Contributed by:
The separation of mixtures of compounds to give the pure components is of great practical importance in chemistry. ~2n~ synthetic reactions give mixtures of products and it is necessary for you to have a reasonably clear idea of
how mixtures of compounds can be separated.
T he separation of mixtures of compounds to give the pure components is of
great practical importance in chemistry. ~ 2 synthetic
n ~ reactions give mix-
tures of products and it is necessary for you to have a reasonably clear idea of
how mixtures of compounds can be separated. Almost all compounds of bio-
chemical interest occur naturally as components of very complex mixtures
from which they can be separated only with considerable difficulty.
Separations can be achieved by differences in physical properties, such
as differences in boiling point, or by chemical means, wherein differences in
physical properties are enhanced by chemical reactions. In this chapter we
will consider some separations of compounds based on differences in physical
2. 258 9 Separat~onand Pur~ficat~on Organ~cCompounds by Spectroscop~cTechn~ques
l d e n t ~ f ~ c a t ~ofo n
properties. Chemical procedures will be discussed elsewhere in connection
with the appropriate classes of compounds.
Identification and structure determination are often closely allied to the
problem of separation. Once a compound is separated, how do we determine
whether it is identical to some previously known compound (identification) or,
if that can't be done, how do we determine its chemical structure? The spectro-
scopic properties of molecules have proven to be extremely informative for
both identification and structure determination and this chapter is mainly con-
cerned with the application of spectroscopy for such purposes. We will give
you now an overview of the spectroscopic properties of the major classes of
organic compounds. In subsequent chapters, spectroscopic properties will be
discussed in the context of the class of compounds under consideration.
The classical criteria for determining the purity of organic compounds are
correct elemental compositions (Section 1- 1A) and sharpness of melting point
or constancy of boiling point. Important though these analytical and physical
criteria are, they can be misleading or even useless. For instance, the analytical
criterion is of no help with possible mixtures of isomers because these mixtures
have the same elemental composition. The simple physical criteria are not
applicable to substances that decompose when one attempts to determine the
melting point or boiling point. Furthermore, boiling points are not very helpful
for liquids that are mixtures of substances with nearly the same boiling point
or are a~eotropes.~ Similar difficulties may be encountered with mixtures of
solid substances that form mixed crystals or are eutectic~.~ Much sharper
criteria for the purity of organic compounds now are provided through use of
"super-separation" methods to see if any contaminants can be separated, or by
spectroscopic techniques, as will be discussed later in this chapter. We begin
here with a brief description of chromatographic methods of separation.
'An azeotrope is a mixture of two or more substances that boils at a constant tempera-
ture, either higher or lower than any of its constituents. Thus an 8.5:l mole mixture of
ethanol and water boils like a pure substance, distilling at 78.2', which is lower than the
boiling point of ethanol (78.5") or of water ( 100"). In contrast, a 1.35: 1 mole mixture
of methanoic (formic) acid and water boils at 107.l0,which is higher than the boiling
points of either methanoic acid ( 100.7") or water (100").
2When solid substances are mixed, the melting point of each normally is depressed. The
eutectic mixture is the mixture of the solids with the lowest melting point.
3. 9-2A Gas-Liqu~dChromatography
9-2A Gas-Liqu id Chromatography
Many separation methods are based on chromatography, that is, separation of
the components of a mixture by differences in the way they become distributed
(or partitioned) between two different phases. To illustrate with an extreme
example, suppose we have a mixture of gaseous methane and ammonia and
contact this mixture with water. Ammonia, being very soluble in water (- 90 g
per 1 0 0 of ~ ~water at 1 atm pressure), will mostly go into the water phase,
whereas the methane, being almost insoluble (- 0.003 g per 100 g of water)
will essentially remain entirely in the gas phase. Such a separation of methane
and ammonia would be a one-stage partitioning between gas and liquid phases
and, clearly, could be made much more efficient by contacting the gas layer
repeatedly with fresh water. Carried through many separate operations, this
partitioning procedure is, at best, a tedious process, especially if the com-
pounds to be separated are similar in their distributions between the phases.
However, partitioning can be achieved nearly automatically by using chromato-
graphic columns, which permit a stationary phase to be contacted by a moving
phase. To illustrate, suppose a sample of a gaseous mixture of ammonia and
methane is injected into a long tube (column) filled with glass beads moistened
with water (the stationary phase), and a slow stream of an inert carrier gas,
such as nitrogen or helium, is passed in to push the other gases through. A
multistage partitioning would occur as the ammonia dissolves in the water and
the resulting gas stream encounters fresh water as it moves along the column.
Carrier gas enriched with methane would emerge first and effluent gas contain-
ing ammonia would come out later. This is a crude description of the method of
gas-liquid chromatography (abbreviated often as glc, GC, or called vapor-phase
chromatography, vpc). This technique has become so efficient as to revolu-
tionize the analysis and separation of almost any organic substance that has
even a slight degree of volatility at some reasonably attainable temperature.
The most modern glc equipment runs wholly under computer control, with
preprogrammed temperatures and digital integration of the detector output. A
wide variety of schemes is available for measuring the concentration of
materials in the effluent carrier gas, and some of these are of such extraordinary
sensitivity that only very small samples are necessary (10-Q, or less).
In the usual glc procedure, a few microliters of an organic liquid to be
analyzed are injected into a vaporizer and carried with a stream of gas (usually
helium) into a long heated column that is p%cked with a porous solid (such as
crushed firebrick) impregnated with a nonvolatile liquid. Gas-liquid partition-
ing occurs, and small differences between partitioning of the components can
be magnified by the large number of repetitive partitions possible in a long
column. Detection often is achieved simply by measuring changes in thermal
conductivity of the effluent gases. A schematic diagram of the apparatus and a
typical separation pattern are shown in Figures 9-1 and 9-2. The method is
extraordinarily useful for detection of minute amounts of impurities provided
4. 260 9 Separation and Purification. Identification of Organic Compounds by Spectroscopic Techniques
pressure-regulated exit
carrler gas supply vapors
1 t /de"c"r Ipacked column
sample injection port
Figure 9-1 Schematic diagram of a gas-l~qu~d chromatography appa-
ratus The detector IS arranged to measure the difference In some property
of the carrler gas alone versus the carrrer gas plus effluent sample at the
e x ~ tDifferences In thermal conductivity are particularly easy to measure
and give reasonably high detect~onsensltivit~es
these are separated from the main peak. Glc also can be used effectively to
purify materials as well as to detect impurities. To do this, the sample size and
the size of the apparatus may be increased, or an automatic system may be
used wherein the products from many small-scale runs are combined.
Figure 9-2 A gas-liquid chromatogram of a mixture of the isomeric
butanols at constant column temperature. A tiny peak on the far left is a
trace of air injected with the sample. The retention times of the various
isomers are in the same order as the boiling points, which are, from left
to right, 82", 99.5", 108", and 11 7". The areas under each peak correspond
to the relative amounts of material present. Raising the column tempera-
ture at a preprogrammed rate while developing the chromatogram speeds
up the removal of the slower-moving components and sharpens their
peaks. Also, by diversion of the gas stream to appropriate cold traps it
is possible to collect pure fractions of each component.
5. 9-2B Liquid-Sol~d Chromatography
9-29 Liquid-Solid Chromatography
Liquid-solid chromatography originally was developed for the separation of
colored substances, hence the name chromatography, which stems from the
Greek word chvoma meaning color. In a typical examination, a colored sub-
stance suspected of containing colored impurities is dissolved in a suitable
solvent and the solution allowed to percolate down through a column packed
with a solid adsorbent, such as alumina or silica, as shown in Figure 9-3. The
"chromatogram" then is "developed" by passing through a suitable solvent
that washes the adsorbate down through the column. What one hopes for, but
may not always find, is that the components of the mixture will be adsorbed
unequally by the solid phase so distinct bands or zones of color appear. The
bands at the top of the column contain the most strongly adsorbed components
and the bands at the bottom the least strongly held components. The zones
may be separated mechanically, or sufficient solvent can be added to wash,
or elute, the zones of adsorbed materials sequentially from the column for
further analysis.
Liquid-solid chromatography in the form just described was developed
first by the Russian biochemist M. S. Tswett, about 1906. In recent years,
many variations have been developed that provide greater convenience, better
separating power, and wider applicability. In thin-layer chromatography,
which is especially useful for rapid analyses, a solid adsorbent containing a
suitable binder is spread evenly on a glass plate, a drop of solution to be ana-
lyzed is placed near one edge and the plate is placed in a container with the
edge of the plate below the spot, dipping into an eluting solvent. The solvent
ascends the plate and the materials in the spot move upward at different rates,
as on a Tswett column. Various detecting means are used- simple visual ob-
servation for colored compounds, differential fluorescence under ultraviolet
light, and spraying of the plate with substances that will give colored materials
with the compounds present. In favorable cases, this form of liquid-solid chro-
matography can be carried out with submicrogram quantities of materials.
solid adsorbent with
bands of separated
sample components
Figure 9-3 A slmple chromatograph~ccolumn for i~qu~d-solid chromatography
6. 262 9 Separation and Puriflcatlon. ldentlfication of Organic Compounds by Spectroscopic Techniques
An extremely important improvement on the Tswett procedure is high-
pressure solid-liquid chromatography. Increasing the input pressure on the
system to 20-70 atmospheres improves the speed of separations by permitting
the use of much smaller \olid particles (with more surface area) than would
be practical for gravity-flow Tswett columns. Automatic monitoring of the
column effluent by ultraviolet spectroscopy (Section 9-9) or by changes in the
refractive index usually provides an effective means of determining how the
separation is proceeding. With such techniques chromatograms similar to
Figure 9-2 are obtained. High-pressure liquid chromatography (hplc) has
great advantages for analysis and separation of high-molecular-weight heat-
sensitive compounds that are unsuitable for glc.
An ingenious variation of solid-liquid chromatography is to use a solid
support to which a material is attached that has a specific affinity for a par-
ticular substance to be separated. The technique is especially useful for sepa-
rating enzymes, and the immobile phase can be constructed from compounds
known to react with, or be complexed by, the enzyme. Some other forms of
chromatography are discussed in Sections 25-48 and 25-7E.
Observation of a single peak in a given chromatographic procedure is
evidence, albeit not definitive evidence, for purity. Contaminants with nearly
the same properties may be very difficult to separate and, if knowing the
degree of purity is highly important, one can run chromatogralns with a variety
of different adsorbents to see if each gives the same result. If they do, the pre-
sumption of purity improves, although it is desirable to determine whether
the spectroscopic techniques to be described in the following section permit
the same conclusion.
The most straightforward way to determine the structures of molecules would
be to "see" how the nuclei are arranged and how the electrons are distributed.
This is not possible with visible light, because the wavelengths of visible light
are very much longer than the usual molecular dimensions. A beam of elec-
trons can have the requisite short wavelengths, but small organic molecules
are destroyed rapidly by irradiation with electrons of the proper wavelengths.
Nonetheless, electron microscopy is a valuable technique for the study of large
molecules, such as D N A , which can be stained with heavy-metal atoms before
viewing, or are themselves reasonably stable to an electron beam (Figures
9-4 and 9-5).
Virtually all parts of the spectrum of electromagnetic radiation, from
x rays lo radio waves, have some practical application for the study of organic
molecules. The use of x-ray difipraction for determination of the structures of
7. 9-3 Some General considerations of Diffraction and Spectroscopic Techniques
Figure 9-4 Electron micrograph (x40,000) of two linked (cat@nated)
cyclic mitochondria1 DNA molecules from a culture of human cells. The
DNA was stained with uranyl acetate, then shadowed wlth platinum and
palladium atoms ~n h ~ g hvacuum to make the molecules eas~lyv ~ s i b l ein
the electron microscope (Photograph suppiled by Dr. B S Hudson and
the late Dr. J. Vinograd.)
Figure 9-5 Electron micrograph ( ~ 1 5 0 , 0 0 0 )of a thin layer of copper
hexadecachlorophthalocyan~nemolecules, The molecules are tilted about
25" from the horizontal plane. (Courtesy JEOL, Ltd.)
8. 264 9 Separation and ~urification.ldentification of Organic Compounds by Spectroscopic Techniques
molecules in crystals is of particular value, and in the past ten years this tech-
nique has become almost routine. Figure 9-6 shows the detailed arrangement
of the carbons, hydrogens, and bromines in 1,8-bis(bromomethyl)naphtkalene,
1, as determined by x-ray diffraction. The apparatus and techniques used are
Figure 9-6 Bond lengths, angles, and arrangement of carbons and bro-
mines in a crystal of 1,8-bis(bromomethyl)naphthalene,1, as determined
by x-ray diffraction. Notice that the preferred conformation in the crystal
has the bromines on opposite sides of the naphthalene ring.
9. 9-3 Some General Considerations of Diffraction and Spectroscopic Techniques 265
highly complex and are not available yet to very many organic lab~ratories.~
Other diffraction methods include electron diffraction, which may be
used to-determine the structures of gases or of volatile liquid substances that
cannot be obtained as crystals suitable for x-ray diffraction, and neutron diffrac-
tion, which has special application for crystals in which the exact location of
hydrogens is desired. Hydrogen does not have sufficient scattering power for
x rays to be located precisely by x-ray diffraction.
The diffraction methods can be used to determine complete structures
of organic molecules, but they are not sufficiently routine to be utilized gen-
erally in practical organic laboratory work. For this reason, in the remainder
of this chapter we will emphasize those forms of spectroscopy that are gener-
ally available for routine laboratory use. As will be seen, these methods are
used by organic chemists in more or less empirical ways. In general, spectro-
scopic methods depend on some form of excitation of molecules by absorption
of electromagnetic radiation and, as we have said, virtually all parts of the
electromagnetic spectrum have utility in this regard. The commonly used span
of the electromagnetic spectrum is shown in Figure 9-7 along with a compar-
ison of the various units that are employed to express energy or wavelength.
The major kinds of spectroscopy used for structural analysis of organic
compounds are listed in Table 9-1. The range of frequencies of the absorbed
radiation are shown, as well as the effect produced by the radiation and spe-
cific kind of information that is utilized in structural analysis. After a brief
account of the principles of spectroscopy, we will describe the methods that
are of greatest utility to practical laboratory work. Nonetheless, it is very
important to be aware of the other, less routine, methods that can be used to
solve special problems, and some of these are discussed in this and in Chap-
ters 19 and 27.
You may have problems with the relationships among the variety of
wavelength and frequency units commonly used in spectroscopy. The rela-
tionship between wavelength, frequency, and velocity should become clear to
you by considering yourself standing on a pier watching ocean waves going by.
Assuming the waves are uniformly spaceq, there will be a uniform distance
between the crests, which is A, the wavelength. The wave crests will pass by
at a certain number per minute, which is v, the frequency. The velocity, c,
3A useful description of how molecular structures can be determined by "x-ray vision"
is given in Chapter XI of Organic Molecules in Action by M. Goodman and F. More-
house, Gordon and Breach, New York, 1973.
10. 266 9 Separat~onand Purificat~onldentif~cationof Organic Compounds by Spectroscopic Techniques
Wavelength Energy Energy in Energy Energy
in nanometers in cm-' kcal mole-' in eV in Hz
x rays
+--visible light
t1 microwave
radio -TV, FM radio
1 -AM r a d ~ o
Figure 9-7 The span of the spectrum of electromagnet~cradiation used
in spectroscopic ~nvesiigationsof organic compounds along with com-
parlson of some of the various units commonly employed for wavelength
and energy of the rad~at~on on log scales
at which the crests move by you is related to A and v by the relationship
c = Av.
This is not really very complicated and it applies equally well to water
waves or electromagnetic radiation. What is almost needlessly complicated
is the variety of units commonly used to express A and v for electromagnetic
radiation. One problem is tradition, the other is the desire to avoid very large
or very small numbers. Thus, as Figure 9-7 shows, we may be interested in
electromagnetic wavelengths that differ by as much as a factor of 1016.Because
the velocity of electromagnetic radiation in a vacuum is constant at 3 x lo8
meters sec-l, the frequencies will differ by the same factor.
Units commonly used for wavelength are meters (m), centimeters (cm),
nanometers (nm), and microns (p). In the past, angstroms (A) and millimi-
crons (mp) also were used rather widely.
Frequency units are in cycles per second (cps) or hertz (Hz), which
are equivalent (radians per second are used widely by physicists).
11. Table 9-1
Principal Spectroscopic Techniques Currently in Use for Analysis of Molecular Structure
Energy range of $
Spectroscop~c absorbed rad~at~on Type of exc~tationproduced %
technique ( ~ wave
n numbers, cm-')" by absorbed rad~at~on lnformat~onobta~ned ol
Ion cyclotron lo-6 to lo-5 Excitation of ions moving in circular Rates and equilibria for reactions of ions with I)
Nuclear magnetic to lo-z
orbits in a magnetic field
Changes in nuclear spin orientations
neutral molecules in the gas phase (Section 27-8)
Chemical shifts and coupling constants; rapid
resonance (nmr)
Electron spin to 1
in a magnetic field
Excitation of unpaired electron-spin
reaction rates (Sections 9-10, 27-1, and 27-2)
Electron distribution in radicals, electron-transfer
resonance (esr) orientations in a magnetic field reactions (Section 27-9) w
Microwave 1 to 100 Rotat~onalexcitation Spacings of rotational energy levels; bond
distances and bond angles (Section 9-6)
Infrared (ir) 100 to 10,000 Rotational-v~brat~onal excltat~on Rotational and vibrational energy levels of 3
molecules (Sectlon 9-7) 2
Rotational-vibrational excitation Rotational and vibrational energy levels of V)
molecules (Section 9-8) u
Visible 5,000 to 25,000 Electronic excitation accompanied Electronic energy levels of molecules (Section
by vibration-rotation changes 9-9)
Ultraviolet 25,000 to 50,000 Electronic excitation accompanied Electronic energy levels of molecules (Sections 2.
by vibration-rotation changes 9-9 and 28-1) -I-I
Photoelectron l o 5 to l o 6 Ejection of an electron from the Ionization energies of valence or inner-shell o
valence or inner shell electrons of molecules (Section 27-5) 2.
Mossbauer l o 7 to l o 9 Excitation of atomic nuclei Electric-field grad~entsat the nucleus produced c
by differences in bond types (Sectlon 27-6)
Mass spectrometry Exc~tat~on produced by Molecular lonlzat~onand Molecular weights; modes of fragmentation
electrons wlth energies fragmentation (Sections 9-1 1 and 27-7)
of about l o 5 cm-I
"These ganges are not meant to be preclse, but to glve you a general Idea of the energy changes ~nvolvedOne wave number (cm-'), 1s equivalent to 2 86
cal mole-' Also see Flgure 9-7 for comparison wlth other commonly used unlts of energy and wavelength
12. 268 9 Separation and Purlflcat~onldentlflcatlon of Organlc Compounds by Spectroscopic Techniques
Frequencies in the electromagnetic spectrum can be seen from Figure
9-7 generally to be large. As a result, it is common to use wave numbers in-
stead of Hz or MHz (megahertz). The frequency in wave number is simply
the frequency v in H z divided by c, the velocity of light in cm. Wave-number
units are cm-I and we can think of the wave number 6 as being the number of
wave crests per centimeter.
1 Hz = MHz = 3.3 x lo-" cm-I
lo6 Hz = 1 MHz = 3.3 X 10-%m-l
3 x 1 0 1 0 H z = 3 x 1 0 4 M H z = 1 cm-I
Exercise 9-1 Suppose you are stand~ngon the end of a p ~ ewatchlng
r the waves and,
between your posltron and a buoy 200 m stra~ghtout, you count 15 wave crests
Further, suppose a wave crest comes by every 15 seconds Calculate v In Hz, h In m,
c In m sec-', and 17 In km-'
Exercise 9-2 Blue light has 17 = 20,800 cm-l. Calculate v in Hz and h in nm
The energies of the hydrogenlike orbitals of various atoms were mentioned in
Chapter 6 and, in particular, we showed a diagram of the most stable state
(1~)~(2s)~(2p)Zof a carbon atom (Figure 6-4). Transfer of one of the 2p elec-
trons to the 3s orbital requires excitation of the atom to a higher energy state
and this can be achieved by absorption of electromagnetic radiation of the
proper wavelength. The usual way that such excitation occurs is by absorption
of a single quantum of radiant energy, and we can say that the absorption of
this amount of energy AE,,, corresponds to excitation of the atom from the
ground state with energy E, to an excited shte of configuration ( 1 ~ ) ~ ( 2 s ) ~
(2p)l(3s)l and energy E,:
E, ( 1~ ) ~ ( 2 ~ ) ~ ( 2 p ) ~ ( 3 s ) l
13. 9-4 Atom~cEnergy States and L ~ n eSpectra
The difference in energy, is related directly to the frequency (v, sec-l)
or wavelength (A, nm)4 of the absorbed quantum of radiation by the equation
AE,, = hv = -
in which h is Planck's constant and c is the velocity of light. The relationship
4 E = hv often is called the Bohr frequency condition.
For chemical reactions, we usually express energy changes in kcal
mole-'.-For absorption of one quantum of radiation by each atom (or each
molecule) in one mole, the energy change is related to A by
AE --
287600kcal mole-l
As defined, AE,, corresponds to one einstein of radiation.
What we have developed here is the idea of a spectroscopic change
being related to a change in energy associated with the absorption of a quan-
tum of energy. Spectra are the result of searches for such absorptions over a
range of wavelengths (or frequencies). If one determines and plots the degree
of absorption by a monoatomic gas such as sodium vapor as a function of
wavelength, a series of very sharp absorption bands or lines are observed,
hence the name line spectra. The lines are sharp because they correspond to
specific changes in electronic configuration without complication from other
possible energy changes.
Exercise 9-3 Calculate the energy in kcal mole-' that corresponds to the absorption
of 1 einstein of light of 589.3 nm (sodium D line) by sodium vapor. Explain how this
absorption of light by sodium vapor may have chemical utility.
Exercise 9-4 a. Use Equations 9-1 and 9-2 to calculate the wavelength in nm and
energy in kcal of an einstein of radiation of radio-frequency energy in the broadcast
band having v = I MHz (1 megahertz) = l o 6 sec-' and knowing that the velocity of
light is approximately 3 x lo8 meters sec-I.
b. In photoelectron spectroscopy, x rays with energies of approximately 1250 elec-
tron volts are used (1 electron volt per mole = 23.05 kcal). What would A (in nm) be
for such x rays?
4See Section 9-3 for discussion of the units of frequency and wavelength.
14. 270 9 Separation-and Purification: Identification of Organic Compounds by Spectroscopic Techniques
The energy states and spectra of molecules are much more complex than those
of isolated atoms. In addition to the energies associated with molecular elec-
tronic states, there is kinetic energy associated with vibrational and rotational
motions. The total energy, E, of a molecule (apart from its translational5 and
nuclear energy) can be expressed as the sum of three terms:
Absorption of electromagnetic radiation by molecules occurs not only by elec-
tronic excitation of the type described for atoms, but also by changes in the
vibrational and rotational energies.
Both rotations and vibrations of molecules are quantized. This means
that only particular values of rotational angular momentum or vibrational
energy are possible. We speak of these permitted values of the energies as
the vibrational and rotational energy levels.
Rotational energy levels normally are very closely spaced so low-energy
radiation, such as is produced by radio transmitters operating in the micro-
wave region, suffices to change molecular rotational energies. Because elec-
tronic and vibrational energy levels are spaced much more widely, and because
changes between them, are induced only by higher-energy radiation, micro-
wave absorptions by gaseous substances can be characterized as essentially
pure "rotational spectra." It is possible to obtain rotational moments of inertia
from microwave spectra, and from these moments to obtain bond angles and
bond distances for simple molecules.
An example of the use of microwave spectra is provided by Figure 9-8,
which shows separate rotational absorptions observed for trans and gauche
conformations of propyl iodide (cf. Section 5-21.
Although microwave spectroscopy, being confined to gases, is not a
routine method in the organic laboratory, it is important to us here in setting
the stage for the consideration of more complex absorptions that occur with
infrared radiation.
jTranslationa1 energy is not very important in connection with spectroscopy and will
not be considered here.
15. 9-6 M~crowaveSpectra Rotational Spectra
gauche -
28 30 32 34 36 38 40
1 1 1 1 11 1 1 1 1 1 1 1 1
- frequency (GHz)-
Figure 9-8 A small part of the microwave spectrum of CH,CH,CH,I at a
pressure of 7 x atm showing absorptions of the trans and gauche
conformations. Notice the regular spacings of the lines for each conforma-
tion. That the spacings are different for the two conformations reflects
their different moments of inertia. The horizontal scale is GHz (giga-
hertz, lo9 Hz) and u = 30 GHz corresponds to h = lo7 nm, which from
Equation 9-1 can be calculated to mean a rotational energy change of
0.0029 kcal mole-'. (Spectrum courtesy of Dr. Howard Harrington, Hewlett-
Packard Corp.)
Exercise 9-5 The microwave spectrum of pure trans-2-butenoic acid (CH3CH=
CHC0,H) shows patterns exactly like those of Figure 9-8, which ~ndicatethe presence
of two different conformations. What are these conformations, and why are there only
two of them? (You may be helped by reviewing Section 6-5.)
At the turn of the nineteenth century 'Sir William Herschel discovered in-
visible radiation beyond the red end of the visible region of the electromagnetic
spectrum. This radiation appropriately is called infrared, meaning "beneath
the red," and it encompasses the wavelength region from lo3 nm to lo6 nm.
You probably are familiar with the common applications of infrared to radiant
heating and photography. In addition to these uses, infrared spectroscopy
has become the most widely used spectroscopic technique for investigating
organic structures.
16. 272 9 Separation and Purification Identification of Organ~cCompounds by Spectroscopic Techniques
Infrared spectroscopy was the province of physicists and physical
chemists until about 1940. At that time, the potential of infrared spectroscopy
as an analytical tool began to be recognized by organic chemists. The change
was due largely to the production of small, quite rugged infrared spectropho-
tometers and instruments of this kind now are virtually indispensable for
chemical analysis. A brief description of the principles and practice of this
spectroscopic method is the topic of this section.
9-7A General Considerations
Absorption of infrared radiation causes transitions between vibrational energy
states of a molecule. A simple diatomic molecule, such as H-CI, has only one
vibrational mode available to it, a stretching vibration somewhat like balls on
the ends of a spring:
stretching vibration
Molecules with three or more atoms can vibrate by stretching and also by
bending of the chemical bonds, as indicated below for carbon dioxide:
stretching vibrations bending vibration
The absorption fretpencies in the infrared spectra of molecules correspond to
changes in the stretching or bending vibrations or both. In general, a polyatomic
molecule with n atoms will have 3n - 6 modes of vibration of which n - 1 are
stretching vibrations and 2n - 5 are bending vibrations. There are circum-
stances, however, where fewer vibrational modes are possible. If the molecule
is linear, like CO,, then there are 3n - 5 possible vibrations, and some of these
vibrations may be equivalent (degenerate vibrations in the language of spectros-
copists). For example, CO, should have 3n - 5 or 4 vibrational modes, two of
which are stretching and two of which are bending modes. However, the two
bending modes are equivalent because the direction in which the molecule
bends is immaterial; in-plane or out-of-plane bending are the same:
in-plane bending out-of-plane bending
Diatomic molecules such as HC1 have one vibrational mode, but it is
important to note that symmetrical diatomic molecules, such as 0 2 , N2, C12,
F,, and H,, do not absorb in the infvared region of the spectrum. This is
17. 9-7A lnfrared Spectroscopy General Cons~derat~ons
because absorption cannot occur if the vibration is electrically symmetrical.
Fortunately, then, the infrared spectra can be recorded in air because the main
components of air, N, and O,, do not interfere.
In practice, infrared spectra can be obtained with gaseous, liquid, or
solid samples. The sample containers (cells) and the optical parts of the instru-
ment are made of rock salt (NaCl) or similar material that transmits infrared
radiation (glass is opaque).
Typical infrared spectra are shown in Figure 9-9 for 2-propanone
(acetone), CH,-CO-CH,, and 2-butanone (methyl ethyl ketone), CH,-
CO-@Hz-CH,. In accord with current practice, the position of absorption
frequency, cm
3600 2800 2000
frequency, cm-'
Figure 9-9 lnfrared absorpt~onspectra of (a) 2-propanone and (b) 2-
butanone In the vapor phase
18. 274 9 ~ e ~ a r a t i oand
; purification: Identification of Organic Compounds by Spectroscopic Techniques
(horizontal scale) is recorded in units of wave numbers (C, cm-l; see Section
9-3). The vertical scale measures the intensity of radiation transmitted through
the sample. Zero transmission means complete absorption of radiation by
the sample as at 1740 cm-I in Figure 9-9. The other absorption bands in
Figure 9-9 that correspond to excitation of stretching or bending vibrations
are not as intense as the absorption at 1740 cm-l.
9-78 Characteristic Stretching Vibrations
What information can we derive about molecular structure from the vibra-
tional bands of infrared spectra? Absorption of radiation in the range of
5000-1250 cm-I is characteristic of the types of bonds present in the molecule,
and corresponds for the most part to stretching vibrations. For example, we
know that the C-H bonds of alkanes and alkyl groups have characteristic
absorption bands around 2900 cm-l; an unidentified compound that shows
absorption in this region will very likely have alkane-type C-H bonds.
More explicitly, the band observed for 2-propanone (Figure 9-9a) at
3050 cm-I arises from absorption of infrared radiation, which causes transitions
between the ground vibrational state (or lowest vibrational energy level) of
a C-H bond and the first excited vibrational energy level for stretching of
that C-H bond. The band at 1740 cm-I corresponds to the infrared absorp-
tion that causes transitions between vibrational energy levels of the C=O
bond. The reason that these are transitions from the vibrational ground state
is because, at room temperature, by far the largest portion of the molecules
are in this state (cf. Exercise 9-9).
Stretching frequencies characteristic of the most important types of
bonds found in organic molecules are given in Table 9-2. You will notice that
the absorption band for each bond type is described by its position within a
more or less broad freqhency range and by its shape (broad, sharp) and inten-
sity (strong, medium, weak).
A qualitative discussion of the factors that determine infrared band position
and band intensities follows. To a first approximation, a chemical bond re-
sembles a mechanical spring that vibrates with a stretching frequency v (cm-"),
in which k is the force constant, and rn, and m, are the masses of the individual
atoms at each end of the bond. The force constant k is a measure of the stiffness
of the bond and usually is related to the bond strength. From Equation 9-3, we
can see that the heavier the bonded atoms, the smaller will be the vibrational
frequency of the bond provided k remains essentially ~ o n s t a n t Thus
.~ if we
increase m, while holding k and m, constant we expect the frequency to de-
crease. This is just what occurs when we change the C-H bond to a C-D
'jRemernber that lower frequency means longer wavelengths and lower energy.
19. 9-7B Characteristic Stretching Vibrations
bond. We also see that the frequency decreases in the order C-H > C-C >
C-N > C-0, which also is in the order of increasing m,, but here matters
are more complicated because k also changes.
Other things being equal, it requires more energy to stretch a bond than to
bend it. Therefore the infrared bands arising from changes in the stretching
vibrations are found at higher frequencies than are those arising from changes
in the bending vibrations.
Another consequence of Equation 9-3 is that if m, and m, remain the same,
the larger the value of k, the higher will be the vibrational frequency. Because
k is expected to run more or less parallel to the bond strength, and because
*multiple bonds are stronger than single bonds, the absorption frequencies of
multiple bonds are higher than for single bonds. Examples are the absorption
of C=C at 2100 cm-l, C=C at 1650 cm-l, and C-C at 1000 cm-l.
Other effects besides mass and bond strength also affect infrared absorption
frequencies. The structural environment of a bond is particularly important.
Thus the absorption frequency of a C-H bond depends on whether it is an
alkyl, alkenyl, alkynyl, or aryl C-H bond (see Table 9-2).
The intensity of an infrared absorption band arising from changes in the
vibrational energy is related to the electrical symmetry of the bond. More
symmetrical, less polarized bonds give weaker absorptions. In fact, if the bond
is completely symmetrical, there is no infrared absorption. In contrast, unsym-
metrical molecules in which the bonds are quite polarized, such as C = O bonds,
show strong infrared absorptions.
Notice in Figure 9-9 that infrared spectra of organic molecules do not
show very sharp absorption lines. This is because changes in rotational ener-
gies can occur together with the vibrational changes. The reason can be seen
more clearly in Figure 9-10, in which each vibrational level, such as El and
E,, of a molecule has associated with it closely spaced rotational levels. Tran-
sitions between El and E, also may involve changes in rotational levels. This
gives a "band" of closely spaced lines for any given vibrational change. For
E2 -- +
vibrational trans~tions
energy levels
Figure 9-10 Schematic vibrational and rotational energy levels. The
arrows correspond to infrared vibrational-rotational transitions of different
20. 276 9 Separation and Purification. Identification of Organic Compounds by Spectroscopic Techniques
Table 9-2
Some Characteristic infrared Absorption Frequencies
Bond Type of compound cm-' Intensity
I 2800-3 100 strong
-C-H alkanes
I - 2200 strong
-C-D alkanes
I 3000-31 00 medium
=C-H al kenes and arenes
zC-H alkynes 3200-3350 strong, sharp
750- 1200" weak to medium
-C-C- alkanes
\ /
alkenes 1600-1 680 variable
-C=C-- alkynes 2050-2260 variable
-C=N n~tr~les 2200-2400 variable
I I I I 980- 1250 strong
-C-0- alcohols -C-OH, ethers -C-0-C-
carboxyl~cacids -C
/P 1350-1 440 weak to medium
\ 1210-1320 strong
1035-1 300 strong (two bands
for unsaturated
\ I1
aldehydes -C-H 1690-1740 strong
1650-1730 strong
21. 9-78 Characteristic Stretching Vibrations 277
Table 9-2 (continued)
Some Characteristic Infrared Absorpt~onFrequencies
Bond Type of compound cm-' Intensity
-0-H alcohols -C-0-H, phenols =C-0-H 3400-3700 variable, sharp
-0-Hb hydrogen-bonded alcohols and 3200-3400 strong, broad
phenols -0-H...O
-0-H alcohols, phenols, acids (bending 1000-1 450 strong
-0-Hb hydrogen-bonded carboxyllc acids 2500-3300 variable, broad
-0-H...O /
-NH2 amines -C-NH, 3200-3600 medium
I (double peak)
3100-3500 medium
(single peak)
"In general, C-C slngle-bond stretch~ngfrequencies are not very useful for ldentlf~catlon
*These bands may not appear at low concentration In solvents where ~ntermolecularhydrogen bondlng
does not occur
complex molecules, particularly in the liquid state, the "rotational fine struc-
ture" of a given vibrational band usually cannot be resolved.
Absorption of infrared radiation over the range from 600 cm-I to 3600
cm-' corresponds to energy-level differences, as in Figure 9-10, of 1.7 kcal
mole-l to 10.3 kcal mole-l.
Exercise 9-6 Use Equation 9-3 and any other pertinent data to predict which com-
pound in each group would absorb in the infrared at the highest frequency for the
changes in the stretching vibration of the specified bond. Give your reasoning.
a. R-CI, R-Br, R-F (carbon-halogen)
b. CH,-NH,, CH,=NH, HC=N (carbon-nitrogen)
22. 278 9 Separation and Purification. Identification of Organ~cCompounds by Spectroscopic Techniques
Exercise 9-7 Whlch compound In each group would have the most Intense Infrared
absorption band corresponding to stretching vibrations of the bonds ~nd~cated?
Give your reasoning
a. (CH,),C=O, (CH,),C=CH, (multiple bond)
b. CH3-CH,, CH3-0-CH, (C-C VS C-0)
c. CH,C=CH, CH,C=CCH, (multiple bond)
d. H-CI, CI-CI
Exercise 9-8* How many vibrational modes are possible for (a) CS, (linear), (b)
BeCI, (linear), and (c) SO, (angular)? Show your reasoning.
Exercise 9-9* Suppose an infrared absorption occurs at 3000 cm-'. Calculate the
corresponding frequency v in sec-'; h in nm, angstroms, and microns, and energy
change in kcal mole-'. Using Equation 4-2 (p. 84) and neglecting AS, calculate
the fraction of the molecules that would be in the ground state and in the first vibra- -
tional excited state (above the ground state by 3000 cm-') at 298°K.
9-7C The Fingerprint Region
Infrared absorption bands between 1250 cm-l and 675 cm-I generally are
associated with complex vibrational and rotational energy changes of the
molecule as a whole and are quite characteristic of particular molecules. This
part of the spectrum is often called the "fingerprint" region and is extremely
useful for determining whether samples are chemically identical. The spectra
of Zpropanone and 2-butanone are seen to be very similar in the region 4000
cm-I to 1250 cm-I but quite different from 1250 cm-' to 675 cm-l. The finger-
print region of the spectrum is individual enough so that if the infrared spectra
of two samples are indistinguishable in the range of frequencies from 3600 cm-I
to 675 cm-I, it is highly probable that the two samples are of the same com-
pound (or the same mixture of compounds).
Characteristic stretching and bending frequencies occur in the finger-
print region, but they are less useful for identifying functional groups, because
they frequently overlap with other bands. This region is sufficiently complex
that a complete analysis of the spectrum is seldom possible.
9-70 Alkanes and Cycloal kanes
The infrared spectra of the alkanes show clearly absorptions corresponding
to the C-H stretching frequencies at 2850 cm-l to 3000 cm-" The C-C
stretching absorptions have variable frequencies and are usually weak. Methyl
(CH,-) and methylene (-CH,-) groups normally have characteristic C-H
bending vibrations at 1400 cm-l to 1470 cm-I. Methyl groups also show a
23. 9-7DAlkanes and Cycloalkanes
weaker band near 1380 cm-l. Two sample infrared spectra that illustrate these
features are given in Figure 9- 11.
The infrared spectra of the cycloalkanes are similar to those of the
alkanes, except that when there are no alkyl substituents the characteristic
frequency, crn-l
frequency, crn-'
Figure 9-11 Infrared spectra of (a) octane and (b) 2,2,4-trirnethylpentane
as pure liquids. Notice the C-H stretching around 2900 crn-l and C-H
bending frequency around 1460 crn-l. The bands near 1370 cm-I for 2,2,4-
trirnethylpentane are characteristic of methyl C-H bending frequencies.
24. 280 9 Separation and Purification, Identification of Organic Compounds by Spectroscopic Techniques
bending frequencies of methyl groups at 1,380 cm-l are absent. A moderately
strong CH, "scissoring" frequency is observed between 1440 cm-I and 1470
cm-l, the position depending somewhat on the size of the ring. These features
of the infrared spectra of cycloalkanes are illustrated in Figure 9-12 using
cyclooctane and methylcyclohexane as examples.
frequency, cm-'
2000 1800 1600 1400 1200 1000 800
frequency, c m - '
Figure 9-12 Infrared spectra of (a) cyclooctane and (b) methylcyclo-
hexane. These spectra can be compared profitably with those in Fig-
ure 9-1 1
25. 9-7E Appl~cationsof lnfrared Spectroscopy to Structure Determination
9-7E Applications of lnfrared Spectroscopy to
Structure Determination
Infrared spectra are very useful both for identification of specific organic
compounds, and for determining types of compounds. For example, Figure
9-13 shows the infrared spectrum of a substance, C4H60,, for which we wish
to determine the compound type and, if possible, the specific structure. The
most informative infrared absorptions for determining the compound type ,
are between 1500 cm-l and 3600 cm-I. Two groups of bands in this region
can be-seen at about 1700 cm-l(s) and 3000 cm-'(s), where (s) means strong;
if we used (m) it would mean medium, and (w) would mean weak. From Table
9-2 we can see that these bands are indicative of C=O (1700 cm-l) and hydro-
gen-bonded O H of carboxylic acids (3000 cm-I). The presumption is that
there is a -CO,H group in the molecule, and we can derive some reassurance
from the fact that the molecular formula C4H60,has enough oxygens to allow
for this possibility.
Table 9-2 also shows that a --C0,H group should have a C-0 ab-
sorption band between 1350 cm-' and 1400 cm-' and 0 - H absorption (bend-
ing frequency) between 1000 cm-' and 1410 cm-', and there is indeed a band
of medium intensity at 1350 cm-' and a strong band at 1240 cm-l. These
absorptions, being in the fingerprint region, do not prove that the compound is a
carboxylic acid; but if there were no absorptions in the 1000 cm-l to 1400
cm-I range, the presence of a -CO,H group would be highly questionable.
frequency, cm '
Figure 9-13 lnfrared spectrum of a compound, C,H,O,
26. 282 9 Separation and Purification, Identification of Organic Compounds by Spectroscopic Techniques
Tentatively, then, we may write a partial structure for C,H,O, as
A propyl group would be C,H,, and C,H, has two hydrogens less, which
indicates the presence of a double bond or a ring. However, Table 9-2 shows
that a double bond should have an absorption of variable intensity at 1600
cm-l to 1680 cm-' and there is no clear sign of such an absorption in Figure
9- 13. The alternative to a double bond would be a ring, which for C,H, has to be
a cyclopropyl ring. The structure that is most compatible with the spectrum is
Final identification may be possible by comparison with an authentic spectrum
of cyclopropanecarboxylic acid, if it is available in one of the several standard
compendia of infrared spectra. A total of about 150,000 infrared spectra are
available for comparison purposes. You should check with the reference sec-
tion of your library to see what atlases of spectral data are available to you.
The foregoing example illustrates the way structures can be determined
from infrared spectral data. For many purposes, the infrared frequencies given
in Table 9-2 are both approximate and incomplete. However, you could be
easily frustrated in interpreting spectral data by being burdened with a very
detailed table in which the unimportant is mixed with the important. The ability
to use extensive tables effectively comes with experience. You should re-
member that tabulated infrared frequencies indicate only the range in which a
given vibrational transition will fall. The exact value for a particular compound
usually is meaningless because it will change depending on whether the spec-
trum is taken of the solid, liquid, or gaseous states, the solvent used, the
concentration, and the temperature. To become familiar with infrared spectra,
we strongly recommend that you work Exercises 9-10 and 9-1 1.
Exercise 9-10 Use Table 9-2 to map the approximate positions and intensities
expected for the characteristic infrared bands corresponding to the stretching vibra-
tions of the various kinds of bonds in the following molecules:
a. l,l,l-trideuteriopropanone H-C-C-C-D b. propyne H-C-C=C-H
(trideuterioacetone) I I
27. 9-7E Appl~cattonsof lnfrared Spectroscopy to Structure Determ~nat~on
H 0 H H H
I II I I \ /H
c. ethyl ethanoate
(ethyl acetate)
d. propenenitrile
(acrylonitrile) F=SC=N
e. 2-oxopropanoic acid H-C-C-C-0-H f. ethanol H-C-C-0-H
(pyruvic acid) I (ethyl alcohol) I I *
(both as pure liquid and as
a dilute solution in CCI,)
Exercise 9-11 The infrared spectra shown in Figure 9-14 are for compounds of
formula C,H,O and C,H,O,. Use the data in Table 9-2 and the molecular formulas to
deduce a structure for each of these substances from its infrared spectrum. Indicate
clearly which lines in the spectra you identify with the groups in your structures.
frequency, cm-'
frequency, cm-'
Figure 9-14 lnfrared spectra for Exercise 9-1 1 . Spectrum (a) corresponds
to C,H,O and Spectrum (b) to C,H,O,.
28. 284 9 Separation and Purification,Identification of Organic Compounds by Spectroscopic Techniques
Raman spectroscopy often is a highly useful adjunct to infrared spectroscopy.
The experimental arrangement for Raman spectra is quite simple in principle.
Monochromatic light, such as from an argon-gas laser, is passed through a
sample, and the light scattered at right angles to the incident beam is analyzed
by an optical spectrometer.
Raman spectra arise as a result o f light photons being "captured" momen-
tarily by molecules in the sample and giving up (or gaining) small increments o f
energy through changes in the molecular vibrational and rotational energies
before being emitted as scattered light. The changes in the vibrational and rota-
tional energies result in changes in wavelength o f the incident light. These
changes are detected as lines falling both above and below the wavelength o f
the incident light. The line positions in Raman spectra always are reported in
wave numbers. Highly efficient laser Raman spectrometers are commercially
Although changes in wavelength in Raman scattering correspond to absorp-
tion or emission o f infrared radiation, infrared and Raman spectra are not always
identical. Indeed, valuable information about molecular symmetry may be
obtained by comparison o f infrared and Raman spectra. When a bond is
electvically symmetvictnl it does not absorb infrared radiation and, for this
reason, symmetrical diatomic molecules such as H, and O,, which are always
electrically symmetrical, do not give infrared absorption spectra. However,
excitation o f symmetrical vibrations does occur in Raman ~cattering.~ In a
molecule such as ethene, CH,=CH,, the double-bond stretching vibration is
symmetrical, because both ends of the molecule are the same. As a result, the
double-bond stretching absorption is not observable in the infrared spectrum
of ethene and is weak in all nearly symmetrically substituted ethenes. None-
theless, this vibration appears strongly in the Raman spectrum o f ethene and
provides evidence for a symmetrical structure for ethene.
As a general conclusion, a molecule has no important symmetry i f all its
infrared bands have counterparts in Raman scattering. To illustrate these
effects,the Raman and infrared spectra o f tetrachloroethene and cyclohexene
are shown in Figures 9-1 5 and 9-1 6. Absorption due to the stretching vibration
o f the double bond in tetrachloroethene ( 1 570 cm-I) is strong in the Raman and
absent in the infrared, whereas that arising from the less symmetrical double
bond of cyclohexene (1658 cm-I) is weak in the infrared and slightly stronger
in the Raman.
tetrachloroethene cyclohexene
7This is in accord with the spectroscopic "selection rules," derived from theoretical
arguments, that predict which transitions between rotational and vibrational energy
levels are "allowed" and which are "forbidden."
29. 9-8 Raman Spectroscopy
frequency, cm '
frequency, cm-'
Figure 9-15 Infrared (top) and Raman spectra (bottom) of tetrachloro-
ethene (notice that the spacings and alignment of the horizontal scales
are not the same). The Raman spectrum was supplied courtesy of the
A ~ p l l e dPhys~csCorporation.
30. 286 9 Separation and Purlflcatlon ldentlficatlon of Organlc Compounds by Spectroscopic Techniques
frequency, cm-'
frequency, cm-'
Figure 9-16 Infrared (top) and Raman spectra (bottom) of cyclohexene (notlce
that the spacings and alignment of the horizontal scales are not the same) The
Raman spectrum was supplied courtesy of the Applled Phys~csCorporation
Exercise 9-12* Classify the following molecules according to the general charac-
teristics expected for their infrared and Raman spectra: (a) HCECH; (b) ICI; (c) CO;
(d) CF,=CH, (double-bond stretch only); (e) (CH,),C=CH, and CH,CH=CHCH,
(double-bond stretch only),
Exercise 9-13* Carbon dioxide gives two infrared absorption bands but only one
Raman line. This Raman line corresponds to a different vibration than the infrared
absorptions. Decide which vibrational modes are infrared active (i.e., make the mole-
cule electrically unsymmetrical during at least part of the vibration) and which is
Raman active (i.e., occurs so the molecule is electrically symmetrical at all times
during the vibration, see Section 9-7A).
31. 9-9A Electron~cSpectra of Organ~cMolecules General Character~st~cs 287
9-9A General Characteristics
A year after Herschel discovered infrared radiation, Johann Ritter discovered
radiation beyond the violet end of the visible spectrum. This radiation came to
be known as ultraviolet and soon was recognized as being especially effective
in causing chemical reactions.
Absorption of light in the ultraviolet and visible regions produces
changes in the electronic energies of molecules associated with excitation of
an electron from a stable to an unstable orbital. Because the energy required
to excite the valence-shell electrons of molecules is comparable to the strengths
of chemical bonds, absorption may lead to chemical reactions. We discussed
this briefly in Chapter 4 in connection with photochemical halogenation of
alkanes; a more detailed account of photochemistry is given in Chapter 28.
The transition of an electron from the ground state, E,, to an excited
electronic state, E,, is accompanied by vibrational and rotational changes in
the molecule, as shown in Figure 9-17. It usually is not possible to resolve the
resulting absorption bands well enough to see the fine structure due to vibra-
tion-rotation transitions. Consequently, absorptions due to electronic excita-
tion are relatively broad.
The ultraviolet spectrum of 2-propanone (acetone) is shown in Figure
9-18. The weak absorption, which peaks (i.e., has ),A at 280 nm, is the result
energy levels
electronic transitions
energy levels
/ f
Figure 9-17 Schematic representation of electronic, vibrational, and
rotational energy levels, The vertical scale is greatly distorted; rotational
energy levels are normally 10-4-10-2 kcal mole-' apart, vibrational energy
levels are 1-10 kcal mole-' apart, while electronic transitions involve
10-1000 kcal mole-'.
32. 9 Separation and Purification. Identification of Organic Compounds by Spectroscopic Techniques
Figure 9-18 The ultrav~olet spectrum of 2-propanone (acetone) In
level. This is called an n -+ IT* (often N -
of excitation of one of the unshared electrons on oxygen to a higher energy
A ) transition, in which n denotes
that the excited electron is one of the unshared n electrons on oxygen and 71."
(pi star) denotes that the excited electron goes to a high-energy antibonding
orbital of the carbon-oxygen double bond (cf. Sections 6-2 and 6-4C). The
same kind of n m-* transition occurs at about the same wavelength and in-
tensity for many simple compounds of the type R,C=O and RCH=O, in
which R is an alkyl group. In a very schematic way, we can write
There also is an absorption of 2-propanone with A,, at 190 nm (the
maximum is not shown in Figure 9-18), which is a different kind of excitation.
This is ascribed to raising an electron in the IT-bondingorbital (Section 6-4C)
of the carbon-oxygen double bond to the T* orbital. Such transitions are
called 71. ---j. IT*,and occur generally for substances with double bonds:
Table 9-3 lists the wavelengths of maximum absorption for some typical
electronic absorption bands of simple molecules. If we remember that ab-
sorptions at longer wavelengths correspond to less energetic transition, it
can be deduced from the A,, values that less energy is required to excite
unshared (nonbonding) electrons than IT electrons in double or triple bonds,
which in turn require less energy than a electrons in single bonds (Figure 9-19).
33. 9-9A Electron~cSpectra of Organ~cMolecules General Character~st~cs
Table 9-3
Some E l e c t r o n ~ cT r a n s ~ t ~ o nofs S ~ m p l eO r g a n ~ cMolecules
Compound TYpe Amax, nm cmaXa Solventb
15 cyclohexane '
CH2=CH2 n- TI*
4 10,000 vapor
CH2=CH-CH=CH, n* 20,900 hexane
CH3-CH=CH-CH=CH-CH, TI- rr* 22,500 hexane
CH2=CH-CH2-CH2-CH=CH2 .ir TI* 20,000 alcohol
CH3-CrCH 450 cyclohexane
7~- TI*
n TI* 24 alcohol
TI- TI* 3,600
strong vapor
strong vapor
weak vapor
200 vapor
365 pentane
150 vapor
2,520 vapor
900 vapor
aThe molar extinction coefficient c is a measure of the absorption efficiency at the wavelength
A,,,. Because, the amount of absorption is proportional to the concentration (c moles liter-')
and thickness of the sample ( I cm), t is obtained from the equation
1 I
t =- log,, 4 or ccl= A
cl I
in which loll is the ratio of intensity of incident light I,to transmitted light I.
The percent trans-
mission of a solution is (lll,) X 100 and the absorb3nce A = log l,/l. Substances for which c 1s
independent of concentration are said to obey Beer's law (or the Beer-Lambert law).
is necessary to specify the solvent because A,, and t,,, vary somewhat with solvent.
dTransitions o -
cThese assignments are not certain.
o * correspond to excitation of a o electron of a single bond to a higher-
"Transitions n -
energy antibonding orbital of the single bond, o* (sigma star).
o * correspond to excitation of an electron of an unshared pair to an anti-
bonding orbital (o*) of a u bond.
34. 290 9 Separation and Purification, Identification of Organic Compounds by Spectroscopic Techniques
L1-- CJ (bonding)
Figure 9-19 Sequence of electronic orbital energies, showing different
kinds of transitions in approximate order of increasing energy, left to right.
The a n*' and + d transitions usually have low transition
probabilities, meaning the bands have low or zero intensities.
Exercise 9-14 L ~ s tthe k~ndsof electronic transltlons that would be expected for
azaethene (methylenelmlne), CH,=NH, In order of lncreaslng energy Use the data
In Table 9-3 to predrct approximately the wavelengths at wh~chthe three lowest-
energy trans~t~onsshould occur
Exercise 9-15 Calculate the percentage of the lncldent l ~ g hthat
t would be absorbed
by an 0 010M solut~onof 2-propanone (acetone) In cyclohexane conta~nedIn a quartz
cell 0 I cm long at 280 nm and at 190 nm (see footnote a of Table 9-3)
Exercise 9-16 Explain why the absorption band at 227.3 nm for trimethylamine,
(CH,),N, disappears in acid solution.
The 7i -
9-9B Effects of Conjugation on Electronic Spectra
T* transition for ethene has , A = 175 nm and E,,, = 10,000. It
would be expected that an alkadiene would give an absorption spectrum similar
to that of ethene but with a larger E , because there are more double bonds per
mole to absorb radiation. This expectation is more or less realized for com-
pounds such as 1,5-hexadiene and 1,3-dimethylenecyclobutane,which have
isolated double bonds, but not for 1,3-butadiene or ethenylbenzene, which
35. 9-9B Effects of Conjugation on Electron~cSpectra
have conjugated double bonds (Section 3-3):
A, = 185 nm,
O C H - C H ,
1,3-butad~ene ethenylbenzene (styrene)
,A = 21 7 nm, t = 21,000 A,, = 244 nm, t = 12,000
In general, conjugated systems of double bonds absorb radiation of
longer wavelengths and with greater intensity than corresponding systems of
isolated double bonds. This means that the difference in energy between the
normal and excited states of conjugated systems is less than for isolated sys-
tems of double bonds. For 1,3-butadiene and 1,5-hexadiene we can calculate
from Equation 9-2 [(28,600)(217 - 185)1(217 X 185)] that the transition
energy is about 23 kcal less for the conjugated system. The ground state of
1,3-butadiene is stabilized by perhaps 3 kcal relative to a nonconjugated sys-
tem of double bonds, which means that the excited state must be much more
stabilized than this if the transition energy is to be 23 kcal less than for 1,5-
Why is the excited state of a conjugated system of double bonds stabi-
lized more, relative to the ground state, than for a nonconjugated system?
Resonance theory provides an explanation (see Section 6-5). Of the several
conventional valence-bond structures that can be written for 1,3-butadiene,
four of which are shown here, 2a-2d, only structure 2a has a low enough
energy to be dominant for the ground state of 1,3-butadiene:
Now, when the molecule is excited to the extent of 132 kcal mole-l by 217 nm
ultraviolet light, its energy is so large that pairing schemes such as 2b, 2c, and
2d, which are too unfavorable to contribute very much to the ground state,
can be very important for the excited state. Thus the stabilization energy of
the excited state, which has a multiplicity of nearly equal-energy pairing
schemes, is expected to be greater than that of the ground state with one
dominant pairing scheme.
36. 292 9 Separation and Purification. Identification of Organic Compounds by Spectroscopic Techniques
The more double bonds in the conjugated system, the smaller the energy
difference between the normal and excited states. The diphenylpolyenes of
formula C,H,+CH=CH+C,H, absorb radiation at progressively longer
wavelengths as n is increased. This is apparent from the colors of the com-
pounds, which range from colorless with n = 1, to orange with n = 2-7, to red
with n = 8, as A,, goes from the ultraviolet into the visible region of the
electromagnetic spectrum.
Similar effects are found with conjugated C=O and C=N double
bonds. For example, the electronic spectra of 2-butanone and 3-buten-2-one
are shown in Figure 9-20.
2-butanone 3-buten-2-one
' (methyl ethyl ketone) (methyl v~nylketone)
The absorption at 277 nm for 2-butanone is an n --+ T* transition, and with
rr -
3-buten-2-one, this absorption shifts to longer wavelengths (324 nm). There is
also an intense absorption band for 3-buten-2-one at 219 nm, which is a
rr* transition. With 2-butanone a corresponding absorption occurs at
185 nm, which is out of the range of the spectrometer used to take the spectra
of Figure 9-20.
Conjugation also can influence infrared spectra. Transitions arising
from C=C and C=O stretching vibrations genei-ally are more intense and
are shifted to slightly lower frequencies (longer wavelengths) for conjugated
compounds relative to nonconjugated compounds. Thus the C=C stretching
of 1-butene occurs at 1650 cm-', whereas that of 1,3-butadiene is observed
at 1597 cm-I.
Alkanes and cycloalkanes have no low-energy electronic transitions
comparable to conjugated systems or molecules with nonbonding electrons.
Therefore alkanes and cycloalkanes show no absorption above 200 nm and
are good solvents to use for electronic spectroscopy.
wavelength, h
Figure 9-20 Electron~cspectra of (a) 2-butanone and (b) 3-buten-2-one
in cyclohexane solution
37. 9-9C Applications of Electron~cSpectroscopy 293
9-9C Applications of Electronic Spectroscopy
How do we use electronic spectroscopy in chemical analysis? The two
principal applications are structure determinations and quantitative analysis.
The position and intensity of an electronic absorption band provides
information as to chemical structure. Such absorptions normally are not as
useful as infrared absorptions because they do not give as detailed information.
For our purposes here, the main points to remember are:
1. A weak absorption ( E = 10-100) suggests an n - T* transition of
an isolated carbonyl group. If this absorption is found in the region 270-350
nm an aldehyde or ketone is probable.
and 260 nm may correspond to n -
2. Somewhat stronger absorptions (E = 100-4000) between 200 nm
cf* transitions.
3. Strong absorptions ( e = 10,000-20,000) usually are characteristic
of T -+ T* transitions. If absorption occurs above 200 nm, a conjugated
system of multiple bonds is indicated. Each additional carbon-carbon double
bond shifts A,, about 30 nm to longer wavelengths and enhances the intensity
of absorption. Conjugation also shifts A,, of n T* transitions to longer
If we are dealing with compounds for which the wavelengths and the
molar intensities of the absorption bands are known, then we can use the
degree of absorption for quantitative analysis with the aid of the Beer-Lambert
law (see Table 9-3 for definitions):
By measuring the absorbance A of a sample of known E in a cell of known path
length 1, the concentration c may be determined. Because changes in absorb-
ance reflect changes in concentration, it is possible to use absorbance mea-
surements to follow rates of chemical reactions, to determine equilibrium
constants (such as the dissociation constants of acids and bases), and to follow
conformational changes in bio-organic molecules such as proteins and nucleic
Exercise 9-17 A compound of formula C,H,O has two absorpt~onbands I r i the ultra-
v~oleth = 320 nm, t = 30 and h = 218 nm, t = 18,000 In ethanol solut~onDraw three
poss~blestructures that are cons~stentw ~ t hthis ~nformat~on
Exercise 9-18 2,4-Pentanedione exists in equilibrium with 4-hydroxy-3-penten-2-one:
38. 294 9 Separation and Purification, Identification of Organic Compounds by Spectroscopic Techniques
The Infrared spectrum of the l ~ q u l dmlxture shows a broad absorptlon band at 3000-
2700 cm-' and an Intense absorptlon band at 1613 cm-' In cyclohexane solutlon, the
substance has, ,A, at 272 nm w ~ t hem,, = 12,000 (a) What can you conclude from this
data as to the magnitude of K, the equ~lrbr~urn constant for the rnterconverslon of the
two forms7 (b) What can you deduce from the fact that the absorptlon at 272 nm 1s
much weaker In aqueous solutlon (pH 7) than ~t1s In cyclohexane?
Exercise 9-19* The electron~cabsorptlon spectrum of 2-n~trobenzenolhas,,A, In
0 1M HCI at 350 nm In 0 1M NaOH, the benzenol 1s largely converted to ~ t s
anlon, and
,,A, shlfts to 41 5 nm
NaOH + QoH d QO' +
~ a @ H20
The ground-state resonance forms of 2-nitrobenzenol and its anion include
2-n~trobenzenol 0
Explaln how the relatlve Importance of these resonance forms to the ground and
exclted states of 2-nltrobenzenol and ~ t sanlon can account for the fact that the anlon
absorbs at longer wavelengths than does 2-nltrobenzenol (Revlew Sectlon 6-5B)
Exercise 9-20* A solutlon contalnlng the two forms of the Important coenzyme
nlcotlnamlde adenlne dlnuclsotlde (abbreviated NAD" and NADH, see Sectlon
15-6C for structures) has an absorbance In a I-cm cell of 0 31 1 at 340 nm and 1 2
at 260 nm Both NADO and NADH absorb at 260 nm, but only NADH absorbs at 340
nrn The molar extlnctlon coefflclents are
Compound 260 nm 340 nrn
NAD@ 18,000 - 0
NADH 15,000 6220
Calculate the proportions of NAD@and NADH in the mixture.
39. 9-10 Nuclear Magnet~cResonance Spectroscopy 295
Nuclear magnetic resonance (nmr) spectroscopy is extremely useful for identi-
fication and analysis of organic compounds. The principle on which this form
of spectroscopy is based is simple. The nuclei of many kinds of atoms act like
tiny magnets and tend t o become aligned in a magnetic field. In nmr spectros-
copy, we measure the energy required to change the alignment of magnetic
nuclei in a magnetic field. To illustrate the procedure with a simple example, '
consider the behavior of a proton (lH) in a magnetic field. There are two
possible alignments of this magnetic nucleus with respect to the direction of
the applied field, as shown in Figure 9-2 1. The nuclear magnets can be aligned
either with the field direction, or opposed to it. The two orientations are not
equivalent, and energy is required to change the more stable alignment to the
less stable alignment.
Figure 9-21 Schematic representation of the possible alignments of a
magnetic nucleus (here hydrogen) in an applied magnetic field. Transi-
tions between the two states constitute the phenomenon of nuclear
magnetic resonance. The arrows through the nuclei represent the average
component of their nuclear magnetic moment in the field direction.
A schematic diagram of an nmr instrument is shown in Figure 9-22.
When a substance such as ethanol, CH,-CH,-OH, the hydrogens of which
have nuclei (protons) that are magnetic, is placed in the transmitter coil and the
magnetic field is increased gradually, at certain field strengths radio-frequency
energy is absorbed by the sample and the ammeter indicates an increase in the
flow of current in the coil. The overall result is a spectrum such as the one
shown in Figure 9-23. This spectrum is detailed enough to serve as a useful
"fingerprint" for ethanol, and also is simple enough that we will be able to
account for the origin of each line. It is the purpose of this section to explain
how the complexities of spectra such as that of Figure 9-23 can be interpreted
in terms of chemical structure.
For what kinds of substances can we expect nuclear magnetic resonance
absorption to occur? Magnetic properties always are found with nuclei of
odd-numbered masses, 'H, I3C, 15N, 170,19F,,IP, and so on, as well as for
40. 296 9 Separation and Pur~flcat~on
ldentlflcatlon of Organlc Compounds by Spectroscopic Techn~ques
nuclei of even mass but odd atomic number, %IH,1°B, "N, and so Nuclei
such as 12C, 1 6 0 , and 32S,which have even mass and atomic numbers, have
no magnetic properties and do not give nuclear magnetic resonance signals.
For various reasons, routine use of nmr spectra in organic chemistry is con-
magnet~cfleld ammeter
of strength H,
Figure 9-22 Essent~alfeatures of a slmple nmr spectrometer
Figure 9-23 Proton nmr spectrum of ethanol (containing a trace of hydro-
chloric acid). Chemical shifts are relative to tetramethylsilane (CH,),Si,
that is, TMS = 0.00 ppm. The stepped line is an integral of the areas under
each of the resonance lines.
8Although the principal isotopes of C1, Br, and 1 have magnetic properties, because of
the special character of all of these isotopes, they act in organic compounds as though
they were nonmagnetic.
41. 9-10A The Relation of NMR to Other K ~ n d sof Spectroscopy
fined to lH, 19F, 13C, and slP. We shall be concerned in this chapter only
with nmr spectra of hydrogen (lH) and of carbon (13C).
The kind of nmr spectroscopy we shall discuss here is limited in its
applications because it can be carried out only with liquids or solutions.
Fortunately, the allowable range of solvents is large, from hydrocarbons to
concentrated sulfuric acid, and for most compounds it is possible to find a
suitable solvent.
Nuclear magnetic resonance spectra may be so simple as to have only a '
single absorption peak, but they also can be much more complex than the
spectrum of Figure 9-23. However, it is important to recognize that no matter
how complex an nmr spectrum appears to be, it involves just three parameters:
chemical shifts, spin-spin splittings, and kinetic (reaction-rate) processes. We
shall have more to say about each of these later. First, let us try to establish
the relationship of nmr spectroscopy to some of the other forms of spectroscopy
we already have discussed in this chapter.
9-10A The Relation of NMR to Other Kinds of Spectroscopy
Nuclear magnetic resonancey spectroscopy involves transitions between the
possible energy levels of magnetic nuclei in an applied magnetic field (see
Figure 9-21). The transition energies are related to the frequency of the
absorbed radiation by the familiar equation A E = hv. An important difference
between nmr and other forms of spectroscopy is that AE is influenced by the
strength of the applied field. This should not be surprising, because if we are
to measure the energy of changing the direction of alignment of a magnetic
nucleus in a magnetic field, then the stronger the field the more energy will be
Nuclear spin (symbolized as I ) is a quantized property that correlates
with nuclear magnetism such that when I is zero the nucleus has no spin and
no magnetic properties. Examples are 12Cand lW. Several nuclei of particular
interest to organic chemists- lH, I T C 15N,
, I9F, and 31P- have spin of l/2. With
I = l/2 there are only two magnetic energy states of the nucleus in a magnetic
field. These states are designated with the spin quantum numbers +I12 and -l/2.
The difference in energy between these states, A E , is given by
in which h is Planck's constant, v is in hertz, y is a nuclear magnetic constant
called the gyromagnetic ratio,1° and H i'3 the magnetic field strength at the
"esonance in the sense used here means that the radio-frequency absorption takes
place at specified "resonance" frequencies. However, you will see that almost all of
the forms of spectroscopy we discuss in this book involve "resonance" absorption in
the same sense.
1°Here, y is in H z per gauss; physicists usually define y in radians per sec per gauss.
42. 298 9 Separation and Purification. Identification of Organic Compounds by Spectroscopic Techniques
nucleus. In general, H will not be exactly equal to H,, the applied magnetic
field and, as we will see, this difference leads to important chemical information.
Each kind of nucleus ('H, I T C ,15N,etc.) has its own y value and, consequently,
will undergo transitions at different frequencies at any particular value of H.
This should become clearer by study of Figure 9-24.
There are several modes of operation of an nmr spectrometer. First and
most common, we hold v constant and vary (or "sweep") H,,. Close to v = yH,
energy is absorbed by the nuclei and the current flow from the transmitter
increases until v is exactly equal to yH. Further increase of Ho makes u < yH,
and the current flow decreases. The form of the energy-absorption curve as a
function of H , when H,, is changed very slowly is shown in Figure 9-25a. The
peak is centered on the point where u = yH. When H , is changed more rapidly,
transient effects are observed on the peak, which are a consequence of the fact
that the nuclei do not revert instantly from the -l/z to +I12 state. The resulting
MHz 2
I -
- - -1 i
v , MHz
10000 I , 20000 30 000 H , gauss
-40 -
14 100 gauss
-60 -
Figure 9-24 Field-frequency diagram that represents the energies (in
frequency units) of the+l/2 and -112 magnetic states of l H and 13Cnuclei
as a function of magnetic field. The vertical scale is of frequency u in
MHz (1 megahertz = 106 Hz = l o 6 cycles per sec) while the horizontal
scale is of magnetic field in gauss. (For comparison, the Earth's magnetic
field is about 0.2 gauss.) The dashed vertical line at 14,100 gauss tells
us that the 'H resonance frequency will be 60.0 MHz and the 73Creso-
nance frequency will be 15.0 MHz at this field strength.