Principal Organic Chemistry, 6th Edition

Organic Chemistry, 6th Edition

,
Año:
2007
Editorial:
Prentice Hall of India
Idioma:
english
Páginas:
1282
ISBN 10:
8120307658
ISBN 13:
9788120307650
File:
PDF, 57.30 MB
Descarga (pdf, 57.30 MB)

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Organic Chemistry, 6th Edition
Organic Chemistry, 6th Edition
13 November 2013 (19:28) 
umberto
This is a fair copy, not great. Some of the text is OCR'd, but most isn't. The text is a bit fuzzy, but it's usable, and the text is complete.
30 August 2016 (05:15) 
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STRUCTURE AND PROPERTID

.

CHAP. 1

shape shown in Fig. 1.9; as with sp and sp2 orbitals, we shall neglect the small back
lobe and represent the front lobe as a sphere.

Figure 1.9 Atomic orbitals: hybrid sp3 orbitals. (a) Cross-section and
approximate shape of a single orbital. It is strongly directed along one axis.
(b) Representation as a sphere, with the small back lobe omitted. (c) Four
orbitals, with the axes directed toward the corners of a tetrahedron.

Now, how are sp3 orbitals arranged in space? The answer is no surprise to us:
in the way that lets them get as far away from each other as possible. They are
directed to the corners of a regular tetrahedron. The angle between any twoorbitals
is the tetrahedral angle 109.5" (Fig. 1.9). Just as mutual repulsion among orbitals
gives two linear bonds or three trigonal bonds, so it gives four tetrahedral bonds.
Overlap of each of the sp3 orbitals of carbon with an s orbital of hydrogen
results in methane: carbon at the center of a regular tetrahedron, and the four
hydrogens at the comers (Fig. 1.10).

Figme 1.10 Bond formation: CH, molecule. (a) Tetrahedral sp3 orbitals.
(b) Predicted shape: H nuclei located for maximum overlap. (c)Shape and
size.

SEC. 1.12

17

UNSHARED P A E 3 OF ELECLaONS

Experimentally, methane has been found to have the highly sjrmmetrical
tetrahedral structure we have assembled. Each carbon-hydrogen bond has sxaply
the same length, 1.10 A; the angle between any pair of bonds is the tetrahedral
angle 109.5". It takes 104 kcal/mol to break one of the bonds of methane.
Thus, in these last three sections, we have seen that there are associated with
covalent bonds not only characteristic bond lengths and bond dissociation energies
but also characteristic bond angles. These bond angles can be conveniently related
to the arrangement of atomic orbitals-including hybrid orbitals-involved in
bond formation; they ultimately go back to the Pauli exclusion principle and the
tendency fo; r unpaired electrons to get as far from each other as possible.
Unlike the ionic bond, which is equally strong in all directions, the covalent
bond is a directed bond. We can begin to see why the chemistry of the covalent bqnd
is so much concerned with molecular size and shape.
Since compounds of carbon are held together chiefly by covalent bonds,
organic chemistry, too, is much concerned with molecular size and shape. To help
us in our study, we should make frequent use of molecular models. Figure 1.11
shows methane as represented by three different kinds of models: stick-and-ball,
framework, and space-filling. These last are made to scale, and reflect accurately
not only bond angles but also relative lengths of bonds and sizes of atoms.

Figure 1.11 Models of methane molecule. (a) Stick-and-ball (Allyn &
Bacon). (b) Framework (Prentice Hall). (c) Space-filling (Corey-PaulingKoltun, CPK); 1.25 cm equals 1.00 A.

.

1.12 Unshared pairs of electrons
Two familiar compounds, ammonia (NH3) and water (H,O), show how
unsharedpairs of electrons can affect molecular structure.

r

N

0

0

7

0 0 0

1Hybridization
sp2

1s

sp3
A

r

N

0

0

0

\

0

0

18

STRUCTURE A ND PROPER~ES

CHAP. 1

In ammonia, nitrogen resembles the carbon of methane. Nitrogen is sp3hybridized, but (Table 1.l) has only three unpaired electrons; they occupy threeof
the sp3 orbitals. Overlap of each of these orbitals with the s orbital of a hydrogen
atom results in ammonia (Fig. 1.12). The fourth sp3 orbital of nitrogen contains a
pair of electrons.

Figure 1.12 Bond formation: NH3 molecule. (a)Tetrahedral sp3 orbitals.
(b) Predicted shape, showing the unshared pair: H nuclei located for

maximum overlap. (c) Shape and size.

If there' is to be maximum overlap and hence maximum bond strength, the
hydrogen nuclei must be located at three comers of a tetrahedron; the fourth corner
is occupied by an unshared pair of electrons. Considering only atomic nuclei, we
would expect ammonia to be shaped like a pyramid with nitrogen at the apex and
hydrogen at the comers of a triangular base. Each bond angle should be the
tetrahedral angle 109.5'.
Experimentally, ammonia is found to have the pyramidal shape calculated by
quantum mechanics. The bond angles are 107", slightly smaller than the predicted
value; it has been suggested that the unshared pair of electrons occupies more
space than any of the hydrogen atoms, and hence tends to compress the bond
angles slightly. The nitrogen-hydrogen bond length is 1.01 A; it takes i03 kcal/
mol to break one of the bonds of ammonia.
The sp3 orbital occupied by the unshared pair of electrons is a region of high
electron density. This region is a source of electrons for electron-seeking atoms
and molecules, and thus gives ammonia its basic properties (Sec. 1.22).
.

-

There are t w n e ~er
conceivableelectronic configurations for ammonia, but neither fits
the facts.
(a) Since nitrogen is bonded to three other atoms, we might have pictured it as using
sp2 orbitals, as boron does in boron trifluoride. But ammonia is not a flat molecule, and so
we must reject this possibility. It is the unshared pair of electrons on nitrogen that makes
the difference between NH, and BF,; these electrons need to stay away from those in the
carbon-hydrogen bonds, and the tetrahedral shape makes this possible.
(b) We might have pictured nitrogen as simply using the p orbitals for overlap, since
they would provide the necessary three unpaired electrons. But this would give bond angles
of 90"-remember, the p orbitals are at right angles to each other-in contrast to the
observed angles of 107". More importantly, the unshared pair would be buried in an s orbitai,
and there is evidence from dipole moments (Sec. 1.16) that this is not so. Evidently the
stability gained by using the highly directed sp3 orbitals for bond formation more than
makes up for raising the unshared pair from an s orbital to the higher-energy sp3orbital.

SEC. 1.12

19

UNSHARED PAIRS OF ELECTRONS

One further fact about ammonia: spectroscopy reveals that the molecule
undergoes inversion, that is, turns inside-out (Fig. 1.13). There is an energy bamer

H

H.. I .H

N'""

',,

It

Figore1.13 Inversion of ammonia.

of only 6 kcal/mol between one pyramidal arrangement and the other, equivalent
one. This energy is provided by molecular collisions, and even at room temperature
the fraction of collisions hard enough to do the job is so large that a rapid
transformation between pyramidal arrangements occurs.
Compare ammonia with methane, which does not undergo inversion. The
unshared pair plays the role of a carbon-hydrogen bond in detetmining the most
stable shape of the molecule, tetrahedral. But, unlike a carbon-hydrogen bond, the
unshared pair cannot maintain aparticular tetrahedralarrangement; the pair points
now in one direction, and the next instant in the opposite direction.
Finally, let us consider water, H20. The situation is similar to that for
ammonia, except that oxygen has only two unpaired electrons, and hence it bonds

0

0

0

0 0 0

sp) Hybridization

with only two hydrogen atoms, which occupy two comers of a tetrahedron. The
other two comers of the tetrahedron are occupied by unshared pairs of electrons
(Fig. 1.14).

Figure 1.14 Bond formation: H,O molecule.(a) Tetrahedral sp3 orbitals.
(8 Predicted shape, showing the unshared pairs: H nuclei located for
maximum overlap. (c) Shape and size.

20

STRUCTURE AND PROPERTIES

CHAP. I

As actually measured, the H U H angle is 105", smaller than the calculated
tetrahedral angle, and even smaller than the angle in ammonia. Here there are two
bulky unshared pairs of electrons compressing the bond angles. The oxygenhydrogen bond length is 0.96 A; it takes 118 kcal/mol to break one of the bonds of
water.
If we examine Fig. 1.15 we can see the fundamental similarity in shape of the
methane, ammonia, and water molecules: a similarity that, by,the approach we
have used, stems from a similarity in bonding.

Flgme 1.15 Models of (a) methane, (b) ammonia, (c) water.

Because of the unshared pairs of electrons on oxygen, water is basic, although
less strongly so than ammonia (Sec. 1.22).
..

,

-

,

,

---.---.

PloMcm 1.'4 Predict the shape of each c)f the follo,wingmoltxules, andI tell how 1YOU
onium ion,i NH,+; (b) the hydronium ion,
arrived at your prediction: (a) the amm~
H,O+ ;(c) methyl alcmhol, CH30H;(d) rnethylamine, CH3NHz.

1.13 Intramolecular forces
We must remember that the particular method of mentally building molecules
that we are learning to use is artificial: it is a purely intellectual process involving
imaginary overlap of imaginary orbitals. There are other, equally artificial ways
that use different mental or physical models. Our method is the one that so far has
seemed to work out best for the organic chemist. Our kit of mental atomic models
will contain just three "kinds " of carbon: tetrahedral (sp3-hybridized), trigonal
(sp2-hybridized), and digonal (sp-hybridized). By use of this kit, we shall find, one
can do an amazingly good job of building hundreds of thousands of organic
molecules.
But, however we amve at it, we see the actual structure of a molecule to be
the net result of a combination of repulsive and attractive forces, which are related
to charge and electron spin.
(a) Repulsiveforces. Electrons tend to stay as far apart as possible because they
have the same charge and also, if they are unpaired, because they have the same
spin (Pauli exclusion principle). The like-charged atomic nuclei, too, repel each
other.

SEC. 1.14

BOND DISSOCIATION ENERGY. HOMOLYSIS AND HETEROLYSIS

21

(b) Attractive forces. Electrons are attracted by atomic nuclei-as are the nuclei
by the electrons-because of their opposite charge, and hence tend to occupy the
region between two nuclei. Opposite spin permits (although, in itself, probably
does not actually encourage) two electrons to occupy the same region.
In methane, for example, the four hydrogen nuclei are as widely separated as
they can be. The distribution of the eight bonding electrons is such that each one
occupies the desirable region near two nuclei-the bond orbital-and yet, except
for its partner, is as far as possible from the other electrons. We can picture each
electron accepting-perhaps reluctantly because of their similar charges-ne
orbital-mate of opposite spin, but staying as far as possible from all other electrons
and even, as it wanders within the loose confines of its orbital, doing its best to
avoid the vicinity of its restless partner.

1.14 Bond dissociation energy. Homolysis and heterolysis
We have seen that energy is liberated when atoms combine to form a molecule.
For a molecule to break into atoms, an equivalent amount of energy must be
consumed. The amount of energy consumed or liberated when a bond is broken or
formed is known as the bond dissociation energy, D. It is characteristicof the particular bond. Table 1.2 lists bond dissociation energies that have been measured for a
number of bonds. As can be seen, they vary widely, from weak bonds like 1-1 (36
kcal/mol) to very strong bonds like H-F (136 kcal/mol). Although the accepted
values may change as experimental methods improve, certain trends are clear.

-

Table 1.2 Hohfo~mcBOND DISSOCIA~ON
ENERGIES,KCAL/MOL
A :B

A.

+ .B

H-H 104
H-F 136
H-Cl 103
H-Br 88
H-I
71

AH = Homolytic bond dissociation energy or D(A-B)

F-F
38
Cl-Cl 58
Br--Br 46
1-1
36

CH3-H
CH3-F
CH3<l
CH3-Br
CH3-I

104
108
84
70
56

.

We must not confuse bonddissociationenergy (D) with another measureof bond strength
called bond energy (E). If one begins with methane, for example, and breaks, successively,
four carbon-hydrogen bonds, one finds four different bond dissociation energies:

CH,

-

CH, + H .

D(CHJ-H) = 104 kcal/mol

CHAP. 1

STRUCIWRE AND PROPERTIES

22

?he carbon-hydrogen bond energy in methane, E(C-H), on the other hand, is a single
avaage value:

CH,

-

C

+ 4H.

AH = 397 kcallmol, E(C-H)

= 39714 = 99 kcal/mol

We shall generally End bond dissociation energies more useful for our purposes.

So far, we have spoken of breaking a molecule into two atoms or into an atom
and a group of atoms. Thus, of the two electrons making up the covalent bond,
one goes to each fragment; such bond-breaking is called homolysis. We shall also
encounter reactions involving bond-breaking of a different kind: heterolysis, in
which both bonding electrons go to the same fragment.
A:B
A:B

-

A.+B-

Homolysis: ont?'electron to eachfragment

+ :B

Heterolysis: both electrons to onefragment

A

(These words are taken from the Greek : homo, the same, and hetero, different;
and lysis, a loosing. To a chemist lysis means "cleavage " as in, for example, hydrolysis, "cleavage by water ".)
The bond dissociation energies given in Table 1.2 are for homolysis, and are
therefore homolytic bond dissociation energies. But bond dissociation energies have
also been measured for heterolysis; some of these heterolytic bond dissociation
energies are given in Table 1.3.

-

BOND DISSOCIATIONENERGIES, KCAL/MOL
Table 1.3 HETEROLY~C
A:B

A+

+ :B-

AH = Heterolytic bond dissociation energy or D(A+-B-)

H-H
H-F
H<i
H-Br
H-I
H- OH

401
370
334
324
315
390

CH3-H
CH3-F
CH3-CI
CH3-Br
CH3-I
CHj-OH

313
256
227
219
212
274

If we examine these values, we see that they are considerably bigger than
those in Table 1.2. Simple heterolysis of a neutral molecule yields, of course, a
positive ion and a negative ion. Separation of these oppositely charged particles
takes a great deal of energy: 100 kcal/mol or so more than separation of neutral
particles. In the gas phase, therefore, bond dissociation generally takes place by
the easier route, homolysis. In an ionizing solvent (Sec. 7 . 9 , on the other hand,
heterolysis is the preferred kind of cleavage.

SEC. 1.16

23

POLARITY OF MOLECULES

1.15 Polarity of bonds
Besides the properties already described, certain covalent bonds have another
property: polarity. Two atoms joined by a covalent bond share electrons; their
nuclei are held by the same electron cloud. But in most cases the two nuclei do not
share the electrons equally; the electron cloud is denser about one atom than the
other. One end of the bond is thus relatively negative and the other end is relatively
positive; that is, there is a negative pole and a positive pole. Such a bond is said to
be a polar b d , or to possess polarity.
We can indicate polarity by using the symbols 6, and 6 - , which indicate
partial and - charges. (We say "delta plus " and "delta minus ".) For example :

+

66, 6-

a+/

H-F

6-

N

0

H

\h

6+/

H

1 '\6+

H

H

8+

Polar bonds

We can expect a covalent bond to be polar if it joins atoms that differ in their
tendency to attract electrons, that is, atoms that differ in electronegativity. Furthermore, the greater the difference in electronegativity, the more polar the bond will
be.
The most electronegative elements are those located in the upper right-hand
comer of the Periodic Table. Of the elements we are likely to encounter in organic
chemistry, fluorine has the highest electronegativity, then oxygen, then nitrogen
and chlorine, then bromine, and finally carbon. Hydrogen does not differ very
much from carbon in electronegativity; rt is not certain whether it is more or less
electronegative.
Electronegativity

F>O>CI,N>Br>C,H

Bond polarities are intimately concerned with both physical and chemical
properties. The polarity of bonds can lead to polarity of molecules, and thus
profoundly affect melting point, boiling point, and solubility. The polarity of a
bond determines the kind of reaction that can take place at that bond, and even
affects reactivity at nearby bonds.

1.1 6

Polarity of molecules

A molecule is polar if the center of negative charge does not coincide with the
center of positive charge. Such a molecule constitutes a dipole: two equal and
opposite charges separated in space. A dipole is often symbolized by u,where
the arrow points from positive to negative. The molecule possesses a dipole
moment, p, which is equal to the magnitude of the charge, e , multiplied by the
distance, d , between the centers of charge :

in
debye
units. D

in
e.s.u.

in
cm

STRUCTURE AND PROPERTIES

CHAP. 1

In a way that cannot be gone into here, it is possible to measure the dipole
moments of molecules; some of the values obtained are listed in Table 1.4. We
shall be interested in the values of dipole moments as indications of the relative
polarities of different molecules.
Table 1.4
H2

0
0
Nz 0
C12 0
Br2 0

DIPOLE MOMENTS,
D

HF
H20
NH,
NF,
BF,

0 2

1.75

1.84
1.46
0.24
O

CH,
CH,CI
CCI,
CO,

0
1.86
0
0

It is the fact that some molecules are polar which has given rise to the
.speculationthat some bonds are polar. We have taken up bond polarity first simply
because it is convenient to consider that the polarity of a molecule is a composite
of the polarities of the individual bonds.
Molecules like H z , O,, N,, Cl,, and Brz have zero dipole moments, that is,
are non-polar. The two identical atoms of each of these molecules have, of course,
the same electronegativity and share electrons equally; e is zero and hence p is
zero, too.
A molecule like hydrogen fluoride has the large dipole moment of 1.75 D.
Although hydrogen fluoride is a small molecule, the very high electronegative
4uorine pulls the electrons strongly; although d is small, e is large, and hence p is
large, too.
Methane and carbon tetrachloride, CCl,, have zero dipole moments. We
certainly would expect the individual bonds-of carbon tetrachloride at least-to
be polar; because of the very symmetrical tetrahedral arrangement, however, they
exactly cancel each other out (Fig. 1.16). In methyl chloride, CH,Cl, the polarity

++

H-F

I

tI

(I

C-H

!I

Hydrogen
fluoride

Methdne

C'a rbon
tetrdchlor~dc

Methyl chlorlde

Figure 1.16 Dipole moments of some molecules. Polarity of bonds and of molecules.

of the carbon-chlorine bond is not canceled, however, and methyl chloride has a
dipole moment of 1.86 D. Thus the polarity of a molecule depends not only upon
the polarity of its individual bonds but also upon the way the bonds are directed,
that is, upon the shape of the molecule.
Ammonia has a dipole moment of 1.46 D. This could be accounted for as a
net dipole moment (a vector sum) resulting from the three individual bond moments,

25

POLARITY OF MOLECULES

SEC. 1.16

and would be in the direction shown in the diagram. In a similar way, we could
account for water's dipole moment of 1.84 D.

H

Dipole moments
expected from
bond moments alone

H

H

Ammonia

Water

Now, what kind of dipole moment would we expect for nitrogen trifluoride,
NF,, which, like ammonia, is pyramidal? Fluorine is the most electronegative
element of all and should certainly pull electrons strongly from nitrogen; the N-F
bonds should be highly polar, and their vector sum should be large-far larger than
for ammonia with its modestly polar N-H bonds.

F
F

Large dipole moment
expectedfrom
bond moments alone

Nitrogen trifluoride
Z

What are the facts? Nitrogen trifluoride has a dipole moment of only 0.24 D. It is
not larger than the moment for ammonia, but rather is much smaller.
How are we to account for this? We have forgotten the unshared pair of
electrons. In NF3 (as in NH3) this pair occupies an sp3 orbital and must contribute
a dipole moment in the direction opposite to that of the net moment of the N-F
bonds (Fig. 1.17); these opposing moments are evidently of about the same size,

p= 1.46 D

I

Ammonia

I

Water

Nitrogen
trifluoride

Figore 1.17 Dipole moments of some molecules. Contribution from unshared pairs. In NF,, the moment due to the unshared pair opposes the
vector sum of the bond moments.

26

STRUCTURE AND PROPERTIES

CHAP. 1

and the result is a small moment, in which direction we cannot say. In ammonia
the observed moment is probably due chiefly to the unshared pair, augmented by
the sum of the bond moments. In a similar way, unshared pairs of electrons must
contribute to the dipole moment of water and, indeed, of any molecules in which
they appear.
Dipole moments can give valuable information about the structure of molecules. For example, any structure for carbon tetrachloride that would result in a
polar molecule can be ruled out on the basis of dipole moment alone. The evidence
of dipole moment thus supports the tetrahedral structure for carbon tetrachloride.
(However, it does not prove this structure, since there are other conceivable
structures that would also result in a non-polar molecule.)
C1, would also have
blem 1.5 Which of the follow.ing concei vable struc
h a chlorirle at each
1
Carbo~
I at the center of a s
ro dipole Imoment? (a)
~ " L a .
.
'. . - .. corner. ( 0 ) arbo on at the~.~
apex or
a pyramla wlrn a- cnlorlne
ar eacn comer
01a square
base.

,. ,.
\

~-

.1

A

.

..

'%Problem
1.6 Suggest a shape for the CO, rnolecule tlhat would account fiBr its zero
dipole momenlt.
Pro1
In Sec. 1.12 we rejected two c:onceivabbe electronic configurations for
If nitroger1 were sp2-hybridize(1, what d imle
~ moment would you expect
amr
.
,
L
_
.
- - ulpoie
>:--amar
u me
moment
of ammonia? (b) If nitrogen usedp orbitals
for ammonla !, .w
for bonding, 1low would you expect the diplole moments of am]monia anci nitrogen
trifluoride to c:ompare? How do they compar,e?
:
..l

---_-:-l

.
d

The dipole moments of most compounds have never been measured. For these
substances we must predict polarity from structure. From our knowledge of
electronegativity, we can estimate the polarity of bonds; from our knowledge of
bond angles, we can then estimate the polarity of molecules, taking into account
any unshared pairs of electrons.

1.17 Structure and physical properties
We have just discussed one physical property of compounds: dipole moment.
Other physical properties-like melting point, boiling point, or solubility in a
particular solvent-are also of concern to us. The physical properties of a new
compound give valuable clues about its structure. Conversely, the structure of a
compound often tells us what physical properties to expect of it.
In attempting to synthesize a new compound, for example, we must plan a
series of reactions to convert a compound that we have into the compound that we
want. In addition, we must work out a method of separating our product from all
the other compounds making up the reaction mixture: unconsumed reactants,
solvent, catalyst, by-products. Usually the isolation and purfication of a product
take much more time and effort than theactual making of it. The feasibility of
isolating the product by distillation depends upon its boiling point and the boiling
points of the contaminants ; isolation by recrystallizationdepends upon its solubility
in various solvents and the solubility of the contaminants. Success in the laboratory
often depends upon making a good prediction of physical properties from structure.
Organic compounds are real substances-not just collections of letters written on
a piece of paper-and we must learn how to handle them.

SEC. 1.18

MELTING POINT

27

We have seen that there are two extreme kinds of chemical bonds: ionic
bonds, formed by the transfer of electrons, and covalent. bonds, formed by the
sharing of electrons. The physical properties of a compound depend largely upon
which kind of bonds hold its atoms together in the molecule.

1.18 Melting point
In a crystalline solid the particles acting as structural units-ions or molecules-are arranged in some very regular, symmetrical way; there is a geometric
pattern repeated over and over within a crystal.
Melting is the change from the highly ordered arrangement of particles in the
crystalline lattice to the more random arrangement that characterizes a liquid (see
Figs. 1.1 8 and 1.19). Melting occurs when a temperature is reached at which the
thermal energy bf the particles is great enough to overcome the intracrystalline
forces that hold them in position.
An ionic compound forms crystals in which the structural units are ions. Solid
sodium chloride, for example, is made up of positive sodium ions and negative
chloride ions alternating in a very regular way. Surrounding each positive ion and

Figure 1.18 ,-Meltingof an ionic crystal. The units are ions.

equidistant from it are six negative ions: one on each side of it, one above and one
below, one in front and one in back. Each negative ion is surrounded in a similar
way by six positive ions. There is nothing that we can properly call a molecule of
sodium chloride. A particular sodium ion does not "belong" to any one chloride
ion; it is equally attracted to six chloride ions. The crystal is an extremely strong,
rigid structure, since the electrostatic forces holding each ion in position are
powerful. These powerful interionic forces are overcome only at. a very high
temperature; sodium chloride has a melting point of 801 "C.
Crystals of other ionic compounds resemble crystals of sodium chloride in
having an ionic lattice, although the exact geometric arrangement may be different.
As a result, these other ionic compounds, too, have high melting points. Many
molecules contain both ionic and covalent bonds. Potassium nitrate, K N 0 3 , for
example, is made up of K + ions and NO3- ions; the oxygen and nitrogen atoms
of the NO3- ion are held to each other by covalent bonds. The physical properties
of compounds like these are largely determined by the ionic bonds; potassium
nitrate has very much the same sort of physical properties as sodium chloride.
A non-ionic compound, one whose atoms are held to each other entirely by
covalent bonds, forms crystals in which the structural units are molecules. It is the

a

STRUCTURE AND PROPERTIES

CHAP. 1

forces holding these molecules to each other that must be overcome for melting to
occur. In general, these intermolecular forces are very weak compared with the

Figure 1.19 Melting of' a non-ionic crystal. The units are molecules.

forces holding ions to each other. To melt sodium chloride we must supply enough
energy to break ionic bonds between Na+ and C1-. To melt methane, CH,, we
do not need to supply enough energy to break covalent bonds between carbon and
hydrogen; we need only supply enough energy to break CH, molecules away from
each other. In contrast to sodium chloride, methane melts at - 183 "C.

1.19.- Intermolecular forces
What kinds of forces hold neutral molecules to each other? Like interionic
forces, these forces seem to be electrostatic in nature, involving attraction of
positive charge for negative charge. There are two kinds of intermolecular forces:
dipole-dipole interactions and van der Wauls forces.
Dipoldpole interaction is the attraction of the positive end of one polar
molecule for the negative end of another polar molecule. In hydrogen chloride, for
example, the relatively positive hydrogen of one molecule is attracted to the
relatively negative chlorine of another:

As a result of dipole-dipole interaction, polar molecules are generally held to each
other more strongly than are non-polar molecules of comparable molecular weight;
this difference in strength of intermolecular forces is reflected in the physical
properties of the compounds concerned.
An especially strong kind of dipoledipole attraction is hydrogen bonding, in
which a hydrogen atom serves as a bridge between two electronegative atoms, holding
one by a covalent bond and the other by purely electrostatic forces. When hydrogen is
attached to a highly electronegative atom, the electron cloud is greatly distorted
toward the electronegative atom, exposing the hydrogen nucleus. The strong
positive charge of the thinly shielded hydrogen nucleus is strongly attracted by the
negative charge of the electronegative atom of a second molecule. This attraction
has a strength of about 5 kcal/mol, and is thus much weaker than the covalent
bond-about 50-100 kcal/mol-that holds it to the first electronegative atom. It is

SEC. 1.19

P

INTERMOLECULAR FORCES

29

much stronger, however, than other dipole4ipole attractions. Hydrogen bonding
is generally indicated in formulas by a broken line :

For hydrogen bonding to be important, both electronegative atoms must come from
the group: F, 0,
N . Only hydrogen bonded to one of these three elements is positive
enough, and only these three elements are negative enough, for the necessary
attraction to exist. These three elements owe their special effectiveness to the
concentrated negative charge on their small atoms.
There must be forces between the molecules of a non-polar compound, since
:ven such compounds can solidify. Such attractions are called van der Waals forces.
The existence of these forces is accounted for by quantum mechanics. We can
roughly visualize them arising in the following way. The average distribution of
charge about, say, a methane molecule is symmetrical, so that there is no net dipole
moment. However, the electrons move about, so that at any instant the distribution will probably be distorted, and a small dipole will exist. This momentary dipole will affect the electron distribution in a second methane molecule
nearby. The negative end of the dipole tends to repel electrons, and the positive
end tends to attract electrons; the dipole thus induces an oppositely oriented dipole
in the neighboring molecule:

Although the momentary dipoles and induced dipoles are constantly changing, the
net result is attraction between the two molecules.
These van der Waals forces have a very short range; they act only between
the portions of different molecules that are in close contact, that is, between the surfaces of molecules. As we shall see, the relationship between the strength of van
der Waals forces and the surface areas of molecules (Sec. 3.12) will help us to
understand the effect of molecular size and shape on physical properties.
With respect to other atoms to which it is not bonded-whether in another molecule or
in another part of the same molecule-very atom has an effective "size", called its van der
Waals radius. As two non-bonded atoms are brought together the attraction between them
steadily increases, and reaches a maximum when they are just "touching'-that is to say,
when the distance between the nuclei is equal to the sum of the van der Waals radii. Now,
if the atoms are forced still closer together, van der Waals attraction is very rapidly replaced
by van der Waals repulsion. Thus, non-bonded atoms welcome each other's touch, but
strongly resist crowding.
We shall find both attractive and repulsive van der Waals forces important to our
understanding of molecular structure.

In Chapter 7, we shall discuss in detail all of these intermolecular forcesthese kinds of secondary bonding.

.

XI
1.20

STRUCTURE AND PROPERTIES

CHAP. 1

Boiling

Although the particles in a liquid are arranged less regularly and are freer to
move about than in a crystal, each particle is attracted by a number of other
particles. Boiling involves the breaking away from the liquid of individual molecules or pairs of oppositely charged ions (see Figs. 1.20 and 1.21). This occurs
when a temperature is reached at which the thermal energy of the particles is
great enough to overcome the cohesive forces that hold them in the liquid.

Figure 1.20 Boiling of an ionic liquid. The units are ions and ion pairs.

In the liquid state the unit of an ionic compound is again the ion. Each ion is
still held strongly by a number of oppositely charged ions. Again there is nothing
we could properly call a molecule. A great deal of energy is required for a pair of
oppositely charged ions to break away from the liquid; boiling occurs only at a
very high temperature. The boiling point of sodium chloride, for example, is
1413 "C.In the gaseousstate we have an ionpair, which can be considered a sodium
chloride molecule.
In the liquid state the unit of a non-ionic compound is again the molecule.
The weak intermolecular forces here-dipole-dipole interactions and van der

Figure 1.21

Boiling of a non-ionic liquid. The units are molecules.

Waals forces-are more readily overcome than the strong interionic forces of ionic
compounds, and boiling occurs at a very much lower temperature. Non-polar
methane boils at - 161.5 "C, and even polar hydrogen chloride boils at only
- 85 "C.
Liquids whose molecules are held together by hydrogen bonds are called
~cwciatedliquids. Breaking these hydrogen bonds takes considerable energy, and
(
;
. 3n associated liquid has a boiling point that is abnormally high for a compound
9 c molecular weight and dipole moment. Hydrogen fluoride, for example, boils
7'

SEC.1.21

31

SOLUBILITY

100 degrees higher than the heavier, non-associated hydrogen chloride; water boils
160ldegrees higher than hydrogen sulfide.
'There are organic compounds, too, that contain hydrogen bonded to oxygen
or nitrogen, and here, too, hydrogen bonding occurs. Let us take, for example,
methane and replace one of its hydrogens with a hydroxyl group, --OH. The
resulting compound, CH,OH, is methanol, the smallest member of the alcohol
family. Structurally, it resembles not only methane. but also water:
H

I

H-C-H

I

H
H-0-H

H
Methane

i

H-C-0-H

I

H
Water

Methanol

Like water, it is an associated liquid with a boiling point "abnormally" high for a
compound of its size and polarity.

The bigger the molecules, the stronger the van der Waals forces. Other things
being equal-polarity, hydrogen bonding-boiling point rises with increasing
molecular size. Boiling points of organic compounds range upward from that of
tiny, non-polar methane, but we seldom encounter boiling points much above
350 "C;at higher temperatures, covalent bohds within the molecules start to break,
and decomposition competes with boiling. It is to lower the boiling point and thus
minimize decomposition that distillation of organic compounds is often carried
out under reduced pressure.

.

.
W e m 1.8 Which of the following organic compounds would you predict to
be associated liquids?Draw structuresto show the hydrogen bonding you would expect.
(a) CH30CH3;(b) CH3F;(c) CH3CI;(d) CH3NH,; (e) (CH3),NH; (f) (CH3),N.

1.21 Solubility
When a solid or liquid didsolves, the structural units-ions or moleculesbecome separated from each other, and the spaces in between become occupied
by solvent molecules. In dissolution, as in melting and boiling, energy must be
supplied to overcome the interionic or intermolecular forces. Where does the
necessary energy come from? The energy required to break the bonds between
solute particles is supplied by the formation of bonds between the solute particles
and the solvent molecules: the old attractive forces are replaced by new ones.
Now, what are these bonds that are formed between solute and solvent? Let
us consider first the case of ionic solutes.
A great deal of energy is necessary to overcome the powerful electrostatic
forces holding together an ionic lattice. Only water or other highly polar solvents

32

STRUCTURE AND PROPERTIES

CHAP. 1

are able to dissolve ionic compounds appreciably. What kinds of bonds are formed
between ions and a polar solvent? By definition, a polar molecule has a positive
end and a xlegative end. Consequently, there is electrostatic attraction between a
positive ion and thenegative end of the solvent molecule, and between a negative
ion and the positive end of the solvent molecule. These attractions are called iondipole bonds. Each ion-dipole borld is relatively weak, but in the aggregate they
supply enough energy to overcome the interionic forces in the crystal. In solution
each ion is surrounded by a cluster of solvent molecules, and is said to be solvated;
if the solvent happens to be water, the ion is said to be hydrated. In solution, as in
the solid and liquid states, the unit of a substance like sodium chloride is the ion,
although in this case it is a solvated ion (see Fig. 1.22).

Figure 1.22 Ion-dipole interactions: a solvated cation and anion.

To dissolve ionic compounds asolvent must also have a high dielectric constant,
thaf is, h b e high insulating properties to lower the attraction between oppositely
charged ions once they are solvated.
Water owes its shperiority as a solvent for ionic substances not only to its
polarity and its high dielectric constant but to another factor as well: it contains
the -OH group and thus can form hydrogen bonds. Water solvates both cations
and 'ixhions: cations, at its negative pole (its unshared electrons, essentially);
anions, through hydrogen bonding.
Now let us turn to the dissolution of non-ionic solutes.
The solubility characteristics of non-ionic compounds are determined chiefly
by their polarity. Non-polar or weakly polar compounds dissolve in non-polar or
weakly polar solvents; highly polar compounds dissolve in highly polar solvents.
"Like dissolves like" is an extremeiy useful rule of thumb. Methane dissolves in
carbon tetrachloride because the forces holding methane molecules to each other
and carbon tetrachloride molecules to each other-van der Waals interactionsare replaced by very similar forces holding methane molecules to carbon tetrachloride molecules.
Neither methane nor carbon tetrachloride is readily soluble in water. The
highly polar water molecules are held to each other by very strong dipole-dipole
interactions-hydrogen bonds; there could be only very weak attractive forces
between water molecules on the one hand and the non-polar methane or carbon
tetrachloride molecules on the other.
In contrast, the highly polar organic compound methanol, CH,OH, is quite .
soluble in water. Hydrogen bonds between water and methanol molecules can
readily replace the very similar hydrogen bonds between different methanol
molecules and different water molecules.
An understanding of the nature of solutions is fundamental to an understanding of organic chemistry. Most organic reactions are camed out in solution and,

U

1

33

ACIDS AND BASES

4

ir is W a g increasingly clear, the solvent does much more than simply bring
different molecules together so that they can react with each other. The solvent is
iradFcd m the reactions that take place in it: just how much it is involved, and in
what ways, is only now being realized. In Chapter 7, when we know a little more
about organic reactions and how they take place, we shall return to this subjectwhich we have barely touched upon here-and examine in detail the role played
by the solvent.

1.22

Acids and bases

Turning from physical to chemical properties, let us review briefly one familiar
topic that is fundamental to the understanding of organic chemistry: acidity and
basicity.
The terms acid and base have been defined in a number of ways, each definition
corresponding to a particular way of looking at the properties of acidity and
basicity. We shall find it useful to look at acids and bases from two of these
viewpoints; the one we select will depend upon the problem at hand.
According to the Lowry-Brensted definition, an acid is a substance t b t gives
up a proton, and a base is a substance that accepts a proton. When sulfuric acid
dissolves in water, the acid H2S04gives up a proton (hydrogen nucleus) to the
base H 2 0 to form the new acid H,O+ and the new base HS04-. When hydrogen
chloride reacts with ammonia, the acid HCl gives up a proton to the base NH, to
form the new acid NH4+ and the new base C1H2S04 +
Stronger
acid

HCI

H20
Stronger
base

+

NH3
Stronger
base

Stronger
acid

H30+ + HS04Weaker
acid

Weaker
5ase

S NH4+ +
Weaker
acid

C1Weaker
base

According to the Lowry-Brmsted definition, the strength of an acid depends
upon its tendency to give up a proton, and the strength of a base depends upon its
tendency to accept a proton. Sulfuric acid and hydrogen chloride are strong acids
since they tend to give up a proton very readily; conversely, bisulfate ion,
HS04-, and chloride ion must necessarily be weak bases since they have little
tendency to hold on to protons. In each of the reactions just described, the
equilibrium favors the formation of the weaker acid and the weaker base.
If aqueous H2S04is mixed with aqueous NaOH, the acid H,O+ (hydronium
ion) gives up a proton to the base OH- to form the new acid H 2 0 and the new
base H20. When aqueous NH,Cl is mixed with aqueous NaOH, the acid NH,'
H30+
Stronger
acid

+

OH-

H20

Stronger
base

Weaker
acid

NH4+ + OHStron r
acir

Stronger
base

H20
Weaker
acid

+

H20
Weaker
base

+

NH3
Weaker
base

34

STRUCl'URE AND PROPERTIES

CHAP. 1

(ammonium ion) gives up a proton to the base OH- to form the new acid H 2 0and
the new base NH3. In each case the strong base, hydroxide ion, has accepted a
proton to form the weak acid H20. If we arrange these acids in the order shown,
we must necessarily arrange the corresponding (conjugate) bases in the opposite
order.
Acid strength

Base strength

> NH4+ > H,O

H2S04

> H,O+

HS04-

< H 2 0 < NH, < OH-

HCI

c1-

Like water, many organic compounds that contain oxygen can act as bases
and accept protons; ethyl alcohol and diethyl ether, for example, form the oxonium
ions I and 11. For convenience, we shall often refer to a structure like I as a
protonated alcohol and a structure like I1 as a protonated ether.
CZH&H
..

+ HtS04

9

C2H50H
.. + H S 0 4 -

H

Ethyl alcohol
I

An oxoniurn ion
Protonated ethyl alcohol

c3

+

( c ~ H ~ ) ~ o : HCI
Diethyl ether

+ ( c ~ H ~ ) ~Ho :+ CI I1

An oxoniurn ion
Protonated diethyl ether

According to the Lewis definition, a base is a substance that can furnish an
electron pair to form a covalent bond, and an acid is a substance that can take up an
electron pair to form a covalent bond. Thus an acid is an electron-pair acceptor and a
base is an electron-pair donor. This is the most fundamental of the acid-base
concepts, and the most general; it includes all the other concepts.
A proton is an acid because it is deficient in electrons, and needs an electron
pair to complete its valence shell. Hydroxide ion, ammonia, and water are bases
because they contain electron pairs available for sharing. In boron trifluoride, BF, ,
boron has only six electrons in its outer shell and hence tends to accept another
pair to complete its octet. Boron trifluoride is an acid and combines with such
bases as ammonia or diethyl ether.
F

I

F-B
I

+

:NH3

F
Acid

F

I

F-B

I
F
Acid

+

97a3

F-B:NH3
I

F
Base

81 a3
..
: o ( c ~ H ~ ) ?3
~
F-B:O(CZHS)Z
I

6
Base

35

ACIDS AND BASES

SEC. 1.22

~luminhm
chloride, AIC13, is ah acid, and for the same reason. In stannic chloride,
SnCl,, tin has a complete octet, but can accept additional pairs of electrons (e.g.,
in SnCl,*-) and hence it is an acid, too.
We write a formal negative charge on boron in these formulas because it has one more
electron-half-interestin the pair shared with nitrogen or oxygen-than is balanced by the
nuclear charge; correspondingly, nitrogen or oxygen is shown with a formal positive charge.
We shall find the Lewis concept of acidity and basicity fundamental to our
understanding of organic chemistry. To make it clear that we are talking about
this kind of acid or base, we shall often use the expression Lewis acid (or Lewis
base), or sometimes acid (or base) in the Lewis sense.
Chemical. properties, like physical properties, depend upon molecular structure. Just what features in a molecule's structure tell us what to expect about its
acidity or basicity? We can try to answer this question in a general way now,
although we shall return to it many times later.
To be acidic in the Lowry-Brsnsted sense, a molecule must, of course, contain
hydrogen. The degree of acidity is determined largely by the kind of atom that
holds the ydrogen and, in particular, by that atom's ability to accommodate the
electron pair left bebind by the departing hydrogen ion. This ability to accommodate
the elt#ctron pair seems to depend upon several factors, including (a) the atom's
electronegativity, and (b) its size. Thus, within a given row of the Periodic Table,
acidity increases as electronegativity increases:

't.

H-CH,

Acidity

< H-NH2 < H--OH < H-F

H-SH < H-CI

And within a given family, acidity increases as the size increases:
H-F < H-CI < H-Br < H-I
H--OH < H-SH < H-SeH

.-

Acidity

Among organic compounds, we can expect appreciable Lowry-Brsnsted acidity
from those containing 0-H, N-H, and S-H groups.
To be acidic in the Lewis sense, a molecule must be electron-deficient; in
oarticular, we would look for an atom bearing only a sextet of electrons.
..
- ..Problem 1.9 Predict the relative acidity of: (a) methyl alcohol (CH30H)and methyl-.
amine (CH,NH,); (b) methyl alcohol (CH,OH) and methar~ethioi(CH3SH);
(c) H30t and N H4+.

.

Problem1.10 \Nhich is tl
r acid of e;ich pair: (a)
- H,Ot
.
. .c~rH20;(b)
-.
a
- - - - - . -- -cxNH,;(c) H,S or ns ;(a) H,U or OH-.! (e) What relationship is there 1
c h g e and

NH,+

<

acidity?

To be basic in either the Lowry-Brmsted or the Lewis sense, a molecule must
tr+t an electron pair available for sharing. The availability of these unshared

-

-e'
is determined largely by the atom that holds them: its electronegativity,
:rs srrt.its charge. The operation of these factors here is necessarily opposite to
k?
w e observed for acidity; the better an atom accommodates the electron pair,
L!X

Iea available the pair is for sharing.

36

STRUCTURE AND PROPERTIES

CHAP. 1

obeach group in order of basici.ty:-.- - Problem 1.1 1 Arrange the rnembers
.. . - -- - (a)F-,OH-,NH2-,CH3-; (b)HF, H20,NH3;(c)Cl-,SH-; ( d ) F - , c l , w , I - ;
(e) OH-, SH-, SeH-.

methyl IFluoride
Problem 1.12 Predict the
alcohol (CH30H),and methylamine (CH3NH2).

I

(1

Problem 1.13 Arrange the rnembers olf each group in order of basicity :
(a) H30+,HZO, OH- ;(b) N H,, NH2'- ;(c) HzS!, HS-, S2-.. (d) What relationship is
there between charge and basicity ?

1.23

Isomerism

Before we start our systematic study of the different kinds of organic
compounds, let us look at one further concept which illustrates especially well the
fundamental importance of molecular structure: the concept of isomerism.
The compound ethyl alcohol is a liquid boiling a/t 78 "C.Analysis (by the
methods described later, Sec. 2.28) shows that it c o p i n s carbon, hydrogen, and
oxygen in the proportions 2C:6H :10. Measurement of its mass spectrum s h o w
that it has a molecular weight of 46. The molecular formula of ethyl alcohol must
therefore be C2H60.Ethyl alcohol is a quite reactive compound. For example, if a
piece of sodium metal is dropped into a test tube containing ethyl alcohol, there is
a vigorous bubbling and the sodium metal is consumed; hydrogen gas is evolved
and there is left behind a compound of formula C2H,0Na. Ethyl alcohol reacts
with hydriodic acid to form water and a compound of formula C2H,I.
The compound dimethyl ether is a gas with a boiling point of - 24 "C. It is
clearly a different substance from ethyl alcohol, differing not only in its physical
properties but also in its chemical properties. It does not react at all with sodium
metal. Like ethyl alcohol, it reacts with hydriodic acid, but it yields a compound of
formula CHjI. Analysis of dimethyl ether sho& that it contains carbon, hydrogen,
and oxygen in the same proportion as ethyl alcohol, 2C:6H: 10. It has the same
molecular weight as ethyl alcohol, 46. We conclude that it has,the same molecular
formula C2H60.
Here we have two substances, ethyl allcohol and d i ~ e t h yether,
l
which have
the same molecular formula, C2H60,and yet quite clearly are different compounds.
How can we account for the existence of these two compounds? The answer is:
they d i / f in molecular structure4 Ethyl alcbhol has the structure =presented by I,
and dimethyl ether the structure represented b t 11. As we shall see, the differences
in physical and chemical properties of thesb two compounds can readily be
accounted for on the basis of the difference in structure.

I

I

H H

I
H

I

H

I

II

Ethyl alcoho!

Dimethyljether

Dijkrmt compounds that have the same molecular formula are called isomers
(Greek: isos, equal; metos, part). They contain the same numbers of the same
kinds of atoms, but the atoms are attached to one another in different ways.
Isomers are different compounds because they have different moleculdr structures.

CHAP. 1

37

PROBLEMS

This difference in molecular structure gives rise to a difference in properties;
it is the difference 41 propprties which tells us that we .are dealing with different

compounds. In some case? the difference in structure-and hence the difference
in properties-is so marked that the isomers are assigned to different chemical
families, as, for example, ethyl alcohol and dimethyl ether. In other cases the
difference in structure is so subtle that it can be described only in terms of threedimensional models. Other kinds of isomerism fall between these two extremes.

PROBLEMS
1. Which of the following would you expect to be ionic, and which non-ionic? Give a
simple electronic structure (Sec. 1.3) for each, showing only valence shell electrons.
(a) MgCIz
(b) CH2Cl2

(c) ICl
(d) NaOCl

(e) KClO,
(f) SiCl,

(g) Bas04
(h) CH3NH2

2. Give a likely simple electronic structure (Sec. 1.3) for each of the following, assuming
them to be completely covalent. Assume that every atom (except hydrogen, of course) has a
complete octet, and that two atoms may share more than one pair of electrons.
(a) N2H4
(b) H2S04
(c) HS0,-

(d) COCI,
(e) HONO
(f) NO2-

(g)

co32"

(h) C2H4
(9 C2H2

( 3C H 2 0
(k) CH202
(1) c3Hs

3. What shape would you expect each of the following to have?
(e) tha amide ion, NH2(a) (CH3)3B
(b) the methyl aniofi, CH3: (f) dimethyl ether
.(g) the fluoroborate ion, BF,(c) the methyl cation, CH3+
( 4 HIS
(h) (CH3)3N
4. In many complex ions, e.g., C O ( N H ~ ) , ~the
+ , bonds to the central atom can be
pictured as utilizing six equivalent sp3d2 (ord2sp3) hybrid orbitals. On the basis of maximum
separation of orbitals, what geometry would you expect these complexes to have?

5. Indicate the direction of the dipole hnoment, ifany, th'at you would expect for each
of the following:
( 4 CH2C12
(g) dimethyl ether
(a) HBr
(b) IC1
(e) CHCI,
(h),(CH3)3N
(c) 12
(f) CH30H
(i) CF2C12

About Working Problems

Working problems is a necessary part of your work for two reasons: it will
guide your study i d the right direction, and, after you have studied a particular
chapter, it will show whether or not you have reached your destination.
You should work all the problems that you can; you should get help with the
ones you cannot work yourself. The first problems in each set are easy, but provide
the drill in drawing formulas, naming compounds, and using reactions that even
the best student needs. The later problems in each set are the kind encountered by
practicing chemists, a1:d test your ability to use what you have learned.
You can check your answers to many of the problems in the answer section in
the back of the book, and by use of the index. You will find more complete answers
to all the problems, together with suggestions about how to approach each type of
problem, in the-study Guide.

38

STRUCTURE AND PROPERTIES

CHAP. 1

6. (a) Although HCl (1.27 A) is a longgr molecule than H F (0.92 A), it has a smaller
dipole moment (1.03 debye compared to 1.75 debye). How do you account for this fact?
(b) The dipole moment of CH3F is 1.847 debyq, and of CD3F, 1.858 debye. (D is 'H,
deuterium.) Compared with the C-H bond, what is the direction of the C-D dipole?
7. What do the differences in properties between lithium acetylacetonate (m.p. very
high, insoluble in chloroform) and beryllium atetylacetonate (m.p. 108 OC, b.p. 270 OC,
soluble in chloroform) suggest about their structures?

8. n-Butyl alcohol (b.p. 118 OC) has a much higher boiling point than its isomer diethyl
ether (b.p. 35 "C), yet both compounds show the same solubility (8 g per 100g) in water.
H H H H

I I I L
H-C-C-C- -0-H
I I I I

H H H H
n-Butyl alcohol
How do you account for these facts?

H H

H H

I I
I
H-C-C-0-C-C-H
I I
I

I
I

H H
H H
Diethyl ether

9. Rewrite the following equations to show the Lowry-Brmsted acids and bases
actually involved. Label each as stronger or weaker, as in Sec. 1.22.
(a) HCl(aq) NaHC03(aq) t- H2C03 + NaCl
(b) NaOH(aq) NaHC03(aq) c- NazC03 + HzO
( 4 NH3(aq) + HN03(aq) t- NH4N03(aq)
(d) NaCN(aq) .-f HCN(aq) + NaOH(aq)
(e) NaH H 2 0
H z NaOH
(f) CaC2 HzO
Ca(OH)2 + C2Hz
Calcium
Acetylene
carbide

+

+
+

+

--

+

10. What is the Lowry-Brmsted acid in (a) HCl dissolved in water; (b) HCI (unionized) dissolved in benzene? (c) Which solution is the more strongly acidic?
11. Account for the fact that nearly every organic compound containing oxygen
dissolves in cold concentrated sulfuric acid to yield a solution from which the compound
can be recovered by dilution with water.

12. How might you account for the following orders of acidity? Be as specific as you

can.
HC104 > HC102 > HC10 a'nd H2SO4> H2S03
13. For each of the following molecular formulas, draw structures like those in Sec.
1.23 (a line for each shared pair, of electrons) for all the isomers you can think of. Assume
that every atom (except hydrogen) has a complete octet, and that two atoms may share more
than one pair of electrons.

c4Hl0
(e) C3H8O
(a) C2H7N
@IC3H8
( 4 C3H7C1
(f) C2H40
14. In ordinary distillation, a liquid is placed in a flask and heated, at ordinary or
reduced pressure, until distillation is complete. In the modification calledjlash distillation,
the liquid is dripped into a heated flask at the same rate that it distills out, so that there is
little liquid in the flask at any time. What advantage might flash distillation have, and under
what conditions might you use it?

Methane
Chapter

Energy of Activation.
Transition State

2.1

Hydrocarbons

Certain organic compounds contain only two elements, hydrogen and carbon,
and hence are known as hydrocarbons. On the basis of structure, hydrocarbons
are divided into two main classes, aliphatic and aromatic. Aliphatic hydrocarbons
are further divided into families: alkanes, alkenes, alkynes, and their cyclic analogs (cycloalkanes, etc.). We shall take up these families in the order given.

Hydrocarbons

Aromatic

Aliphatic

Alkanes

Alkenes

Alkynes

Cyclic
aliphatic

The

simplest

member of the alkane

family and, indeed, one of the simplest of

organic compounds is methane, CH 4 We shall sUidy this single compound at
some length, since most of what we learn about it can be carried over with minor
all

.

modifications to any alkane.

2.2

Structure of methane

As we discussed in the previous chapter (Sec. 1.11), each of the four hydrogen
atoms is bonded to the carbon atom by a covalent bond, that is, by the sharing of a
pair of electrons. When carbon is bonded to four other atoms, its bonding orbitals
(sp* orbitals, formed by the mixing of one s and three/? orbitals) are directed to the
corners of a tetrahedron (Fig. 2. la). This tetrahedral arrangement is the one that
permits the orbitals to be as far apart as possible. For each of these orbitals to

40

METHANE

CHAP. 2

2.2 structure of methane
As we discussed in the previous chapter (Sec. 1.1 l), each of the four hydrogen
atoms is bonded to the carbon atom by a covalent bond, that is, by the sharing of
a pair of electrons. When carbon is bonded to four other atoms, its bonding orbitals
(sp3 orbitals, formed by the mixing of ones and threep orbitals) are directed to the
comers of a tetrahedron (Fig. 2. la). This tetrahedral arrangement is the one that
permits the orbitals to be as far apart as possible. For each of these orbitals to
overlap most effectively the spherical s orbital of a hydrogen atom, and thus to
form the strongest bond, each hydrogen nucleus must be located at a comer of this
tetrahedron (Fig. 2. lb).

Figure 2.1 Methane molecule. (a) Tetrahedral sp3 orbitals. (b) Predicted
shape: H nuclei located for maximum overlap. (c) Shape and size.

The tetrahedral structure of methane has been verified by electron diffraction
(Fig. 2. lc), which shows beyond question the arrangement of atoms in such simple
molecules. Later on, we shall examine some of the evidence that led chemists to
accept this tetrahedral structure long before quantum mechanics or electron
diffraction was known.
We shall ordinarily write methane with a dash to represent each pair of
electrons shared by carbon and hydrogen (I). To focus our attention on individual
electrons, we may sometimes indicate a pair of electrons by a pair of dots (11).
Finally, when we wish to represent the actual shape of the molecule, we shall use
a simple three-dimensional formula like I11 or IV.
H

I

H-C-H

H
H:C:H

y
H-c- H

HI
..C.~

In three-dimensional formulas of this kind, a solid wedge represents a bond
coming toward us out of the plane of the paper; a broken wedge, a bond going
away from us behind the plane of the paper; and an ordinary line, a bond lying in
the plane of the paper. Thus formulas I11 and IV represent methane as in Fig. 2.22
and Fig. 2.26, respectively.

SEC.

PHYSICAL PROPERTIES

2.3

41

overlap most effectively the spherical s orbital of a hydrogen atom, and thus to form
the strongest bond, each hydrogen nucleus must be located at a corner of this

tetrahedron

(Fig.- 2. 1 b).

(c)

Figure 2.1. Methane molecule, (a) Tetrahedral sp* orbitals. (b) Predicted
nuclei located for maximum overlap, (c) Shape and size.
shape:

H

The

methane has been verified by electron diffraction
which
shows
the arrangement of atoms in such simple
beyond
question
(Fig. 2.1c),
molecules. Later on, we shall examine some of the evidence that led chemists to
accept this tetrahedral structure long before quantum mechanics or electron
diffraction was known.
We shall ordinarily write methane with a dash to represent each pair of electrons shared by carbon and hydrogen (I). To focus our attention on individual
electrons, we may sometimes indicate a pair of electrons by a pair of dots (II).
Finally,

tetrahedral structure of

when we wish

to consider the actual shape of the molecule,

we

shall use a

simple three-dimensional picture (HI).

H

H-C H

H:C:H

H

H
H

23

Physical properties

As we discussed in
compound, whether

the previous chapter (Sec. 1.18), the unit of such a non-ionic
Because the methane

solid, liquid, or gas, is the molecule.

highly symmetrical, the polarities of the individual carbon-hydrogen
result, the molecule itself is non-polar.
Attraction between such non-polar molecules is limited to van der Waals

molecule

is

bonds cancel out; as a

forces; for such small molecules, these attractive forces

must be tiny compared

with the enormous forces between, say, sodium and chloride ions. It is not surprising, then, that these attractive forces are easily overcome by thermal energy,
so that melting and boiling occur at very low temperatures: m.p
-183, b.p.
- 101.5. (Compare these values with the corresponding ones for sodium chloride:

m.p. 801. b.p. 1413.)
tures.

As a consequence, methane

is

a gas at ordinary tempera-

METHANE

42

Methane

is

when

colorless and,

liquefied,

.

is

less

CHAP.

dense than water (sp.gr.

2

0.4).

In agreement with the rule of thumb that "litfe dissolves like," it is only slightly
soluble in water, but very soluble in organic liquids such as gasoline, ether, and
alcohol. In its physical properties methane sets the pattern for the other members

of the alkane family.

Source

2.4

Methane
that

is,

is an end product of the anaerobic ("without air") decay of plants,
of the breakdown of certain very complicated molecules. *As such, it is the

major constituent (up to 97

)

of natural gas.

It is

the dangerousy/m/ow/? of the

coal mine, and can be seen as marsh gas bubbling to the surface of swamps.
If methane is wanted in very pure form, it can be separated from the other
constituents of natural gas (mostly other alkanes) by fractional distillation. Most

of course, is consumed as fuel without purification.
According to one theory, the origins of life go back to a primitive earth
surrounded by an atmosphere of methane, water, ammonia, and hydrogen.
Energy radiation from the sun, lightning discharges- -broke these simple moleof

it,

cules into reactive fragments (free radicals, Sec. 2.12); these combined to form
larger molecules which eventually yielded the enormously complicated organic

compounds

that

make up

living organisms. (Recent detection of organic

cules in space has even led to the speculation that "organic seeds for

have existed

life

molecould

in interstellar clouds/')

Evidence that this could have happened was found in 1953 by the Nobel Prize
winner Harold C. Urey and his student Stanley Miller at the University of Chicago.
They showed that an electric discharge converts a mixture of methane, water,

animonia, and hydrogen into a large number of organic compounds, including
amino acids, the building blocks from which proteins, the "stuff of life*' (Chap. 36),
are made. (It is perhaps appropriate that we begin this study of organic chemistry
with methane and its conversion .into free radicals.)

The methane generated in the
be the very substance from which
**.

.

2.5

.

final

decay of a once-living organism

in the final analysis

ear tlf to earth, ashes to ashes, dust to dust.

.

.

may

well

the organism was derived.

."

Reactions

its chemical properties as in its physical properties, methane sets the pattern for the alkane family (Sec. 3.18). Typically, it reacts only with highly reactive
substances or under very vigorous conditions, which, as we shall see, amounts to

In

the same thing. At this point
halogens, and even by water.

2.6

we

shall take

up only

its

oxidation: by oxygen, by

Oxidation. Heat of combustion

Combustion to carbon dioxide and water is characteristic of organic compounds; under special conditions it is used to determine their content of carbon
and hydrogen (Sec. 2.26).
Combustion of methane is the principal reaction taking place during the

SEC.

CHLORINATION: A SUBSTITUTION REACTION

2.7

43

REACTIONS OF METHANE
1.

Oxidation

CH 4 + 2O 2 -5^> CO 2 + 2H 2 O +

heat (213 kcal/mole)

2HC-CH + 2CO +

10H 2

Combustion

Discussed in Sec.

8.5.

Acetylene

CO + 3H 2
2.

Halogenation

HX

HX

HX

^ CHX

CH 4 -^-> CH X ^> CH 2X 2
3

Reactivity of

X2 F2 >

C1 2

HX

> Br 2 (>

-^-> CX 4

3

light required

I 2)

Unreactive

burning of natural gas.
areas where natural gas

It is
is

hardly necessary to emphasize

available; the important product

is

its importance in the
not carbon dioxide or

water but heat.

Burning of hydrocarbons takes place only
for example, by a flame or a spark.

Once

at high temperatures, as provided,

started,

however, the reaction gives off

often sufficient to maintain the high temperature and to permit
burning to continue. The quantity of heal evolved when one mole of a hydrocarbon
is burned to carbon dioxide and water is called (he heat of combustion; for methane

heat which

its

value

is

is

213 kcal.

Through controlled

partial oxidation of

methane

methane and the high-temperature

an increasingly important source of
products other than heat: of hydrogen, used in the manufacture of ammonia;
of mixtures of carbon monoxide and hydrogen, used in the manufacture of

catalytic reaction with water,

is

methanolznd other alcohols; and of acetylene
large-scale production of

(Sec. 8.5), itself the starting point of

many

organic compounds.
of particular interest to us

Oxidation by halogens is
it than the other reactions of methane

more about

partly because

and, in one

way

we know

or another,

is

the topic of discussion throughout the remainder of this chapter.

2.7

Chlorination: a substitution reaction

Under the influence of ultraviolet light or at a temperature of 250-400 a
mixture of the two gases, methane and chlorine, reacts vigorously to yield hydrogen
chloride and a compound of formula CH 3 C1. We say that methane has undergone
chlorination, and we call the product, CH 3 C1, chloromethane or methyl chloride

(CH 3 =

methyl).

Chlorination
as substitution.

A

is

a typical example of a broad class of organic reactions known
atom has been substituted for a hydrogen atom of

chlorine

METHANE

44

CHAP. 2

methane, and the hydrogen atom thus replaced
atom of chlorine.

is

H

found combined with a second

H
Ii8htorheat

>

H--C-C1 + H-C1

A
M

Chlorine

Ji
w

Methane

Hydrogen
chloride

Methyl chloride
(Chloromethane)

The methyl chloride can itself undergo further substitution to form more
hydrogen chloride and CH 2 C1 2 dichloromethane or methylene chloride (CH 2 =
,

methylene).

H

H
1

H C

Cl

+

H

Cl

I

I

H

H
Methylene chloride
(Dichloromethane)

In a similar way, chlorination may continue to yield CHCIa trichloromethane
or chloroform, and CC1 4 tetrachloromethane or carbon tetrachloride. These last
,

,

two compounds are already familiar

an anesthetic, and carbon
and the fluid in certain fire

to us, chloroform as

tetrachloride as a non-flammable cleaning agent
extinguishers.

HC1

+

CH 4 ^L> CH 3 C1
Methane

Methyl
chloride

2.8

Control of chlorination

Chlorination of methane may yield any one of four organic products, depending upon the stage to which the reaction is carried. Can we control this reaction
so that methyl chloride is the principal organic product? That is, can we limit
the reaction to the

first

stage,

wowochlorination ?

We might at first expect

to accomplish this by pronaively, as it turns out
viding only one mole of chlorine for each mole of methane. But let us see what
happens if we do so. At the beginning of the reaction there is only methane for
the chlorine to react with, and consequently only the first stage of chlorination

takes place. This reaction, however, yields methyl chloride, so that as the reaction
proceeds methane disappears and methyl chloride takes its place.

As

the proportion of methyl chloride grows, it competes with the methane for
By the time the concentration of methyl chloride exceeds

the available chlorine.

that of methane, chlorine

is

more likely to attack methyl chloride than methane, and

A

the second stage of chlorination becomes more important than the first.
large
amount of methylene chloride is formed, which in a similar way is chlorinated to

chloroform and
finally

work up

this, in turn, is

chlorinated to carbon tetrachloride.

chlorinated methanes together

When we

we find that it is a mixture of
with some unreacted methane.

the reaction product,

all

four

RELATIVE REACTIVITY

SEC. 2.10

The

45

reaction may, however, be limited almost entirely to monochlorination

if

use a large excess of methane. In this case, even at the very end of the reaction
unreacted methane greatly exceeds methyl chloride. Chlorine is more likely to

we

attack methane than methyl chloride, and thus the

first stage of chlorination is the
principal reaction.
Because of the great difference in their boiling points, it is easy to separate the
excess methane (b.p. -161.5) from the methyl chloride (b.p. -24) so that the

methane can be mixed with more chlorine and put through the process again.
While there is a low conversion of methane into methyl chloride in each cycle, the
yield of methyl chloride based on the chlorine consumed is quite high.
The use of a large excess of one reactant is a common device of the organic
chemist when he wishes to limit reaction to only one of a number of reactive sites
in the

2.9

molecule of that reactant.

Reaction with other halogens: halogenation

Methane

under the
bromomethanes: methyl

reacts with bromine, again at high temperatures or

influence of ultraviolet light, to yield the corresponding

bromide, methylene bromide, bromoform, and carbon tetrabromide.

HBr
+

C H 4 -"^ CH 3 Br
Methane

Methyl
bromide

Bromination takes place somewhat

Methane does not

less readily

react with iodine at

all.

than chlorination.

With

fluorine

it

reacts so vigor-

ously that, even in the dark and at room temperature, the reaction must be carefully controlled the reactants, diluted with an inert gas, are mixed at low pressure.
:

We

can, therefore, arrange the halogens in order of reactivity.

Reactivity of halogens

F2 >

C1 2

> Br2

O

I 2)

This same order of reactivity holds for the reaction of the halogens with other
alkanes and, indeed, with most other organic compounds. The spread of reactivities is so great that only chlorination and bromination proceed at such rates as
to be generally useful.

2.10

Relative reactivity

Throughout our study of organic chemistry, we shall constantly be interested
We shall compare the reactivities of various reagents toward
the same organic compound, the reactivities of different organic compounds toward the same reagent, and even the reactivities of different sites in an organic
molecule toward the same reagent.
in relative reactivities.

It should be understood that when we compare reactivities we compare rates
of reaction. When we say that chlorine is more reactive than bromine toward

methane, we mean that under the same conditions (same concentration, same
temperature, etc.) chlorine reacts with methane faster than does bromine. From
another point of view, we mean that the bromine reaction must be carried out under

METHANE

46

CHAP.

2

more vigorous conditions (higher concentration or higher temperature) if it is to
take place as fast as the chlorine reaction. When we say that methane and iodine
do not react at all, we mean that the reaction is too slow to be significant.

We

want

shall

whenever

possible,

know
how to

to

not only what these relative reactivities are, but also,
account for them. To see what factors cause one

reaction to be faster than another,

we

however, we must understand a

2.11

little

up in more detail this matter of the
toward methane. Before we can do this,
more about the reaction itself.

shall take

different reactivities of the halogens

Reaction mechanisms
important for us to

It is

but also

how

it

happens, that

know
is,

to

not only what happens in a chemical reaction
not only the facts but also the theory.

know

For example, we know that methane and chlorine under the influence of heat
or light form methyl chloride and hydrogen chloride. Just how is a molecule of
methane converted into a molecule of methyl chloride? Does this transformation
involve

more than one

step, and. if so,

what are these steps? Just what

tion of heat or light ?
The answer to questions like these, that

is,

is

the func-

the detailed, step-by-step description

of a chemical reaction, is called a mechanism. It is only a hypothesis; it is advanced
to account for the facts. As more facts are discovered, the mechanism must also
account for them, or else be modified so that it does account for them; it may even
be necessary to discard a mechanism and to propose a new one.
It would be difficult to say that a mechanism had ever been proved. If, however, a mechanism accounts satisfactorily for a wide variety of facts; if we make
predictions based upon this mechanism and find these predictions borne out; if the
mechanism is consistent with mechanisms for other, related reactions; then the

mechanism

is

said to be well established,

and

it

becomes part of the theory of

organic chemistry.

Why

are

we

interested in the

mechanisms of reactions? As an important part
make up the framework on which

of the theory of organic chemistry, they help

we hang

the facts

we

learn.

An

understanding of mechanisms will help us to see a

We

shall
pattern in the complicated and confusing multitude of organic reactions.
find that many apparently unrelated reactions proceed by the same or similar

mechanisms, so that most of what we have already learned about one reaction
be applied directly to

many new

may

ones.

By knowing how a reaction takes place, we can make changes in the experimental conditions not by trial and error, but logically that will improve the
yield of the product we want, or that will even alter the course of the reaction
completely and give us an entirely different product. As our understanding of
reactions grows* so does our power to control them.

2.12

Mechanism of
It will

in

chlorination. Free radicals

be worthwhile to examine the mechanism of chlorination of methane

some

tion,

detail. The same mechanism holds for bromination as well as chlorinaand for other alkanes as well as methane; it even holds for many compqunds

which, while not alkanes, contain alkane-like portions in their molecules. Closely

MECHANISM OF CHLORINATION

SEC. 2.12

mechanisms are involved

related

More

alkanes.

this

important,

in oxidation

mechanism

47

(combustion) and other reactions of
illustrates certain general principles

wide range of chemical reactions. Finally, by studying
we can learn something of how a chemist
finds out what goes on during a chemical reaction.
Among the facts that must be accounted for are these: (a) Methane and
that can be carried over to a

the evidence that supports the mechanism,

chlorine

do not

readily,

however,

react in thr dark at

room

of ultraviolet light at

many

room

temperature, (b) Reaction takes place
(c) under the influence

dark at temperatures over 250, or

in the

temperature, (d)

When

the reaction

is

induced by

light,

(several thousand) molecules of methyl chloride are obtained for each

photon of light that is absorbed by the system, (e) The presence of a small amount
of oxygen slows down the reaction for a period of time, after which the reaction
proceeds normally; the length of this period depends upon how much oxygen is
present.

The mechanism
generally accepted,

that accounts for these facts

is

shown

C1 2

(1)

Cl-

(2)

-

3

then

heatorlight

satisfactorily,

2C1-

>

HC1 +

>

CH 3C1 +

C1 2

and hence

is

:

>

+ CH 4

CH +

(3)

most

in the following equation

CH

-

3

Ci-

(2), (3), (2), (3), etc.

The first step is the breaking of a chlorine molecule into two chlorine atoms.
Like the breaking of any bond, this requires energy, the bond dissociation energy,
and in Table 1.2 (p. 21) we find that in this case the value is 58 kcal/mole. The
energy

is

supplied as either heat or light.
energy

The

+

:CI:C1:

>

:C1-

+

-Cl:

chlorine molecule undergoes homolysis (Sec. 1.14): that is, cleavage of
bond takes place in a symmetrical way, so that each atom

the chlorine-chlorine

one electron of the pair that formed the covalent bond. This odd electron
not paired as are all the other electrons of the chlorine atom; that is, it does not
have a partner of opposite spin (Sec. 1.6). An atom or group of atoms possessing an
odd (unpaired) electron is called a free radical. In writing the symbol for a free
retains
is

we generally include a dot to represent the odd electron just as we include
a plus or minus sign in the symbol of an ion.
Once formed, what is a chlorine atom most likely to do? Like most free
radicals, it is extremely reactive because of its tendency to gain an additional electron
radical,

and thus have a complete

from another point of view, energy was supplied
the cleavage of the chlorine molecule, and this
energy-rich particle tends strongly to lose energy by the formation of a new chemical
to each chlorine

octet;

atom during

bond.

To form a new chemical bond, that is, to react, the chlorine atom must collide
with some other molecule or atom. What is it most likely to collide with? Obviously, it is most likely to collide with the particles that are present in the" highest
concentration: chlorine molecules and methane molecules. Collision with another

METHANE

48
chlorine

atom

CHAP.

2

quite unlikely simply because there are very few of these reactive,
around at any time. Of the likely collisions, that with a

is

short-lived particles

no net change; reaction may occur, but
only in the exchange of one chlorine atom for another:

chlorine molecule causes

:

+

Cl-

:

Cl Cl

>

:

.

:

Cl Cl
:

:

+

:

it

can result

Collision probable but not productive

Cl-

atom with a methane molecule is both probable and
The chlorine atom abstracts a hydrogen atom, with one electron, to

Collision of a chlorine

productive.

form a molecule of hydrogen chloride

:

H

H
H C H +
H

Cl

:

:

>

:

H

Cl

:

:

+ H C

-

:

Collision probable

and productive

H

Methane

Methyl radical

Now the methyl group is left with an odd, unpaired electron; the carbon atom has
only seven electrons in its valence shell. One free radical, the chlorine atom, has
been consumed, and a new one, the methyl radical,
3 -, has been formed in its

CH

place. This

is

Now, what
it is

mechanism.
methyl radical most likely to do? Like the chlorine atom,
and for the same reason: the tendency to complete its octet,

step (2) in the
this

is

extremely reactive,

to lose energy by forming a new bond. Again, collisions with chlorine molecules
or methane molecules are the probable ones, not collisions with the relatively

scarce chlorine atoms or methyl radicals. But collision with a methane molecule
could at most result only in the exchange of one methyl radical for another:

H
H C H
H
:

:

The

H
H C
H

H
+ C

:

H

>

:

H
-f

H C H
:

:

J

H
collision

Collision probable but not productive

H

of a methyl radical with a chlorine molecule

is,

then, the importhe bonding

The methyl radical abstracts a chlorine atom, with one of
electrons, to form a molecule of methyl chloride:
tant one.

H
H C
H
:

-

-f

:

Cl Cl
:

:

Methyl

>

H
H C
H
:

:

Cl

:

+

:

Cl

Collision probable

and productive

Methyl chloride

radical

The other product

is a chlorine atom. This is step (3) in the mechanism.
Here again the consumption of one reactive particle has been accompanied by
the formation of another. The new chlorine atom attacks methane to form a
methyl radical, which attacks a chlorine molecule to form a chlorine atom, and so
the sequence is repeated over and over. Each step produces not only a new reactive

particle but also a molecule of product: methyl chloride or

hydrogen chloride.

This process cannot, however, go on forever. As we saw earlier, union of two
r^^rt \\*, a A folotlyglv crarrf* nartirh*c is not Itkelv but CVerV SO often it dOCS
happen
!

INHIBITORS

SEC. 2.14

and when
are

49

does, this particular sequence of reactions stops. Reactive particles

it

consumed but not generated.
:C1-

+

>

:C1:C1:

3

>

CH :CH

-Cl:

>

CH

-Cl:

CH 3 + -CH
-

CH
It is clear,

on page 47

:

then,

how

the

either light or heat

-

3

+

many

3

is

3

:CJ:

mechanism accounts

and (d)
and form

for facts (a), (b), (c),

required to cleave the chlorine molecule

atom may eventually bring about

the initial chlorine atoms; once formed, each

formation of

3

the

molecules of methyl chloride.

Chain reactions

2.13

is an example of a chain reaction, a reaction that
of steps, each of which generates a reactive substance that brings
about the next step. While chain reactions may vary widely in their details, they all
have certain fundamental characteristics in common.

The

chlorination of methane

a

involves

series

heatorli8ht

C1 2

(1)

Cl-

(2)

+ CH 4

CH r +

(3)

then

Chain-initiating step

HC1 +
>

C1 2

CH

CH 3 C1 +

-

3

1
>

Cl-

J

Chain-propagating steps

(2), (3), (2), (3), etc., until finally:

Cl-

(4)

2C1

>

+

>

C1 2

3

>

CH CH

-CI

>

-Cl

or
(5)

CH 3 + -CH
-

3

3

}>

Chain-terminating steps

or

CH

(6)

3

+

First in the chain

of reactions

CH
is

3

C1

a chain-initiating step, in which energy

absorbed and a reactive particle generated;

in the present reaction

it is

is

the cleavage

of chlorine into atoms (step 1).
There are one or more chain-propagating steps, each of which consumes a
reactive particle and generates another; here they are the reaction of chlorine atoms
with methane (step

2),

and of methyl radicals with chlorine

(step 3).

Finally, there are chain-terminating steps, in which reactive particles are consumed but not generated; in the chlorination of methane these would involve the

union of two of the reactive

particles, or the

capture of one of them by the walls

of the reaction vessel.

Under one set of conditions, about 10,000 molecules of methyl chloride are
formed for every quantum (photon) of light absorbed. Each photon cleaves one
chlorine molecule to form two chlorine atoms, each of which starts a chain. On
the average, each chain consists of 5000 repetitions of the chain-propagating cycle
before

it

2.14

Inhibitors

is finally

Finally.

h^ w

stopped.

^^ the mechanism of chlorination account for fact

(e),

that a

METHANE

50

CHAP.

2

amount of oxygen slows down the reaction for a period of time, which depends upon the amount of oxygen, after which the reaction proceeds normally?
Oxygen is believed to react with a methyl radical to form a new free radical:

small

CH

CH 3 OO-

The

radical

-

3

much

is

+

CH -O-0-

>

2

less reactive

3

than the

CH 3

-

radical,

and can do

little

By combining with a methyl radical, one oxygen molecule
breaks a chain, and thus prevents the formation of thousands of molecules of
methyl chloride; this, of course, slows down the reaction tremendously. After all
to continue the chain.

combined with methyl radicals* the reaction is
normal rate.
A substance that slows down or stops a reaction even though present in small
amount is called an inhibitor. The period of time during which inhibition lasts, and
the oxygen molecules present have
free to

proceed at

its

after which the reaction proceeds normally,

by a

reactions

we

is

called the inhibition period. Inhibition

amount of an added material is quite characteristic of chain
of any type, and is often one of the clues that first leads us to suspect that

relatively small

are dealing with a chain reaction.

It is

could prevent the reaction of so many.

oxygen to

hard to see

(We

how

shall frequently

else

a few molecules

encounter the use of

inhibit free-radical reactions.)

Heat of reaction

2.15

In our consideration of the chlorination of methane, we have so far been concerned chiefly with the particles involved molecules and atoms- and the changes
that they undergo. As with any reaction, however, it is important to consider also
the energy changes involved, since these changes determine to a large extent how

whether it will take place at all.
dissociation energies given in Table 1.2 (p. 21),
we can calculate the energy changes that take place in a great number of reactions.
In the conversion of methane into methyl chloride, two bonds are broken,
3

fast the reaction will go, and, in fact,

the values of

By using

bond

CH -H

and CI Cl, consuming 104 f 58, or a total of 162 kcal/mole. At the same time
two new bonds are formed, CH 3 - Cl and H Cl, liberating 84 f 103, or a total
of 187 kcal/mole. The result is the liberation of 25 kcal of heat for every mole of

CH -H +
3

Cl -CI

104

>

58

methane that

3

84

162

tion.

CH -C! + H-C1
103.
187

A//

= -25

kcal

converted into methyl chloride; this is, then, an exothermic reacwe note, does not depend on our knowing the mechanism

is

(This calculation,

of the reaction.)

When

heat

is

liberated, the heat content (enthalpy), //,

of the molecules them-

must decrease; the change in heat content, A//, is therefore given a negative*
(In the case of an endothermic reaction, where heat is absorbed, the increase

selves
sign.

in heat content

Problem
(a)

of the molecules

2.1

is

indicated by a positive A//.)

Calculate A// for the corresponding reaction of methane with:

bromine, (b) iodine,

(c) fluorine.

ENERGY OF ACTIVATION

SEC. 2.16

51

The value of -25 kcal that we have just calculated is the netkH for the overall
A more useful picture of the reaction is given by the A//'s of the indi-

reaction.

vidual steps. These are calculated below:

C1-C1

(1)

Atf=+58kcal

*

2C\>

>

CH 3 + H

(58)

Cl-

(2)

+ CH 3

H

CH 3 +

>

CH 3-C1 +

A//

Cl-

= -26

(84)

(58)
It is clear

+1

(103)

C1-C1

-

(3)

A# =

Cl

-

(104)

why this reaction, even though exothermic,

ture (in the absence of light).

The

occurs only at a high temperawhich reaction

chain-initiating step, without

is highly endothermic, and takes place (at a significant rate) only at a
high temperature. Once the chlorine atoms are formed, the two exothermic chainpropagating steps occur readily many times before the chain is broken. The

cannot occur,

difficult

cleavage of chlorine

is

the barrier that must be surmounted before the

subsequent easy steps can be taken.
Problem 2.2 Calculate A/7 for the corresponding steps
ane with: (a) bromine, (b) iodine, (c) fluorine.

in the reaction of

meth-

We have assumed so far that exothermic reactions proceed readily, that is, are
reasonably fast at ordinary temperatures, whereas endothermic reactions proceed
with difficulty, that is, are slow except at very high temperatures. This assumed
relationship between A// and rate of reaction is a useful rule of thumb when other
is not available; it is not, however, a necessary relationship, and there
exceptions to the rule. We shall go on, then, to a discussion of another
energy quantity, the energy of activation* which is related in a more exact way to

information
are

many

rate of reaction.

2.16

Energy of activation

To

see

closely at
Cl-

a
-I-

what

happens during a chemical reaction, let us look more
example, the attack of chlorine atoms on methane:

actually

specific

CH 3 H

>

(104)

H

Cl

4-

CH 3

-

\H =

+

1

kcal

E^

4 kcal

(103)

is comparatively simple: it occurs in the gas phase, and is thus not
complicated by the presence of a solvent; it involves the interaction of a single atom
and the simplest of organic molecules. Yet from it we can learn certain principles

This reaction

that apply to any reaction.
Just what must happen if this reaction

is

to occur? First of

all,

a chlorine

atom and a methane molecule must collide. Since chemical forces are of extremely
short range, a hydrogen-chlorine bond can form only when the atoms are in close
contact.

Next, to be

must provide a certain minimum amount of
Cl bond liberates 103 kcal/mole; breaking the

effective, the collision

energy. Formation of the

H

METHANE

52

CHAP.

2

CHj H bond

requires 104 kcal/mole. We might have expected that only 1 kcal/
mole additional energy would be needed for reaction to occur; however, this is
not so. Bond-breaking and bond-making evidently are not perfectly synchronized,
and the energy liberated by the one process is not completely available for the
other. Experiment has shown that if reaction is to occur, an additional 4 kcal/mole
of energy must be supplied.
The minimum amount of energy that must be provided by a collision for reaction
"

to occur is called the energy of activation,

the

moving

and are

particles.

Most

acl

.

Its

source

is

the kinetic energy of

collisions provide less than this

fruitless, the original particles

minimum

quantity
simply bouncing apart. Only solid collisions

between particles one or both of which are moving unusually fast are energetic
enough to bring about reaction. In the present example, at 275, only about one
collision in 40 is sufficiently energetic.
Finally, in addition to being sufficiently energetic, the collisions must occur
when the particles are properly oriented. At the instant of collision, the methane
molecule must be turned in such a way as to present a hydrogen atom to the full
force of the impact. In the present example, only about one collision in eight is
properly oriented.
In genera], then, a chemical reaction requires collisions of sufficient energy
(a c t) an d of proper orientation. There is an energy of activation for nearly every
reaction where bonds are broken, even for exothermic reactions, in which bond-

making liberates more energy than is consumed by bond-breaking.
The attack of bromine atoms on methane is more highly endothermic, with
a A// of +16 kcal.
'*

Br-

+ CH 3 H

*<

H

>

Br

+ CH 3

-

A/f

= +16

kcaJ

Eact =

18 kcal

(88)

(104)

CH 3 H bond, as before, requires 104 kcal/mole, of which only 88
provided by formation of the H Br bond. It is evident that, even if this
88 kcal were completely available for bond-breaking, at least an additional 16
Rr

liking the

kcal

is

kcal/mole would have to be supplied by the collision. In other words, the act
of an endothermic reaction must be at least as large as the A//. As is generally
true, the
&ct of the present reaction (18 kcal) is actually somewhat larger than
the A//.

Progress of reaction: energy changes

2.17

and

These energy relationships can be seen more clearly in diagrams like Figs. 2.2
2.3. Progress of reaction is represented by horizontal movement from reac-

on the

left to products on the right. Potential energy (that is, all energy except
any stage of reaction is indicated by the height of the curve.
Let us follow the course of reaction in Fig. 2.2. We start in a potential energy
valley with a methane molecule and a chlorine atom. These particles are moving,
and hence possess kinetic energy in addition to the potential energy shown. The
exact amount of kinetic energy varies with the particular pair of particles, since
some move faster than others. They collide, and kinetic energy is converted into
potential energy. With this increase in potential energy, reaction begins, and we

tants

kinetic) at

move up
of the

hill

the energy

and

start

hill.

down

If

enough

kinetic energy

the far side.

is

converted,

we reach

the top

SEC.

2.17

PROGRESS OF REACTION: UN ERG Y CHANGES

A//-

CHh-

4-1 kcai

53

+ HCI

CHi-f-CI-

CHi+Cl-

+HC1

>

Progress of reaction

Potential energy changes during progress of reaction: the
methane-chlorine atom reaction.

Figure 2.2.

is converted back into kinetic energy,
the level of the products. The products contain a little more potential energy than did the reactants, and we find ourselves in a slightly higher valley
than the one we left. With this net increase in potential energy there must be a

During the descent, potential energy

until

we reach

corresponding decrease in kinetic energy. The new particles break apart, and sin<
they are moving more slowly than the particles from which they were for*

CHr + HBr
act

=I8kcal

+ Br

CH3 + HBr
-

Progress of reaction

Figure 2.3.

Potential energy changes during progress of reaction
reaction.

methane-bromine atom

:

the

METHANE

54

CHAP.

2

we observe

a drop in temperature. Heat will be taken up from the surroundings.
bromine reaction, shown in Fig. 2.3, we climb a much higher hill and
a much higher valley. The increase in potential energy and the corre-

In the

end up in
sponding decrease

in kinetic energy
is much larger than in the chlorine reaction;
be taken up from the surroundings.
An exothermic reaction follows much the same course. (Take, for example,
the reverse of the bromine reaction: that is, read from right to left in Fig. 2.3.)

more heat

will

In this case, however, the products contain less potential energy than did the reacwe end up in a lower valley than the one we left. Since this time the

tants so that

new particles contain more kinetic energy than the particles from which they
were formed, and hence move faster, we observe a rise in temperature. Heat will
be given offio the surroundings.
In any reaction there are
to reach the top of the

hill.

many

collisions that provide too

These collisions are

fruitless,

little

and we

energy for us
back to our

slide

Many collisions provide sufficient energy, but take place when the
molecules are^improperly oriented. We then climb an energy hill, but we are off
the road; we may climb very high without finding the pass that leads over into the
original valley.

next valley.

The

difference in level between the

two

valleys,

is,

of course, the A//; the

dif-

ference in level between the reactant valley and the top of the hill is the
act
are concerned only with these differences, and not with the absolute height at
any stage of the reaction. We are not even concerned with the relative levels of the
.

We

and bromine reactions. We need only to know that
we climb a hill 4 kcal high and end up in a valley kcal
point; and that in the bromine reaction we climb a hill 18

reactant valleys in the chlorine
in the chlorine reaction

higher than our starting

and end up

1

higher than our starting point.
the height of the hill, the act that determines the rate of
reaction, and not the difference in level of the two valleys, A//. In going to a lower
or even non-existent.
valley, the hill might be very high, but could be very low.
kcal hign

As we

shall see,

in a valley 16 kcal

it is

In climbing to a higher valley, however, the hill can be no lower than the valley to
which we are going; that is to say, in an endothermic reaction the ACl must be at least

E

as large as the A//.

An
because

energy diagram of the sort shown in Figs. 2.2 and 2.3 is particularly useful
it tells us not only about the reaction we are
considering, but also about the

reverse reaction. Let us

move from

right to left in Fig. 2.2, for example.

We

see

that the reaction

CH 3 + H-C1
-

>

(103)

CH 3 H +

Cl-

A# =

-1,

aot

=

3

(104)

has an energy of activation of 3 kcal, since in this case we climb the hill from the
1 kcal.
higher valley. 'fUfcis, of course, an exothermic reaction with a A// of
In the

same way we can

CH 3 + H
.

Br

(88)

see
>

from

Fig. 2.3 that the reaction

CH 3^~H +

Br

A//

=-16,

;<*

=

2

(104)

has an energy of activation of 2 kcal, and is exothermic with a A// of - 16 kcal.
(We notice that, even though exothermic, these last two reactions have energies of
activation.)

RATE OF REACTION

SEC. 2.18

55

Reactions like the cleavage of chlorine into atoms
Cl

>

Cl

Cl-

A#=

+ -Cl

fall

into a special category:

+58,

*

=

58

(58)
C.

a bond

is

broken but no bonds are formed. The reverse of

this reaction, the union"

of chlorine atoms, involves no bond-breaking and hence would be expected to
Cl-

+

Cl-Cl

-Cl

= -58,

E

ft

ct

=

(58)

take place very easily, in fact, with no energy of activation at all. This is considered to be generally true for reactions involving the union of two free radicals.
If there is no hill to climb in going from chlorine atoms to a chlorine molecule,
but simply a slope to descend, the cleavage of a chlorine molecule must involve
simply the ascent of a slope as shown in Fig. 2.4. The act for the cleavage of a

chlorine molecule, then, must equal the A//, that is, 58 kcal. This equality of ^. t
and A/7 is believed to hold generally' for reactions in which molecules dissociate
into radicals.

Progress of reaction

>

Potential energy changes during progress of reaction: simple

Figure 2.4.
dissociation.

Rate of reaction

2.18

A

chemical reaction

orientation.

The

is

the result of collisions of sufficient energy and proper
must be the rate at which these effec-

rate of reaction, therefore,

tive collisions occur, the

number of

effective collisions, let us say, that

occur

during each second within each cc of reaction space. We can then express the
rate as the product of three factors. (The number expressing the probability that a

METHANE

CHAP.

2

have the proper orientation is commonly called the probability factor.)
Anything that affects any one of these factors affects the rate of reaction.
collision will

The

collision frequency

depends upon

(a)

how

closely the particles are

crowded

together, that is, concentration or pressure; (b) how large they are; and (c) how
fast they are moving, which in turn depends upon their weight and the temperature.
We can change the concentration and temperature, and thus change the rate.

We

are familiar with the fact that an increase in concentration causes

in rate;

an increase
does so, of course, by increasing the collision frequency. A rise in

it

temperature increases the collision frequency; as we shall
energy factor,

and

see,

this latter effect is so great that the effect

it

also increases the

of temperature en

collision frequency

is by comparison unimportant.
and weight of the particles are characteristic of each reaction and
cannot be changed. Although they vary widely from reaction .to reaction, this

The

size

A

heavier weight makes
variation does not affect the collision frequency greatly.
the particle move more slowly at a given temperature, and hence tends to decrease
the collision frequency. A heavier particle is, however, generally a larger particle,
and the larger size tends to increase the collision frequency. These two factors thus

tend to cancel out.

The

probability factor

of reaction that

is

depends upon the geometry of the particles and the kind
it does not
vary

taking place. For closely related reactions

widely.
Kinetic energy of the moving molecules is not the only source of the energy needed
energ> can also be provided, for example, from vibrations among the various
atoms within the molecule. Thus the probability factor has to do not only with what
for reaction

atoms
in the

;

in the molecule suffer the collision, but also with the alignment of the other atoms
molecule at the time of collision.

By

far the

most important factor determining

rate

is

the energy factor: the

fraction of collisions that are sufficiently energetic. This factor depends upon the
temperature, which we can control, and upon the energy of activation, which is
characteristic of each reaction.

At a
velocity
in fact,

vidual

given* temperature the molecules of a particular compound have an average
and hence an average kinetic energy that is characteristic of this system;
the temperature is a measure of this average kinetic energy. But the indi-

n, )lecules

do not

all

travel with the

same

velocity,

some moving

faster

than the average and some slower. The distribution of velocities is shown in
Fig. 2,5 by the familiar bell-shaped curve that describes the distribution among
individuals of so
Jife

expectancy.

many

qualities, for

example, height, intelligence, income, or even

The number of molecules with a

particular velocity

is

greatest

RATE OF REACTION

SEC. 2.18

57

Collisions with energy

>

i

Collisions with energy

>

2

Average
*>

Energy

>

Energy

Figure 2.6. Distribution of kinetic energy among collisions.

Figure 2.5. Distribution of kinetic energy among molecules.

for a velocity near the average
smaller than the average.

and decreases as the

becomes

velocity

larger or

The distribution of collision energies, as we might expect, is described by a
similar curve, Fig. 2.6. Let us indicate collisions of a particular energy,
act , by a
vertical line. The number of collisions with energy equal to or greater than
ftct
is

indicated by the shaded area under the curve to the right of the vertical line.
number of collisions that have this minimum energy,

fraction of the total
is

then the fraction of the total area that

the value

The

is

shaded.

It is

The
'

act

,

evident that the greater

0/ act , the smaller the fraction of collisions that possess that energy.
exact relationship between energy of activation and fraction of collisions

with that energy

is:

e~ E *ct /RT

=

fraction of collisions with energy greater than

where
e =
R=
T=

2.7 1 8 (base of natural logarithms)

1.986 (gas constant)
absolute temperature.

Using P for the probability factor and
the rate equation

Z for

the collision frequency,

we

arrive at

:

rate

=

This exponential /relationship is important to us in that it indicates that a
small difference in act has a large effect on the fraction of sufficiently energetic
collisions,

and hence on the

rate of reaction.

For example,

at

275, out of every

million collisions, 10,000 provide sufficient energy if act = 5 kcal, 100 provide
= 10 kcal, and only one provides sufficient energy if
sufficient energy if
act
=
15
kcal.
This
that (all other things being equal) a reaction with
means
act
acl

=

go 100 times as fast as one with
as one with aot = 15 kcal.

5 kcal will

times as fast

ftct

=

10 kcal, and 10,000

METHANE

58

CHAP.

2

have so far considered a system held at a given temperature. A rise in
temperature, of course, increases the average kinetic energy and average velocities,
and hence shifts the entire curve to the right, as shown in Fig. 2.7. For a given

We

energy of activation, then, a rise in temperature increases the fraction of sufficiently
and hence increases the rate, as we already know.

energetic collisions,

The exponential relationship again leads to a large change in rate, this time for
a small change in temperature. For example, a rise from 250 to 300, which is
increase in absolute temperature, increases the rate by 50",', if act =
only a 10
5 kcal, doubles the rate

if

act

=

10 kcal,

As this example shows, the greater the

and

trebles the rate if

act

=

15 kcal.

the greater the effect of a given change in
E IRT
temperature; this follows from the e~ *"
relationship. Indeed, it is from the
relationship between rate and temperature that the act of a reaction is determined:
the rate

act

,

measured at different temperatures, and from the results act is calculated.
have examined the factors that determine rate of reaction. What we have

is

We

learned may be used in many ways. To speed up a particular reaction, for
example, we know that we might raise the temperature, or increase the concentration of reactants, or even (in ways that we shall take up later) lower the
acL
.

Of immediate interest, however, is the matter of relative reactivities.
therefore, how our knowledge of reaction rates can help us to account

see,

fact that

one reaction proceeds

two reactions are

faster than another,

Let us
for the

even though conditions for the

identical.

Collisions at T\ with energy

EAct

with energy

& et

Collisions at

T-i

15

3

act

>

Energy
Figure 2.7.

Change

in collision energies

with change in temperature.

Relative rates of reaction

2.19

We have s