Download Infrared Spectroscopy

CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Infrared Spectroscopy
Infrared (IR) spectroscopy is one of the most common spectroscopic techniques used by organic and
inorganic chemists. Simply, it is the absorption measurement of different IR frequencies by a sample
positioned in the path of an IR beam. The main goal of IR spectroscopic analysis is to determine the
chemical functional groups in the sample. Different functional groups absorb characteristic frequencies of
IR radiation. Hence it is used to identify chemical compounds or monitor changes occurring in the course
of a chemical reaction. For example, the spectra of two related molecules, 1-propanol (an alcohol), and
propanoic acid (a carboxylic acid) are given below. Their spectra are considerably different and tell about
important features of a molecule. Using various sampling accessories, IR spectrometers can accept a wide
range of sample types such as gases, liquids, and solids. Thus, IR spectroscopy is an important and
popular tool for structural elucidation and compound identification.
Figure 1. IR spectra of 1-propanol (an alcohol), and propanoic acid
Infrared refers to that part of the electromagnetic spectrum between the visible and microwave regions.
Electromagnetic spectrum refers to the seemingly diverse collection of radiant energy, from cosmic rays
to X-rays to visible light to microwaves, each of which can be considered as a wave or particle traveling
at the speed of light. These waves differ from each other in the length and frequency as shown bellow-
Figure 2. The electromagnetic spectrum
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Infrared radiation spans a section of the electromagnetic spectrum having wave numbers from roughly
13,000 to 10 cm–1, or wavelengths from 0.78 to 1000 µm. It is bound by the red end of the visible region
at high frequencies and the microwave region at low frequencies.
Of greatest practical use to the organic chemist is the limited portion of this region, the mid IR range
4000–400 cm–1. An increase in wave number corresponds to an increase in energy. This is a convenient
relationship for the organic chemist.
Figure 3. The IR regions of the electromagnetic spectrum
Infrared radiation in the range from about 10000-100 cm-1 is absorbed by organic molecules and
converted into energy of molecular vibration. This absorption is quantized but vibrational spectra appear
as bands rather than as lines because a single vibrational energy change is accompanied by a number of
rotational energy changes. it is with these vibrational-rotational bands, particularly those occurring
between 4000-400 cm-1 that an organic chemist concerned. In IR spectroscopy, an organic molecule is
exposed to infrared radiation. When the radiant energy matches the energy of a specific molecular
vibration, absorption occurs. The frequency of absorption depends on the1. Relative masses of atoms
2. Force constants of the bonds and
3. Geometry of the molecule
In a typical IR spectrum the band positions are presented as wave numbers. The wave number, plotted on
the X-axis, is proportional to energy; therefore, the highest energy vibrations are on the left. Band
intensities can be expressed either as transmittance (T) or absorbance (A). Transmittance, T, is the ratio of
radiant power transmitted by the sample (I) to the radiant power incident on the sample (I0). Absorbance
(A) is the logarithm to the base 10 of the reciprocal of the transmittance (T). The percent transmittance
(%T) is plotted on the Y-axis. Absorption of radiant energy is therefore represented by a “trough” in the
curve. Zero transmittance corresponds to 100% absorption of light at that wavelength. Organic chemist
usually report intensity in semi-quantitative terms (s = strong, m = medium, w = weak).
Except at very, very low temperatures, all molecules are in motion in some manner. Molecules translate
(move from place to place), they rotate in space, and, importantly for this experiment, they vibrate. At
temperatures above absolute zero, all the atoms in molecules are in continuous vibration with respect to
each other. When the frequency of a specific vibration is equal to the frequency of the IR radiation
directed on the molecule, the molecule absorbs the radiation.
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
You may have come to think of a molecule as having rigid bond lengths and bond angles, this is not the
actual case, since bond lengths and angles represent the average positions about which atoms vibrate. A
molecule is not a rigid assemblage of atoms. A molecule can be said to resemble a system of balls of
varying masses corresponding to atoms and springs of varying strengths (force constants ) corresponding
to the chemical bonds of a molecule.
There are two types of molecular vibrations, stretching and bending. A stretching vibration is a
rhythmical movement along the bond axis such that the inter-atomic distance is increasing or decreasing.
A bending vibration may consist of a change in bond angle between bonds with a common atom or the
movement of a group of atoms with respect to the remainder of the molecule without movement of the
atom in the group with respect to one another.
Only those vibrations that result in a rhythmical change in the dipole moment of the molecule are
observed in the IR. Various stretching and bending vibration of a molecule occur at certain quantized
frequencies. When IR light of that same frequency is incident on the molecule, energy is absorbed and the
amplitude of that vibration is increased. The frequency of the vibration remains unchanged. When the
molecule reverts from the excited state to ground stale the absorbed energy released as heat.
Each atom has three degrees of freedom, corresponding to motions along any of the three cartesian
coordinate axes (x, y, z). A polyatomic molecule of n atoms has 3n total degrees of freedom. However, 3
degrees of freedom are required to describe translation, the motion of the entire molecule through space.
Additionally, 3 degrees of freedom correspond to rotation of the entire molecule. Therefore, the
remaining 3n – 6 degrees of freedom are true, fundamental vibrations for nonlinear molecules. Linear
molecules possess 3n – 5 fundamental vibrational modes because only 2 degrees of freedom are sufficient
to describe rotation.
The net number of fundamental vibrations for nonlinear and linear molecules is therefore:
Fundamental vibrations involved no change in the center of gravity of the molecule. Among the 3n – 6
(or 3n – 5) fundamental vibrations (also known as normal modes of vibration), those that produce a net
change in the dipole moment may result in an IR activity and those that give polarizability changes may
give rise to Raman activity. Naturally, some vibrations can be both IR- and Raman-active.
Calculation reveals that a simple molecule such as propane, C3H8, has 27 fundamental vibrations, and
therefore, you might predict 27 bands in an IR spectrum! (The actual number is sometimes different as
discussed in part later.) The fundamental vibrations for water, H2O, are given in Figure 4. Water, which is
nonlinear, has three fundamental vibrations.
Figure 4. Vibrational modes for H2O
Carbon dioxide, CO2, is linear and hence has four fundamental vibrations (Figure 5). The symmetrical
stretch of CO2 gives a strong band in the IR at 2350 cm-1. You may notice this band in samples which you
run on the instruments in the teaching labs, since CO2 is present in the atmosphere. The two scissoring or
bending vibrations are equivalent and therefore, have the same frequency and are said to be degenerate,
appearing in an IR spectrum at 666 cm–1.
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Figure 5. Vibrational modes for CO2
The symmetrical stretch of CO2 is inactive in the IR because this vibration produces no change in the
dipole moment of the molecule. In order to be IR active, a vibration must cause a change in the dipole
moment of the molecule. (The reason for this involves the mechanism by which the photon transfers its
energy to the molecule, which is beyond the scope of this discussion.) Of the following linear molecules,
carbon monoxide and iodine chloride absorb IR radiation, while hydrogen, nitrogen, and chlorine do not.
In general, the larger the dipole change, the stronger the intensity of the band in an IR spectrum.
Only two IR bands (2350 and 666 cm –1) are seen for carbon dioxide, instead of four corresponding to the
four fundamental vibrations. Carbon dioxide is an example of why one does not always see as many
bands as implied by our simple calculation. In the case of CO2, two bands are degenerate, and one
vibration does not cause a change in dipole moment..
The stretching and bending vibrations for the important organic group, -CH2-, are illustrated in Figure 6.
(The 3n–6 rule does not apply since the -CH2- group represents only a portion of a molecule.) Note that
bending vibrations occur at lower frequencies than corresponding stretching vibrations.
Figure 6. Stretching and bending vibrational modes for a CH2 group
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
The theoretical number of fundamental vibrations will seldom be observed because of the following
reasons –
1. The following phenomenon reduces the theoretical number of bandsa. Overtones (multiple of a given frequency) and
b. Combination tones (sum of two different vibrations)
2. The following will reduces the theoretical number of bands
a. Fundamental frequencies that fall outside of the 400-400 cm-1 region.
b. Fundamental bands that are too weak to be observed.
c. Fundamental vibrations that are too close to each other to be resolved on the instrument.
d. The occurrence of a degenerate band from several absorptions of the same frequency in
highly symmetrical molecules.
e. The failure of certain fundamental vibrations to appear in the IR because of the lack of
change in molecular dipole.
Stretching Vibrations
The stretching frequency of a bond can be approximated by Hooke’s Law. In this approximation, two
atoms and the connecting bond are treated as a simple harmonic oscillator composed of 2 masses (atoms)
joined by a spring The energy curve for a simple harmonic oscillator is illustrated in Figure 15.7.
According to Hooke’s law, the frequency of the vibration of the spring is related to the mass and the force
constant of the spring, k, by the following formula:
Where
k is the force constant
m is the mass
ν is the frequency of the vibration
In the classical harmonic oscillator, E = 1/2kx2 = hν, where x is the displacement of the spring. Thus, the
energy or frequency is dependent on how far one stretches or compresses the spring, which can be any
value. If this simple model were true, a molecule could absorb energy of any wavelength.
However, vibrational motion is quantized: it must follow the rules of quantum mechanics, and the only
transitions which are allowed fit the following formulaE = (n +1/2) hν
Where
ν is the frequency of the vibration and n is the vibrational quantum number (0, 1, 2, 3, . . . )
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
The lowest energy level is E0 = 1/2 hν, the next highest is E1 = 3/2 hν. According to the selection rule,
only transitions to the next energy level (∆n = ± 1) are allowed; therefore molecules will absorb an
amount of energy equal to (3/2)hν – (1/2) hν or hν. This rule is not inflexible, and occasionally transitions
of 2hν, 3hν, or higher are observed. These correspond to bands called overtones in an IR spectrum. They
are of lower intensity than the fundamental vibration bands.
A molecule is not just two atoms joined on a spring, of course. A bond can come apart, and it cannot be
compressed beyond a certain point. A molecule is actually an anharmonic oscillator. As the interatomic
distance increases, the energy reaches a maximum, as seen in Figure 8. Note how the energy levels
become more closely spaced with increasing interatomic distance in the anharmonic oscillator. The
allowed transitions, hν, become smaller in energy. Therefore, overtones can be lower in energy than
predicted by the harmonic oscillator theory.
The following formula has been derived from Hooke’s law. For the case of a diatomic molecule, (ν has
been substituted for ν,
Where
k is the force constant
µ is the reduced mass [1/µ= (1/m1 +1/m2 )]
ν is the frequency of the vibration and
c is the velocity of light (3x1010 cm/s)
Above equation shows the relationship of bond strength and atomic mass to the wave number at which a
molecule will absorb IR radiation. As the force constant increases, the vibrational frequency
(wavenumber) also increases. Approximate value of force constants for bonds are:
Single bond Double bond Triple bond-
5 x 105 dyne/cm
10 x 105 dyne/cm
15 x 105 dyne/cm
As the mass of the atoms increases, the vibration frequency decreases. Using the following mass values
C, carbon 12/6.02 x 1023 and H, hydrogen 1/6.02 x 1023 ν for a C–H bond is calculated to be 3032 cm–1.
(Try this calculation!) The actual range for C–H absorptions is 2850–3000 cm–1. The region of an IR
spectrum where bond stretching vibrations are seen depends primarily on whether the bonds are single,
double, or triple or bonds to hydrogen. The following table shows where absorption by single, double,
and triple bonds are observed in an IR spectrum.
Bond
C–C, C–O, C–N
C=C, C=O, C=N, N=O
C≡C, C≡N
C–H, N–H, O–H
Absorption region, (cm–1)
800–1300
1500–1900
2000–2300
2700–3800
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Although a useful approximation, the motion of two atoms in a large molecule cannot be isolated from
the motion of the rest of the atoms in the molecule. In a molecule, two oscillating bonds can share a
common atom. When this happens, the vibrations of the two bonds are coupled. As one bond contracts,
the other bond can either contract or expand, as in asymmetrical and symmetrical stretching. In general,
when coupling occurs, bands at different frequencies are observed, instead of superimposed (or
degenerate) bands as you might expect from two identical atoms in a bond vibrating with an identical
force constant. In the case of the –CH2– group in Figure 6, you note there are two bands in the region for
C—H bonds: 2926 cm–1 and 2853 cm–1.
Table of Characteristic IR Absorptions
Frequency, cm–1
Bond
Functional Group
3640–3610 (s, sh)
3500–3200 (s,b)
3400–3250 (m)
3300–2500 (m)
3330–3270 (n, s)
3100–3000 (s)
3100–3000 (m)
3000–2850 (m)
2830–2695 (m)
2260–2210 (v)
2260–2100 (w)
1760–1665 (s)
1760–1690 (s)
1750–1735 (s)
1740–1720 (s)
1730–1715 (s)
1715 (s)
1710–1665 (s)
1680–1640 (m)
1650–1580 (m)
1600–1585 (m)
1550–1475 (s)
1500–1400 (m)
1470–1450 (m)
1370–1350 (m)
1360–1290 (m)
1335–1250 (s)
1320–1000 (s)
1300–1150 (m)
1250–1020 (m)
1000–650 (s)
950–910 (m)
910–665 (s, b)
900–675 (s)
850–550 (m)
725–720 (m)
700–610 (b, s)
690–515 (m)
O–H stretch, free hydroxyl
O–H stretch, H–bonded
N–H stretch
O–H stretch
–C≡C–H: C–H stretch
C–H stretch
=C–H stretch
C–H stretch
H–C=O: C–H stretch
C≡N stretch
–C≡C– stretch
C=O stretch
C=O stretch
C=O stretch esters,
C=O stretch aldehydes,
C=O stretch
C=O stretch ketones,
C=O stretch
–C=C– stretch
N–H bend
C–C stretch
N–O asymmetric stretch
C–C stretch (in–ring)
C–H bend
C–H rock
N–O symmetric stretch
C–N stretch
C–O stretch
C–H wag (–CH2X)
C–N stretch
=C–H bend
O–H bend
N–H wag
C–H
C–Cl stretch
C–H rock
–C≡C–H: C–H bend
C–Br stretch
alcohols, phenols
alcohols, phenols
1°, 2° amines, amides
carboxylic acids
alkynes (terminal)
aromatics
alkenes
alkanes
aldehydes
nitriles
alkynes
carbonyls (general)
carboxylic acids
saturated aliphatic
saturated aliphatic
α, β–unsaturated esters
saturated aliphatic
α, β –unsaturated aldehydes, ketones
alkenes
1° amines
(in–ring) aromatics
nitro compounds
aromatics
alkanes
alkanes
nitro compounds
aromatic amines
alcohols, carboxylic acids, esters, ethers
alkyl halides
aliphatic amines
alkenes
carboxylic acids
1°, 2° amines
“oop” aromatics
alkyl halides
alkanes
alkynes
alkyl halides
m=medium, w=weak, s=strong, n=narrow, b=broad, sh=sharp
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Examples
Alkanes
The spectra of simple alkanes are characterized by absorptions due to C–H stretching and bending (the C–C
stretching and bending bands are either too weak or of too low a frequency to be detected in IR spectroscopy). In
simple alkanes, which have very few bands, each band in the spectrum can be assigned.
-1
• C–H stretch from 3000–2850 cm
-1
• C–H bend or scissoring from 1470-1450 cm
-1
• C–H rock, methyl from 1370-1350 cm
-1
• C–H rock, methyl, seen only in long chain alkanes, from 725-720 cm
The IR spectrum of octane is shown below. Note the strong bands in the 3000-2850 cm-1 region due to C-H stretch.
The C-H scissoring (1470), methyl rock (1383), and long-chain methyl rock (728) are noted on this spectrum. Since
most organic compounds have these features, these C-H vibrations are usually not noted when interpreting a
routine IR spectrum.
The region from about 1300-900 cm-1 is called the fingerprint region. The bands in this region originate in
interacting vibrational modes resulting in a complex absorption pattern. Usually, this region is quite complex and
often difficult to interpret; however, each organic compound has its own unique absorption pattern (or fingerprint)
in this region and thus an IR spectrum be used to identify a compound by matching it with a sample of a known
compound.
Alkenes
Alkenes are compounds that have a carbon-carbon double bond, –C=C–. The stretching vibration of the C=C bond
usually gives rise to a moderate band in the region 1680-1640 cm-1.Stretching vibrations of the –C=C–H bond are
of higher frequency (higher wavenumber) than those of the –C–C–H bond in alkanes.The strongest bands in the
spectra of alkenes are those attributed to the carbon-hydrogen bending vibrations of the =C–H group. These bands
are in the region 1000-650 cm-1 (Note: this overlaps the fingerprint region).
-1
• C=C stretch from 1680-1640 cm
-1
• =C–H stretch from 3100-3000 cm
-1
• =C–H bend from 1000-650 cm
The IR spectrum of 1-octene is shown below. Note the band greater than 3000 cm-1 for the =C–H stretch and the
several bands lower than 3000 cm-1 for –C–H stretch (alkanes). The C=C stretch band is at 1644 cm-1. Bands for C–
H scissoring (1465) and methyl rock (1378) are marked on this spectrum; in routine IR analysis, these bands are not
specific to an alkene and are generally not noted because they are present in almost all organic molecules (and they
are in the fingerprint region).
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Note: Absorptions above 3000 cm-1 is a very useful tool for interpreting IR spectra: Only alkenes and aromatics
show a C-H stretch slightly higher than 3000 cm-1. Compounds that do not have a C=C bond show C-H stretches
only below 3000 cm-1.
Alkynes
Alkynes are compounds that have a carbon-carbon triple bond (–C≡C–). The –C≡C– stretch appears as a weak
band from 2260-2100 cm-1. This can be an important diagnostic tool because very few organic compounds show an
absorption in this region. A terminal alkyne (but not an internal alkyne) will show a C–H stretch as a strong, narrow
band in the range 3330-3270 cm-1. (Often this band is indistinguishable from bands resulting from other functional
groups on the same molecule which absorb in this region, such as the O-H stretch.). A terminal alkyne will show a
C–H bending vibration in the region 700-610 cm-1.
-1
• –C≡C– stretch from 2260-2100 cm
-1
• –C≡C–H: C–H stretch from 3330-3270 cm
-1
• –C≡C–H: C–H bend from 700-610 cm
The spectrum of 1-hexyne, a terminal alkyne, is shown below. Note the C–H stretch of the C–H bond adjacent to
the carbon-carbon triple bond (3324), the carbon-carbon triple bond stretch (2126), and the C–H bend of the C-H
bond adjacent to the carbon-carbon triple bond (636). The other bands noted are C–H stretch, scissoring, and
methyl rock bands from the alkane portions of the molecule.
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Ketones
The carbonyl stretching vibration band C=O of saturated aliphatic ketones appears at 1715 cm-1. Conjugation of the
carbonyl group with carbon-carbon double bonds or phenyl groups, as in alpha, beta-unsaturated aldehydes and
benzaldehyde, shifts this band to lower wavenumbers, 1685-1666 cm-1.
• C=O stretch:
-1
o aliphatic ketones 1715 cm
-1
o α, β-unsaturated ketones 1685-1666 cm
The spectrum of 2-butanone is shown below. This is a saturated ketone, and the C=O band appears at 1715. Note
the C–H stretches (around 2991) of alkyl groups. It's usually not necessary to mark any of the bands in the
fingerprint region (less than 1500 cm-1).
Alcohols
Alcohols have characteristic IR absorptions associated with both the O-H and the C-O stretching vibrations. When
run as a thin liquid film, or "neat", the O–H stretch of alcohols appears in the region 3500-3200 cm-1 and is a very
intense, broad band. The C–O stretch shows up in the region 1260-1050 cm-1.
-1
• O–H stretch, hydrogen bonded 3500-3200 cm
-1
• C–O stretch 1260-1050 cm (s)
The spectrum of ethanol is shown below. Note the very broad, strong band of the O–H stretch (3391) and the C–O
stretches (1102, 1055).
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
Aromatics
The =C–H stretch in aromatics is observed at 3100-3000 cm-1. Note that this is at slightly higher frequency than is the
–C–H stretch in alkanes. Aromatic hydrocarbons show absorptions in the regions 1600-1585 cm-1 and 1500-1400 cm-1
due to carbon-carbon stretching vibrations in the aromatic ring. Bands in the region 1250-1000 cm-1 are due to C–H
in-plane bending, although these bands are too weak to be observed in most aromatic compounds. Besides the C–H
stretch above 3000 cm-1, two other regions of the infrared spectra of aromatics distinguish aromatics from organic
compounds that do not have an aromatic ring:
-1
• 2000-1665 cm (weak bands known as "overtones")
-1
• 900-675 cm (out-of-plane or "oop" bands)
Not only do these bands distinguish aromatics, but they can be useful if you want to determine the number and
positions of substituents on the aromatic ring. The pattern of overtone bands in the region 2000-1665 cm-1 reflect
the substitution pattern on the ring. The pattern of the oop C–H bending bands in the region 900-675 cm-1 are also
characteristic of the aromatic substitution pattern. Details of the correlation between IR patterns in these two
regions and ring substitution are available in the literature references linked in the left frame (especially the books
by Shriner and Fuson, Silverstein et. al., and the Aldrich Library of IR Spectra).
In some instances, it is useful to remember that aromatics in general show a lot more bands than compounds that do
not contain an aromatic ring. If you are presented with two spectra and told that one is aromatic and one is not, a
quick glance at the sheer multitude of bands in one of the spectra can tell you that it is the aromatic compound.
-1
• C–H stretch from 3100-3000 cm
-1
• overtones, weak, from 2000-1665 cm
-1
• C–C stretch (in-ring) from 1600-1585 cm
-1
• C–C stretch (in-ring) from 1500-1400 cm
-1
• C–H "oop" from 900-675 cm
The spectrum of toluene is shown below. Note the =C–H stretches of aromatics (3099, 3068, 3032) and the –C–H
stretches of the alkyl (methyl) group (2925 is the only one marked). The characteristic overtones are seen from about
2000-1665. Also note the carbon-carbon stretches in the aromatic ring (1614, 1506, 1465), the in-plane C–H bending
(1086, 1035), and the C–H oop (738).
Aldehydes
The carbonyl stretch C=O of saturated aliphatic aldehydes appears from 1740-1720 cm-1. As in ketones, if the
carbons adjacent to the aldehyde group are unsaturated, this vibration is shifted to lower wave numbers, 1710-1685
cm-1. Another useful diagnostic band for aldehydes is the O=C–H stretch. This band generally appears as one or
two bands of moderate intensity in the region 2830-2695 cm-1. Since the band near 2830 cm-1 is usually
indistinguishable from other C–H stretching vibration bands (recall that the C–H stretches of alkanes appear from
3000-2850 cm-1), the presence of a moderate band near 2720 cm-1 is more likely to be helpful in determining
whether or not a compound is an aldehyde.
DR. JITENDRA SINGH, IET-DAVV, INDORE
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CHEMISTRY - BE-I Year (July’ 2014), Sem -A
H–C=O stretch 2830-2695 cm-1
• C=O stretch:
-1
o aliphatic aldehydes 1740-1720 cm
-1
o alpha, beta-unsaturated aldehydes 1710-1685 cm
The spectra of benzaldehyde and butyraldehyde are shown below. Note that the O=C stretch of the alpha, betaunsaturated compound -- benzaldehyde -- is at a lower wavenumber than that of the saturated butyraldehyde.
Note the O=C–H stretches in both aldehydes in the region 2830-2695 cm-1, especially the shoulder peak at 2725
cm-1 in butyraldehyde and 2745 cm-1 in benzaldehyde.
•
General Uses of IR spectroscopy:
• Identification of all types of organic and many types of inorganic compounds
• Determination of functional groups in organic materials
• Determination of the molecular composition of surfaces
• Identification of chromatographic effluents
• Quantitative determination of compounds in mixtures
• Nondestructive method
• Determination of molecular conformation (structural isomers) and stereochemistry (geometrical isomers)
• Determination of molecular orientation (polymers and solutions)
Common Applications:
• Identification of compounds by matching spectrum of unknown compound with reference spectrum
(fingerprinting)
• Identification of functional groups in unknown substances
DR. JITENDRA SINGH, IET-DAVV, INDORE
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