Infrared Spectrophotometry.ppt

analysis ppt

1. Chapter 3: Infrared Spectroscopy
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2. 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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3. Introduction
Infraredspectroscopydealswiththeinteractionofinfraredlightwith
matter.
Infraredradiation(IR)refersbroadlytothatpartoftheelectromagnetic
spectrumbetweenthevisibleandmicrowaveregions.
IRdoesnothavesufficientenergytocauseexcitationofelectronsbutit
doescauseatomsandgroupsofatomsoforganiccpdstovibratefaster
aboutthecovalentbondsthatconnectthem.
–ThequantummechanicalenergylevelsobservedinIRspectroscopyarethose
ofmolecularvibration
•Nondestructive
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4. 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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5. 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)NearIRregion: whichextendsfrom0.8to2.5μm(waven.about
12500to4000cm-1)
–2)MiddleIR(Vibration-Rotationregion):whichextendsfrom2.5to50μm
(waven.about4000to200cm-1)
–3)FarIR(Rotationregion):whichextendsfrom50to1000μm(wavenabout
200to10cm-1)
–ThemainregionforanalyticalpurposesisthemiddleIRregion.
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6. What happen when a molecule absorbs infrared radiation?
Thecovalentbondbetweentwoatomsactslikeaspring.
Whenacpdisbombardedwithradiationofafrequencythatexactly
matchesthefrequencyofoneofitsvibrations(naturalvibrational
frequencies),themoleculewillabsorbenergy.
•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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7. When does IR absorption occur?
1.IRabsorptiononlyoccurswhenIRradiationinteractswithamolecule
undergoingachangeindipolemomentasitvibratesorrotates.
2.InfraredabsorptiononlyoccurswhentheincomingIRphotonhas
sufficientenergyforthetransitiontothenextallowedvibrationalstate
•Note:Ifthe2rulesabovearenotmet,noabsorptioncanoccur
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8. When does IR absorption occur?
•NOTALLbondsinamoleculearecapableofabsorbingIRenergy.
•Onlythosebondsthathavechangeindipolemomentarecapableto
absorbIRradiation.
•Thelargerthedipolechange,thestrongertheintensityofthebandinan
IRspectrum.
-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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9. When does IR absorption occur?
•Inheteronucleardiatomicmolecule,becauseofthedifferencein
electronegativitiesofthetwoatoms,oneatomacquiresasmall
positivecharge(q+),theotheranegativecharge(q-).
•Thismoleculeisthensaidtohaveadipolemomentwhose
magnitude,
•Manymoleculeshavesuchdipolemomentsduetonon-uniform
distributionsofpositiveandnegativechargesonthevariousatoms
μ = q x r
μ =dipole moment (Coulomb ·meters)
q =magnitude of charges
r =vector
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10. 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
–Vibrationoftwosimilaratomsagainsteachother(e.g.O2orN2
molecules)willnotresultinachangeofelectricalsymmetryordipole
momentofthemoleculesandsuchmoleculeswillnotabsorbenergy
intheIRregion(butotherbondsandvibrationalmodesinthese
moleculesdoabsorbIRlight).CCRR
R
R R
R
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11. 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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12. 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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13. 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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14. Molecular vibration
divided
into
stretching bending
back & forth
movement
involves change in
bond angles
symmetrical
asymmetrical
scissoring
rockingtwisting
wagging
in-plane
vibration
outofplane
vibration
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15. 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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16. 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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17. IR Spectrum in Absorption Mode
•TheIRspectrumisbasicallyaplotoftransmitted(orabsorbed)
frequenciesvs.intensityofthetransmission(orabsorption).
•Frequenciesappearinthex-axisinunitsofinversecentimeters
(wavenumbers),andintensitiesareplottedonthey-axisinpercentage
units.
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The graph above shows a spectrum in absorptionmode.
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18. 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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19. AN IR SPECTRUM IN TRANSMISSION MODE
•The IR spectrum of octane, plotted as transmission (left)
and absorbance (right).
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20. 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 vibrationto 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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22. The Position of Absorption Band
•Frequencies (position) of normal modes of vibration of two
atoms held together by a chemical bond depend on:
–Themassesofthetwovibratingatoms;Lightatomsvibrateathigher
frequenciesthantheheavierones.
–Theforceconstantofthebondbetweenthemandbondorder:
(relativestiffnessofthebond).
•Triplebondsarestiffer(vibrateathigherfrequencies)thandouble
bonds,anddoublebondsarestifferthansinglebonds.
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23. 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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24. 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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25. The Position of Absorption Bands…
•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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26. The Position of Absorption Bands…
•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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27. The Position of Absorption Bands…
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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28. The Position of Absorption Bands…
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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29. 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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30. The Intensity of Absorption Bands…
•Numberofbonds:theintensityofanabsorptionbandalsodependson
thenumberofbondsresponsiblefortheabsorption.
•Forexample,theabsorptionbandfortheC–Hstretchwillbemore
intenseforacompoundsuchasoctyliodide,whichhas17C–H
bonds,thanformethyliodide,whichhasonlythreeC–Hbonds.
•Concentration:concentratedsampleshavegreaternumbersof
absorbingmoleculesand,therefore,moreintenseabsorptionbands.
•Intensitiesarereferredasstrong(s),medium(m),weak(w),broad,
andsharp.
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31. Infrared Instrumentation
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Sample
compartm
ent
IR
Source
Detecto
r
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32. Infrared Instrumentation
•Theinstrumentthatdeterminestheabsorptionspectrumforacompound
iscalledaninfraredspectrometeror,moreprecisely,a
spectrophotometer.
•Twotypesofinfraredspectrometersareincommonuse:dispersiveand
Fouriertransform(FT)instruments.
•Bothofthesetypesofinstrumentsprovidespectraofcompoundsinthe
commonrangeof4000to400cm-1.
•Insimpleterms,IRspectraareobtainedbydetectingchangesin
transmittance(orabsorption)intensityasafunctionoffrequency.
•Basic components include
–Radiation sources
–Monochromators/Interferometers/
–Sample compartment
–Detector
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33. Infrared Instrumentation
•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.
•Globarsource : is a silicon carbide rod (5mm diameter, 50mm long)
which is electrically heated to about 1500 oC.
•Nichromewire,: electrically heated to 1100 oC.It produces a lower
intensity of radiation than the Nernst or Globarsources, but has a longer
working life.
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34. Infrared Instrumentation
•Monochromators
–Themonochromatorisadeviceusedtodisperseabroadspectrumof
radiationandprovideacontinuousseriesofelectromagneticenergybandsof
determinablewavelengthorfrequencyrange.
–Prismsorgratingsarethedispersivecomponentsusedinconjunctionwith
variable-slitmechanismsandmirrors.
–UsedonlyindispersiveIRspectrometer.
•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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35. Infrared Instrumentation
•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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36. Infrared Instrumentation
•Interferometers…
–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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37. INSTRUMENTATION…
•Interferometers…
–Theinterferogramatimedomainspectrumandrecordsthedetector
responsechangesversustimewithinthemirrorscan.
–Ifthesamplehappenstoabsorbatthisfrequency,theamplitudeofthe
sinusoidalwaveisreducedbyanamountproportionaltotheamountof
sampleinthebeam.
–AmathematicaloperationknownasFouriertransformationconvertsthe
interferogram(atimedomainspectrumdisplayingintensityversustime
withinthemirrorscan)tothefinalIRspectrum,whichisthefamiliar
frequencydomainspectrumshowingintensityversusfrequency.
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38. School of Pharmacy, CHS, AAU
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39. School of Pharmacy, CHS, AAU
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40. Fourier Transform
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41. Infrared Instrumentation
•Detectors
–MostdetectorsusedindispersiveIRspectrometerscanbecategorizedinto
twoclasses:thermaldetectorsandphotondetectors.
–Thermaldetectorsincludethermocouples,thermistors,andpneumaticdevices
(Golaydetectors).
–Theymeasuretheheatingeffectproducedbyinfraredradiation.
–Avarietyofphysicalpropertychangesarequantitativelydetermined:
expansionofanonabsorbinggas(Golaydetector),electricalresistance
(thermistor),andvoltageatjunctionofdissimilarmetals(thermocouple).
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42. Infrared Instrumentation
•Detectors…
–PhotondetectorsrelyontheinteractionofIRradiationandasemiconductor
material.
•Non-conductingelectronsareexcitedtoaconductingstate.Thus,a
smallcurrentorvoltagecanbegenerated.
–Theresponsetimesofmanydetectors(forexample,thermocoupleand
thermistor)usedindispersiveIRinstrumentsaretooslowfortherapidscan
times(1secorless)oftheinterferometer.
–ThetwomostpopulardetectorsforaFTIRspectrometeraredeuterated
triglycinesulfate(DTGS)andmercurycadmiumtelluride(MCT).
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43. 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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44. Dispersive IR spectrometer
•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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45. Dispersive IR spectrometer
•Fouriertransform-IRspectrometer
–Itwasdevelopedinordertoovercomethelimitations
encounteredwithdispersiveinstruments.
•Themaindifficultywastheslowscanningprocess.
–Amethodformeasuringalloftheinfraredfrequencies
simultaneously,ratherthanindividually,wasneeded.
–Asolutionwasdevelopedwhichemployedaverysimpleoptical
devicecalledaninterferometer
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46. Dispersive IR spectrometer
•Fouriertransform-IRspectrometer(FTIR)
•TheIRradiationfromabroadbandsourceisfirstdirectedintoan
interferometer,
–whereitisdividedandthenrecombinedafterthesplitbeams
traveldifferentopticalpathstogenerateconstructiveand
destructiveinterference.
•Next,theresultingbeampassesthroughthesamplecompartmentand
reachestothedetector.
•MostFTIRspectrometersaresingle-beaminstruments
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47. 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
transformedinto a spectrum using
a Fourier Transform (FT).
FTIR
Comparison Between Dispersion Spectrometer and FTIR
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48. Dispersive IR spectrometer
•Advantages of FT-IR
–Better speed
•Acompletespectrumcanbeobtainedduringasinglescanofthe
movingmirror,whilethedetectorobservesallfrequencies
simultaneously.
•AninterferometerinanFT-IRinstrumentdoesnotseparateenergyinto
individualfrequenciesformeasurementoftheinfraredspectrum.
•Incontrast,everywavelengthacrossthespectrummustbemeasured
individuallyinadispersivespectrometer.
•Thisisaslowprocess,andtypicallyonlyonemeasurementscanofthe
sampleismadeinadispersiveinstrument
•FT-IRadvantageisthatmanyscanscanbecompletedandcombinedon
anFT-IRinashortertimethanonescanonadispersiveinstrument.
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49. Dispersive IR spectrometer
•Better sensitivity and Resolution
–AnFT-IRinstrumentdoesnotuseaslittolimittheindividualfrequency
reachingthesampleanddetectorasadispersiveinstrumentdoes.
–TherearealsofewermirrorsurfacesinanFT-IRspectrometer,sothereare
lessreflectionlossesthaninadispersivespectrometer.
–Overall,moreenergyreachesthesampleandhencethedetectorinanFT-IR
spectrometerthaninadispersivespectrometer.
–Highersensitivityandclearspectrum
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50. Dispersive IR spectrometer
–Simplermechanicaldesign
•Thereisonlyonemovingpart,themovingmirror,resultingin
betterreliability.
–Eliminationofstraylightandemissioncontributions
•TheinterferometerinFTIRmodulatesallthefrequencies.The
unmodulatedstraylightandsampleemissions(ifany)arenot
detected.
–Powerfuldatastation
•ModernFTIRspectrometersareusuallyequippedwithapowerful,
computerizeddatasystem.
•Itcanperformawidevarietyofdataprocessingtaskssuchas
Fouriertransformation,baselinecorrection,andlibrarysearching.
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51. 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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52. 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;
–Adropofaliquidorganiccompoundisplacedbetweenapairofpolished
sodiumchlorideorpotassiumbromideplates,referredtoassaltplates.
–Whentheplatesaresqueezedgently,athinliquidfilmformsbetweenthem.
–Aspectrumdeterminedbythismethodisreferredtoasaneatspectrum
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53. Preparation of Samples…
•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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54. Preparation of Samples…
•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.
–Underpressure,thepotassiumbromidemeltsandsealsthecompoundintoa
matrix.
–TheresultisaKBrpelletthatcanbeinsertedintoaholderinthe
spectrometer.
–Themaindisadvantageofthismethodisthatpotassiumbromideabsorbs
water,whichmayinterferewiththespectrumthatisobtained.
–Ifagoodpelletisprepared,thespectrumobtainedwillhavenointerfering
bandssincepotassiumbromideistransparentdownto400cm-1
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55. Preparation of Samples…
•Thesecondmethod,aNujolmull,involvesgrindingthecompoundwith
mineraloil(Nujol)tocreateasuspensionofthefinelygroundsample
dispersedinthemineraloil.
–Thethicksuspensionisplacedbetweensaltplates.
–Themaindisadvantageofthismethodisthatthemineraloilobscuresbands
thatmaybepresentintheanalyzedcompound.
–Nujolbandsappearat2924,1462,and1377cm-1.
•Thethirdcommonmethodusedwithsolidsistodissolvetheorganic
compoundinasolvent,mostcommonlycarbontetrachloride(CCl4).
–some regions of the spectrum are obscured by bands in the solvent.
–Althoughitispossibletocanceloutthesolventfromthespectrumby
computerorinstrumentaltechniques,theregionaround785cm-1isoften
obscuredbythestrongC-Clstretchthatoccursthere.
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2019/20 A.Y.
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56. Con’t…
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57. Preparation of Samples…
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58. 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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59. Sample cells…
•Gases-thegassamplecellissimilartothecellforliquidsamples.
•Gaseshavedensitieswhichareseveralordersofmagnitudelessthan
liquids,andhencepathlengthsmustbecorrespondinglygreater,usually
10cmorlonger.
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School of Pharmacy, CHS, AAU
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60. 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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61. Absorption Regions
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62. 1.Absorptionofaninfraredphotonresultsintheexcitationofthe
moleculetoahighervibrationalquantumstate.
2.Forastretchingvibration,aphotonisabsorbedandthemolecule
excitedtoahighervibrationalquantumstateonlyifthatvibration
resultsinachangeinbonddipole.
•Thebonddipoleisaproductofbondlengthandchargedifferenceof
thebondedatoms.
Basic Concept in IR
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63. Basic Concept in IR
•Asabondvibrates,thebondlengthchanges,sothiscriterionismet.
Thechargedifferenceisdeterminedbytheelectronegativityofthe
bondedatoms.
•Iftheseatomsarenotidenticalthentheyhaveadifferencein
electronegativityandthusadifferenceincharge.
•Iftheseatomsareidentical,theyhaveequalelectronegativityandthus
nochargedifference.
•Inthiscase,theproductofbondlengthchangeandchargedifference
(zero)iszero,sonophotonisabsorbed.Iftheatomsareevenslightly
different,thenasmallchangeinbonddipoleoccurs,alongwiththe
correspondingabsorptionofaninfraredphoton.
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64. 5.E.g.allbondstretchesofH
2C=OshowupontheIRspectrumbecause
allofthebondsconsistofunequalatoms.ForH
2C=CH
2,theC–Hbond
stretchesgiveIRbands,buttheC=Cbond(whichismadeupoftwo
identicalcarbonatoms)doesnot.TheC=CbondsofH
2C=CH–CH=CH
2
showupintheIRbecausethecarbonsarenotequal.Onecarbonisa
CH
2andtheotherisaHC–CH.
6.Insummary,asymmetricalbondsgiveIRabsorptionswhereas
symmetricalbondsdonot.
7.Theenergyofthephotonnecessarytoexcitethemoleculetoahigher
vibrationalquantumleveliscontrolledbythemassesoftheattached
atomsandthebondstrength.Similarfunctionalgroupsaremadeupof
similaratoms.Thusmoleculeswithsimilarfunctionalgroupsabsorb
photonsofsimilarenergies.ThisiswhywecanuseIRspectroscopyto
determinewhatfunctionalgroupsarepresentorabsentinacompound
bysimplylookingatthepositionsoftheIRpeaks.
Basic Concept in IR….
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65. 8.WedividetheIRspectrumintofivezones,basedupontheenergyof
photonsabsorbedbytwelvecommonfunctionalgroups.Thezonesand
functionalgroupsthereinaresummarizedbelow:
ZoneFiveAnalysis
Basic Concept in IR….
65
Zone1(3700–3200cm1)alcoholO–H,amide,oramineN–H,
terminalalkyne≡C–H
Zone2(3200–2700cm-1)alkylC–H,arylorvinylC–H,aldehydeC–
H,carboxylicacidO–H
Zone3(2300–2000cm-1)alkyneandnitriletriplebonds
Zone4(1850–1650cm-1)carbonylfunctionalgroups
Zone5(1680–1450cm-1)alkene,benzenering
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2019/20 A.Y.

66. Basic Concept in IR….
9.-Alcohol,Amine:theO–HandN–Hstretchesareusuallybroadeneddue
tohydrogenbonding.
-Carboxylicacid:musthavetwobands:broadO–Hstretchinzone2
andC=OstretchinZone4.
-Aldehyde:musthavetwosetsofbands:2900and2700cm
-1
inzone
2(2900cm
-1
oftenhiddenbyalkylC–Hstretches),andC=Ostretch
inzone4.
-Alkyne:apeakcausedbythetriplebondstretchisnotobserved
whenthealkyneissymmetricornearlyso.
-Carbonyl:oftenthemostintensepeakintheentirespectrum.
-Benzenering:musthavebandsatabout1600cm
-1
(maybetwo
bandsclosetogether)andanotheratabout1500cm
-1
.
•Sincemostbenzeneringcompoundshaveatleastonehydrogen
atomattachedtothebenzenering,ansp2C–Hstretchpeakisalso
seeninZone2
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67. 10.Theregionfromabout1450to400cm-1iscalledthefingerprint
region.
•Thisregionisoftencomplexbecauseitcontainsmanyabsorptions.
•ItisusuallyignoredinroutineIRanalysisbecauseitcanbedifficulttomake
preciseassignments.
•Forexample,apeakat1360cm
-1
couldbeaCH
2bendoraC–Nstretch.
•Withoutpreciseassignments,itcanbedifficulttodeterminetheabsenceor
presenceoffunctionalgroupsfromthefingerprintregion.
•However,thefingerprintregionsoftwoIRspectraofthesamecompoundare
identical,inthesamewaythattwofingerprintsleftbythesamepersonin
differentplacesareidentical.
•Thusthefingerprintregionofanunknowncompoundcanbecomparedwith
thefingerprintregionofaknownsample,andiftheymatchexactly,the
unknownandknownareidentical
Basic Concept in IR….
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68. 68
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69. 69
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70. How to Approach the Analysis of a Spectrum
1.Isacarbonylgrouppresent?
–TheC=Ogroupgivesrisetoastrongabsorptionintheregion
at1870to1550cm–1.Thepeakisoftenthestrongestinthe
spectrumandofmediumwidth.(Youcan'tmissit)
•2.IfC=Oispresent,checkthefollowingtypes(ifabsent,goto
3).
–ACIDS-presenceofO-Hstretching
–broadabsorptionnear3400-2400cm–1
–AMIDES-presenceofN-Hstretching–mediumabsorptionnear
3500cm–1
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71. How to Approach the Analysis of a Spectrum
•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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72. How to Approach the Analysis of a Spectrum
•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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73. How to Approach the Analysis of a Spectrum
•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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74. How to Approach the Analysis of a Spectrum
•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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75. Thespectrumoftolueneisshownbelow.
Notethe=C–Hstretchesofaromatics(3099,3068,3032)and
the–C–Hstretchesofthealkyl(methyl)group(2925istheonlyone
marked).
Thecharacteristicovertonesareseenfromabout2000-1665.
Alsonotethecarbon-carbonstretchesinthearomaticring(1614,1506,
1465),thein-planeC–Hbending(1086,1035),andtheC–Hoop(738).
IR spectrum of tolune School of Pharmacy, CHS, AAU
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76. PracticeProblems
1.Severalfunctionalgroupshaveabsorptionsinmorethanoneofthe
fivezones.Toconcludethatthefunctionalgroupispresent,both
absorptionsmustbeseen.Prepareatablelistingthesefunctional
groupsandthecorrespondingabsorptions.
Hint:Thereareatleastsixsuchfunctionalgroups.Someofthemrequire
threedistinctabsorptions.
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77. FunctionalGroupAbsorptions
terminalalkyne3300cm-1(C–Hstretch) -z1
2260–2000cm-1(C≡Cstretch) -z3
aromaticring3100–3000cm-1(sp2C–Hstretch) -z2
~1600cm-1and1500–1450cm-1(benzeneringstretch)-z5
aldehyde~2900and~2700cm-1(C–Hstretch) -z2
1740–1720cm-1(C=Ostretch) -z4
carboxylicacid3000–2500cm-1(O–Hstretch) -z2
1725–1700cm-1(C=Ostretch) -z4
arylketone1700–1680cm-1(C=Ostretch) -z4
~1600cm-1and1500–1450cm-1(benzeneringstretch)-z5
enone1685–1665cm-1(C=Ostretch) -z4
1680–1620cm-1(C=Cstretch) -z5
Thereareafewuncommonexceptionstothistable.Forexample,acetyleneisa
terminalalkynethatdisplaysazone1≡C–Hstretch,butnotazone3C≡C
stretch,duetothesymmetryofthetriplebond.Abenzeneringinwhichallsix
ringhydrogenshavebeensubstituted(e.g.,hexmethylbenzene)haszone5peaks
butnotzone2sp2C–Hstretches.Atertiaryamide(onewithoutanN–Hbond)
hasaC=Ostretchinzone4butnoN–Hstretchinzone1.
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78. 2. What is the effect of conjugation on the energy of an infrared absorption
band?
Conjugationshiftsabandtolowerenergy.
Forexample,aketonetypicallyabsorbsintheregionof1750–1705cm-1,
whereasanenone(alkene-carbonylconjugationO=C–C=C;alsocalledan
α,β-unsaturatedketone)absorbsat1685–1665cm
-1
,andanarylketone
(ketoneconjugatedwithabenzenering)absorbsat1700–1680cm
-1
.This
shifttoalowerenergyisseenwhenanyfunctionalgroupisconjugated.
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79. 79
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80. 80
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81. 81
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82. An Alkane IR Spectrum (n-hexane)
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83. An Alkene IR Spectrum (1-hexene)
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84. Alkyne---
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85. A Ketone IR Spectrum
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86. n-Hexane: CH
3CH
2CH
2CH
2CH
2CH
3
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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87. Exercise 2 Identify the functional group that present in the molecule
and predict the possible structure of the IR spectrum given below
1-Hexene: CH
2=CHCH
2CH
2CH
2CH
3
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School of Pharmacy, CHS, AAU
2019/20 A.Y.

88. 3-heptanone: CH
2CH
2
H
3C
O
CH
2
CH
2
CH
3
Exercise 3 Identify the functional group that present in the molecule
and predict the possible structure of the IR spectrum given below
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89. Exercise 4 Identify the functional group that present in the molecule
and predict the possible structure of the IR spectrum given below
1-Hexanol: CH
3CH
2CH
2CH
2CH
2CH
2OH
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90. Summary of IR Absorptions
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91. 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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92. School of Pharmacy, CHS, AAU
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93. School of Pharmacy, CHS, AAU
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94. School of Pharmacy, CHS, AAU
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95. 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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2019/20 A.Y.
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96. 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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97. Thank You!
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