Infrared Spectrophotometry.ppt

Chapter 3: Infrared Spectroscopy
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School of Pharmacy, CHS, AAU
2019/20 A.Y.

Learning Outcomes
• By the end of this topic, students should be able to:
– Explain the principles and the working mechanism of infrared (IR)
absorption spectroscopy
– Identify the molecular species that absorb IR radiation
– Interpret IR spectrum
– Differentiate between a dispersive IR instrument and a FTIR
spectrometer
– Determine unknown qualitatively using IR absorption
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Introduction
 Infrared spectroscopy deals with the interaction of infrared light with
matter.
 Infrared radiation (IR) refers broadly to that part of the electromagnetic
spectrum between the visible and microwave regions.
 IR does not have sufficient energy to cause excitation of electrons but it
does cause atoms and groups of atoms of organic cpds to vibrate faster
about the covalent bonds that connect them.
– The quantum mechanical energy levels observed in IR spectroscopy are those
of molecular vibration
• Non destructive
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Introduction…
• General Uses
– Identification of functional groups on a molecule important tool in
organic chemistry
– IR spectra gives you better structural information
than UV/Vis spectra
– Spectral matching can be done by computer
software and library spectra
– Since absorbance follows Beer’s Law, can do quantitative analysis
(A=E C l)
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Introduction…
• Absorption of IR radiation corresponds to energy changes on the order of
8 to 40 kJ/mole.
– Radiation in this energy range corresponds to stretching and bending
vibrational frequencies of the bonds in most covalent molecules.
• IR Spectrum, has three regions;
– 1) Near IR region : which extends from 0.8 to 2.5 μm (wave n. about
12500 to 4000 cm-1)
– 2) Middle IR (Vibration-Rotation region): which extends from 2.5 to 50 μm
(wave n. about 4000 to 200 cm-1)
– 3) Far IR (Rotation region): which extends from 50 to 1000 μm (wave n about
200 to 10 cm-1)
– The main region for analytical purposes is the middle IR region.
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What happen when a molecule absorbs infrared radiation?
 The covalent bond between two atoms acts like a spring.
 When a cpd is bombarded with radiation of a frequency that exactly
matches the frequency of one of its vibrations (natural vibrational
frequencies ), the molecule will absorb energy.
• The energy absorbed will increase the amplitude of the vibrational
motions of the bonds in the molecule.
– This allows the bonds to stretch and bend a bit more.
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When does IR absorption occur?
1.IR absorption only occurs when IR radiation interacts with a molecule
undergoing a change in dipole moment as it vibrates or rotates.
2. Infrared absorption only occurs when the incoming IR photon has
sufficient energy for the transition to the next allowed vibrational state
• Note: If the 2 rules above are not met, no absorption can occur
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• NOT ALL bonds in a molecule are capable of absorbing IR energy.
• Only those bonds that have change in dipole moment are capable to
absorb IR radiation.
• The larger the dipole change, the stronger the intensity of the band in an
IR spectrum.
- Dipole moment (μ) - a measure of net molecular polarity in a chemical
bond or molecule, equal to the product of one charge and the distance
between the charges (μ = Q x r).
• Many molecules have such dipole moments due to non-uniform
distributions of positive and negative charges on the various atoms
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• In heteronuclear diatomic molecule, because of the difference in
electronegativities of the two atoms, one atom acquires a small
positive charge (q+), the other a negative charge (q-).
• This molecule is then said to have a dipole moment whose
magnitude,
• Many molecules have such dipole moments due to non-uniform
distributions of positive and negative charges on the various atoms
μ = q x r
μ =dipole moment (Coulomb ·meters)
q =magnitude of charges
r =vector
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Molecular Species that Absorb Infrared Radiation
• Can a vibration change the dipole moment of a molecule?
• Asymmetrical stretching and asymmetrical bending change the dipole
moment of a molecule.
– Asymmetrical stretching/bending are IR active.
– Symmetrical stretching/bending are not IR active
• E.g Symmetrically substituted alkenes and alkynes
– Vibration of two similar atoms against each other (e.g. O2 or N2
molecules) will not result in a change of electrical symmetry or dipole
moment of the molecules and such molecules will not absorb energy
in the IR region (but other bonds and vibrational modes in these
molecules do absorb IR light).
C C R
R
R
R R
R
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Number of Vibrational modes
• A molecule consisting of n atoms has a total of 3n degrees of freedom,
corresponding to the Cartesian coordinates of each atom in the
molecule.
• In a nonlinear molecule, 3 of these degrees are rotational and 3 are
translational and the remaining correspond to fundamental vibrations.
• In a linear molecule, 2 degrees are rotational and 3 are translational.
• The net number of fundamental vibrations for nonlinear and linear
molecules is therefore:
• 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.
Molecules Degree of freedoms
Non linear 3n-6
Linear 3n-5
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Number of Vibrational modes…
• But this number may be increased or decreased.
• Reasons for decrease
– Two bands might overlap (Degenerate)
– An absorption might not be in the 4000–400 cm–1 range
– An absorption might be too weak to be observed
– Absorptions might be too close to each other to be resolved on the
instrument.
– Lack of dipole change
• Reasons for increase
– Overtones –multiples of the fundamental vibrations
– Coupling-splitting of vibrational modes
– Combination tones-two vibrations abs. Simultaneously
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Mode of Vibration
• Molecules can vibrate in a variety of ways, but generally
bond vibration modes are divided into two types:
• Stretching Vibrations and
• Bending (deformation) vibrations.
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Molecular vibration
divided
into
stretching bending
back & forth
movement
involves change in
bond angles
symmetrical asymmetrical
scissoring
rocking twisting
wagging
in-plane
vibration
out of plane
vibration
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Mode of Vibration…
A. Stretching Vibrations: Where the atoms move back and forth as if they
are joined by a spring along the bond axis
– It affects the bond length
– More energetic than the bending vibrations
– Divided into two types; Symmetric and asymmetric stretching
vibrations.
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Mode of Vibration…
B. Bending (Deformation) Vibrations:
• They are displacements occurring at right angles to the bond axis.
• Affect the bond angle & require lower energy than stretching vibrations.
• Divided to: In-Plane bending vibrations & Out-of-Plane bending
vibrations
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IR Spectrum in Absorption Mode
• The IR spectrum is basically a plot of transmitted (or absorbed)
frequencies vs. intensity of the transmission (or absorption).
• Frequencies appear in the x-axis in units of inverse centimeters
(wavenumbers), and intensities are plotted on the y-axis in percentage
units.
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The graph above shows a spectrum in absorption mode.
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IR Spectrum in Transmission Mode
 The graph below shows a spectrum in transmission mode.
 This is the most commonly used representation and the one found in
most chemistry and spectroscopy books.
 The transmittance spectra provide better contrast between intensities
of strong and weak bands.
 Therefore we will use this representation.
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AN IR SPECTRUM IN TRANSMISSION MODE
• The IR spectrum of octane, plotted as transmission (left)
and absorbance (right).
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The Position of Absorption Band
• Amount of energy required to stretch a bond depends on the strength
of the bond and the masses of the bonded atoms (Hooke’s Law).
• The equation relates the wavenumber of the stretching vibration to the
force constant of the bond ( f ) and the masses of the atoms (in grams)
joined by the bond and the force constant is a measure of the strength
of the bond.
• The stronger the bond, the greater the energy required to stretch it,
because a stronger bond corresponds to a tighter spring.
• Frequency of the vibration is inversely related to the mass of the atoms
attached to the spring, so heavier atoms vibrate at lower frequencies.
• Approximate wave number of an absorption can be calculated from the
equation derived from Hooke’s law, which describes motion of a
vibrating spring:
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The Position of Absorption Band
• Frequencies (position) of normal modes of vibration of two
atoms held together by a chemical bond depend on:
– The masses of the two vibrating atoms; Light atoms vibrate at higher
frequencies than the heavier ones.
– The force constant of the bond between them and bond order:
(relative stiffness of the bond).
• Triple bonds are stiffer (vibrate at higher frequencies) than double
bonds, and double bonds are stiffer than single bonds.
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Carbon-Carbon Bond Stretching
• Stronger bonds absorb at higher frequencies:
– C-C 1200 cm-1
– C=C 1660 cm-1
– CC 2200 cm-1 (weak or absent if internal)
• Conjugation lowers the frequency:
– isolated C=C 1640-1680 cm-1
– conjugated C=C 1620-1640 cm-1
– aromatic C=C approx. 1600 cm-1
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The Position of Absorption Bands…
Carbon-Hydrogen Stretching
• Hybridization affects the force constant K,
• Bonds are stronger in the order sp > sp2 > sp3,
– Bonds with more s character absorb at a higher frequency
– sp3 C-H, just below 2900 cm-1
– sp2 C-H, just above 3100 cm-1
– sp C-H, at 3300 cm-1
• Bending motions occur at lower energy (lower frequency) than the
typical stretching motions because of the lower value for the bending
force constant K.
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• Resonance also affects the strength and length of a bond and hence its
force constant K.
• Thus, whereas a normal ketone has its C=O stretching vibration at 1715
cm-1, a ketone that is conjugated with a C=C double bond absorbs at a
lower frequency, near 1675 to 1680 cm-1.
• That is because resonance lengthens the C=O bond distance and gives it
more single-bond character:
• Resonance has the effect of reducing the force constant K, and the
absorption moves to a lower frequency.
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• The electronegativity of the two vibrating atoms (Dipole moment
effects) also affects the position of absorption band.
• Examples:
– C – H 2850 – 2960 cm-1 N – H 3300 – 3500 cm-1
– O – H 3590 – 3650 cm-1
– C – N Around 1350 cm-1 C – O Around 1400 cm-1
– C = C 1620 – 1680 cm-1 C = N 1620 – 1690 cm-1 C = O
1630 – 1780 cm-1
– C = C 2100 – 2260 cm-1 C = N 2220 – 2260 cm-1
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Hydrogen Bonding: alters both Stretching and bending vibrations:
(a) Stretching vibrations: cause vibrations to move to lower frequencies.
– To longer wavelength with increased intensity
– Cause band widening
(b) Bending vibrations: cause shift to shorter wavelengths. Less
pronounced than stretching vibrations.
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Hydrogen Bonding have two types:
(a) Inter-molecular H.B.: Involves association of two or more molecules of
the same or different compounds
(b) Intera-molecular H.B : Formed when the proton donor and acceptor are
present in a single molecule
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The Intensity of Absorption Bands…
• The intensity of an absorption band depends on the size of the change in
dipole moment associated with the vibration, number of bonds
responsible for absorption and concentration.
• Dipole moment: the greater the change in dipole moment, the more
intense the absorption.
– The stretching vibration of an O-H bond will be associated with a greater change in
dipole moment than that of an N-H bond because the bond is more polar.
– Consequently, the stretching vibration of the O-H bond will be more intense.
– Likewise, the stretching vibration of an N-H bond is more intense than that of a C-H
bond because the N-H bond is more polar.
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O - H > N - H > C - H
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The Intensity of Absorption Bands…
• Number of bonds: the intensity of an absorption band also depends on
the number of bonds responsible for the absorption.
• For example, the absorption band for the C – H stretch will be more
intense for a compound such as octyl iodide, which has 17 C – H
bonds, than for methyl iodide, which has only three C – H bonds.
• Concentration: concentrated samples have greater numbers of
absorbing molecules and, therefore, more intense absorption bands.
• Intensities are referred as strong (s), medium (m), weak (w), broad,
and sharp.
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Infrared Instrumentation
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Sample
compartm
ent
IR
Source
Detecto
r
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• The instrument that determines the absorption spectrum for a compound
is called an infrared spectrometer or, more precisely, a
spectrophotometer.
• Two types of infrared spectrometers are in common use: dispersive and
Fourier transform (FT) instruments.
• Both of these types of instruments provide spectra of compounds in the
common range of 4000 to 400 cm-1.
• In simple terms, IR spectra are obtained by detecting changes in
transmittance (or absorption) intensity as a function of frequency.
• Basic components include
– Radiation sources
– Monochromators/Interferometers/
– Sample compartment
– Detector
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• Radiation sources
• An inert solid is electrically heated to a temperature in the range 1500 to
2000 oC. The heated material will then emit infra red radiation.
• Three popular types of sources.
• Nernst glower : is a cylinder (1-2 mm diameter, approximately 20 mm
long) of rare earth oxides. The Nernst glower can reach temperatures of
2200 oC .
• Globar source : is a silicon carbide rod (5mm diameter, 50mm long)
which is electrically heated to about 1500 oC.
• Nichrome wire,: electrically heated to 1100 oC. It produces a lower
intensity of radiation than the Nernst or Globar sources, but has a longer
working life.
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• Monochromators
– The monochromator is a device used to disperse a broad spectrum of
radiation and provide a continuous series of electromagnetic energy bands of
determinable wavelength or frequency range.
– Prisms or gratings are the dispersive components used in conjunction with
variable-slit mechanisms and mirrors.
– Used only in dispersive IR spectrometer.
• Interferometers
– An FT-IR instrument uses a system called an interferometer to collect a
spectrum.
– It consists of three active components: a moving mirror, a fixed mirror, and a
beam splitter.
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• Interferometers…
– The energy goes from the source to the beams splitter which splits the
beam into two parts.
– One part is transmitted to a moving mirror; one part is reflected o a
fixed mirror.
– The moving mirror moves back and forth at a constant velocity.
– The two beams are reflected from the mirrors and recombined at the
beam splitter.
– The beam from the moving mirror has traveled a different distance than
the beam from the fixed mirror.
– When the beams are combined an interference pattern is created, since
some of the wavelengths recombine constructively and some
destructively.
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– If the movable mirror moves away from the beam splitter at a constant
speed, radiation reaching the detector goes through a steady sequence
of maxima and minima as the interference alternates between
constructive and destructive phases.
– This interference pattern is called interferogram.
– The resulting beam then passes through the sample and is eventually
focused on the detector.
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INSTRUMENTATION…
– The interferogram a time domain spectrum and records the detector
response changes versus time within the mirror scan.
– If the sample happens to absorb at this frequency, the amplitude of the
sinusoidal wave is reduced by an amount proportional to the amount of
sample in the beam.
– A mathematical operation known as Fourier transformation converts the
interferogram (a time domain spectrum displaying intensity versus time
within the mirror scan) to the final IR spectrum, which is the familiar
frequency domain spectrum showing intensity versus frequency.
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Fourier Transform
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• Detectors
– Most detectors used in dispersive IR spectrometers can be categorized into
two classes: thermal detectors and photon detectors.
– Thermal detectors include thermocouples, thermistors, and pneumatic devices
(Golay detectors).
– They measure the heating effect produced by infrared radiation.
– A variety of physical property changes are quantitatively determined:
expansion of a non absorbing gas (Golay detector), electrical resistance
(thermistor), and voltage at junction of dissimilar metals (thermocouple).
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• Detectors…
– Photon detectors rely on the interaction of IR radiation and a semiconductor
material.
• Non-conducting electrons are excited to a conducting state. Thus, a
small current or voltage can be generated.
– The response times of many detectors (for example, thermocouple and
thermistor) used in dispersive IR instruments are too slow for the rapid scan
times (1 sec or less) of the interferometer.
– The two most popular detectors for a FTIR spectrometer are deuterated
triglycine sulfate (DTGS) and mercury cadmium telluride (MCT).
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Dispersive IR spectrometer
• The instrument produces a beam of infrared radiation from a hot wire
and, by means of mirrors, divides it into two parallel beams of equal-
intensity radiation.
• The sample is placed in one beam, and the other beam is used as a
reference.
• The beams then pass into the monochromator, which disperses each
into a continuous spectrum of frequencies of infrared light.
• The monochromator consists of a rapidly rotating sector (beam chopper)
that passes the two beams alternately to a diffraction grating (a prism in
older instruments)
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• The slowly rotating diffraction grating varies the frequency or
wavelength of radiation reaching the thermocouple detector.
• The detector senses the ratio between the intensities of the reference
and sample beams.
• In this way, the detector determines which frequencies have been
absorbed by the sample and which frequencies are unaffected by the
light passing through the sample.
• After the signal from the detector is amplified, the recorder draws the
resulting spectrum of the sample on a chart.
• It is important to realize that the spectrum is recorded as the frequency
of infrared radiation changes by rotation of the diffraction grating.
• Dispersive instruments are said to record a spectrum in the frequency
domain.
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• Fourier transform-IR spectrometer
– It was developed in order to overcome the limitations
encountered with dispersive instruments.
• The main difficulty was the slow scanning process.
– A method for measuring all of the infrared frequencies
simultaneously, rather than individually, was needed.
– A solution was developed which employed a very simple optical
device called an interferometer
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• Fourier transform-IR spectrometer (FTIR)
• The IR radiation from a broadband source is first directed into an
interferometer,
– where it is divided and then recombined after the split beams
travel different optical paths to generate constructive and
destructive interference.
• Next, the resulting beam passes through the sample compartment and
reaches to the detector.
•Most FTIR spectrometers are single-beam instruments
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To separate IR light, a grating is used.
Grating
Light source
Detector
Sample
Slit
To select the specified IR light,
A slit is used.
Dispersion
Spectrometer
Fixed CCM
B.S.
Moving CCM
IR Light source
Sample
Detector
An interferogram is
first made by the
interferometer using
IR light.
The interferogram is calculated and
transformed into a spectrum using
a Fourier Transform (FT).
FTIR
Comparison Between Dispersion Spectrometer and FTIR
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• Advantages of FT-IR
– Better speed
• A complete spectrum can be obtained during a single scan of the
moving mirror, while the detector observes all frequencies
simultaneously.
• An interferometer in an FT-IR instrument does not separate energy into
individual frequencies for measurement of the infrared spectrum.
• In contrast, every wavelength across the spectrum must be measured
individually in a dispersive spectrometer.
• This is a slow process, and typically only one measurement scan of the
sample is made in a dispersive instrument
• FT-IR advantage is that many scans can be completed and combined on
an FT-IR in a shorter time than one scan on a dispersive instrument.
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• Better sensitivity and Resolution
– An FT-IR instrument does not use a slit to limit the individual frequency
reaching the sample and detector as a dispersive instrument does.
– There are also fewer mirror surfaces in an FT-IR spectrometer, so there are
less reflection losses than in a dispersive spectrometer.
– Overall, more energy reaches the sample and hence the detector in an FT-IR
spectrometer than in a dispersive spectrometer.
– Higher sensitivity and clear spectrum
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– Simpler mechanical design
• There is only one moving part, the moving mirror, resulting in
better reliability.
– Elimination of stray light and emission contributions
• The interferometer in FTIR modulates all the frequencies. The
unmodulated stray light and sample emissions (if any) are not
detected.
– Powerful data station
• Modern FTIR spectrometers are usually equipped with a powerful,
computerized data system.
• It can perform a wide variety of data processing tasks such as
Fourier transformation, baseline correction, and library searching.
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Preparation Of Samples For Infrared Spectroscopy
• It is possible to analyze samples in the liquid, solid or gaseous forms.
• The material used to contain the sample must always be transparent to
IR radiation.
• Cells must be constructed of ionic substances; typically sodium chloride
or potassium bromide.
– Glass and plastics absorb strongly throughout the infrared region of the
spectrum.
– Potassium bromide plates are more expensive than sodium chloride plates
but have the advantage of usefulness in the range of 4000 to 400 cm-1.
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Preparation of Samples…
– Sodium chloride plates are used widely because of their relatively low cost.
– The practical range for their use in spectroscopy extends from 4000 to 650
cm-1
– Sodium chloride begins to absorb at 650 cm-1, and any bands with
frequencies less than this value will not be observed.
– Since few important bands appear below 650 cm-1, sodium chloride plates
are in most common use for routine infrared spectroscopy.
• Liquid Sample;
– A drop of a liquid organic compound is placed between a pair of polished
sodium chloride or potassium bromide plates, referred to as salt plates.
– When the plates are squeezed gently, a thin liquid film forms between them.
– A spectrum determined by this method is referred to as a neat spectrum
since no solvent is used.School of Pharmacy, CHS, AAU
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• Salt plates break easily and are water soluble.
– Organic compounds analyzed by this technique must be free of water.
– The pair of plates is inserted into a holder that fits into the spectrometer.
– A compensating cell, containing pure solvent is placed in the reference
beam of the instrument.
– The choice of solvent depends on the solubility of the sample and its own
minimal absorption in IR region.
– Carbon tetrachloride, chloroform and carbon disulfide are
preferred solvents.
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• Solid Sample
• There are at least three common methods for preparing a solid sample
for spectroscopy.
• The first method mixing the finely ground solid sample with powdered
potassium bromide and pressing the mixture under high pressure.
– Under pressure, the potassium bromide melts and seals the compound into a
matrix.
– The result is a KBr pellet that can be inserted into a holder in the
spectrometer.
– The main disadvantage of this method is that potassium bromide absorbs
water, which may interfere with the spectrum that is obtained.
– If a good pellet is prepared, the spectrum obtained will have no interfering
bands since potassium bromide is transparent down to 400 cm-1
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• The second method, a Nujol mull, involves grinding the compound with
mineral oil (Nujol) to create a suspension of the finely ground sample
dispersed in the mineral oil.
– The thick suspension is placed between salt plates.
– The main disadvantage of this method is that the mineral oil obscures bands
that may be present in the analyzed compound.
– Nujol bands appear at 2924, 1462, and 1377 cm-1.
• The third common method used with solids is to dissolve the organic
compound in a solvent, most commonly carbon tetrachloride (CCl4).
– some regions of the spectrum are obscured by bands in the solvent.
– Although it is possible to cancel out the solvent from the spectrum by
computer or instrumental techniques, the region around 785 cm-1 is often
obscured by the strong C-Cl stretch that occurs there.
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Con’t…
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BP methods of preparation of substance
KBr disc Methods
• Acetazolamide tablet
• Cimetidine HCl
• Warfarin
• Predisolone
• Chloropromazine
• Spironolactone
Mineral oil
• Clindamycin phosphate
• Procarbazone
• Tolazoline
• Claritromycin‘
• Erythromycin
• etc
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Sample cells…
• Gases- the gas sample cell is similar to the cell for liquid samples.
• Gases have densities which are several orders of magnitude less than
liquids, and hence path lengths must be correspondingly greater, usually
10 cm or longer.
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Analytical Information
• The combination of the fundamental vibrations or rotations of various
functional groups and the subtle interactions of these functional groups
with other atoms of the molecule
– Results in the unique, generally complex IR spectrum for each individual
compound.
• IR spectroscopy is mainly used in two ways: structural elucidation and
compound identification.
• It has been found that many functional groups give characteristic IR
absorption at specific, narrow frequency ranges.
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Absorption Regions
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1. Absorption of an infrared photon results in the excitation of the
molecule to a higher vibrational quantum state.
2. For a stretching vibration, a photon is absorbed and the molecule
excited to a higher vibrational quantum state only if that vibration
results in a change in bond dipole.
• The bond dipole is a product of bond length and charge difference of
the bonded atoms.
Basic Concept in IR
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Basic Concept in IR
• As a bond vibrates, the bond length changes, so this criterion is met.
The charge difference is determined by the electronegativity of the
bonded atoms.
• If these atoms are not identical then they have a difference in
electronegativity and thus a difference in charge.
• If these atoms are identical, they have equal electronegativity and thus
no charge difference.
• In this case, the product of bond length change and charge difference
(zero) is zero, so no photon is absorbed. If the atoms are even slightly
different, then a small change in bond dipole occurs, along with the
corresponding absorption of an infrared photon.
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5. E.g. all bond stretches of H2C=O show up on the IR spectrum because
all of the bonds consist of unequal atoms. For H2C=CH2, the C–H bond
stretches give IR bands, but the C=C bond (which is made up of two
identical carbon atoms) does not. The C=C bonds of H2C=CH–CH=CH2
show up in the IR because the carbons are not equal. One carbon is a
CH2 and the other is a HC–CH.
6. In summary, asymmetrical bonds give IR absorptions whereas
symmetrical bonds do not.
7. The energy of the photon necessary to excite the molecule to a higher
vibrational quantum level is controlled by the masses of the attached
atoms and the bond strength. Similar functional groups are made up of
similar atoms. Thus molecules with similar functional groups absorb
photons of similar energies. This is why we can use IR spectroscopy to
determine what functional groups are present or absent in a compound
by simply looking at the positions of the IR peaks.
Basic Concept in IR….
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8. We divide the IR spectrum into five zones, based upon the energy of
photons absorbed by twelve common functional groups. The zones and
functional groups therein are summarized below :
Zone Five Analysis
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Zone 1 (3700–3200 cm1) alcohol O–H, amide, or amine N–H,
terminal alkyne ≡C–H
Zone 2 (3200–2700 cm-1) alkyl C–H, aryl or vinyl C–H, aldehyde C–
H, carboxylic acid O–H
Zone 3 (2300–2000 cm-1) alkyne and nitrile triple bonds
Zone 4 (1850–1650 cm-1) carbonyl functional groups
Zone 5 (1680–1450 cm-1) alkene, benzene ring
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9. -Alcohol, Amine: the O–H and N–H stretches are usually broadened due
to hydrogen bonding.
- Carboxylic acid: must have two bands: broad O–H stretch in zone 2
and C=O stretch in Zone 4.
- Aldehyde: must have two sets of bands: 2900 and 2700 cm-1 in zone
2 (2900 cm-1 often hidden by alkyl C–H stretches), and C=O stretch
in zone 4.
- Alkyne: a peak caused by the triple bond stretch is not observed
when the alkyne is symmetric or nearly so.
- Carbonyl: often the most intense peak in the entire spectrum.
- Benzene ring: must have bands at about 1600 cm-1 (may be two
bands close together) and another at about 1500 cm-1.
• Since most benzene ring compounds have at least one hydrogen
atom attached to the benzene ring, an sp2 C–H stretch peak is also
seen in Zone 2
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10. The region from about 1450 to 400 cm-1 is called the fingerprint
region.
• This region is often complex because it contains many absorptions.
• It is usually ignored in routine IR analysis because it can be difficult to make
precise assignments.
• For example, a peak at 1360 cm-1 could be a CH2 bend or a C–N stretch.
• Without precise assignments, it can be difficult to determine the absence or
presence of functional groups from the fingerprint region.
• However, the fingerprint regions of two IR spectra of the same compound are
identical, in the same way that two fingerprints left by the same person in
different places are identical.
• Thus the fingerprint region of an unknown compound can be compared with
the fingerprint region of a known sample, and if they match exactly, the
unknown and known are identical
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How to Approach the Analysis of a Spectrum
1. Is a carbonyl group present?
– The C=O group gives rise to a strong absorption in the region
at 1870 to 1550 cm–1. The peak is often the strongest in the
spectrum and of medium width.(You can't miss it)
• 2. If C=O is present, check the following types (if absent, go to
3).
– ACIDS-presence of O-H stretching
– broad absorption near 3400-2400 cm–1
– AMIDES-presence of N-H stretching– medium absorption near
3500 cm–1
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• 2. If C=O is present, check the following types…
– ESTERS- presence of C-O stretching – strong intensity absorptions
near 1300-1000 cm–1
– ANHYDRIDES - have two C=O bands near 1810 and 1760 cm–1
– ALDEHYDES- presence of aldehydic C-H stretching (2900 to 2700
cm–1)
• Two weak absorptions near 2850 and 2750 cm–1 on the right-
hand side of CH band
– KETONES -the above 5 choices have been eliminated
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• 3. If C=O is absent
– ALCOHOLS Check for O-H (3600)
– PHENOLS – check for O-H (3550-3500)
• confirm this by finding C–O near 1300-1000 cm–1
– AMINES Check for N-H medium absorptions near 3500 cm–1
– ETHERS Check for C–O (and absence of OH) near 1300-1000 cm–1
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• 4- Double Bonds and/or Aromatic Rings
– C=C is a weak absorption near 1650 cm–1
– medium to strong absorptions in the region 1650-1450 cm–1 often
imply an aromatic ring
– confirm the above by consulting the C-H region; aromatic and vinyl
C-H occurs to the left of 3000 cm–1 (aliphatic CH occurs to the right
of this value)
• 5. Triple Bonds
– – C≡N is a medium, sharp absorption near 2250 cm–1
– – C≡C is a weak but sharp absorption near 2150 cm–1
– Check also for acetylenic C-H near 3300 cm–1
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• 6. Nitro Groups
– Two strong absorptions at 1660 - 1500 cm–1 and 1390-1260
cm–1
• 7. Hydrocarbons
– None of the above are found
– Major absorptions are in C-H region near 3000 cm–1
– Very simple spectrum, only other absorptions near 1450 cm–1
and 1375 cm–1 (bending vibrations)
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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).
IR spectrum of tolune School of Pharmacy, CHS, AAU
2019/20 A.Y.
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Practice Problems
1. Several functional groups have absorptions in more than one of the
five zones. To conclude that the functional group is present, both
absorptions must be seen. Prepare a table listing these functional
groups and the corresponding absorptions.
Hint: There are at least six such functional groups. Some of them require
three distinct absorptions.
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Functional Group Absorptions
terminal alkyne 3300 cm-1 (C–H stretch) - z1
2260–2000 cm-1 (C≡C stretch) - z3
aromatic ring 3100–3000 cm-1 (sp2 C–H stretch) -z2
~1600 cm-1 and 1500–1450 cm-1 (benzene ring stretch) - z5
aldehyde ~2900 and ~2700 cm-1 (C–H stretch) - z2
1740–1720 cm-1 (C=O stretch) - z4
carboxylic acid 3000–2500 cm-1 (O–H stretch) -z2
1725–1700 cm-1 (C=O stretch) -z4
aryl ketone 1700–1680 cm-1 (C=O stretch) -z4
~1600 cm-1 and 1500–1450 cm-1 (benzene ring stretch) -z5
enone 1685–1665 cm-1 (C=O stretch) -z4
1680–1620 cm-1 (C=C stretch) -z5
There are a few uncommon exceptions to this table. For example, acetylene is a
terminal alkyne that displays a zone 1 ≡C–H stretch, but not a zone 3 C≡C
stretch, due to the symmetry of the triple bond. A benzene ring in which all six
ring hydrogens have been substituted (e.g., hex methylbenzene) has zone 5 peaks
but not zone 2 sp2 C–H stretches. A tertiary amide (one without an N–H bond)
has a C=O stretch in zone 4 but no N–H stretch in zone 1.
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2. What is the effect of conjugation on the energy of an infrared absorption
band?
Conjugation shifts a band to lower energy.
For example, a ketone typically absorbs in the region of 1750–1705 cm-1,
whereas an enone (alkene-carbonyl conjugation O=C–C=C; also called an
α,β-unsaturated ketone) absorbs at 1685–1665 cm-1, and an aryl ketone
(ketone conjugated with a benzene ring) absorbs at 1700–1680 cm-1. This
shift to a lower energy is seen when any functional group is conjugated.
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An Alkane IR Spectrum (n-hexane)
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An Alkene IR Spectrum (1-hexene)
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Alkyne---
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A Ketone IR Spectrum
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n-Hexane: CH3CH2CH2CH2CH2CH3
Exercise 1. Identify the functional group that present in the molecule
and predict the possible structure of the IR spectrum given below
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Exercise 2 Identify the functional group that present in the molecule
1-Hexene: CH2=CHCH2CH2CH2CH3
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3-heptanone:
CH2
CH2
H3C
O
CH2
CH2 CH3
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1-Hexanol: CH3CH2CH2CH2CH2CH2OH
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Summary of IR Absorptions
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1. Fundamental chemistry
Determination of molecular structure/geometry.
e.g. Determination of bond lengths, bond angles of
gaseous molecules
2. Qualitative analysis – simple, fast, nondestructive
Main uses of IR spectroscopy:
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Quantitative application of IR
• IR more difficult than UV-Vis because
– narrow bands
– complex spectra
– weak incident beam
– solvent absorption
• The basis for quantitative analysis of absorption
spectrometry is the Beer’s law. A=abc
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Quantitative…
• For quantification of a single component, a strong absorption band,
which is relatively free of overlapping or interference, is selected
from the IR spectrum.
• The percent transmittance is plotted against the wave number.
• The intensity of the incident radiation (I0) and the intensity of the
transmitted radiation (I) can be measured by the base– line method
illustrated below.
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Thank You!
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Infrared Spectrophotometry.ppt

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