infrared spectroscopy. spectroscopy is an instrumentally aided studies of the interactions between...
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Infrared SpectroscopyInfrared Spectroscopy
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• Spectroscopy is an instrumentally aided studies of the interactions between matter (sample being analyzed) and energy (any portion of the electromagnetic spectrum, EMS)
• Chemists can use portions of the EMS to selectively manipulate the energies contained within a molecule, to uncover detailed evidence of its chemical structure and bonding.
• EMS 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.
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=>
EMSEMS and Molecular Effects and Molecular Effects
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• Energy (E) E = h = hc/ – where h is Planck’s constant, c is the speed of
light, is frequency or the number of vibrations per secondand is the wavelength
• Wavenumber (’) ’ = 1/ – given in cm-1
• Period (P) P = 1/– the time between a vibration
= hc’
Energy, frequency, and wavenumber are directly proportional to each other.
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Method Abbrev.
Energy used
Units
Ultraviolet-Visible Spectroscopy
UV-Vis ultraviolet-visible
nm
Infrared Spectroscopy
IR infrared m or cm-1
Nuclear Magnetic Resonance
NMR radio frequencie
s
Hz
Mass Spectroscopy MS electron volts
amu
The four most common spectroscopic methods used in organic analysis are:
What actually happens to the sample during an analysis?{How do the sample and energy “interact”?}
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• What happens when a sample absorbs UV/Vis energy?
excitation of ground state electrons(typically p and n electrons)
Eelectronic increases momentarily
• What happens when a sample absorbs IR energy?
stretching and bending of bonds (typically covalent bonds)
Evibration increases momentarily
UV/Vis
*
sample *
transition
IR-O-H -O
(3500 cm-1)—H
(200 nm)
Matter/Energy Interactions
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• Infrared spectroscopy (IR) measures the bond vibration frequencies in a molecule and is used to determine the functional group
• Mass spectrometry (MS) fragments the molecule and measures the masses
• Nuclear magnetic resonance (NMR) spectroscopy detects signals from hydrogen atoms and can be used to distinguish isomers
• Ultraviolet (UV) spectroscopy uses electron transitions to determine bonding patterns
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• Just below red in the visible region
• Wavelengths usually 2.5-25 m
• More common units are wavenumbers, or cm-1, the reciprocal of the wavelength in centimeters (104/m = 4000-400 cm-1)
• Wavenumbers are proportional to frequency and energy
The IR Region
• The IR region is divided into three regions: the near, mid, and far IR. The mid IR region is of greatest practical use to the organic chemist.
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Molecular Vibrations and IR Spectroscopy
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Molecules are made up of atoms linked by chemical Molecules are made up of atoms linked by chemical bonds. The movement of atoms and chemical bonds bonds. The movement of atoms and chemical bonds like spring and balls (vibration)like spring and balls (vibration)
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What is a vibration in a molecule?
Any change in shape of the molecule- stretching of bonds, bending of bonds, or internal rotation around single bonds
Vibrations
There are two main vibrational modes :1. Stretching - change in bond length
(higher frequency)
Stretching vibration
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Stretching Types
Symmetric Asymmetric
Bending Types
In-plane (Scissoring)
2. Bending - change in bond angle (lower frequency)
Out-plane (Twisting)
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Modes of vibrationsModes of vibrations
Stretching: change in bond distance. Stretching: change in bond distance. Occurs at higher energy: 4000-1250 Occurs at higher energy: 4000-1250 cmcm11..
-CH2-
H2O
Bending: change in bond angle.Bending: change in bond angle. Occurs at lower energy: 1400-666 cmOccurs at lower energy: 1400-666 cm11..
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More complex types of stretching and bending are possible
Can a vibration change the dipole moment of a molecule?
Infrared active vibrations (those that absorb IR radiation) must result in a change of dipole moment
Asymmetrical stretching/bending and internal rotation change the dipole moment of a molecule. Asymmetrical stretching/bending are IR active.
Symmetrical stretching/bending does not. Not IR active
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Fundamental Vibrations (Absorption Frequencies)A molecule has as many as degrees of freedom as the total degree of freedom of its individual atoms.• Each atom has 3 degree of freedom (x,y,z)• A molecule of n atoms therefore has 3n degrees of freedom. • Non linear molecules (e.g. H2O) Vibrational degrees of freedom or Fundamental Vibrations = 3n – 6
HO
H HO
H HO
H
SymmetricalStretching (υs OH)
3652 cm-1
AsymmetricalStretching (υas OH)
3756 cm-1
Scissoring (δs HOH)1596 cm-1
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• For linear molecule (e.g. CO2) :Vibrational degrees of freedom or Fundamental
Vibrations = 3n – 5
O C O O C O
SymmetricalStretching (υs CO2)
1340 cm-1
AsymmetricalStretching (υas CO2)
2350 cm-1
O C O O C O
Scissoring (bending outof the plane of the paper)
(δs CO2) 666 cm-1
Scissoring (bending in the plane of the paper)
(δs CO2) 666 cm-1
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O=C=O
H-O-HH-O-H
“BACKGROUND”
Carbon Dioxide and WaterCarbon Dioxide and Water
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The theoretical no. of fundamental vibrations will seldom be observed because overtones (multiples of a given frequencies) and combination tones (sum of two other vibrations) increase the no. of bands.
Other phenomena reduce the no. of bands including: 1. Fundamental frequencies that fall outside the 4000-400 cm-1 region.
2. Fundamental bands that are too weak to be observed.
3. Fundamental bands that are so close that they coalesce.
4. The occurrence of a degenerate band from several absorptions of the same frequency in highly symmetrical molecules.
5. The failure of certain fundamental vibrations to appear in the IR because of the lack of change in molecular dipole.
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Which of the following atoms or molecules will absorb IR radiation:
I—Cl H2 N2 Cl2
Why?
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IR-Active and InactiveIR-Active and Inactive
• A polar bond is usually IR-active.• A nonpolar bond in a symmetrical
molecule will absorb weakly or not at all.
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How does the mass influence the vibration?H2 I2
MM =2 g/mole MM =254 g/mole
The greater the mass - the lower the wavenumber (ύ)
For a vibration at 4111 cm-1 (the stretch in H2), how many vibrations occur in a second?
120 x 1012 vibrations/sec or a vibration every 8 x 10-15 seconds!
120 trillion vibration per second!!!!
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Hooke’s Law
ύ = The vibration frequency (cm-1)c = Velocity of light (cm/s)f = force constant of bond (dyne/cm)
M1 and M2 are mass (g) of atom M1 and M2
M1 M2
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The relative contributions of bond strength is also considered in vibrational frequencies.
In general functional groups that have a strong dipole give rise to strong absorptions in the IR.
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=1
2c K
=m1 m2
m1 + m2
= frequencyin cm-1
c = velocity of light
K = force constant in dynes/cm
m = atomic masses
SIMPLE HARMONIC OSCILLATOR
C CC CC C > >multiple bonds have higher K’s
=reduced mass
( 3 x 1010 cm/sec )
THE EQUATION OF A
This equation describes the vibrations of a bond.
where
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=1
2c K
larger K,higher frequency
larger atom masses,lower frequency
constants
2150 1650 1200
C=C > C=C > C-C=
C-H > C-C > C-O > C-Cl > C-Br3000 1200 1100 750 650
increasing K
increasing
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• Frequency decreases with increasing atomic mass.• Frequency increases with increasing bond energy.
Stretching FrequenciesStretching Frequencies
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wave number = ύ1
ύ= ------- * ((k(m1 + m2))/(m1 m2))1/2
2c 1ύ = ---------------------- cm-1
2(3.14)(3.0E10)* ((5.0E5(12+16)(6.02E23))/(12*16))1/2
EXAMPLE: Calculate the fundamental frequency expected in the infrared absorption spectrum for the C - O stretching frequency. The value of the force constant is 5.0 X 105 dynes/cm.
ύ = 1112 cm-1
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150020002500300035004000
0%50
%10
0%
Tra
nsm
issi
on
Wavenumber (cm-1)
Vibrational Spectroscopy
4500
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IR Correlation DiagramTra
nsm
itta
nce
(%
) 100
80
60
40
20
04000 3500 3000 2500 2000 1500 1000
2.5 3.0 4.0 5.0 6.0 10.0
Frequency (cm-1)
Region I3600-2700 cm-1
Region II1800-1600 cm-1
/ Wavelength (microns, m)
O-H N-H C-Hbond stretching
alcohols phenols carboxylic acids
aminesamides
alkynes alkenes alkanes
C=O
acid chloridesanhydrides
estersketones
aldehydescarboxylic acids
amides
FingerprintRegion
(below 1500 cm-1)
C-H=C-H -C-H
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InstrumentationInstrumentation
schematic diagram of a double beam double-grating infrared spectrophotometer
1. Radiation source 2. Monochromator3. Solvents, sample cells, samples 4. Readout / Recorder
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Solvents1. Must be transparent in the region studied: no single solvent is
transparent throughout the entire IR region
2. Water and alcohols are seldom employed to avoid O-H band of
water .
3. Must be chemically inert (does not react with substance or cell
holder).
CCl4, CS2, or CHCl3; may be used but we should consider its IR
spectrum
SOLVENTS, CELLS, SAMPLES
Cells- NaCl or KCl cells may be used (moisture from air and sample
should be avoided: even with care, their surfaces eventually
become fogged due to absorption of moisture)
- Very thin (path length = 0.1 to 1.0 mm)
- Sample concentration = about 0.1 – 10%
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Samples
1. Solid KBr disk (1 mg solid sample + 100 mg KBr pressed into a
disk)
Mull: 1 mg solid sample suspended in Nujol (heavy liquid
hydrocarbon)
2. Liquid Neat (thin film of liquid between two NaCl plates
solution in CCl4 and put in special NaCl cells.
3. Gas IR spectrum is obtained directly by permitting the sample
to expand into an evacuated special cells.
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Interferogram The interferogram at the right
displays the interference pattern and contains all of the spectrum information.
A Fourier transform converts the time domain to the frequency domain with absorption as a function of frequency.
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• Identification of functional groups on a molecule – this is a very important tool in organic chemistry
• Spectral matching can be done by computer software and library spectra
• Since absorbance follows Beer’s Law, can do quantitative analysis
Use of IR spectraUse of IR spectra
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• An IR spectrum is a plot of per cent transmittance (or absorbance) against wavenumber (frequency or wavelength). A typical infrared spectrum is shown below.
• A 100 per cent transmittance in the spectrum implies no absorption of IR radiation. When a compound absorbs IR radiation, the intensity of transmitted radiation decreases. This results in a decrease of per cent transmittance and hence a dip in the spectrum. The dip is often called an absorption peak or absorption band.
FEATURES OF AN IR SPECTRUMFEATURES OF AN IR SPECTRUM
• Different types of groups of atoms (C-H, O-H, N-H, etc…) absorb infrared radiation at different
characteristic wavenumbers.
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• No two molecules will give exactly the same IR spectrum (except enantiomers)
• Simple stretching: 1600-3500 cm-1
• Complex vibrations: 400-1400 cm-1, called the “fingerprint region”
Baseline
Absorbance/Peak
IR SpectrumIR Spectrum
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Infrared Spectrum of Propanal (CH3CH2CHO)
4600 3800 3000 2200 1800 1400
0
10
20
30
40
50
60
70
80
90
100
1000 800 600 400
C =O
Per
cent
tran
smitt
ance
W avenum ber (cm )-1
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Describing IR AbsorptionsIR absorptions are described by their frequency
and appearance.• Frequency () is given in wavenumbers (cm-1)• Appearance is qualitative: intensity and
shape• conventional abbreviations:
vs very strong
s strongm medium w weakbr broadsh sharp OR shoulder
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• In general, the IR spectrum can be split into four regions for interpretation:
• 4000 2500 cm-1: Absorption of single bonds formed by hydrogen and other elements e.g. OH, NH, CH
• 2500 2000 cm-1: Absorption of triple bonds e.g. C≡C, C≡N
• 2000 1500 cm-1: Absorption of double bonds e.g. C=C, C=O
• 1500 400 cm-1: This region often consists of many different, complicated bands. This part of the spectrum is unique to each compound and is often called the fingerprint region. It is rarely used for identification of particular functional groups.
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Summary of IR AbsorptionsSummary of IR Absorptions
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BASE VALUESBASE VALUES(+/- 10 cm-1)
These arethe minimumnumber ofvalues tomemorize.
O-H 3600N-H 3400C-H 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C O ~1100 large range
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O-H STRETCHO-H STRETCH
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Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
O-H2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550
650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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• O-H 3600 cm-1 (alcohol, free)• O-H 3300 cm-1 (alcohols & acids, H-bonding)
3600 3300
H-BONDEDFREE
broadens
shifts
The O-H stretching region
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OH
R
O
H
R
R OH
R
OH
HR O
R HO
HYDROGEN-BONDED HYDROXYLHYDROGEN-BONDED HYDROXYL
Many kinds of OHbonds of differentlengths and strengths This leads to a broad absorption.
Longer bonds are weaker and lead to lower frequency.
Hydrogen bonding occurs in concentrated solutions ( for instance, undiluted alcohol ).
“Neat” solution.
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““FREE” HYDROXYLFREE” HYDROXYL
R OH
CCl4
CCl4
CCl4CCl4
CCl4
Distinct bond has a well-defined length and strength.
Occurs in dilute solutions of alcohol in an “inert” solvent like CCl4.
Solvent molecules surround but do nothydrogen bond.
The “free” hydroxyl vibrates without interference from any other molecule.
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CyclohexanolCyclohexanol
OH O-HH-bond
C-H
C-O
CH2
ALCOHOL
neat solution
,
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Butanoic AcidButanoic Acid
CH3 CH2 CH2 C OH
O
O-HH-bond
C-H C=O
CH2
C-O
CARBOXYLIC ACID
neat solution
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C
O
OHRC
O
O HR
CARBOXYLIC ACID DIMERCARBOXYLIC ACID DIMER
Strong hydrogen bonding in the dimer weakens the OHbond and leads to a broad peak at lower frequency.
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N-H STRETCHN-H STRETCH
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Typical Infrared Absorption RegionsTypical Infrared Absorption Regions
N-H2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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The N-H stretching region
• Primary amines give two peaks
• Secondary amines give one peak• Tertiary amines give no peak
NH
HN
H
Hsymmetric asymmetric
N-H 3300 - 3400 cm-1
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CH3 CH2 CH2 CH2 NH2
NH2
NH2
scissor
CH2
CH3
PRIMARY AMINEaliphatic
1-Butanamine1-Butanamine
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NH2
CH3
NH2
Ar-H
-CH3
benzeneAr-H
PRIMARY AMINEaromatic
3-Methylbenzenamine3-Methylbenzenamine
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NH CH2 CH3
NH
benzeneAr-H
CH3
SECONDARY AMINE
N N -Ethylbenzenamine-Ethylbenzenamine
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N
CH3
CH3
no N-H
benzene
CH3
Ar-H
Ar-H
-CH3
TERTIARY AMINE
N,N N,N -Dimethylaniline-Dimethylaniline
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C-H STRETCHC-H STRETCH
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Typical Infrared Absorption RegionsTypical Infrared Absorption Regions
C-H2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
We will look at this area first
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•C-H aldehyde, two peaks (both weak) ~ 2850 and 2750 cm-1
3000 divides
UNSATURATED
SATURATED
•C-H sp stretch ~ 3300 cm-1
•C-H sp2 stretch > 3000 cm-1
•C-H sp3 stretch < 3000 cm-1
The C-H stretching regionBASE VALUE = 3000 cm-1
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3000
-C-H=C-H
31003300
=C-H=
2900 2850 2750
-CH=O(weak)
increasing CH Bond Strength
sp3-1ssp2-1ssp-1s
increasing frequency (cm-1)
aldehyde
increasing s character in bond
increasing force constant K
STRONGER BONDS HAVE LARGER FORCE CONSTANTSAND ABSORB AT HIGHER FREQUENCIES
CH BASE VALUE = 3000 cm-1
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HexaneHexane
CH3 CH2 CH2 CH2 CH2 CH3
CHstretchingvibrations
ALKANE
includesCH3 sym and asym
CH2 sym and asym
CH bending vibrationsdiscussed shortly
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C-H BENDINGC-H BENDING
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• CH2 bending ~ 1465 cm-1
• CH3 bending (asym) appears near
the CH2 value ~ 1460 cm-1
• CH3 bending (sym) ~ 1375 cm-1
THE C-H BENDING REGIONTHE C-H BENDING REGION
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C
H
H
C
H
HC
H
H
C
H
H
C
HH
C
HH
Scissoring Wagging
Rocking Twisting
BendingVibrations
~1465 cm-1
~720 cm-1
~1250 cm-1
~1250 cm-1
in-plane out-of-plane
METHYLENE GROUP BENDING VIBRATIONSMETHYLENE GROUP BENDING VIBRATIONS
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CH3CH2
1465 1460 1375
asym sym
METHYLENE AND METHYL BENDING VIBRATIONSMETHYLENE AND METHYL BENDING VIBRATIONS
these two peaks frequently overlap
and are not resolved
C-H Bending, look near 1465 and 1375 cm-1
CH
H
H
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CH3CH2
1465 1460 1375
asym sym
13701380
13701390
CCH3
CH3
C CH3
CH3
CH3
C CH3
METHYLENE AND METHYL BENDING VIBRATIONSMETHYLENE AND METHYL BENDING VIBRATIONS
geminal dimethyl
t-butyl
(isopropyl)two peaks
two peaks
The sym methyl peaksplits when you havemore than one CH3 attached to a carbon.
ADDITIONAL DETAILS FOR SYM CH3
one peak
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HexaneHexane
CH3 CH2 CH2 CH2 CH2 CH3
CHstretch
CH2 bend
CH3
bend
CH2
rocking
ALKANE
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1-Hexene1-Hexene
CH2 CH CH2 CH2 CH2 CH3
=CH
CH
C=C CH2 CH3
bend
CH
ALKENE
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TolueneToluene
CH3 C=Cbenzene
CH3
Ar-H
Ar-H
AROMATIC
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1-Hexyne1-Hexyne
CH C CH2 CH2 CH2 CH3
C=C=
=C-H= C-H
CH2, CH3
ALKYNE
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C10H22
C12H26
SimilarBut NotIdentical
Fingerprinting
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C N AND C C STRETCH
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Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
C=N
C=C
=
=2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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The triple bond stretching region
• C N 2250 cm-1 • C C 2150 cm-1=
==
=
The cyano group often gives a strong, sharp peak due to its large dipole moment.
The carbon-carbon triple bond gives a sharp peak, but it is often weak due to a lack of a dipole. This isespecially true if it is at the center of a symmetricmolecule.
R C C R
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PropanenitrilePropanenitrile
CH3 CH2 C N
C=N=
NITRILE
BASE = 2250
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1-Hexyne1-Hexyne
CH C CH2 CH2 CH2 CH3
C=C==C-H=
ALKYNE
BASE = 2150
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C=O STRETCHINGC=O STRETCHING
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Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
C=O
2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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This region stretches from about 1800 to 1650 cm-1 - RIGHT IN THE MIDDLE OF THE SPECTRUM
The base value is 1715 cm-1 (ketone)
The bands are very strong !!! due to the large C=O dipole moment.
C=O is often one of the strongest peaks in the spectrum
THE CARBONYL STRETCHING REGIONTHE CARBONYL STRETCHING REGION
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2-Butanone2-Butanone
CH3 C CH2 CH3
O
KETONE
C=O
C-H
overtone2x C=O
CH bend
BASE = 1715
1715
C=O
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CR
O
H
CR
O
O C R
O
CR
O
ClCR
O
OR'CR
O
RCR
O
NH2CR
O
OH
169017101715172517351800
1810 and 1760
BASEVALUE
acid chloride ester aldehyde
carboxylicacid amideketone
anhydride
( two peaks )
EACH DIFFERENT KIND OF C=O COMES AT A DIFFERENT FREQUENCY
C=O IS SENSITIVE TO ITS ENVIRONMENTC=O IS SENSITIVE TO ITS ENVIRONMENT
THESE VALUES ARE WORTH LEARNING all are +/- 10 cm-1
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1.225 A 1.231 A 1.235 A 1.248 A
acidchloride
ester ketone amide
C=O BOND LENGTHS IN CARBONYL COMPOUNDSC=O BOND LENGTHS IN CARBONYL COMPOUNDS
shorter longer
1780 cm-1 1735 cm-1 1715 cm-1 1680 cm-1
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Ketones are at lower frequency than Aldehydes because ofthe second electron-donating alkyl group.
Acid chlorides are at higher frequency than ketones because of the electron-withdrawing halide.
Esters are at higher frequencies than ketones due to the electron-withdrawing oxygen atom. This is more important than resonance with the electron pair on the oxygen.
Amides are at lower frequencies than ketones due to resonance involving the unshared pair on nitrogen. The electron-withdrawing effect of nitrogen is less important than the resonance.
SUMMARYSUMMARY
Note the electronegativity difference, O versus N, weights thetwo factors (resonance/ e-withdrawal) differently in esters than in amides.
Acids are at lower frequency than ketones due to H-bonding.
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CH3 C CH2 CH3
O
KETONE
C=O
C-H
overtone
CH bend
BASE = 1715
2-Butanone2-Butanone1719 x 2 = 3438overtone of strong C=O peak
3438
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CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C H
O
ALDEHYDE
C=O
CHO
CH bend
BASE = 1725
NonanalNonanal
3460
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CCl
O
CH3 (CH2)10
ACID CHLORIDE
C=OC-H
CH bend
BASE = 1800Dodecanoyl ChlorideDodecanoyl Chloride
3608
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CO
O
CH2 CH2 CH3CH2CH3
ESTER
C=O
C-OC-H
BASE = 1735
Ethyl ButanoateEthyl Butanoate
3482
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COH
O
CHCH3
CH3
CARBOXYLIC ACID
C=O
O-H
C-HC-O
BASE = 1710
2-Methylpropanoic Acid2-Methylpropanoic Acid
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C
O
OHRC
O
O HR
CARBOXYLIC ACID DIMERCARBOXYLIC ACID DIMER
Strong hydrogen-bonding in the dimer weakens the O-H andC=O bonds and leads to broad peaks at lower frequencies.
lowersfrequencyof C=O
and also of O-H
RECALL
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CNH2
O
CH2CH3
AMIDE
C=O
NH2
C-H
CH bend
BASE = 1690two peakssym / asymPropanamidePropanamide
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C=C STRETCHING
ALKENESAROMATICS
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Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
C=C2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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The C=C stretching region
• C=C double bond at 1650 cm-1 is often weak or not even seen.
• C=C benzene ring shows peak(s) near 1600 and 1400 cm-
1 , one or two at each value - CONJUGATION LOWERS THE VALUE.
• When C=C is conjugated with C=O it is stronger and comes at a lower frequency.
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1-Hexene1-Hexene
CH2 CH CH2 CH2 CH2 CH3
ALKENE
oops
C=C=C-H
C-Haliphatic
C-Hbend
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TolueneToluene
CH3
AROMATIC
benzene
oops
benzeneC-H
C=C
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Typical Infrared Typical Infrared AbsorptionAbsorption
RegionsRegionsC-O
2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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C-O STRETCHING
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The C-O stretching region
• The C-O band appears in the range of 1300 to 1000 cm-1
• Look for one or more strong bands appearing in this range!
• Ethers, alcohols, esters and carboxylic acids have C-O bands
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CH3 CH2 CH2 CH2O CH2 CH2 CH2 CH3
ETHER
C-O
BASE = 1100
CH2 CH3
bendingC-H
Dibutyl EtherDibutyl Ether
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O CH3
AROMATIC ETHER
benzene oops
C-O
C-Haromatic
BASE = 1100
AnisoleAnisole
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OH
ALCOHOL
BASE = 3600BASE = 1100
OH
C-O
CH2
bend
C-H
CyclohexanolCyclohexanol
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COH
O
CHCH3
CH3
CARBOXYLIC ACID
OH
CH C=OC-O
2-Methylpropanoic Acid2-Methylpropanoic Acid
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CO
O
CH2 CH2 CH3CH2CH3
ESTER
CH
C=O
C-O
Ethyl ButanoateEthyl Butanoate
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N=O STRETCHING
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Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
N-O2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 18001650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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The N=O stretching region
• N=O stretching -- 1550 and 1350 cm-1
asymmetric and symmetric stretchings
• Often the 1550 cm-1 peak is stronger than the other one
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CH3
CHCH3
NO2
NITROALKANENITROALKANE
N=O
N=O
C-H
gem-dimethyl
2-Nitropropane2-Nitropropane
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Typical Infrared Typical Infrared AbsorptionAbsorption
RegionsRegionsC-Cl
2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O
C=NVery
fewbands
C=C
C-ClC-O
C-NC-CX=C=
Y(C,O,N,S)
C N
C C
N=O N=O*
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The C-X stretching regionThe C-X stretching region
• C-Cl 785 to 540 cm-1, often hard to find amongst the fingerprint bands!!
• C-Br and C-I appear outside the useful range of infrared
spectroscopy.
• C-F bonds can be found easily, but are not that common.
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CCl Cl
Cl
Cl
Often used as a solvent for IR spectra.When it is used, spectra show C-Cl absorptions.
C-Cl
Carbon TetrachlorideCarbon Tetrachloride
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Cl
C-Cl
oops
benzeneC=C
benzene ringcombinationbands
ChlorobenzeneChlorobenzene
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=C-H OUT OF PLANE BENDING
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H
H
H
OUT-OF-PLANE BENDINGOUT-OF-PLANE BENDING
above
below
PLANE
(OOPS)
H
ALKENES
BENZENES
also with
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Monosubstituted
cis-1,2-
trans-1,2-
1,1-
Trisubstituted
Tetrasubstituted
C CH
R H
H
C CH
R R
H
C CR
H R
H
C CR
R R
R
C CR
R R
H
C CR
R H
H
Disubstituted
10 11 12 13 14 15
1000 900 800 700 cm-1
s s
m
s
s
s
=C-H OUT OF PLANE BENDING
ALKENES
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10 11 12 13 14 15
1000 900 800 700 cm-1
Monosubstituted
Disubstituted
ortho
meta
para
Trisubstituted
1,2,4
1,2,3
1,3,5
BENZENES
s s
s
s s
s
s
s
s
m
m
m
m
OOPSRING H’s
combination bands