Most of us are quite familiar with infrared radiation. We have
seen infrared lamps keep food hot and often associate infrared
radiation with heat. While the generation of heat is a probable
event following the absorption of infrared radiation, it is important
to distinguish between the two. Infrared is a form of radiation
that can travel through a vacuum while heat is associated with
the motion and kinetic energy of molecules. The concept of heat
in a vacuum has no meaning because of the lack of molecules and
molecular motion. Infrared spectroscopy is the study of how molecules
absorb infrared radiation and ultimately convert it to heat. By
examining how this occurs, we will not only learn about how infrared
radiation is absorbed, but we will also learn about molecular
structure and how the study of infrared spectroscopy can provide
information about the structure of organic molecules. An infrared
spectrum of a chemical substance, is very much like a photograph
of a molecule. However, unlike a normal photograph which would
reveal the position of nuclei, the infrared spectrum will only
reveal a partial structure. It is the purpose of this narrative
to provide you with the tools necessary to interpret infrared
spectra, successfully. In some respects, this process is similar
to reading an X-ray of the chest. While most of us could easily
identify the gross structural features of the chest such as the
ribs, most of us would need some guidance in identifying those
features on the X-ray film associated with disease.
In order to interpret infrared spectra, having some idea or model
of the physical process involved when a molecule interacts with
infrared radiation would be useful. You may recall in introductory
chemistry, the discussion of how atoms interact with electromagnetic
radiation led to the development of quantum theory and the introduction
of quantum numbers. The interaction of infrared radiation with
molecules requires a similar treatment. While the use of quantum
theory is necessary to explain this interaction, most of us live
in a world that appears continuous to us and we do not have much
experience discussing phenomena that occur in discrete steps.
The discussion that follows will attempt to develop a model of
how molecules interact with infrared radiation that is based as
much as possible on classical physics. When necessary, we will
insert the modifications required by quantum mechanics. This model,
while perhaps oversimplified, will contain the physical picture
that is useful to understand the phenomena and will be correct
from a quantum mechanical standpoint.
Let's begin first by considering two isolated atoms, a hydrogen and a bromine atom moving toward each other from a great distance. What do you suppose will happen once the atoms approach each other and can feel each others presence? The potential energy curve for the H-Br molecule is shown in Figure 1. As the two atoms approach each other notice that the potential energy drops. If we recall that energy must be conserved, what must happen to the kinetic energy? The two atoms must attract each other and accelerate toward each other, thereby increasing their kinetic energy. The change in kinetic energy is illustrated by the dotted line in the figure. At some point they will "collide" as indicated by the part of the potential energy curve that rises steeply at small interatomic distances and then the atoms will begin to move away from each other. At this point, we might ask, "Will the molecule of HBr survive the collision"? Unless some energy from this system is lost, say by emission of a photon of light or collision by a third body to remove some energy, these are two ships passing in the night. The kinetic energy resulting from the coulombic attraction of the two atoms will exactly equal the drop in potential energy and the two atoms will
Figure 1. The potential (solid line) and kinetic energy
(dotted line) of HBr as a function of the separation of the two
nuclei. The kinetic energy at every point illustrated by the dotted
line is equal to the potential energy plus the small amount of
kinetic energy associated with initial motion of the two nuclei
when separated at large distances.
fly apart. The spontaneous emission of a photon of light is improbable, so this mechanism is unlikely to drop the HBr molecule into the well. Most probable from a physical perspective, is the collision of our HBr with a third body which will remove some energy and result in the trapping of the HBr molecule in the well. Though very excited, this molecule will now survive until other collisions with less energetic molecules leads to an HBr molecule at the bottom of the well and the generation of heat (kinetic energy) that would be experienced in the exothermic reaction of hydrogen and bromine atoms to form hydrogen bromide. Let us now consider a hydrogen bromide molecule that has lost a little kinetic energy by collision and has been trapped in the potential energy well of Figure 1. We might ask, "How would a molecule that does not have enough kinetic energy to escape the well behave in this well? A molecule with some kinetic energy below this threshold value (total energy slightly less than 0 in Fig. 1) will be able to move within this well. The internuclear separation will vary within the limits governed by the available kinetic energy. Since this motion involves a stretching or compression of the internuclear distance it is usually described as a vibration. Additional collisions with other molecules will eventually lead to the dissipation of the energy associated with formation of the hydrogen bromide bond. At this point we might ask the following question. If we remove all the excess kinetic energy from HBr, what will be its kinetic and potential energy? Alternatively we might ask, "Will the hydrogen bromide molecule reside at the very bottom of the well when it is cooled down to absolute zero Kelvin?" Before we answer this question, let's digress for a little and discuss the relative motions of the hydrogen and bromine atoms in terms of the physics of everyday objects. Once we learn how to describe the classical behavior of two objects trapped in a potential energy well, we will return to the question we have just posed.
One model we can use to describe our hydrogen bromide molecule
is to consider our HBr molecule to be made up of balls of uneven
mass connected to each other by means of a spring. Physicists
found many years ago some interesting properties of such a system
which they referred to as a harmonic oscillator. Such a system
repeatedly interconverts potential and kinetic energy, depending
on whether the spring is exerting a force on the balls or the
momentum of the balls is causing the spring to be stretched or
compressed. The potential energy of this system (PE) is given
by the parabola,
PE = k(x-xo)2 1
where x-xo is the displacement of the balls from their equilibrium
condition when the system is at rest and k is a measure of the
stiffness of the spring. While this simple equation does not apply
to molecules, please notice how similar the potential energy surface
of the parabola (Figure 3) is to the bottom of the surface of
Figure 1. The constant k is used to describe chemical bonds and
is referred to as the force constant. As you might imagine,
it is a measure of the stiffness of the chemical bond.
Several other relationships were observed that do carry over in
describing molecular systems. For example, they found that when
a ball was suspended on a spring from a horizontal wall, the frequency
of vibration or oscillation, n, depended
only on the mass of the ball and the stiffness of the spring.
The term A is a constant of the proportionality. By varying the
mass of the ball and the stiffness of the spring, they were able
to uncover the following simple relationship between frequency,
mass and force constant:
Suspending a ball and spring from a horizontal surface is a special case of the more general situation when you have two more comparable masses attached to each other. Under these circumstances, when two similar masses are attached to a spring, the relationship between frequency of vibration, mass and force constant is given by:
where m, represents the product of
the masses divided by their sum (m1m2)/(m1+m2). This latter term
is found in other physical relationships and has been given the
name, the reduced mass. It can easily be seen that
equation 2 is a special case of the more general relationship
given by equation 3. If we consider m1to be much larger than m2,
the sum of m1+ m2 m1
and substituting this approximation into (m1m2)/(m1+m2)
m2. Substituting m2 into
equation 3 where m2 is the smaller of the two masses gives us
exactly the same relationship as we had above when the ball was
suspended from a horizontal wall. The horizontal wall is much
more massive than the ball so that the vibration of a smaller
ball has very little effect on the wall. Despite their simplicity,
equations 2 and 3 play an important role in explaining the behavior
of molecular systems. However, before we discuss the important
role these equations play in our understanding of infrared spectroscopy,
we need to review some of the properties of electromagnetic radiation,
particularly radiation in the infrared range.
The electromagnetic spectrum is summarized in Figure 2. On the
extreme right we find radiowaves and scan from right to left we
encounter terms which have become familiar to us; microwave, infrared,
visible ultraviolet and X-rays. All of these forms of electromagnetic
1010 108 106
2x105 1x105 4000
650 12 5 x10-2 10-3 10-6
10-6 10-4 10-2
5x10-2 10-1 2.5
15.4 830 4x105 107 1010
Figure 2. The electromagnetic spectrum.
are related to each other in a simple and obvious way. First let us discuss why we refer to these different forms of light as electromagnetic radiation. Simply stated, all these forms of radiation have an electric and magnetic field associated with them that varies as shown for the standing wave in Figure 3. Only the electric field is shown in this figure. If we were to include the magnetic field it would look exactly as the electric field but would be rotated 90 ° out of the plane of the paper and would oscillate above and below the plane of the paper like a sin or cos wave. In
infrared spectroscopy, only the electric field associated with
the electromagnetic radiation is important and we will limit our
present discussion to how this field varies with time. We called
the light wave associated with Figure 3 a standing wave because
this is how the electric field would
Figure 3. The electric field of light associated with a
standing wave with a fixed wavelength.
vary if we took a picture of the wave. One of the properties of all electromagnetic radiation is that it travels in a vacuum at the speed of 3 x 1010 cm/sec. Therefore, if we were to turn this standing wave "on" we would observe this oscillating field rapidly passing us by. If we examine the electric field (or the magnetic field which is not shown), we observe that the field is repetitive, varying as a cos or sin wave. The length of the repeat unit along the x axis is called the wavelength, l, and it is this property which varies continuously from 106 cm (1010 microns) for radio waves down to 10-13 cm (10-6 microns) for cosmic radiation. A unit of length that is frequently used in infrared spectroscopy is the micron. A micron is equivalent to 10-4 cm. If we were to "stand on the corner and watch all the wavelengths go by", since all electromagnetic radiation would be traveling at 3 x 1010 cm/sec, the frequency, n, at which the shorter wavelengths would have to pass by would have to increase in order to keep up with the longer wavelengths. This relationship can be described in the following mathematical equation:
ln = c; (c = 3 x 1010 cm/sec). 4
The frequency of the light times the wavelength of the light must equal the speed at which the light is traveling.
In addition to having wave properties such as the ones we have
been discussing, electromagnetic radiation also has properties
we would normally attribute to particles. These "particle
like" properties are often referred to as characteristics
of photons. We can discuss the wave properties of photons
by referring to the wavelength (eqn. 4) and frequency associated
with a photon. The energy of a single photon is a measure of a
property we would normally associate with a particle. The relationship
which determines the energy associated with a single photon of
light, E, and the total energy incident on a surface by monochromatic
light, ET, is given by:
E = h n (or equivalently, E = h c/
l, from equation 4),
ET = n h n 6
where h is Planck's constant and is numerically equal to 6.6 x
10-27 erg s and n is the number of photons. Equations 4 and 5
tell us that photons with short wavelengths, in addition to having
higher frequencies associated with them, also carry more punch!
The energy associated with a photon of light is directly proportional
to its frequency.
At this point we are ready to return to a discussion of how infrared
radiation interacts with molecules. Following our discussion of
balls and springs, you have probably figured that infrared spectroscopy
deals with the vibration of molecules. Actually, both rotation
and vibration of molecules is involved in the absorption of infrared
radiation, but since molecular rotation is not usually resolved
in most infrared spectra of large organic molecules, we will ignore
this additional consideration. In order to derive the relationship
between vibrational energy and molecular structure, it is necessary
to solve the Schoedinger equation for vibrational-rotational interactions.
Since solution of this equation is beyond the scope of this treatment,
we will simply use the relationship that is derived for a harmonic
oscillator from this equation. As you see, the quantum mechanical
solution of a harmonic oscillator, equation 7, is remarkably simple
and very similar to the relationship we obtained from considering
the classical model of balls and springs.
Before discussing the implications of equation 7, let's take a moment to see how similar it is to equations 3 and 5. From equation 5, we see that substituting equation 3 for n results in equation 7 except for the (n + 1/2) term. However we should point out that we have substituted the vibrational frequency of two masses on a spring for a frequency associated with the number of wave maxima (or minima, null points. etc.) passing a given point (or street corner) per unit time. We are able to do this because of the presence of the (n +1/2) term. Let's discuss the significance of the (n + 1/2) term before we returning to answer this question. The previous time you encountered the Schroedinger equation was probably when studying atomic spectra in Introductory Chemistry. An important consequence of this encounter was the introduction of quantum numbers, at that time the principle quantum number, N, the azimuthal quantum number, l, the magnetic, ml, and spin quantum number, s. This time is no exception. Meet n, the vibrational quantum number. These numbers arise in a very similar manner. The Schroedinger equation is a differential equation which vanishes unless certain terms in it have very discrete values. For n, the allowed values are 0,1,2,... Let us now consider the energy of vibration associated with a molecule in its lowest energy of vibration, n = 0. According to equation 7, the energy of vibration is given by, when n = 0, the zero point energy. This equation allows us to answer the question posed earlier about what would happen to the vibrational energy of a molecule at absolute zero. According to quantum theory the molecule would continue to vibrate. From the relationship E = hn, we can evaluate the vibrational frequency as , the same as found by classical physics for balls and springs. This equation states that the vibrational frequency of a given bond in a molecule depends only on the stiffness of the chemical
Figure 4. The potential energy surface for a HBr molecule
illustrating how the vibrational energy levels vary in energy
with increasing vibrational quantum number.
bond and the masses that are attached to that bond. Similarly,
according to equation 7, once the structure of a molecule is defined,
the force constants and reduced mass are also defined by the structure.
This also defines the vibrational frequencies and energy of absorption.
Stated in a slightly different manner, a molecule will not absorb
vibrational energy in a continuous fashion but will do so only
in discrete steps as determined by the parameters in equation
7 and illustrated for the HBr molecule in Figure 4. We have pointed
out that the vibrational quantum number can have positive integer
values including a value of zero. Upon absorption of vibration
energy, this vibrational quantum number can change by +1 unit.
At room temperature, most molecules are in the n = 0 state.
Figure 4 illustrates the real vibrational levels for HBr. Notice
that equation 7 predicts that the energy level spacings should
all be equal. Notice according to Figure 4, the spacings actually
converge to a continuum for large values of n. For small values
of n, n = 0, 1, 2, equation 7 gives a good approximation of the
vibrational energy levels for HBr. Equation 7 was derived from
the approximation that the potential energy surface is like a
parabola. Near the minimum of this surface, around the zero point
energy, this is a good approximation. As you go up from the minimum,
the resemblance decreases and the assumptions made in solving
the Schroedinger equation no longer are valid.
Let us now return and question the wisdom of substituting the vibrational frequency of a molecule for the frequency of electromagnetic radiation in equation 5. I hope at this point of the discussion this does not seem so absurd. If the vibrational frequency of the molecule, as determined by the force constant and reduced mass, equals the frequency of the electromagnetic radiation, then this substitution makes good sense. In fact, this gives us a mechanism by which we can envision why a molecule will absorb only distinct frequencies of electromagnetic radiation. It is known that symmetrical diatomic molecules like nitrogen, oxygen and hydrogen, do not absorb infrared radiation, even though their vibrational frequencies are in the infrared region. These homonuclear diatomic molecules have no permanent dipole moment and lack a mechanism by which they can interact with the electric field of the light. Molecules like HBr and HCl which have a permanent dipole, resulting from an unequal sharing of the bonding electrons, have a dipole which oscillates as the bond distance between the atoms oscillate. As the frequency of the electric field of the infrared radiation approaches the frequency of the oscillating bond dipole and the two oscillate at the same frequency and phase, the chemical bond can absorb the infrared photon and increase its vibrational quantum number by +1. This is illustrated in Figure 5. Of course, some HBr molecules may not be correctly oriented toward the light to interact and these molecules will not absorb light. Other factors will also influence the intensity and shape of the absorption. However, when the frequency of the electromagnetic radiation equals the vibrational frequency of a molecule, absorption of light does occur and this leads to an infrared spectrum that is characteristic of the structure of a molecule.
Up to now we have discussed molecules changing their vibrational quantum number by +1. A change of -1 is also equally possible under the influence of infrared radiation. This would lead to emission of infrared radiation. The reason why we have not discussed this possibility is that most molecules at room temperature are in the ground vibrational level (n=0) and cannot go any lower. If we could get a lot of molecules, let's say with n = 1, use of infrared could be used to stimulate emission. This is how an infrared laser works.
Figure 5. An HBr molecule interacting with electromagnetic
radiation. In order for this interaction to occur successfully,
the frequency of the light must equal the natural vibrational
frequency of the HBr and the electric field must be properly orientated.
We have previously discussed the infrared region of the electromagnetic
spectrum in terms of the wavelength of the light that is involved,
2.5-15 m ((4000-650 cm-1)
(Figure 3). According to equation 4, we can also express this
region of the electromagnetic spectrum in terms of the frequency
of the light. There is an advantage to discussing the absorption
of infrared radiation in frequency units. According to equation
5, energy is directly proportional to frequency. The energy associated
with an absorption occurring at twice the frequency of another
can be said to require twice the energy. Occasionally, weak bands
occur at twice the frequency of more intense bands. These are
called overtones and they result when the vibrational quantum
number changes by +2. While these transitions are weak and are
theoretically forbidden (i.e. they occur with an intensity of
less than 5 % of the same transition that involves a change of
+1 in the vibrational quantum number) they are easy to identify
when units of frequency are used. Sometimes absorption bands involving
a combination of frequencies occur. There is no physical significance
to adding together wavelengths - there is a physical significance
to the addition of frequencies since they are directly proportional
to energy. To convert wavelength to frequency according to equation
4, we need to multiply the speed of light by the reciprocal of
wavelength. Since the speed of light is a universal constant,
the curious convention of simply using the reciprocal of wavelength
has evolved. Thus a peak at 5 m would
be expressed as 1/(5x10-4 cm) or 2000 cm-1.
You will note that 2000 cm-1 is not a true
frequency. A true frequency would have units of cycles/sec. To
convert 2000 cm-1 to a true frequency one
would need to multiply by the speed of light (cm/sec). However,
2000 cm-1 is proportional to frequency and
this is how frequency units in infrared spectroscopy are expressed.
What would be the frequency of light with a wavelength of 10 m?
Analysis of IR Spectra
At this point we are ready to leave diatomic molecules and start
talking about complex organic molecules. Before doing so, it should
be pointed out that the discussion that follows is an oversimplification
of the true vibrational behavior of molecules. Many vibrational
motions of molecules are motions that involve the entire molecule.
Analysis of such motions can be very difficult if you are dealing
with substances of unknown structure. Fortunately, the infrared
spectrum can be divided into two regions, one called the functional
group region and the other the fingerprint region. The functional
group region is generally considered to range from 4000 to approximately
1500 cm-1 and all frequencies below 1500 cm-1
are considered characteristic of the fingerprint region. The fingerprint
region involves molecular vibrations, usually bending motions,
that are characteristic of the entire molecule or large fragments
of the molecule. Hence the origin of the term. Used together,
both regions are very useful for confirming the identity of a
chemical substance. This is generally accomplished by a comparison
of the spectrum of an authentic sample. As you become more proficient
in analyzing infrared spectra, you may begin to assign bands in
this region. However, if you are just beginning to interpret spectra
of organic molecules, it is best to focus on identifying the characteristic
features in the functional group region. The functional group
region tends to include motions, generally stretching vibrations,
that are more localized and characteristic of the typical functional
groups found in organic molecules. While these bands are not very
useful in confirming identity, they do provide some very useful
information about the nature of the components that make up the
molecule. Perhaps most importantly, the frequency of these bands
are reliable and their presence or absence can be used confidently
by both the novice and expert interpreter of infrared spectra.
The discussion which follows focuses primarily on the functional
group region of the spectrum. Some functional groups are discussed
in more detail than others. You will find that all this information
is summarized in Table 1 which should prove useful to you when
you try to interpret an unknown spectrum. Finally, you should
bear in mind that although we have developed a model that can
help us understand the fundamental processes taking place in infrared
spectroscopy, interpretation of spectra is to a large extent an
empirical science. Information about the nature of a compound
can be extracted not only from the frequencies that are present
but also by peak shape and intensity. It is very difficult to
convey this information in Table form. Only by examining real
spectra will you develop the expertise to accurately interpret
the information contained within. Be sure to examine the spectra
contained in this handout carefully. Whenever you interpret a
spectrum and extract structural information, check your assignments
by examining the spectrum of a known substance that has similar
Table 1. A summary of the principle infrared bands and their assignments.
R is an aliphatic group.
|C-H||sp3 hybridized||R3C-H||2850-3000||M(sh)||6, 18, 22|
|sp2 hybridized||=CR-H||3000-3250||M(sh)||7, 13, 42|
|aldehyde C-H||H-(C=O)R||2750, 2850||M(sh)||14, 15|
|N-H||primary amine, amide||RN-H2, RCON-H2||3300, 3340||S,S(br)||18, 19|
|secondary amine, amide||RNR-H, RCON-HR||3300-3500||S(br)||20, 21|
|tertiary amine, amide||RN(R3), RCONR2||none||22, 23|
|O-H||alcohols, phenols||free O-H||3620-3580||W(sh)||17, 24, 25|
|hydrogen bonded||3600-3650||S(br)||24, 25, 28|
|carboxylic acids||R(C=O)O-H||3500-2400||S(br)||26, 27, 29, 30|
|anhydrides||R(CO2CO)R||1820, 1750||S, S(sh)||36|
|C=C||olefins||R2C=CR2||1680-1640||W(sh)||10, 39, 40|
|-NO2||nitro groups||RNO2||1550, 1370||S,S(sh)||28|
Factors Influencing the Location and Number of Peaks
Before beginning a detailed analysis of the various peaks observed
in the functional group region, it might be useful to mentioned
some of the factors that can influence the location and number
of peaks we observe in infrared spectroscopy. Theoretically, the
number of fundamental vibrations or normal modes available
to a polyatomic molecule made up of N atoms is given by 3N-5 for
a totally linear molecule and 3N-6 for all others. By a normal
mode or fundamental vibration, we mean the simple independent
bending or stretching motions of two or more atoms, which when
combined with all of normal modes associated with the remainder
of the molecule will reproduce the complex vibrational dynamics
associated with the real molecules. Normal modes are determined
by a normal coordinate analysis (which will not be discussed in
this presentation). If each of these fundamental vibrations were
to be observed, we would expect either 3N-5 or 3N-6 infrared bands.
There are some factors which decrease the number of bands observed
and others that cause an increase in this number. Let's discuss
the latter first.
We have already mentioned overtones, which are absorption
of energy caused by a change of 2 rather than 1 in the vibrational
quantum number. While overtones are usually forbidden transitions
and therefore are weakly absorbing, they do give rise to more
bands than expected. Overtones are easily identified by the presence
of a strongly absorbing fundamental transition at slightly more
than half the frequency of the overtone. On occasion, combination
bands are also observed in the infrared. These bands, as their
name implies, are absorption bands observed at frequencies such
as 1 + 2 or 1 - 2,
where 1 and 2 refer to fundamental frequencies.
Other combinations of frequencies are possible. The symmetry properties
of the fundamentals play a role in determining which combinations
are observed. Fortunately, combination bands are seldom observed
in the functional group region of most polyatomic molecules and
the presence of these bands seldom cause a problem in identification.
Another cause of splitting of bands in infrared is due to a phenomena
called Fermi Resonance. While a discussion of Fermi Resonance
is beyond the scope of this presentation, this splitting can be
observed whenever two fundamental motions or a fundamental and
combination band have nearly the same energy (i.e. 1
and 22 or 1 and 2 + 3).
In this case, the two levels split each other. One level increases
while the other decreases in energy. In order to observe Fermi
Resonance, in addition to the requirement that a near coincidence
of energy levels occurs, other symmetry properties of these vibrations
must also be satisfied. As a consequence, Fermi Resonance bands
are not frequently encountered.
There are also several factors which decrease the number of infrared
bands observed. Symmetry is one of the factors that can significantly
reduce the number of bands observed in the infrared. If stretching
a bond does not cause a change in the dipole moment, the vibration
will not be able to interact with the infrared radiation and the
vibration will be infrared inactive. Other factors include the
near coincidence of peaks that are not resolved by the spectrometer
and the fact that only a portion of the infrared spectrum is usually
accessed by most commercial infrared spectrometers.
This concludes the general discussion of infrared spectroscopy.
At this point we are ready to start discussing some real spectra.
Carbon-Hydrogen Stretching Frequencies
Let's take one more look at equation 7 and consider the carbon-hydrogen stretching frequencies. Since k and mH are the only two variables in this equation, if we assume that all C-H
stretching force constants are similar in magnitude, we would
expect the stretching frequencies of all C-H bonds to be similar.
This expectation is based on the fact that the mass of a carbon
atom and whatever else is attached to the carbon is much larger
the mass of a hydrogen. The reduced mass for vibration of a hydrogen
atom would be approximately the mass of the hydrogen atom which
is independent of structure. All C-H stretching frequencies are
observed at approximately 3000 cm-1, exactly
as expected. Fortunately, force constants do vary some with structure
in a fairly predictable manner and therefor it is possible to
differentiate between different types of C-H bonds. You may recall
in your study of organic chemistry, that the C-H bond strength
increased as the s character of the C-H bond increased. Some typical
values are given below in Table 2 for various hydridization states
of carbon. Bond strength and bond stiffness measure different
properties. Bond strength measures the depth of the potential
energy well associated with a C-H. Bond stiffness is a measure
of how much energy it takes to compress or stretch a bond. While
these are different properties, the stiffer bond is usually associated
with a deeper potential energy surface. You will note in Table
2 that increasing the bond strength also increases the C-H bond
Table 2. Carbon Hydrogen Bond Strengths as a Function of Hybridization
Type of C-H bond Bond Strength IR Frequency
|sp3 hybridized C-H||CH3CH2CH2-H||99||<3000|
|sp2 hybridized C-H||CH2=CH-H||108||>3000|
|sp hybridized C-H||HCC-H||128||3300|
C-H sp3 hybridization
Methyl groups, methylene groups and methine hydrogens on sp3 carbon atoms all absorb between 2850 and 3000 cm-1. While it is sometimes possible to differentiate between these types of hydrogen, the beginning student should probably avoid this type of interpretation. It should be
pointed out however, that molecules that have local symmetry,
will usually show symmetric and asymmetric stretching frequencies.
Take, for example, a CH2 group. It is not
possible to isolate an individual frequency for each hydrogen.
These two hydrogens will couple and will show two stretching frequencies,
a symmetric stretching frequency in which stretching and compression
of both hydrogens occurs simultaneously, and an asymmetric stretching
frequency in which stretching of one hydrogen is accompanied by
compression of the other. While these two motions will occur at
different frequencies, both will be found between the 2850-3000
cm-1 envelope. Similarly for a CH3
group, symmetric and asymmetric vibrations are observed. This
behavior is found whenever this type of local symmetry is present.
We will find other similar examples in the functional groups we
will be discussing. Some examples of spectra containing only sp3
hybridization can be found in Figures 5-6, and located at the
end of this discussion. These peaks are usually sharp and of medium
intensity. Considerable overlap of several of these bands usually
results in absorption that is fairly intense and broad in this
C-H sp2 hybridization
Hydrogens attached to sp2 carbons absorb at
3000-3250 cm-1. Both aromatic and vinylic carbon hydrogen
bonds are found in this region. An example of a molecule that
contains only sp2 hybridization can be found in Figure
7. Other examples of molecules that contain sp2 C-H
bonds along with other functional groups include Figures 13, 25
and 37. Examples of hydrocarbons that contain both sp2and
sp3 hybridization can be found in Figures 8-12. These
peaks are usually sharp and of low to medium intensity.
C-H sp hybridization
Hydrogens attached to sp carbons absorb at 3300 cm-1.
An example of a spectrum that contains sp hybridization can be
found in Figure 13. These peaks are usually sharp and of medium
to strong intensity.
Before concluding the discussion of the carbon hydrogen bond,
one additional type of C-H stretch can be distinguished, the C-H
bond of an aldehyde. The C-H stretching frequency appears as a
doublet, at 2750 and 2850 cm-1. Examples of
spectra that contain a C-H stretch of an aldehyde can be found
in Figures 14 and 15. You may (should) question why the stretching
of a single C-H bond in an aldehyde leads to the two bands just
described. The splitting of C-H stretching frequency into a doublet
in aldehydes is due to the phenomema we called "Fermi Resonance".
It is believed that the aldehyde C-H stretch is in Fermi resonance
with the first overtone of the C-H bending motion of the aldehyde.
The normal frequency of the C-H bending motion of an aldehyde
is at 1390 cm-1. As a result of this interaction, one
energy level drops to ca. 2750 and the other increases
to ca. 2850 cm-1. Only one C-H stretch is observed
for aldehydes that have the C-H bending motion of an aldehyde
significantly shifted from 1390 cm-1.
In summary, it is possible to identify the type of hydrogen based
on hybridization by examining the infrared spectra in the 3300
to 2750 cm-1 region. Before concluding, we
should also mention some exceptions to the rules we just outlined.
Cyclopropyl hydrogens which are formally classified as sp3
hybridized actually have more s character than 25 %. Carbon-hydrogen
frequencies greater than 3000 cm-1 are observed
for these stretching vibrations. Halogen substitution can also
affect the C-H stretching frequency. The C-H stretching frequencies
of hydrogens attached to a carbon also bearing halogen substitution
can also be shifted above 3000 cm-1. This
is illustrated in Figure 16. The last exception we will mention
is an interesting case in which the force constant is increased
because of steric interactions. The infrared spectrum of tri-t-butylcarbinol
is given in Figure 17. In this case, the hydrogens are sp3
hybridized but stretching the C-H bonds leads to increased crowding
and bumping, and this is manifested by a steeper potential energy
surface and an increase in k, the force constant in equation 6.
Nitrogen Hydrogen Stretching Frequencies
Much of what we have discussed regarding C-H stretching frequencies is also applicable here. There are three major differences between the C-H and N-H stretching frequencies. First, the force constant for N-H stretching is stronger, there is a larger dipole moment associated with the N-H bond, and finally, the N-H bond is usually involved in hydrogen bonding. The stronger force constant leads to a higher frequency for absorption. The N-H stretching frequency is usually observed from 3500-3200 cm-1. The larger dipole moment leads to a stronger absorption and the presence of hydrogen bonding has a definite influence on the band shape and frequency position. The presence of hydrogen bonding has two major influences on spectra. First, its presence causes a shift toward lower frequency of all functional groups that are involved in hydrogen bonding and second, the peaks are generally broadened. Keep these two factors in mind as you examine the following spectra, regardless of what atoms and functional groups are involved in the hydrogen bonding.
The N-H stretching frequency is most frequently encountered in
amines and amides. The following examples will illustrate the
behavior of this functional group in a variety of circumstances.
Primary amines and amides derived from ammonia
The N-H stretching frequency in primary amines and in amides derived
from ammonia have the same local symmetry as observed in CH2.
Two bands, a symmetric and an asymmetric stretch are observed.
It is not possible to assign the symmetric and asymmetric stretches
by inspection but their presence at approximately 3300 and 3340
cm-1 are suggestive of a primary amine or
amide. These bands are generally broad and a third peak at frequencies
lower than 3300 cm-1, presumably due to hydrogen
bonding, is also observed. This is illustrated by the spectra
in Figures 18 and 19 for n-butyl amine and benzamide.
Secondary amines and amides
Secondary amines and amides show only one peak in the infrared.
This peak is generally in the vicinity of 3300 cm-1. This is illustrated
in Figures 20 and 21. Again notice the effect of hydrogen bonding
on the broadness of the N-H peak.
Tertiary amines and amides
Tertiary amines and amides from secondary amines have no observable
N-H stretching band as is illustrated in Figures 22 and 23.
N-H bending motions
You may recall that we will be ignoring most bending motions because
these occur in the fingerprint region of the spectrum. One exception
is the N-H bend which occurs at about 1600 cm-1.
This band is generally very broad and relatively weak. Since many
other important bands occur in this region it is important to
note the occurrence of this absorption lest it be mistakenly interpreted
as another functional group. Figure 18 illustrates the shape and
general intensity of the bending motion. Most other functional
groups absorbing in this region are either sharper or more intense.
The hydroxyl stretch is similar to the N-H stretch in that it
hydrogen bonds but does so more strongly. As a result it is often
broader than the N-H group. In those rare instances when it is
not possible to hydrogen bond, the stretch is found as a relative
weak to moderate absorption at 3600-3650 cm-1.
In tri-t-butylmethanol where steric hindrance prevents hydrogen
bonding, a peak at 3600 cm-1 is observed as
shown in Figure 17. Similarly for hexanol, phenol, and hexanoic
acid, Figures 24, 25, and 26, gas phase and liquid phase spectra
illustrate the effect of hydrogen bonding on both the O-H stretch
and on the rest of the spectrum. In should be pointed out that,
in general, while gas phase spectra are usually very similar,
frequencies are generally shifted to slightly higher values in
comparison to condensed phase spectra. Gas phase spectra that
differ significantly from condensed phase spectra are usually
taken as evidence for the presence of some sort of molecular association
in the condensed phase.
The hydroxyl group in phenols and alcohols usually is found as
a broad peak centered at about 3300 cm-1 in
the condensed phase as noted above and in the additional examples
of Figures 24, 28, and 29. The O-H of a carboxylic acid, so strongly
associated that the O-H absorption in these materials, is often
extended to approximately 2500 cm-1. This
extended absorption is clearly observed in Figures 26, 27, and
29 and serves to differentiate the O-H stretch of a carboxylic
acid from that of an alcohol or phenol. In fact, carboxylic acids
associate to form intermolecular hydrogen bonded dimers both in
the solid and liquid phases.
The nitrile group
The nitrile group is another reliable functional group that generally
is easy to identify. There is a significant dipole moment associated
with the CN bond which leads to a significant change when it interacts
with infrared radiation usually leading to an intense sharp peak
at 2200-2280 cm-1. Very few other groups absorb
at this region with this intensity. The spectrum in Figure 31
illustrates the typical behavior of this functional group. If
another electronegative atom such as a halogen is attached to
the same carbon as the nitrile group, the intensity of this is
The carbon-carbon triple bond
The CC bond is not considered to be a very reliable functional
group. This stems in part by considering that the reduced mass
in equation 7 is likely to vary. However it is characterized by
a strong force constant and because this stretching frequency
falls in a region where very little else absorbs, 2100-2260 cm-1,
it can provide useful information. The terminal carbon triple
bond (CC-H) is the most reliable and easiest to identify. We have
previously discussed the C-H stretching frequency; coupled with
a band at 3300 cm-1, the presence of a band
at approximately 2100 cm-1 is a strong indication
of the -CC-H group. The spectrum in Figure 13 illustrates the
presence of this group.
An internal -CC- is more difficult to identify and is often missed.
Unless an electronegative atom such as nitrogen or oxygen is directly
attached to the sp hybridized carbon, the dipole moment associated
with this bond is small; stretching this bond also leads to a
very small change. In cases where symmetry is involved, such as
in diethyl acetylenedicarboxylate, Figure 32, there is no change
in dipole moment and this absorption peak is completely absent.
In cases where this peak is observed, it is often weak and difficult
to identify with a high degree of certainty.
The carbonyl group
The carbonyl group is probably the most ubiquitous group in organic
chemistry. It comes in various disguises. The carbonyl is a polar
functional group that frequently is the most intense peak in the
spectrum. We will begin by discussing some of the typical acyclic
aliphatic molecules that contain a carbonyl group. We will then
consider the effect of including a carbonyl as part of a ring
and finally we will make some comments of the effect of conjugation
on the carbonyl frequency.
Acyclic aliphatic carbonyl groups
Esters, aldehydes, and ketones
Esters, aldehydes, and ketones are frequently encountered examples
of molecules exhibiting a C=O stretching frequency. The frequencies,
1735, 1725, 1715 cm-1 respectively, are too close to allow a clear
distinction between them. However, aldehydes can be distinguished
by examining both the presence of the C-H of an aldehyde (2750,
2850 cm-1) and the presence of a carbonyl group. Examples of an
aliphatic aldehyde, ester, and ketone are given in Figures 14,
34, 36, and 35, respectively.
Carboxylic acids, amides and carboxylic acid anhydrides
Carboxylic acids, amides and carboxylic acid anhydrides round out the remaining carbonyl groups frequently found in aliphatic molecules. The carbonyl frequencies of these molecules, 1700-1730 (carboxylic acid), 1640-1670 (amide) and 1800-1830, 1740-1775 cm-1 (anhydride), allow for an easy differentiation when the following factors are also taken into consideration.
A carboxylic acid can easily be distinguished from all the carbonyl containing functional groups by noting that the carbonyl at 1700-1730 cm-1 is strongly hydrogen bonded and broadened as a result. In addition it contains an O-H stretch which shows similar hydrogen bonding as noted above. Spectra which illustrate the effect of hydrogen bonding include Figures 27, and 29.
Amides are distinguished by their characteristic frequency which is the lowest carbonyl frequency observed for an uncharged molecule, 1640-1670 cm-1(Amide I). In addition, amides from ammonia and primary amines exhibit a weaker second band (Amide II) at 1620-1650 cm-1 and 1550 cm-1 respectively, when the spectra are run on the solids. Amides from secondary amines do not have a hydrogen attached at nitrogen and do not show an Amide II band. The Amide I band is mainly attributed to the carbonyl stretch. The Amide II involves several atoms including the N-H bond. We will return to the frequency of the amide carbonyl when we discuss the importance of conjugation and the effect of resonance on carbonyl frequencies. The spectra of benzamide, a conjugated amide (Figure 19), and N-methyl acetamide (Figure 21) clearly identify the Amide I and II bands. The spectrum of N,N dimethyl acetamide (Figure 23) illustrates an example of an amide from a secondary amine.
Anhydrides can be distinguished from other simple carbonyl containing
compounds in that they contain and exhibit two carbonyl frequencies.
However, these frequencies are not characteristic of each carbonyl.
Rather they are another example of the effects of local symmetry
similar to what we have seen for the CH2 and
NH2 groups. The motions involved here encompass
the entire anhydride (-(C=O)-O-(O=C-) in a symmetric and asymmetric
stretching motion of the two carbonyls. The two carbonyl frequencies
often differ in intensity. It is not possible to assign the peaks
to the symmetric or asymmetric stretching motion by inspection
nor to predict the more intense peak. However, the presence of
two carbonyl frequencies and the magnitude of the higher frequency
(1800 cm-1) are a good indication of an anhydride. Figure 36 contains
a spectrum of an aliphatic anhydride.
Cyclic aliphatic carbonyl containing compounds
The effect on the carbonyl frequency as a result of including
a carbonyl group as part of a ring is usually attributed to ring
strain. Generally ring strain is believed to be relieved in large
rings because of the variety of conformations available. However
as the size of the ring gets smaller, this option is not available
and a noticeable effect is observed. The effect of increasing
ring stain is to increase the carbonyl frequency, independent
of whether the carbonyl is a ketone, part of a lactone(cyclic
ester), anhydride or lactam (cyclic amide). The carbonyl frequencies
for a series of cyclic compounds is summarized in Table 3.
Table 3. The Effect of Ring Strain on the Carbonyl Frequencies
of Some Cyclic Molecules
|Ring Size||ketone: cm-1||lactones: cm-1||lactams: cm-1|
|4||cyclobutanone: 1775||b-propiolactone: 1840|
|5||cyclopentanone: 1751||g-butyrolactone: 1750||g-butyrolactam: 1690|
|6||cyclohexanone: 1715||d-valerolactone: 1740||d-valerolactam: 1668|
|7||cycloheptanone: 1702||e-caprolactone: 1730||e-caprolactam: 1658|
Carbon carbon double bond
Like the CC bond, the C=C bond stretch is not a very reliable
functional group. However, it is also characterized by a strong
force constant and because of this and because the effects of
conjugation which we will see can enhance the intensity of this
stretching frequency, this absorption can provide useful and reliable
In simple systems, the terminal carbon carbon double bond (C=CH2)
is the most reliable and easiest to identify since the absorption
is of moderate intensity at 1600-1675 cm-1. We have previously
discussed the C-H stretching frequency of an sp2
hybridized C-H. The spectrum in Figure 9 illustrates the presence
of this group. In addition the terminal C=CH2
is also characterized by a strong band at approximately 900 cm-1.
Since this band falls in the fingerprint region, some caution
should be exercised in its identification.
An internal non-conjugated C=C is difficult to identify and can
be missed. The dipole moment associated with this bond is small;
stretching this bond also leads to a very small change. In cases
where symmetry is involved, such as in 4-octene, Figure 10, there
is no change in dipole moment and this absorption peak is completely
absent. In cases where this peak is observed, it is often weak.
In 2,5-dihydrofuran, Figure 39, it is difficult to assign the
C=C stretch because of the presence of other weak peaks in the
vicinity. The band at approximately 1670 cm-1 may be the C=C stretch.
In 2,5-dimethoxy-2,5-dihydrofuran, Figure 40, the assignment at
1630 cm-1 is easier but the band is weak.
There is one circumstance that can have a significant effect on the intensity of both internal and terminal olefins and acetylenes. Substitution of a heteroatom directly on the unsaturated carbon to produce, for example, a vinyl or acetylenic ether, or amine leads to a significant change in the polarity of the C=C or CC bond and a substantial increase in intensity is observed. The C=C in 2,3-dihydrofuran is observed at 1617.5 cm-1 and is one of the most intense bands in the spectrum (Figure 41). Moving the C=C bond over one carbon gives 2,5-dihydrofuran attenuates the effect and results in a weak absorption (Figure 39).
Aromatic ring breathing motions
Benzene rings are encountered frequently in organic chemistry.
Although we may write benzene as a six membered ring with three
double bonds, most are aware that this is not a good representation
of the structure of the molecule. The vibrational motions of a
benzene ring are not isolated but involve the entire molecule.
To describe one of the fundamental motions of benzene, consider
imaginary lines passing through the center of the molecule and
extending out through each carbon atom and beyond. A symmetric
stretching and compression of all the carbon atoms of benzene
along each line is one example of what we might describe as a
ring breathing motion. Simultaneous expansions and compressions
of these six carbon atoms lead to other ring breathing motions.
These vibrations are usually observed between 1450 and 1600 cm-1
and often lead to four observable absorptions of variable intensity.
As a result of symmetry, benzene, Figure 7, does not exhibit these
bands. However most benzene derivatives do and usually 2 or 3
of these bands are sufficiently separate from other absorptions
that they can be identified with a reasonable degree of confidence.
The least reliable of these bands are those observed at approximately
1450 cm-1 where C-H bending motions are observed.
Since all organic molecules that contain hydrogen are likely to
have a C-H bond, absorptions observed at 1450 cm-1
are not very meaningful and should usually be ignored. Two of
the four bands around 1600 cm-1 are observed
in ortho and meta xylene, identified by the greek
letter and a third band at about 1500 cm-1
is assigned (Figure 11 and 12). We will return to a discussion
of these bands when we discuss the effects of conjugation on the
intensities of these motions.
The final functional group we will include in this discussion is the nitro group. In addition to being an important functional group in organic chemistry, it will also begin our discussion of the importance of using resonance to predict effects in infrared spectroscopy. Let's begin by drawing a Kekule or Lewis structure for the nitro group. You will find that no matter what you do, it will be necessary to involve all 5 valence
electrons of nitrogen and use them to form the requisite number
of bonds to oxygen. This will lead to a positive charge on nitrogen
and a negative charge on one oxygen. As a result of resonance,
we will delocalize the negative charge on both oxygens and as
shown, this leads to an identical structure. Since the structures
are identical, we would expect the correct structure to be a resonance
hybrid of the two. In terms of geometry, we would expect the structure
to be a static average of the two geometric structures both in
terms of bond distances and bond angles. Based on what we observed
for the CH2 and NH2 stretch,
we would expect a symmetric and an asymmetric stretch for the
N-O bond in the nitro group halfway between the N=O and N-O stretches.
Since both of those functional groups are not covered in this
discussion, we will need to assume for the present that this is
correct. Two strong bands are observed, one at 1500-1600 cm-1
and a second between 1300-1390 cm-1, Figure 28.
Effect of resonance and conjugation on infrared frequencies
Let's continue our discussion of the importance of resonance but shift from the nitro group to the carboxylate anion. The carboxylate anion is represented as a resonance hybrid by the following figure:
Unlike the nitro group which contained functional groups we will
not be discussing, the carboxyl group is made up of a resonance
hybrid between a carbon oxygen single bond and a carbon oxygen
double bond. According to resonance, we would expect the C-O bond
to be an average between a single and double bond or approximately
equal to a bond and a half. We can use the carbonyl frequency
of an ester of 1735 cm-1 to describe the force
constant of the double bond. We have not discussed the stretching
frequency of a C-O single bond for the simple reason that it is
quite variable and because it falls in the fingerprint region.
However the band is known to vary from 1000 to 1400 cm-1.
For purposes of this discussion, we will use an average value
of 1200 cm-1. The carbonyl frequency for a
bond and a half would be expected to fall halfway between 1735
and 1200 or at approximately 1465 cm-1. The
carboxyl group has the same symmetry as the nitro and CH2
groups. Both a symmetric and asymmetric stretch should be observed.
The infrared spectrum of sodium benzoate is given in Figure 42.
An asymmetric and symmetric stretch at 1410 and 1560 cm-1
is observed that averages to 1480 cm-1, in good agreement with
the average frequency predicted for a carbon oxygen bond with
a bond order of 1.5. While this is a qualitative argument, it
is important to realize that the carboxylate anion does not show
the normal carbonyl and normal C-O single bond stretches (at approximately
1700 and 1200 cm-1) suggested by each of the
static structures above.
In the cases of the nitro group and the carboxylate anion, both
resonance forms contribute equally to describing the ground state
of the molecule. We will now look at instances where two or more
resonance forms contribute unequally to describing the ground
state and how these resonance forms can effect the various stretching
Most carbonyl stretching frequencies are found at approximately 1700 cm-1. A notable exception is the amide carbonyl which is observed at approximately 1600 cm-1. This suggests that the following resonance form makes a significant contribution to describing the ground state of amides:
You may recall that resonance forms that lead to charge separation are not considered to be very important. However the following information support the importance of resonance in amides. X-ray crystal structures of amides show that in the solid state the amide functional group is planar. This suggests sp2 hybridization at nitrogen rather than sp3. In addition the barrier to rotation about the carbon nitrogen bond has been measured. Unlike the barrier of rotation of most aliphatic C-N bonds which are of the order of a few kcal/mol, the barrier to rotation about the carbon nitrogen bond in dimethyl formamide is approximately 18 kcal/mol. This suggests an important contribution of the dipolar structure to the ground state of the molecule and the observed frequency of 1600 cm-1, according to the arguments given above for the carboxylate anion, is consistent with more C-O single bond character than would be expected otherwise.
Conjugation of a carbonyl with a C=C bond is thought to lead to an increase in resonance interaction. Again the resonance forms lead to charge separation which clearly de-emphasizes their importance.
However this conjugative interaction is useful in interpreting several features of the spectrum. First it predicts the small but consistent shift of approximately 10 cm-1 to lower frequency, observed when carbonyls are conjugated to double bonds or aromatic rings. This feature is summarized in Table 4 for a variety of carbonyl groups. Next, the dipolar resonance form suggests a more polar C=C than that predicted for an unconjugated C=C. In terms of the change in dipole moment, contributions from this structure suggest that the intensity of infrared absorption of a C=C double bond would increase relative to an unconjugated system. Comparison of Figures 9, 10 and 35 with Figures 43, and 44-47 shows this to be the case. Conjugation is associated with an increase in intensity of the C=C stretching frequency. Finally, examination of Figures 43-46 reveals an intricacy not previously observed with simple non-conjugated carbonyls. The carbonyls of Figures 43-46 which are all conjugated appear as multiplets while those unconjugated carbonyls such as those in Figures 14 and 35 appear as single frequencies. Note however that not all conjugated carbonyls appear as multiplets (Figures 15 and 47. Resolution of this additional complicating feature can be achieved if we consider that conjugation requires a fixed conformation. For most conjugated carbonyls, two or more conformations are possible. The s-cis form is shown above and the s-trans form is shown below.
If the resonance interaction in these two forms differ, the effect
of resonance on the carbonyl will differ leading to similar but
different frequencies. The presence of multiple carbonyl frequencies
is a good indication of a conjugated carbonyl. In some conjugated
systems such as benzaldehyde
Table 4. The effect of conjugation on carbonyl frequencies.
|2-butanone||1717||methyl vinyl ketone||1700,
|propanoic acid||1715||propenoic acid||1702||benzoic acid||1688|
|ethyl propionate||1740||ethyl acrylate||1727||ethyl benzoate||1718|
and benzyl 4-hydroxyphenylketone (Figures 15 and 47), only one
conformation by symmetry is possible and conjugation does not
lead to any additional carbonyl frequencies. It should also be
noted that in many of the examples given above, cis-trans
isomerization about the carbon-carbon double bond is also possible.
Some of the observed bands may also be due to the presence of
these additonal isomers. Since the intensity of the peak is determined
by the change in dipole moment, the presence of a small amount
of geometric isomer can still lead to a detectable peak.
Experimental infrared spectra
Up to now we have been focusing in on theory and interpretation
of infrared spectra. At this point we should spend some time discussing
the practical aspects of how infrared spectra are obtained and
the factors to take into consideration when trying to interpret
the results. Let's first start by considering gas phase spectra.
Cells and gas phase spectra
These type of spectra were more a curiosity and of theoretical
interest until the introduction of the combined techniques of
gas chromatography-Fourier transfer infrared spectroscopy (GC-FTIR).
The major advantages of this method is that spectra can be obtained
on micrograms of material and the spectra do not show the effects
of interactions between molecules characteristic of condensed
phase spectra. These spectra are usually obtained at elevated
temperatures. Condensed phase spectra however will continue to
be important because of the fact that many compounds do not survive
injection into a gas chromatograph. Currently, most frequency
correlations for various functional groups are reported for the
condensed phase. Frequencies observed in the gas phase are usually
slightly higher than those observed for the same functional group
in the condensed phase.
Gas phase spectra can also be taken at room temperature. All that
is needed is a sample with a vapor pressure of several millimeters
and a pathlength of about a decimeter (10 cm). Cells with NaCl
or KBr windows are commercially available or can be built easily.
Crystals of KBr are transparent from 4000-250 cm-1
and are perfectly acceptable for most uses. They have the disadvantage
of being hydroscopic and must be stored in a desiccator. Cells
of sodium chloride are transparent from 4000-600 cm-1,
less expensive and less hydroscopic. These cells are also acceptable
for routine spectra.
Cells and condensed phase spectra
Condensed phase spectra can be taken as a solid or as a liquid.
Comparison of the same sample in the liquid and solid phase will
differ. However the major differences observed will be in the
fingerprint region. In cases where infrared spectroscopy is used
as a criteria of identity, the spectra under comparison should
be obtained under identical experimental conditions. Liquid phase
spectra are the easiest to obtain. All that is needed are two
polished disks of NaCl or KBr, both commercially available. A
thin film is prepared by depositing a drop of the liquid between
the two plates and mounting them in the beam of the spectrometer.
This is referred to as a neat liquid. Glass is not a useful material
in infrared spectroscopy because of the strong absorptions due
to the Si-O group. The infrared spectrum of quartz is shown in
Spectra of solids can be obtained in a variety of ways. The method
of choice varies depending on the physical properties of the material
under consideration. We will list several methods that can be
used satisfactorily along with the limitations and advantages
Neat Spectra (thin film)
In order to obtain an infrared spectrum of a solid, it is necessary
to get light, mainly infrared, through the sample. This can be
achieved in various ways and we will outline some that have proven
successful in the past. A thin layer of a solid deposited as a
solution on an infrared cell and the solvent allowed to evaporate
has proven successful with many compounds. Solvents such as CHCl3,
CH2Cl2 and CCl4
have been frequently been used. The solid sample should have an
appreciable solubility in one of these solvents. A drop of a solution
left to evaporate will deposit a thin film of crystal that will
often transmit sufficient light to provide an acceptable infrared
spectrum. This method suffers from the disadvantage that a spectrum
of the solvent must also be run to determine whether all of the
solvent has evaporated.
A mull is a suspension of a solid in a liquid. Under these conditions, light can be transmitted through the sample to afford an acceptable infrared spectrum. The commercial sample of Nujol, or mineral oil, which is a long chain hydrocarbon is often used for this purpose. Most solids do not dissolve in this medium but can be ground up in its presence. A small mortar and pestle is used for this purpose. If the grinding process gives rise to small particles of solid with diameters roughly the same as the wavelength of the infrared radiation being used, 2-5 microns, these particles will scatter rather than transmit the light. The effect of poor grinding is illustrated in Figures 29 and 30 for a sample of benzoic acid. If you find this type of band distortion with either a Nujol mull or a KBr pellet (discussed below), simply continue grinding the sample up until the particles become finer.
The major disadvantage of using a Nujol mull is that the information
in the C-H stretching region is lost because of the absorptions
of the mulling agent. A spectrum of Nujol is shown in Figure 5.
To eliminate this problem, it may be necessary to run a second
spectrum in a different mulling agent that does not contain any
C-H bonds. Typical mulling agents that are used for this purpose
are perfluoro- or perchlorohydrocarbons. Examples include perchlorobutadiene,
perfluorokerosene or a perfluorohydrocarbon oil (Figure 48).
A KBr pellet is a dilute suspension of a solid in a solid. It
is usually obtained by first grinding the sample in anhydrous
KBr at a ratio of approximately 1 part sample to 100 parts KBr.
Although it is best to weigh the sample (1 mg) in the KBr (100mg),
with some experience it is possible to use your judgment in assigning
proportions of sample to KBr. The mixture is then ground up in
an apparatus called a Wiggle-Bug, frequently used by dentists
to prepare amalgams. The ground up sample mixture is then placed
on a steel plate containing a paper card with a hole punched in
it. The sample is placed in the hole, making sure that some sample
also overlaps the paper card. Paper the thickness and consistency
of a postcard is usually used and the hole is positioned on the
card so that it will lie in the infrared beam when placed on the
spectrometer. A second steel plate is placed over the sample and
card and the steel sandwich is placed in a hydraulic press and
subjected to pressures of 15000 psi for about 20 seconds. Removal
of the paper card following decompression usually results in a
KBr pellet that is reasonably transparent both to visible light
and infrared radiation. Some trial and error may be necessary
before quality pellets can be obtained routinely. Samples that
are not highly crystalline sometimes prove difficult and do not
produce quality pellets. However good quality spectra can be obtained
on most samples. The only limitation of KBr is that it is hydroscopic.
Because of this, it is usually a good idea to obtain a spectrum
run as a Nujol mull on your sample as well. The two spectra should
be very similar and since Nujol is a hydrocarbon and has no affinity
for water, any absorption in Nujol between 3400-3600 cm-1 can
be attributed to the sample and not to the absorption of water
PE 1600 FT Infrared Spectrometer
The operation of the Perkin Elmer 1600 FTIR, the instrument that you will be using in this laboratory, will be demonstrated. However before learning how to use it you should familiarize yourself with some of the general operating features of the instrument and its capabilities and limitation. In addition this brief tutorial will serve as a useful reminder once you have learned how to use the instrument. A discussion of the performance of a Fourier Transfer infrared spectrometer is beyond the scope of this publication. However the following will summarize some of the essential features of the PE 1600. To begin with the PE 1600 is a single beam instrument. Unlike a double beam instrument that simultaneously corrects for absorptions due to atmospheric water vapor and carbon dioxide, most FTIR spectrometers correct for the background absorption by storing an interferogram and background spectrum before recording your spectrum. An interferogram contains the same information as a regular spectrum, frequency vs. intensity, but this information is contained in the form of intensity vs. time. The Fourier Tranform is the mathematical process which coverts the information from intensity and time to intensity and frequency. One of the major advantages of a FTIR instrument is that is takes on the order of a second to record an entire spectrum. This makes it very convenient to record the same spectrum a number of times and to display an averaged spectrum. Since the signal to noise ratio varies as the square root of the number of scans averaged, it is easy to obtain good signal to noise on this type of instrument, even if you have very little sample. On a typical spectrum you should average at least four spectra. Be sure you average a similar number of background spectra.
Looking at the keyboard of the instrument you will find keys with
a permanent function. Two functions are defined on some keys and
these functions can be accessed by pressing the key directly or
by a combination of the shift + function key. A number of other
keys, those directly under the monitor, are defined by the screen
and their function may vary depending on the screen. The instrument
has four active memory sites where spectra can be stored and retrieved.
These are called the background, x, y and z. Once the instrument
has been turned on and has passed the self tests, it is usually
a good idea to allow the infrared source to warm up for 5 min.
With the sample pathway empty , pressing "scan + background
+ the key under the monitor prompt consistent with the number
of scans you which to average" will produce a background
interferogram. Placing your sample in the beam and pressing "scan
x, y, or z + the number of scans you desire" will produce
an interferogram with the requisite number of scans that will
be corrected for the background and displayed on the screen. You
may wish to record a spectrum, then make some adjustments to see
if they improve the quality of the spectrum. Storing the second
spectrum in a different memory region allows you to evaluate the
adjustments. If the first spectrum was stored in x and the second
in y, pressing x or y allows you to retrieve either.
To plot your spectrum on the HP plotter, simply turn the plotter
on, load paper on the plotter, and make sure the plotter is equipped
with at least one pen. Press "plot" and the monitor
will tell you the number of peaks the instrument has detected
that satisfy the current settings. The frequency of each peak
identified as meeting these criteria will be printed out. If the
number of peaks is too large or too few peaks are identified,
it may be necessary to change the current setting on the instrument.
Simply press "cancel + plot (the key is located on the monitor)".
To change the peak threshold, press the following keys located
on the monitor: press "setup", "view", "peaks".
You can change the threshold value using the number pad on the
left of the console. Increasing the threshold will decrease the
number of peaks while decreasing the existing value will include
more peaks. Next press "execute"; and "exit".
Additional details and instructions are generally available at
When you are first learning to make KBr pellets, you will be using
a paper punch to produce a hole on a paper card. You will need
to adjust the size and position of the hole properly so that it
will allow sufficient infrared light to pass through the card.
To monitor how much radiation is reaching the detector, press
"shift + monitor and then the energy key under the monitor".
This will let you know how much energy is reaching the detector.
With nothing else in the beam except the paper card with the hole
punched in, at least 60-70% of the energy should reach the detector.
If you observe a reading much less than this, you may either have
to adjust the position of the aperture or enlarge it or both.
The cancel key under the monitor will return you to normal mode.
Interpretation of Infrared Spectra
We have just concluded a discussion of a large number of frequencies
and the functional groups that are generally associated with these
frequencies. At this point you may be asking yourself how to begin
to interpret these frequencies with regards to obtaining information
of molecular structure. There are a number of different approaches
that can be used and often the best approach to use depends on
the nature of the information you would like to obtain from your
infrared spectrum. For example, if you are repeating a synthesis
in the laboratory and you wish to determine whether you have successfully
isolated the material you intended to prepare, you may be able
to compare your spectrum to an infrared spectrum of an authentic
sample. In this case, you are using infrared analysis for establishing
the identity of your sample. Assuming that your spectrum has been
run under the same conditions, as your reference, i.e. neat sample,
KBr pellet, etc., you should be able to reproduce the spectrum
of the reference material, peak for peak. The presence of some
additional peaks in your spectrum may indicate a contamination
with solvent, starting material or an impurity that has not been
removed. The presence of fewer peaks than your reference is of
more concern. This generally indicates a failure to obtain the
If the structure of the material of interest is unknown, then
a more systematic analysis of your spectrum will be necessary.
You should be aware that it is not usually possible to determine
molecular structure from the infrared spectrum alone. Usually,
some supplemental spectroscopic and/or structural information
(such as molecular formula) is also necessary. For the unknowns
in this course, you will generally be using infrared spectroscopy
to differentiate between a few possible compounds. Frequently,
this can be achieved by an analysis of the functional groups in
your spectrum. The discussion which follows, uses a more generalized
approach to analyze spectra. This approach should be applicable
in a variety of different circumstances. If a portion of the discussion
is not relevant to you, simply skip it and continue until it does
The Degree of Unsaturation
Once the molecular formula of an unknown is known, it is a simple matter to determine the degree of unsaturation. The degree of unsaturation is simply the sum of the number of carbon-carbon double bonds and rings. Each reduces the number of hydrogens or any other element with a valance of one by two. Although there is a general formula that can be memorized and used, a much simpler procedure is to note the number of carbon atoms and any other elements in your molecular formula and simply draw a molecule that contains the requisite number of carbons atoms and any other elements that are present. Make sure there are no rings or carbon-carbon double bonds in your structure. Each carbon, nitrogen, sulfur and oxygen should have four, three, two and two single bonds, respectively. Use as many hydrogens as you need to make up the appropriate number of bonds for each element. Be sure to include all halogens and any other elements in your structure as well. Count the number of hydrogens in your structure and subtract this number from the number in your original molecular formula. The difference, divided by two equals the degree of unsaturation, the sum of the number of rings and double bonds.
Consider C6Cl6 as an example: CH2Cl-CHCl-CHCl-CHCl-CHCl-CH2Cl. The number of hydrogens in my sample molecule is 8; there are none in the original molecular formula. The difference, divided by two is four. The degree of unsaturation is four. An unsaturation factor of four is quite common and characteristic of benzene and its derivatives.
Application of the degree of unsaturation to the interpretation
of an infrared spectrum is quite straightforward. Clearly some
functional groups can be eliminated by composition. Amines, amides,
nitriles and nitro groups can be eliminated if the molecule does
not contain any nitrogen. Alternatively everything but amines
can be eliminated if the molecular formula contains nitrogen and
no degrees of unsaturation. The following steps should serve as
a general protocol to follow and should prove useful regardless
of the structure of your unknown or whether the degree of unsaturation
1. Examine the C-H stretching frequencies at 3000 cm-1. Absorption bands bands at frequencies slightly larger than 3000 cm-1 are indicative of vinyl or/and aromatic hydrogens. The presence of these peaks should be consistent with the degree of unsaturation of your molecule. The absence of absorption above 3000 cm-1 but the presence of some unsaturation in the molecular formula are consistent with a cyclic compound.
If your degree of unsaturation is 4 or greater, look for 2 to
4 absorption peaks between 1600-1450 cm-1
and weak peaks at 2000-1667 cm-1. These are
characteristic of aromatic compounds.
2. Next look for a doublet at 2750 and 2850 cm-1 characteristic
of an aldehyde. The presence of these two bands should also be
accompanied by a strong absorption at approximately 1700 cm-1.
Most spectra display strong absorption in the 1800-1700 cm-1 region.
If your spectrum does, check to see if the carbonyl is a closely
spaced doublet or multiplet. Closely spaced multiplicity in the
carbonyl region accompanied by C-H absorption at 3000-3100 cm-1
is frequently characteristic of an a,
b- unsaturated carbonyl compounds. Check to make sure that
the carbonyl frequency is consistent with conjugation.
3. If you unknown contains broad absorption from 3600-3000, your
molecule could have an O-H or N-H stretch. Your molecular formula
may allow you to differentiate. Check the multiplicity of this
peak. A doublet is characteristic of a primary amine or and amide
derived from ammonia. Check the carbonyl region at around 1650-1600
cm-1. Two bands in this region are consistent of an amide from
ammonia or a primary amine. Remember a broad and relatively weak
band at about 1600 cm-1 is characteristic of N-H bending. Usually
you will only see this band in amines, since that carbonyl group
of the amide will interfere. Be sure to look for the effect of
hydrogen bonding which usually results in a general broadening
of the groups involved.
4. If the broad band starting at 3600 expands to nearly 2400 cm-1,
look for the presence of a broad carbonyl at approximately 1700
cm-1. This extremely broad OH band is only observed in carboxylic
acids and enols from b-diketones. The
presence of a relatively intense but broad band at approximately
1700 cm-1 is good evidence for a carboxylic acid.
5. Don't try to over-interpret your spectrum. Often, it is not
possible to arrive at a unique structure based on infrared analysis
alone. You should use your infrared analysis much like you would
use other classification tests. You can learn a great deal about
your unknown from your spectrum but be sure to use other important
physical data such as melting point, boiling point and solubility
characteristics of your unknown to assist you in narrowing down
the different structural possibilities.