom
H
2-Methylpropane
CH
3
CCH
3
CH
3
H
H9004H° H11005 H11001397 kJ
(95 kcal)
H11001
Propane
CH
3
CHCH
3
H
Isopropyl
radical
(secondary)
CH
3
CHCH
3
Hydrogen
atom
H
Energy
Propyl radical (primary)
Isopropyl radical (secondary)
CH
3
CH
2
CH
2
H11001 H?
CH
3
CH
2
CH
3
Propane
CH
3
CHCH
3
H11001 H?
13 kJ/mol
(3 kcal/mol)
?
?
410 kJ/mol
(98 kcal/mol)
397 kJ/mol
(95 kcal/mol)
FIGURE 4.16 Diagram show-
ing how bond dissociation
energies of methylene and
methyl C±H bonds in
propane reveal a difference
in stabilities between two
isomeric free radicals. The
secondary radical is more sta-
ble than the primary.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.18 Mechanism of Methane Chlorination 153
PROBLEM 4.16 Carbon–carbon bond dissociation energies have been measured
for alkanes. Without referring to Table 4.3, identify the alkane in each of the fol-
lowing pairs that has the lower carbon–carbon bond dissociation energy, and
explain the reason for your choice.
(a) Ethane or propane
(b) Propane or 2-methylpropane
(c) 2-Methylpropane or 2,2-dimethylpropane
SAMPLE SOLUTION (a) First write the equations that describe homolytic car-
bon–carbon bond cleavage in each alkane.
Cleavage of the carbon–carbon bond in ethane yields two methyl radicals,
whereas propane yields an ethyl radical and one methyl radical. Ethyl radical is
more stable than methyl, and so less energy is required to break the carbon–car-
bon bond in propane than in ethane. The measured carbon–carbon bond dissoci-
ation energy in ethane is 368 kJ/mol (88 kcal/mol), and that in propane is 355
kJ/mol (85 kcal/mol).
Like carbocations, most free radicals are exceedingly reactive species—too reac-
tive to be isolated but capable of being formed as transient intermediates in chemical
reactions. Methyl radical, as we shall see in the following section, is an intermediate in
the chlorination of methane.
4.18 MECHANISM OF METHANE CHLORINATION
The generally accepted mechanism for the chlorination of methane is presented in Fig-
ure 4.17. As we noted earlier (section 4.16), the reaction is normally carried out in the
gas phase at high temperature. The reaction itself is strongly exothermic, but energy must
be put into the system in order to get it going. This energy goes into breaking the weak-
est bond in the system, which, as we see from the bond dissociation energy data in Table
4.3, is the Cl±Cl bond with a bond dissociation energy of 242 kJ/mol (58 kcal/mol).
The step in which Cl±Cl bond homolysis occurs is called the initiation step.
Each chlorine atom formed in the initiation step has seven valence electrons and
is very reactive. Once formed, a chlorine atom abstracts a hydrogen atom from methane
as shown in step 2 in Figure 4.17. Hydrogen chloride, one of the isolated products from
the overall reaction, is formed in this step. A methyl radical is also formed, which then
attacks a molecule of Cl
2
in step 3. Attack of methyl radical on Cl
2
gives chloromethane,
the other product of the overall reaction, along with a chlorine atom which then cycles
back to step 2, repeating the process. Steps 2 and 3 are called the propagation steps of
the reaction and, when added together, give the overall equation for the reaction. Since
one initiation step can result in a great many propagation cycles, the overall process is
called a free-radical chain reaction.
PROBLEM 4.17 Write equations for the initiation and propagation steps for the
formation of dichloromethane by free-radical chlorination of chloromethane.
Ethane
CH
3
CH
3
Two methyl radicals
H11001CH
3
CH
3
Propane
CH
3
CH
3
CH
2
H11001
Ethyl radical
CH
3
CH
2
Methyl radical
CH
3
The bond dissociation energy
of the other reactant,
methane, is much higher. It
is 435 kJ/mol (104 kcal/mol).
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
154 CHAPTER FOUR Alcohols and Alkyl Halides
In practice, side reactions intervene to reduce the efficiency of the propagation
steps. The chain sequence is interrupted whenever two odd-electron species combine to
give an even-electron product. Reactions of this type are called chain-terminating steps.
Some commonly observed chain-terminating steps in the chlorination of methane are
shown in the following equations.
Combination of a methyl radical with a chlorine atom:
Combination of two methyl radicals:
CH
3
CH
3
EthaneTwo methyl radicals
CH
3
CH
3
Chloromethane
CH
3
Cl
Chlorine atom
Cl
Methyl radical
CH
3
Cl
(a) Initiation
Step 1: Dissociation of a chlorine molecule into two chlorine atoms:
Chlorine molecule Two chlorine atoms
2[
(b) Chain propagation
Step 2: Hydrogen atom abstraction from methane by a chlorine atom:
Chlorine atom Methane Hydrogen chloride Methyl radical
H11001 CH
3
Step 3: Reaction of methyl radical with molecular chlorine:
Chlorine molecule Methyl radical
H11001
Chlorine atom Chloromethane
H11001
(c) Sum of steps 2 and 3
Methane Chlorine Chloromethane Hydrogen
chloride
]Cl Cl
HHCH
3
Cl H11001 Cl
Cl Cl CH
3
ClCH
3
Cl
±￡
±￡
CH
4
H11001 Cl
2
CH
3
Cl H11001 HCl
FIGURE 4.17 Equations de-
scribing the initiation and
propagation steps in the
free-radical mechanism for
the chlorination of methane.
Together the two propaga-
tion steps give the overall
equation for the reaction.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.18 Mechanism of Methane Chlorination 155
FROM BOND ENERGIES TO HEATS OF REACTION
Y
ou have seen that measurements of heats of
reaction, such as heats of combustion, can pro-
vide quantitative information concerning the
relative stability of constitutional isomers (Section
2.15) and stereoisomers (Section 3.12). The box in
Section 2.15 described how heats of reaction can be
manipulated arithmetically to generate heats of for-
mation (H9004H°
f
) for many molecules. The following ma-
terial shows how two different sources of thermo-
chemical information, heats of formation and bond
dissociation energies (Table 4.3), can reveal whether
a particular reaction is exothermic or endothermic
and by how much.
Consider the chlorination of methane to
chloromethane. The heats of formation of the reac-
tants and products appear beneath the equation.
These heats of formation for the chemical com-
pounds are taken from published tabulations; the
heat of formation of chlorine, as it is for all elements,
is zero.
The overall heat of reaction is given by
H9004H° H11005 (heats of formation of products) H11002
(heats of formation of reactants)
H9004H° H11005 (H1100281.9 kJ H11002 92.3 kJ) H11002 (H1100274.8 kJ) H11005H1100299.4 kJ
Thus, the chlorination of methane is calculated to be
an exothermic reaction on the basis of heat of forma-
tion data.
The same conclusion is reached using bond dis-
sociation energies. The following equation shows the
bond dissociation energies of the reactants and prod-
ucts taken from Table 4.3:
435
CH
4
BDE:
(kJ/mol)
242
Cl
2
H11001H11001
349
CH
3
Cl
431
HCl
H20858
H20858
H1100274.8
CH
4
H9004H°
f
:
(kJ/mol)
0
Cl
2
H11001H11001
H1100281.9
CH
3
Cl
H1100292.3
HCl
Because stronger bonds are formed at the expense of
weaker ones, the reaction is exothermic and
H9004H° H11005 (BDE of bonds broken) H11002
(BDE of bonds formed)
H9004H° H11005 (435 kJ H11001 242 kJ) H11002
(349 kJ H11001 431 kJ) H11005H11002103 kJ
This value is in good agreement with that obtained
from heat of formation data.
Compare chlorination of methane with iodina-
tion. The relevant bond dissociation energies are
given in the equation.
H9004H° H11005 (BDE of bonds broken) H11002
(BDE of bonds formed)
H9004H° H11005 (435 kJ H11001 150 kJ) H11002
(234 kJ H11001 297 kJ) H11005H1100154 kJ
A positive value for H9004H° signifies an endothermic re-
action. The reactants are more stable than the prod-
ucts, and so iodination of alkanes is not a feasible re-
action. You would not want to attempt the
preparation of iodomethane by iodination of
methane.
A similar analysis for fluorination of methane
gives H9004H° H11005H11002426 kJ for its heat of reaction. Fluori-
nation of methane is four times as exothermic as
chlorination. A reaction this exothermic, if it also oc-
curs at a rapid rate, can proceed with explosive vio-
lence.
Bromination of methane is exothermic, but less
exothermic than chlorination. The value calculated
from bond dissociation energies is H9004H° H11005H1100230 kJ. Al-
though bromination of methane is energetically fa-
vorable, economic considerations cause most of the
methyl bromide prepared commercially to be made
from methanol by reaction with hydrogen bromide.
H20858
H20858
435
CH
4
BDE:
(kJ/mol)
150
I
2
H11001H11001
234
CH
3
I
297
HI
H20858
H20858
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
156 CHAPTER FOUR Alcohols and Alkyl Halides
Combination of two chlorine atoms:
Termination steps are, in general, less likely to occur than the propagation steps.
Each of the termination steps requires two free radicals to encounter each other in a
medium that contains far greater quantities of other materials (methane and chlorine mol-
ecules) with which they can react. Although some chloromethane undoubtedly arises via
direct combination of methyl radicals with chlorine atoms, most of it is formed by the
propagation sequence shown in Figure 4.17.
4.19 HALOGENATION OF HIGHER ALKANES
Like the chlorination of methane, chlorination of ethane is carried out on an industrial
scale as a high-temperature gas-phase reaction.
As in the chlorination of methane, it is often difficult to limit the reaction to monochlo-
rination, and derivatives having more than one chlorine atom are also formed.
PROBLEM 4.18 Chlorination of ethane yields, in addition to ethyl chloride, a
mixture of two isomeric dichlorides. What are the structures of these two dichlo-
rides?
In the laboratory it is more convenient to use light, either visible or ultraviolet, as
the source of energy to initiate the reaction. Reactions that occur when light energy is
absorbed by a molecule are called photochemical reactions. Photochemical techniques
permit the reaction of alkanes with chlorine to be performed at room temperature.
Methane, ethane, and cyclobutane share the common feature that each one can give
only a single monochloro derivative. All the hydrogens of cyclobutane, for example, are
equivalent, and substitution of any one gives the same product as substitution of any
other. Chlorination of alkanes in which all the hydrogens are not equivalent is more com-
plicated in that a mixture of every possible monochloro derivative is formed, as the chlo-
rination of butane illustrates:
CH
3
CH
2
CH
2
CH
3
Butane
H11001CH
3
CH
2
CH
2
CH
2
Cl
1-Chlorobutane (28%)
(n-butyl chloride)
CH
3
CHCH
2
CH
3
Cl
W
2-Chlorobutane (72%)
(sec-butyl chloride)
Cl
2
hH9263, 35°C
H11001 H11001
Hydrogen
chloride
HCl
Cyclobutane Chlorine
Cl
2
Chlorocyclobutane (73%)
(cyclobutyl chloride)
Cl
hH9263
H11001H11001CH
3
CH
3
Ethane
Cl
2
Chlorine
CH
3
CH
2
Cl
Chloromethane (78%)
(ethyl chloride)
HCl
Hydrogen chloride
420°C
Chlorine molecule
Cl
2
Two chlorine atoms
ClCl
Photochemical energy is indi-
cated by writing “light” or
“hH9263” above the arrow. The
symbol hH9263 is equal to the en-
ergy of a light photon and
will be discussed in more de-
tail in Section 13.1.
The percentages cited in this
equation reflect the compo-
sition of the monochloride
fraction of the product mix-
ture rather than the isolated
yield of each component.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
These two products arise because in one of the propagation steps a chlorine atom may
abstract a hydrogen atom from either a methyl or a methylene group of butane.
The resulting free radicals react with chlorine to give the corresponding alkyl chlorides.
Butyl radical gives only 1-chlorobutane; sec-butyl radical gives only 2-chlorobutane.
If every collision of a chlorine atom with a butane molecule resulted in hydrogen abstrac-
tion, the n-butyl/sec-butyl radical ratio and, therefore, the 1-chloro/2-chlorobutane ratio,
would be given by the relative numbers of hydrogens in the two equivalent methyl groups
of CH
3
CH
2
CH
2
CH
3
(six) compared with those in the two equivalent methylene groups
(four). The product distribution expected on a statistical basis would be 60% 1-chlorobu-
tane and 40% 2-chlorobutane. The experimentally observed product distribution, how-
ever, is 28% 1-chlorobutane and 72% 2-chlorobutane. sec-Butyl radical is therefore
formed in greater amounts, and n-butyl radical in lesser amounts, than expected statisti-
cally.
The reason for this behavior stems from the greater stability of secondary com-
pared with primary free radicals. The transition state for the step in which a chlorine
atom abstracts a hydrogen from carbon has free-radical character at carbon.
A secondary hydrogen is abstracted faster than a primary hydrogen because the transi-
tion state with secondary radical character is more stable than the one with primary rad-
ical character. The same factors that stabilize a secondary radical stabilize a transition
state with secondary radical character more than one with primary radical character.
Hydrogen atom abstraction from a CH
2
group occurs faster than from a CH
3
group. We
can calculate how much faster a single secondary hydrogen is abstracted compared with
a single primary hydrogen from the experimentally observed product distribution.
Transition state for abstraction
of a secondary hydrogen
Transition state for abstraction
of a primary hydrogen
Cl CH
2
CH
2
CH
2
CH
3
H
H9254 H9254
Cl CHCH
2
CH
3
CH
3
H
H9254 H9254
H11001H11001
n-Butyl radical
CH
3
CH
2
CH
2
CH
2
1-Chlorobutane
(n-butyl chloride)
CH
3
CH
2
CH
2
CH
2
Cl ClCl
2
Cl
2
H11001H11001
sec-Butyl radical
CH
3
CHCH
2
CH
3
Cl
2-Chlorobutane
(sec-butyl chloride)
CH
3
CHCH
2
CH
3
Cl
H11001H11001
n-Butyl radical
CH
3
CH
2
CH
2
CH
2
Butane
CH
3
CH
2
CH
2
CH
2
H Cl HCl
H11001H11001
sec-Butyl radical
CH
3
CHCH
2
CH
3
Cl HCl
Butane
CH
3
CHCH
2
CH
3
H
4.19 Halogenation of Higher Alkanes 157
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
158 CHAPTER FOUR Alcohols and Alkyl Halides
H11005
H11005H11003H11005
A single secondary hydrogen in butane is abstracted by a chlorine atom 3.9 times as fast
as a single primary hydrogen.
PROBLEM 4.19 Assuming the relative rate of secondary to primary hydrogen
atom abstraction to be the same in the chlorination of propane as it is in that of
butane, calculate the relative amounts of propyl chloride and isopropyl chloride
obtained in the free-radical chlorination of propane.
A similar study of the chlorination of 2-methylpropane established that a tertiary
hydrogen is removed 5.2 times faster than each primary hydrogen.
In summary then, the chlorination of alkanes is not very selective. The various
kinds of hydrogens present in a molecule (tertiary, secondary, and primary) differ by only
a factor of 5 in the relative rate at which each reacts with a chlorine atom.
Bromine reacts with alkanes by a free-radical chain mechanism analogous to that
of chlorine. There is an important difference between chlorination and bromination, how-
ever. Bromination is highly selective for substitution of tertiary hydrogens. The spread
in reactivity among primary, secondary, and tertiary hydrogens is greater than 10
3
.
In practice, this means that when an alkane contains primary, secondary, and tertiary
hydrogens, it is usually only the tertiary hydrogen that is replaced by bromine.
H11001H11001
2-Methylpentane
CH
3
CCH
2
CH
2
CH
3
CH
3
H
W
W
Bromine
Br
2
2-Bromo-2-methylpentane
(76% isolated yield)
CH
3
CCH
2
CH
2
CH
3
CH
3
Br
W
W
Hydrogen
bromide
HBr
hH9263
60°C
Relative rate (bromination)
(tertiary)
1640
R
3
CH H11022H11022
(secondary)
82
R
2
CH
2
(primary)
1
RCH
3
Relative rate (chlorination)
(tertiary)
5.2
R
3
CH H11022H11022
(secondary)
3.9
R
2
CH
2
(primary)
1
RCH
3
H11001
2-Methylpropane
CH
3
CCH
3
CH
3
H
W
W
1-Chloro-2-methylpropane
(63%) (isobutyl chloride)
CH
3
CCH
2
Cl
CH
3
H
W
W
2-Chloro-2-methylpropane (37%)
(tert-butyl chloride)
CH
3
CCH
3
CH
3
Cl
W
W
Cl
2
hH9263, 35°C
3.9
1
6
4
72
28
Rate of secondary H abstraction
Rate of primary H abstraction
rate of secondary H abstraction H11003 4 secondary hydrogens
rate of primary H abstraction H11003 6 primary hydrogens
72% 2-chlorobutane
28% 1-chlorobutane
The yield cited in this reac-
tion is the isolated yield of
purified product. Isomeric
bromides constitute only a
tiny fraction of the crude
product.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.20 Summary 159
PROBLEM 4.20 Give the structure of the principal organic product formed by
free-radical bromination of each of the following:
(a) Methylcyclopentane (c) 2,2,4-Trimethylpentane
(b) 1-Isopropyl-1-methylcyclopentane
SAMPLE SOLUTION (a) Write the structure of the starting hydrocarbon, and
identify any tertiary hydrogens that are present. The only tertiary hydrogen in
methylcyclopentane is the one attached to C-1. This is the one replaced by
bromine.
This difference in selectivity between chlorination and bromination of alkanes
needs to be kept in mind when one wishes to prepare an alkyl halide from an alkane:
1. Since chlorination of an alkane yields every possible monochloride, it is used only
when all the hydrogens in an alkane are equivalent.
2. Bromination is normally used only to prepare tertiary alkyl bromides from alkanes.
Selectivity is not an issue in the conversion of alcohols to alkyl halides. Except
for certain limitations to be discussed in Section 8.15, the location of the halogen sub-
stituent in the product corresponds to that of the hydroxyl group in the starting alcohol.
4.20 SUMMARY
Chemical reactivity and functional group transformations involving the preparation of
alkyl halides from alcohols and from alkanes are the main themes of this chapter.
Although the conversions of an alcohol or an alkane to an alkyl halide are both classi-
fied as substitutions, they proceed by very different mechanisms.
Section 4.1 Alcohols and alkyl halides may be named using either substitutive or
functional class nomenclature. In substitutive nomenclature alkyl halides
are named as halogen derivatives of alkanes. The parent is the longest
continuous chain that bears the halogen substituent, and in the absence
of other substituents the chain is numbered from the direction that gives
the lowest number to the carbon that bears the halogen. The functional
class names of alkyl halides begin with the name of the alkyl group and
end with the halide as a separate word.
Section 4.2 The substitutive names of alcohols are derived by replacing the -e end-
ing of an alkane with -ol. Functional class names of alcohols begin with
the name of the alkyl group and end in the word ”alcohol.”
Substitutive name: 2-Hexanol
Functional class name: 1-Methylpentyl alcohol
CH
3
CHCH
2
CH
2
CH
2
CH
3
OH
W
Substitutive name: 2-Bromohexane
Functional class name: 1-Methylpentyl bromide
CH
3
CHCH
2
CH
2
CH
2
CH
3
Br
W
Br
2
light
H
CH
3
Methylcyclopentane
Br
CH
3
1-Bromo-1-methylcyclopentane
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
Section 4.3 Alcohols (X H11005 OH) and alkyl halides (X H11005 F, Cl, Br, or I) are classified
as primary, secondary, or tertiary according to the degree of substitution
at the carbon that bears the functional group.
Section 4.4 The halogens (especially fluorine and chlorine) and oxygen are more
electronegative than carbon, and the carbon–halogen bond in alkyl
halides and the carbon–oxygen bond in alcohols are polar. Carbon is the
positive end of the dipole and halogen or oxygen the negative end.
Section 4.5 Dipole/induced-dipole and dipole–dipole attractive forces make alcohols
higher boiling than alkanes of similar molecular weight. The attractive
force between ±OH groups is called hydrogen bonding.
Hydrogen bonding between the hydroxyl group of an alcohol and water
makes the water-solubility of alcohols greater than that of hydrocarbons.
Low-molecular-weight alcohols [CH
3
OH, CH
3
CH
2
OH, CH
3
CH
2
CH
2
OH,
and (CH
3
)
2
CHOH] are soluble in water in all proportions. Alkyl halides
are insoluble in water.
Section 4.6 Br?nsted acids are proton donors; Br?nsted bases are proton acceptors.
Strong acids transfer protons to alcohols to form alkyloxonium ions. An
alkyloxonium ion is the conjugate acid of an alcohol.
Section 4.7 Proton transfer from a Br?nsted acid to the oxygen of water is a single-
step process and is very fast. It is a bimolecular, concerted process.
Section 4.8 See Table 4.4
Section 4.9 Secondary and tertiary alcohols react with hydrogen halides by a mech-
anism that involves formation of a carbocation intermediate in the rate-
determining step.
(1)
Alcohol
ROH H11001H11001
Hydrogen
halide
HX
H11001
Alkyloxonium
ion
ROH
2
Halide
anion
X
H11002
fast
H11001H11001
R
H
O
Alcohol
(base)
H
Alkyloxonium ion
(conjugate acid)
R
H
HO
H11001
Cl Cl
H11002
RR
H
OOH
Primary
RCH
2
X
Secondary
RCHRH11032
X
W
Tertiary
RCRH11032
X
RH11033
W
W
160 CHAPTER FOUR Alcohols and Alkyl Halides
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.20 Summary 161
TABLE 4.4 Conversions of Alcohols and Alkanes to Alkyl Halides
Reaction (section) and comments
Reactions of alcohols with hydrogen halides (Section
4.8) Alcohols react with hydrogen halides to yield
alkyl halides. The reaction is useful as a synthesis of
alkyl halides. The reactivity of hydrogen halides
decreases in the order HI H11022 HBr H11022 HCl H11022 HF. Alcohol
reactivity decreases in the order tertiary H11022 secondary
H11022 primary H11022 methyl.
Reaction of alcohols with phosphorus tribromide
(Section 4.14) As an alternative to converting alco-
hols to alkyl bromides with hydrogen bromide, the
inorganic reagent phosphorus tribromide is some-
times used.
Free-radical halogenation of alkanes (Sections 4.15
through 4.19) Alkanes react with halogens by substi-
tution of a halogen for a hydrogen on the alkane.
The reactivity of the halogens decreases in the order
F
2
H11022 Cl
2
H11022 Br
2
H11022 I
2
. The ease of replacing a hydrogen
decreases in the order tertiary H11022 secondary H11022 pri-
mary H11022 methyl. Chlorination is not very selective and
so is used only when all the hydrogens of the alkane
are equivalent. Bromination is highly selective,
replacing tertiary hydrogens much more readily than
secondary or primary ones.
General equation and specific example(s)
Reaction of alcohols with thionyl chloride (Section
4.14) Thionyl chloride is a synthetic reagent used to
convert alcohols to alkyl chlorides.
Alcohol
ROH H11001H11001
Hydrogen
halide
HX
Alkyl
halide
RX
Water
H
2
O
Alkane
RH H11001H11001
Halogen
X
2
Alkyl
halide
RX
Hydrogen
halide
HX
Alcohol
ROH H11001H11001H11001
Thionyl
chloride
SOCl
2
Alkyl
chloride
RCl
Sulfur
dioxide
SO
2
Hydrogen
chloride
HCl
Alcohol
3ROH H11001H11001
Phosphorus
tribromide
PBr
3
Alkyl
bromide
3RBr
Phosphorous
acid
H
3
PO
3
HCl
OH
CH
3
1-Methylcyclopentanol
Cl
CH
3
1-Chloro-1-
methylcyclopentane (96%)
1-Pentanol
CH
3
CH
2
CH
2
CH
2
CH
2
OH
1-Chloropentane (80%)
CH
3
CH
2
CH
2
CH
2
CH
2
Cl
SOCl
2
pyridine
2-Pentanol
CH
3
CHCH
2
CH
2
CH
3
OH
W
2-Bromopentane (67%)
CH
3
CHCH
2
CH
2
CH
3
Br
W
PBr
3
2,2,3-Trimethylbutane
(CH
3
)
2
CHC(CH
3
)
3
2-Bromo-2,3,3-
trimethylbutane (80%)
(CH
3
)
2
CC(CH
3
)
3
Br
W
Br
2
hH9263
Cyclodecane Cyclodecyl chloride (64%)
Cl
Cl
2
hH9263
(2)
(3)
Section 4.10 Carbocations contain a positively charged carbon with only three atoms
or groups attached to it. This carbon is sp
2
-hybridized and has a vacant
2p orbital.
H11001
Alkyl halide
RX
Halide ion
X
H11002
Carbocation
R
H11001
fast
H11001
Carbocation
R
H11001
H11001
Alkyloxonium ion
ROH
2
Water
H
2
O
slow
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
162 CHAPTER FOUR Alcohols and Alkyl Halides
Carbocations are stabilized by alkyl substituents attached directly to the
positively charged carbon. Alkyl groups are electron-releasing sub-
stituents. Stability increases in the order:
(least stable) CH
3
H11001
H11021 R
2
CH
H11001
H11021 R
3
C
H11001
(most stable)
Carbocations are strongly electrophilic (Lewis acids) and react with
nucleophiles (Lewis bases).
Section 4.11 The conversion of an alcohol to an alkyl halide on reaction with a hydro-
gen halide is a nucleophilic substitution. Nucleophilic substitutions (S
N
)
are classified as S
N
1 or S
N
2 according to whether the rate-determining
step is unimolecular or bimolecular.
Section 4.12 The rates at which alcohols are converted to alkyl halides depends on the
rate of carbocation formation: tertiary alcohols are most reactive; primary
alcohols and methanol are least reactive.
Section 4.13 Primary alcohols do not react with hydrogen halides by way of carboca-
tion intermediates. The nucleophilic species (Br
H11002
) attacks the alkyloxo-
nium ion and “pushes off” a water molecule from carbon in a bimolec-
ular step. This step is rate-determining, and the mechanism is S
N
2.
Section 4.14 See Table 4.4
Section 4.15 See Table 4.4
Section 4.16 Methane reacts with Cl
2
to give chloromethane, dichloromethane,
trichloromethane, and tetrachloromethane.
Section 4.17 Chlorination of methane, and halogenation of alkanes generally, proceed
by way of free-radical intermediates. Alkyl radicals are neutral and have
an unpaired electron on carbon.
Like carbocations, free radicals are stabilized by alkyl substituents. The
order of free-radical stability parallels that of carbocation stability.
Section 4.18 The elementary steps (1) through (3) describe a free-radical chain mech-
anism for the reaction of an alkane with a halogen.
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(1)
(2)
(3)
Section 4.19 See Table 4.4
PROBLEMS
4.21 Write structural formulas for each of the following alcohols and alkyl halides:
(a) Cyclobutanol (e) 2,6-Dichloro-4-methyl-4-octanol
(b) sec-Butyl alcohol (f) trans-4-tert-Butylcyclohexanol
(c) 3-Heptanol (g) 1-Cyclopropylethanol
(d) trans-2-Chlorocyclopentanol (h) 2-Cyclopropylethanol
4.22 Name each of the following compounds according to substitutive IUPAC nomenclature:
(a) (CH
3
)
2
CHCH
2
CH
2
CH
2
Br (f)
(b) (CH
3
)
2
CHCH
2
CH
2
CH
2
OH (g)
(c) Cl
3
CCH
2
Br (h)
(d) (i)
(e) CF
3
CH
2
OH
4.23 Write structural formulas, or build molecular models for all the constitutionally isomeric
alcohols of molecular formula C
5
H
12
O. Assign a substitutive and a functional class name to each
one, and specify whether it is a primary, secondary, or tertiary alcohol.
4.24 A hydroxyl group is a somewhat “smaller” substituent on a six-membered ring than is a
methyl group. That is, the preference of a hydroxyl group for the equatorial orientation is less pro-
nounced than that of a methyl group. Given this information, write structural formulas or build
molecular models for all the isomeric methylcyclohexanols, showing each one in its most stable
conformation. Give the substitutive IUPAC name for each isomer.
4.25 By assuming that the heat of combustion of the cis isomer was larger than the trans, struc-
tural assignments were made many years ago for the stereoisomeric 2-, 3-, and 4-methylcyclo-
hexanols. This assumption is valid for two of the stereoisomeric pairs but is incorrect for the other.
For which pair of stereoisomers is the assumption incorrect? Why?
OH
Cl
2
CHCHBr
W
Cl
Br
CH
3
CH
3
OH
OH
Halogen
molecule
X
2
Alkyl
halide
RX
Alkyl
radical
R
Halogen
atom
XH11001H11001(propagation step)
Halogen
atom
X
Alkyl
radical
R
Alkane
RH
Hydrogen
halide
HXH11001H11001(propagation step)
Two halogen atoms
2X
Halogen molecule
X
2
(initiation step)
Problems 163
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
4.26 (a) Menthol, used to flavor various foods and tobacco, is the most stable stereoisomer of 2-
isopropyl-5-methylcyclohexanol. Draw or make a molecular model of its most stable
conformation. Is the hydroxyl group cis or trans to the isopropyl group? To the methyl
group?
(b) Neomenthol is a stereoisomer of menthol. That is, it has the same constitution but dif-
fers in the arrangement of its atoms in space. Neomenthol is the second most stable
stereoisomer of 2-isopropyl-5-methylcyclohexanol; it is less stable than menthol but
more stable than any other stereoisomer. Write the structure, or make a molecular model
of neomenthol in its most stable conformation.
4.27 Each of the following pairs of compounds undergoes a Br?nsted acid–base reaction for
which the equilibrium lies to the right. Give the products of each reaction, and identify the acid,
the base, the conjugate acid, and the conjugate base.
(a) HI H11001 HO
H11002
BA (e) (CH
3
)
3
CO
H11002
H11001 H
2
O BA
(b) (f) (CH
3
)
2
CHOH H11001 H
2
N
H11002
BA
(c) HF H11001 H
2
N
H11002
BA (g) F
H11002
H11001 H
2
SO
4
BA
(d)
4.28 Transition-state representations are shown for two acid–base reactions. For each one, write
the equation for the reaction it represents in the direction for which the equilibrium lies to the
right. Label the acid, the base, the conjugate acid, and the conjugate base, and use curved arrows
to show the flow of electrons.
(a) (b)
4.29 Calculate K
a
for each of the following acids, given its pK
a
. Rank the compounds in order
of decreasing acidity.
(a) Aspirin: pK
a
H11005 3.48
(b) Vitamin C (ascorbic acid): pK
a
H11005 4.17
(c) Formic acid (present in sting of ants): pK
a
H11005 3.75
(d) Oxalic acid (poisonous substance found in certain berries): pK
a
H11005 1.19
4.30 The pK
a
’s of methanol (CH
3
OH) and methanethiol (CH
3
SH) are 16 and 11, respectively.
Which is more basic, KOCH
3
or KSCH
3
?
4.31 Write a chemical equation for the reaction of 1-butanol with each of the following:
(a) Sodium amide (NaNH
2
) (d) Phosphorus tribromide
(b) Hydrogen bromide, heat (e) Thionyl chloride
(c) Sodium bromide, sulfuric acid, heat
4.32 Each of the following reactions has been described in the chemical literature and involves
an organic starting material somewhat more complex than those we have encountered so far. Nev-
ertheless, on the basis of the topics covered in this chapter, you should be able to write the struc-
ture of the principal organic product of each reaction.
O
HCH
3
H O
H
H9254H11001H9254H11002H9254H11002 H9254H11002
CH
3
H
CH
3
H BrC O
HClH11001CH
3
CO
H11002
X
O
CH
3
CH
2
O
H11002
H11001 CH
3
COH
X
O
164 CHAPTER FOUR Alcohols and Alkyl Halides
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(a)
(b)
(c)
(d)
(e)
4.33 Select the compound in each of the following pairs that will be converted to the corre-
sponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason
for your choice.
(a) 1-Butanol or 2-butanol
(b) 2-Methyl-1-butanol or 2-butanol
(c) 2-Methyl-2-butanol or 2-butanol
(d) 2-Methylbutane or 2-butanol
(e) 1-Methylcyclopentanol or cyclohexanol
(f) 1-Methylcyclopentanol or trans-2-methylcyclopentanol
(g) 1-Cyclopentylethanol or 1-ethylcyclopentanol
4.34 Assuming that the rate-determining step in the reaction of cyclohexanol with hydrogen bro-
mide to give cyclohexyl bromide is unimolecular, write an equation for this step. Use curved
arrows to show the flow of electrons.
4.35 Assuming that the rate-determining step in the reaction of 1-hexanol with hydrogen bro-
mide to give 1-bromohexane is an attack by a nucleophile on an alkyloxonium ion, write an equa-
tion for this step. Use curved arrows to show the flow of electrons.
4.36 Two stereoisomers of 1-bromo-4-methylcyclohexane are formed when trans-4-methylcy-
clohexanol reacts with hydrogen bromide. Write structural formulas or make molecular models of:
(a) trans-4-Methylcylohexanol
(b) The carbocation intermediate in this reaction
(c) The two stereoisomers of 1-bromo-4-methylcyclohexane
4.37 Basing your answers on the bond dissociation energies in Table 4.3, calculate which of the
following reactions are endothermic and which are exothermic:
(a) (CH
3
)
2
CHOH H11001 HF ±￡ (CH
3
)
2
CHF H11001 H
2
O
C
10
H
15
Br
Br
2
, light
100°C
heat
HOCH
2
CH
2
CH
2
CH
2
OH H11001 2HBr
HCl
Br
C
CH
3
CH
3
OH
CH
3
OH
COCH
2
CH
3
O
SOCl
2
pyridine
CH
2
CH
2
OH
PBr
3
pyridine
Problems 165
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
(b) (CH
3
)
2
CHOH H11001 HCl ±￡ (CH
3
)
2
CHCl H11001 H
2
O
(c) CH
3
CH
2
CH
3
H11001 HCl ±￡ (CH
3
)
2
CHCl H11001 H
2
4.38 By carrying out the reaction at H1100278°C it is possible to fluorinate 2,2-dimethylpropane to
yield (CF
3
)
4
C. Write a balanced chemical equation for this reaction.
4.39 In a search for fluorocarbons having anesthetic properties, 1,2-dichloro-1,1-difluoropropane
was subjected to photochemical chlorination. Two isomeric products were obtained, one of which
was identified as 1,2,3-trichloro-1,1-difluoropropane. What is the structure of the second com-
pound?
4.40 Among the isomeric alkanes of molecular formula C
5
H
12
, identify the one that on photo-
chemical chlorination yields
(a) A single monochloride (c) Four isomeric monochlorides
(b) Three isomeric monochlorides (d) Two isomeric dichlorides
4.41 In both the following exercises, assume that all the methylene groups in the alkane are
equally reactive as sites of free-radical chlorination.
(a) Photochemical chlorination of heptane gave a mixture of monochlorides containing 15%
1-chloroheptane. What other monochlorides are present? Estimate the percentage of each
of these additional C
7
H
15
Cl isomers in the monochloride fraction.
(b) Photochemical chlorination of dodecane gave a monochloride fraction containing 19%
2-chlorododecane. Estimate the percentage of 1-chlorododecane present in that fraction.
4.42 Photochemical chlorination of 2,2,4-trimethylpentane gives four isomeric monochlorides.
(a) Write structural formulas for these four isomers.
(b) The two primary chlorides make up 65% of the monochloride fraction. Assuming that
all the primary hydrogens in 2,2,4-trimethylpentane are equally reactive, estimate the
percentage of each of the two primary chlorides in the product mixture.
4.43 Photochemical chlorination of pentane gave a mixture of three isomeric monochlorides. The
principal monochloride constituted 46% of the total, and the remaining 54% was approximately a
1
:
1 mixture of the other two isomers. Write structural formulas for the three monochloride iso-
mers and specify which one was formed in greatest amount. (Recall that a secondary hydrogen is
abstracted three times faster by a chlorine atom than a primary hydrogen.)
4.44 Cyclopropyl chloride has been prepared by the free-radical chlorination of cyclopropane.
Write a stepwise mechanism for this reaction.
4.45 Deuterium oxide (D
2
O) is water in which the protons (
1
H) have been replaced by their heav-
ier isotope deuterium (
2
H). It is readily available and is used in a variety of mechanistic studies
in organic chemistry and biochemistry. When D
2
O is added to an alcohol (ROH), deuterium
replaces the proton of the hydroxyl group.
ROH H11001 D
2
O BA ROD H11001 DOH
The reaction takes place extremely rapidly, and if D
2
O is present in excess, all the alcohol is con-
verted to ROD. This hydrogen–deuterium exchange can be catalyzed by either acids or bases. If
D
3
O
H11001
is the catalyst in acid solution and DO
H11002
the catalyst in base, write reasonable reaction mech-
anisms for the conversion of ROH to ROD under conditions of (a) acid catalysis and (b) base
catalysis.
166 CHAPTER FOUR Alcohols and Alkyl Halides
Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
