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Physical Properties
Melting Point
Boiling Point
Density
Solubility
Refractive Index
Chemical Tests
Hydrocarbons
Alkanes
Alkenes
Alkynes
Halides
Alcohols
Aldehydes
Ketones
Spectroscopy
Mass
(Molecular Weight)
Ultraviolet/Visual
(Conjugation, Carbonyl)
Infrared
Functional Groups
NMR
(Number, Type, Location of protons)
Gas Chromatography
(Identity, Mole %)
SpectroscopyBuilding A Toolset
ForThe Identification of Organic Compounds
04/21/23 1
Spectroscopy
04/21/23 2
Spectroscopy Tools
Spectroscopy Spectroscopy
The Absorption of Electromagnetic Radiation and the use of the Resulting Absorption Spectra to Study the Structure of Organic Molecules.
When continuous radiation passes through a transparent material, some of the radiation can be absorbed.
If the portion that is not absorbed is passed through a prism, a spectrum with gaps is produced.
This is called an:
ABSORPTION SPECTRUM04/21/23 3
Spectroscopy Energy States
Energy absorption by transparent materials in any portion of the electromagnetic spectrum causes atoms or molecules to pass from a state of low energy (ground state) to a state of higher energy (excited state).
There are 3 types of Energy States:
Electronic
Vibrational
Spin04/21/23 4
Spectroscopy Electromagnetic Spectrum
Cosmic (Gamma) X-Ray
Ultraviolet Visible Infrared
Microwave Radio
Energy States and the Electromagnetic Spectrum
Electronic – Ultraviolet
Vibrational – Infrared
Spin – Radio
04/21/23 5
MicrowaveInfraredX-RayVacuum
UV
VisibleNear Ultraviolet
VibrationalInfrared
NuclearMagnetic
Resonance
Radio Frequency
400 nm200 nm 800 nm 2.5 15
1 m 5 m
Blue Red
Cosmic&
Ray
0.01 nm
3 x 1019 Hz 3 x 1016 Hz 2 x 1013 Hz
10 nm 30 cm
1 x109cm-1
0.002 cm-1
10 cm-1 3 cm-1 0.01 cm-1
1 mm
Frequency ()
Energy (E)
High
High Low
Low
Wavelength ()Short Long
1 x107cm-1 5 x104cm-1
2.5 x104cm-1
1.25 x104cm-1
667cm-1
4 x103cm-1
6 x 107 Hz
3 x 108 Hz
1.5 x 1015 Hz 1 x 109 Hz3 x 1011 Hz
1.2 x 1014 HzFrequency
Wave Number
Wavelength
Spectroscopy
04/21/23 6
E = hc /
E = [E (excited) – E (ground)] = h
= Frequency (Hz) c = Velocity of Light (cm/sec) = Wavelength (cm) h = Planck’s Constant
= c /
Where:
SpectroscopyQuantization
The excitation process is quantized, in which only selected frequencies of energy are absorbed representing the energy difference (E) between the excited and ground states.
04/21/23 7
SpectroscopySpectroscopy Types:
Mass Spectrometry (MS) – Hi-Energy Electron Bombardment
Use – Molecular Weight, Presence of Nitrogen, Halogens
Ultraviolet Spectroscopy (UV) – Electronic Energy States
Use –Conjugated Molecules; Carbonyl Group, Nitro Group
Infrared Spectroscopy (IR) – Vibrational Energy States
Use – Functional Groups; Compound Structure
Nuclear Magnetic Resonance (NMR) – Nuclear Spin States
Use – The number, type, and relative position of protons (Hydrogen nuclei) and Carbon-13 nuclei
04/21/23 8
Mass Spectroscopy High energy electrons bombard organic molecules breaking
some or all of the original molecules into fragments.
The process usually removes a single electron to produce a positive ion (cation radical) that can be separated in a magnetic field on the basis of the mass / charge ratio.
Removal of the single electron produces a charge of +1 for the cation.
Thus, the cation represents the Molecular Weight of the original compound or any of the fragments that are produced.
The mass spectrum produced is a plot of relative abundance of the various fragments (positively charged cation radicals) versus the Mass / Charge (M/Z) ratio.
The most intense peak is called the “Base Peak”, which is arbitrarily set to 100% abundance; all other peaks are reported as percentages of abundance of “Base Peak.”
04/21/23 9
M + e- M+ + 2e-
Molecule High EnergyElectron
Molecular Ion(Radical Cation)
TypicalMass
Spectrum
Molecular Ion Peak (M+ 88)
M - H2O
M - (H2O and CH3)
M - (H2O and CH2 – CH2)
Base Peak
CH2OH
1-Pentanol - MW 88
CH3(CH2)3 – CH2OH
Mass Spectroscopy
04/21/23 10
Mass Spectroscopy Molecular Ion Peak (M+)
Largest mass/charge ratio Always the last peak on the right side of
spectrum May or may not be the base peak (usually
not)! Abundance can be quite small, i.e., very
small peaks The Molecular Ion Peak represents the
Molecular Weight of the Compound
04/21/23 11
Mass Spectroscopy
04/21/23 12
Methyl Propyl Ketone (C5H10O) (CAS 107-87-9)
M+
– 15
(CH3) lost
M+
86
M+
– 28
(CH2CH2) lost
M+
– 43
(C2C2CH3) lostPropyl Group
Molecular Ion Peak
Mass Spectroscopy The Presence of Nitrogen in the Compound
If the Mass / Charge (m/z) ratio for the Molecular Ion peak is “Odd”, then the molecule contains an Odd number of Nitrogen atoms, i.e., 1, 3, 5, etc.
Note: An “Even” value for the Mass / Charge ratio could represent a compound with an even number of Nitrogen atoms, i.e., 0, 2, 4 etc.
The actual presence of Nitrogen in the compound is not explicitly indicated as it is with an “Odd” value for the ratio.
04/21/23 13
Mass Spectroscopy Halogens in Organic Compounds
Most elements exist in several isotopic forms:Ex. 1H1, 2H1, 12C6, 13C6, 35Cl17, 37Cl17, 79Br35, 81Br35
“Average Molecular Weight”
The average molecular weight of “All” isotopes of a given element relative to the abundance of the each isotope in nature
“Integral Molecular Weight”
The Number of Protons and Neutrons in a specific isotope
Each fragment represented in a Mass Spectrum produces several peaks each representing a particular isotopic mixture of the elements in the compound, i.e., an “integral molecular weight.
04/21/23 14
Mass Spectroscopy The Presence of Chlorine in a Compound
The two (2) principal Chlorine Isotopes in nature areCl-35 and Cl-37 (2 additional Neutrons in Cl-37)
The relative abundance ratio of Cl-35 to Cl-37 is:
100 : 32.6 or 75.8 : 24.2 or 3 : 1
Therefore, a Molecule containing a single Chlorine atom will show two Mass Spectrum Molecular Ion peaks, one for Cl-35 (M+) and one for Cl-37 (M+2)
Note: M+2 denotes 2 more neutrons than M+
Based on the natural abundance ratio of 100 / 32.6 (about 3:1), the relative intensity (peak height) of theCl-35 peak will be 3 times the intensity of the Cl-37 peak
04/21/23 15
1-Chloropropane
Molecule contains 1 Chlorine atom resulting in two Molecular Ion Peaks of about 3:1 relative intensity,
based on the 3:1 natural abundance ratio ofCl-35 / Cl-37
Molecular Ion PeaksM+ 78: M+2 80
very small
Mass Spectroscopy The Presence of Chlorine in a Compound (Con’t)
04/21/23 16
Mass Spectroscopy The Presence of Bromine in a Compound
The two (2) principal Bromine Isotopes in nature areBr-79 and Br-81 (2 additional Neutrons in Br-81)
The relative abundance ratio of Br-79 to Br-81 is
100 : 97.1 or 50.5 : 49.5 or 1 : 1
Molecules containing a single Bromine atom will also show two molecular ion peaks one for Br-79 (M+) and one for Br-81 M+2
Based on the natural abundance ratio of 100 / 97.1 (about 1:1), the relative intensity of the Br-79 peak will be about the same as the Br-81 peak
04/21/23 17
3-Bromo-1-Propene
Molecular Ion PeaksM+ 120; M+2 122
Molecule contains 1 Bromine atom resulting in two Molecular Ion Peaks of about 1:1 relative intensity, based on the 50.5:49.5 (1:1) natural
abundance ratio ofBr-79 / Br-81
Mass Spectroscopy The Presence of Bromine in a Compound (Con’t)
04/21/23 18
Mass SpectroscopyThe Presence of Fluorine in a Compound
Fluorine exists in nature principally as a single isotope
19F9
A compound containing any number of Fluorine atoms will have a single Molecular Ion peak (assuming no other Halogens present)
04/21/23 19
Mass Spectroscopy Multiple Halogens in a Compound
Compounds containing two (2) Chlorine atoms will produce three (3) Molecular Ion peaks representing the 3 possible isotope combinations of Chlorine:
35Cl17 35Cl17 (Rel Peak Intensity - 100.0)
35Cl17 37Cl17 (Rel Peak Intensity - 65.3)
37Cl17 37Cl17 (Rel Peak Intensity - 10.6)
04/21/23 20
Mass Spectroscopy Multiple Halogens in a Compound
Compounds containing three (3) Chlorine atoms will produce four (4) Molecular Ion peaks representing the 4 possible isotope combinations for Chlorine:
35Cl17 35Cl17 35Cl17 (Rel Peak Intensity - 100.0)
35Cl17 35Cl17 37Cl17 (Rel Peak Intensity - 97.8)
35Cl17 37Cl17 37Cl17 (Rel Peak Intensity - 31.9)
37Cl17 37Cl17 37Cl17 (Rel Peak Intensity - 3.5)
04/21/23 21
Mass Spectroscopy & Molecular Formula
Information from the Mass Spectrum can used to determine the Molecular Formula of a compound
Ex. Molecular Ion Peaks – M+ 94; M+2 96 (95)
2 Molecular Ion Peaks (3:1) suggest: 1 Chlorine atom
Partial Analysis: C – 25.4%; H – 3.2 %
Use 95 as average molecular weight
Carbon: 95 x 0.254 = 24.1 / 12 = 2 C atoms
Hydrogen: 95 x 0.032 = 3.0 / 1 = 3 H atoms
95 – (24 + 3) = 68 unresolved mass
(Use oxygen, nitrogen, halides (Cl or Br) to resolve mass)
2 Oxygen (16 + 16) + 1 Chlorine (35.5) 68
Molecular Formula - C2H3O2Cl
04/21/23 22
Mass Spectroscopy Summary
Fragmentation of Organic Molecules by high energy electrons
Base Peak – 100 % abundance
Molecular Ion Peak – Highest Mass/Charge ratio
Molecular Ion Peak – Last peak(s) on right side of chart
Molecular Ion Peak – Represents Molecular Weight of compound
Molecular Ion Peak – If value is “Odd” the compound contains an odd number of “Nitrogen” atoms
Molecular Ion Peak – If two peaks occur and the relative abundance ratio is 3:1, then the compound contains a single Chlorine atom.
Molecular Ion Peak – If two peaks occur and the relative abundance ration is 1:1, then the compound contains a single Bromine Atom
04/21/23 23
Ultraviolet/Visible (UV) Spectroscopy UV-Visible Spectrum : 190 nm – 800 nm
In Ultraviolet and Visible Spectroscopy, the energy absorption transitions that occur are between electronic energy levels of valence electrons, that is, orbitals of lower energy are excited to orbitals of higher energy
Thus, UV / Visible spectra often called Electronic Spectra All organic compounds absorb Ultraviolet light to some
degree, but in many cases at such short wavelengths to make its utility of very limited value in organic chemistry
For the purpose of this course, the primary use of UV/Vis will be to confirm: The presence of conjugated molecules, either aliphatic
alkene structures or aromatic ring structures To a lesser degree, the presence of the Carbonyl group
and the Nitro group
04/21/23 24
Ultraviolet/Visible (UV) Spectroscopy When a molecule absorbs radiation a valence electron
is generally excited from its highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO)
For most molecules, the lowest energy orbitals are thesigma () orbitals ( - bonds)
The electrons of sigma bonds are too tightly bound to be promoted by radiation in the 200-700 nm region.
Therefore alkanes, saturated alcohols, simple alkenes show no or very little UV absorption.
The orbitals occupy somewhat higher energy levels. Orbitals that hold unshared pairs of electrons, the
nonbonding (n) orbitals, lie at even higher energies. Unoccupied or antibonding orbitals (* and *) have
the highest energies.
04/21/23 25
Ultraviolet/Visible (UV) Spectroscopy Chromophores
The absorption of Ultraviolet radiation results from the excitation of electrons from ground to excited state
The Nuclei in molecules, however, determine the strength with which electrons are bound to the molecule, thus, influencing the spacing between ground and excited states
The characteristic energy of a transition and the wavelength of radiation absorbed are properties of a group of atoms rather than the electrons themselves.
The group of atoms producing such an absorption is called a Chromophore.
As the structure of the molecule (alkane, alkene, alkyne, alcohol, amine, nitrile, carbonyl, etc.) changes, the energy and intensity of the Ultraviolet absorption will change accordingly
04/21/23 26
(sigma)
(pi)
n (non-bonding)
C::C
C : H Sigma & pi bonds have antibonding (unocuupied)orbitals associatedwith them * & *
Ultraviolet/Visible (UV) Spectroscopy Radiation in the range 190nm – 800nm causes
valence electrons (those that participate in bonding) to be excited to a higher energy level.
The ground state of an organic molecule can contain valence electrons in three principal types of molecular orbitals:
04/21/23 27
Energy Transition Examplesn * in carbonyl compoundsn * in oxygen, nitrogen, sulfur, and halogen compounds * in alkenes, alkynes, carbonyl and azo compounds * in carbonyl compounds * in alkanesIn
crea
sin
g
En
erg
y
**
n
Antibonding (single bonds)Antibonding (double bonds)
Nonbonding (as in lone electron pairs or the propenyl (allyl) radical
Bonding (double bonds)Bonding (single bonds)
Incr
easi
ng
En
erg
y
Note:Electronic energy levels in aromatic molecules are more complicated than are presented here.
Ultraviolet/Visible (UV) Spectroscopy
04/21/23 28
Ultraviolet/Visible (UV) Spectroscopy Atoms produce sharp lines of absorption
Molecules have many excited modes of vibration and rotation at room temperature. The rotational and vibrational levels are superimposed on the electronic levels
Electron transitions may occur from any of several vibrational and rotational states of one electronic level to any of several vibrational and rotational states of a higher electronic level.
Thus, the UV spectrum of a molecule consists of a broad band of absorption centered near the wavelength of the major transition, i.e. where the radiation has its maximum absorption (max).
04/21/23 29
Ultraviolet/Visible (UV) Spectroscopy The Ultraviolet-Visible
spectrum is generally recorded as a plot of absorbance versus wavelength; but the plot is more often presented with the Absorptivity () or log plotted on the ordinate (y-axis) and the wavelength plotted on the abscissa (x-axis)
Ex: Cyclohexane
(A Conjugated Aromatic Molecule)
Wavelength of Maximum Absorbance
max – 230 nm
************************
Molar Absorptivity
– 15,000 cm-1
Log – 4.204/21/23 30
Ultraviolet/Visible (UV) Spectroscopy The Beer-Lambert Law
The Ultraviolet/Visible Spectrum is a plot of the Wavelength () in nanometers (nm) over the entire Ultraviolet / Visible region versus the Absorbance (A) of the radiation at each wavelength.
A = log (Ir / Is) = C L
Is = Intensity of light through sample solution
Ir = Intensity of incident light passing through
Reference cell
= Molar Absorptivity (Molar Extinction Coefficient) A measure of the strength or intensity of the absorption.
Units – l/(mol • cm) (m2 x 10-2 /mol) (mmol/dm3)
C = Concentration of solute (mol/L) or (g/L) if molecular mass is unknown
L = Length of cell (cm)
04/21/23 31
Ultraviolet/Visible (UV) Spectroscopy
A = • C • l
= A / (C • l )
Values of are usually expressed as Log
Aliphatic (single
band) = 10,000 –
20,000 (Log = 4.0 – 4.3)
Aromatic (two bands
= 1,000 –
10,000 (Log = 3.0 – 4.0)
Carbonyl compounds
= 10 – 100
(Log = ~ 2)
Nitro compounds= 10 (Log = ~
1)04/21/23 32
Example Transitionmax
(nm)max
Log
n-Butyl Iodide n * 257 486 2.7
Acetone n * 279 15 1.2
Acrolein * 210 11,500 4.1
(C=C & C=O) n * 315 14 1.1
1,3-Butadiene * 217 21,000 4.3
1,3,5-Hexadiene * 258 35,000 4.5
Benzene(2 transitions)
Ar * Ar *
ca 200
255
8,000
215
3.9
2.3
Ultraviolet/Visible (UV) Spectroscopy Typical Transitions & Associated Wavelengths of Maximum Absorption and Molar Absorptivities
04/21/23 33
Ultraviolet/Visible (UV) Spectroscopy
Typical Transitions and Absorptions
of Simple IsolatedChromophores
04/21/23 34
Class Transition max (nm) Log
R-OH n* 180 2.5
R-O-R n* 180 3.5
R-NH2 n* 190 3.5
R-SH n* 210 3.0
R2C=CR2 * 175 3.0
R-C=C-R * 170 3.0
R-CN n* 160 <1.0
R-N=N-R n* 340 1.0
R-NO2 n* 271 <1.0
R-CHO * 190 2.0
R-CHO n* 290 1.0
R2CO * 180 3.0
R2CO n* 280 1.5
RCOOH n* 205 1.5
RCOOR’ n* 205 1.5
RCONH2 n* 210 1.5
Ultraviolet/Visible (UV) Spectroscopy
Computation Example:
An -unsaturated ketone of relative molecular weight 110 has an absorption band with max at 215 nm and = 10,000 (l / mol • cm)
A solution of this ketone showed absorbance A = 2.0 with a 1 cm cell. Calculate the concentration of the ketone in this solution expressed in grams per liter.
Ans: A = c L
c = A / L
c = 2.0 / ((10,000 l/mol • cm) * 1.0 cm)
c = 2 x 10-4 mol/l
c = 2 x 10-4 mol/l x 110 g/mol
c = 2.20 x 10-2 g/l04/21/23 35
Ultraviolet/Visible (UV) Spectroscopy
Computation Example:
Calculate the Molar Absorptivity, , for a solution containing 1.0 mmol dm-3 (1.0 x 10-3 moles per liter) of solute, when the absorbance of a 1 cm cell was 1.5.
Ans: A = c L
= A / c L
= 1.5 / (1 x 10-3 mol / L) (1 cm)
= 1.5 x 103 L / mol • cm
What would be the Absorbance for a solution of double this concentration?
Ans: A = 1.5 x 103 L / mol • cm • 2.0 x 10-3 moles / L • 1 cm
A = 3.004/21/23 36
Ultraviolet/Visible (UV) Spectroscopy
Alkanes
Contain single sigma bonds resulting in only * transitions which absorb ultraviolet radiation at wavelengths generally too short for use in UV spectroscopy.
Utility: None
Alcohols, Ethers, Amines, Sulfur Compounds
The n * transitions absorb UV radiation within the experimentally accessible range (>180 nm).
Utility: Very little
04/21/23 37
Ultraviolet/Visible (UV) Spectroscopy
Alkenes and Alkynes
Absorb UV radiation in the range < 180 nm.
“Cumulated” alkenes ( * transitions), which have one or more “” sigma bonds between the double bounds usually have absorption maxima below 200 nm.
Utility: Very little
Compounds with Oxygen double bonds
Unsaturated molecules containing oxygen or nitrogen structures such as Carbonyl (C=O) and Nitro (NO2) have both n * (280 - 290 nm) and * transitions (188 nm).
Utility: Moderate
04/21/23 38
Ultraviolet/Visible (UV) Spectroscopy Conjugated unsaturated systems are molecules with
two or more double or triple () bonds each alternating with a single or sigma bond ().
Conjugated unsaturated systems have delocalized bonds, i.e., a p-orbital on an atom adjacent to a double bond producing * transitions. Single electron as in the allyl radical (CH2=CHCH2•) Vacant p orbital as in allyl cation (CH2=CHCH2
+) P orbital of another double bond
(CH2=CHCH=CH2
Conjugated systems include the Aliphatic Alkenes as well as the Aromatic ring structures.
Compounds whose molecules contain conjugated multiple bonds absorb strongly in the UV / Visible portion of the electromagnetic spectrum (> 200 nm).
Utility: Good04/21/23 39
The Wavelength of Maximum Absorption ( max ) is obtained from the Absorption Spectrum
Wavelength of Maximum Absorbance (max) – 242.5 nm
Molar Absorptivity ( ) – 13,100 M-1 cm-1 (Log = 4.1)
2,5-Dimethyl-2,4-Hexadiene (in Methanol)
Ultraviolet/Visible (UV) Spectroscopy
Conjugated Unsaturated Systems
Conjugated systems consist of alternating sigma () bonds and pi () bonds) and the Ultraviolet absorptions show large values of
04/21/23 40
Ultraviolet/Visible (UV) Spectroscopy
Conjugated Unsaturated Systems (Con’t)
, - Unsaturated ketones, Dienes, Polyenes
Transitions - *
High Intensity Bands
= 10,000 to 20,000 (log = 4.0 - 4.3)
max > 210 nm
Aromatic Conjugated Systems
Transitions - *
2 Medium Intensity Bands
= 1000 - 60,000 (log = 3.0 - 4.8)
max both bands > 200 nm
Note: Substitution on ring increases Molar Absorptivity above 10,000
04/21/23 41
Ultraviolet/Visible (UV) Spectroscopy Carbonyl (C=O), Nitro Group (NO2) (Resonance
effects on substituted benzene)
Transitions n - * & *
Single Low Intensity Band = 10 (log = 1) to = 300 (log =
2.5)
max (250 - 360 nm)
Nitro (NO2) log (~1.0)
Carbonyl (C=O) log (~2.0)
The presence of these functional groups should be used only as confirmations of species identified in the IR Spectra.
04/21/23 42
Ultraviolet/Visible (UV) Spectroscopy
Practical Approach to Interpreting UV/Vis Information
If the problem you are working on provides an UV/Vis spectrum and it indicates “No” absorption in the 200 – 700 nm range, the following conclusions are applicable:
The compound is not conjugated, i.e., it does not contain alternating double/single bonds (including Benzene ring.)
The compound probably does not contain “Carbonyl” or “Nitro” groups (confirm with IR).
If the problem provides Log Absorptivity values (Log ) the following possibilities exist:
Log (> 4.0) - Conjugated , - Unsaturated ketones, Dienes, Polyenes
Log (3.0 – 4.0) - Aromatic ring (Check IR, NMR)
Log (1.5 – 2.5) - C=O (Check IR)
Log (1.0 – 1.5) - NO2 (Check IR)04/21/23 43
Infrared Spectroscopy Infrared Spectroscopy References
Pavia, et al - pp. 851 - 886 Solomon’s et al - pp. 79 - 84; 821 – 822
Infrared Radiation That part of the electromagnetic spectrum
between the visible and microwave regions
0.8 m (12,500 cm-1) to 50 m (200 cm-
1). Area of Interest in Infrared Spectroscopy
The Vibrational portion of infrared spectrum
2.5 m (4,000 cm-1) to 25 m (400 cm-1) Radiation in the vibrational infrared region is
expressed in units called wavenumbers ( )
04/21/23 44
Infrared Spectroscopy Wavenumbers are expressed in units of reciprocal
centimeters (cm-1) i.e. the reciprocal of the wavelength () expressed in centimeters.
(cm-1) = 1 / (cm)
Wave Numbers can be converted to a frequency () by multiplying them by the speed of light (c) in cm/sec
(Hz) = c = c / (cm /sec /cm = 1/sec)
Recall: E = h c /
Thus, wavenumbers are directly proportional to energy
04/21/23 45
Infrared Spectroscopy Polar Covalent Bonds & Dipole Moments
Organic compounds are organized into families of compounds on the basis of certain groupings of atoms, i.e., Functional Groups.
The Electrons between atoms in an organic compound are shared forming “Covalent bonds.”
Covalent bonds between atoms with different electronegativities have an unequal sharing of the bond electrons setting up an electrostatic charge difference between the atoms.
The atom with the greater Electronegativity pulls the electrons closer to it forming a “Polar Covalent Bond.”
04/21/23 46
Infrared Spectroscopy Polar Covalent Bonds & Dipole Moments (Con’t)
The relative strength of the Polar Covalent Bond impacts the ability of the molecule, i.e., a Functional Group, to attract or repel other polar entities (functional groups).
The separation of the positive and negative charges in a Polar Covalent Bond is referred to as a Dipole.
A dipole has a Dipole Moment defined as the product of the magnitude of the partial charges (in electrostatic units, esu) times the distance (in cm) of separation.
Only those Covalent bonds with Dipole Moments are capable of absorbing Infrared Radiation.
04/21/23 47
Infrared Spectroscopy The Radiation (Energy) Absorption Process
The absorption of Infrared Radiation by a Polar Covalent Bond raises the molecule to a higher energy state.
This is a Quantized process in which only selected frequencies are absorbed dependent on the relative masses of the atoms, the force constants of the bond (electronegativity), and the geometry of the atoms.
Covalent Bonds possess Rotational and Vibrational frequencies.
Every type of bond has a natural frequency of vibration.
The same bond in different compounds has a slightly different frequency of vibration.
04/21/23 48
Infrared Spectroscopy When the frequencies of Infrared Radiation match the
natural vibrational frequencies of a bond with a Dipole Moment, the radiation is absorbed increasing the amplitude of the vibrational motions of the covalent bonds.
Infrared radiation is absorbed and converted by organic molecules with polar covalent bonds and dipole moments into energy of molecular rotation and molecular vibration.
Rotation - Less than 100 cm-1 (Spectrum is lines)
Vibration - 10,000 cm-1 to 100 cm-1 (Spectrum is bands)
The vibrational bands appears because each vibrational energy change is accompanied by a number of rotational changes
Infrared Spectroscopy is concerned only with the vibrational spectrum (4,000 cm-1 to 400 cm-1)
04/21/23 49
Infrared Spectroscopy Molecular Vibrations
Absorption of infrared radiation corresponds to energy changes on the order of 8-40 KJ/mole (2-10 Kcal/mole
The frequencies in this energy range correspond to the stretching and bending frequencies of the covalent bonds with dipole moments.
Stretching (requires more energy than bending) Symmetrical Asymmetrical
Bending Scissoring (in-plane bending) Rocking (in-plane bending) Wagging (out-of-plane bending) Twisting (out of plane bending)
04/21/23 50
C—H HC H
HC H
SymmetricStretch
(2853 cm-1)
AsymmetricStretch
(2926 cm-1)
Infrared Spectroscopy Stretching – A rhythmical movement along the
bond axis such that the interatomic distance is increasing or decreasing.
In any group of three or more atoms – at least two of which are identical - there are two modes of stretching or bending: Symmetric and Asymmetric
For the Methylene Group (CH2):
04/21/23 51
H H C
H H C
HC H
HC H
Scissoring
~1450 cm-1
(In Plane)
Rocking
~750 cm-1
(In Plane)
Wagging
~1250 cm-1
(Out of Plane)
Twisting
~1250 cm-1
(Out of Plane)
Infrared Spectroscopy Bending – A change in bond angle between bonds
with a common atom or
A movement of a group of atoms with respect to the remainder of the molecule
04/21/23 52
Infrared Spectroscopy Thus, no two molecules of different structure will have
exactly the same natural frequency of vibration, each will have a unique infrared absorption pattern or spectrum.
Two Uses: IR can be used to distinguish one compound from
another. Absorption of IR energy by organic compounds will
occur in a manner characteristic of the relative strengths of the Polar Covalent Bonds in the Functional Groups present in the compound; thus, an Infrared Spectrum gives structural information about the functional groups present in a molecule.
The absorptions of each type of bond (N–H, C–H, OH, C–X, C=O, C–O, C–C, C=C, C≡C, C≡N, etc.) are regularly found only in certain small portions of the vibrational infrared region, greatly enhancing analysis possibilities.
04/21/23 53
The split beams pass into a Monochromator, which consists of a rapidly rotating sector that passes each beam to a diffraction grating or prism.
The slowly rotating diffraction grating varies the wavelength of radiation reaching the detector.
The detector senses the ratio in intensity between the reference (air) and sample beams and records the differences on a chart.
Detector
Slit
Monochromator
IR Source
Recorder
SplitBeams Air
LenzSample
Infrared Spectroscopy Instrumentation
Dispersive (Double Beam) IR Spectrophotometer
04/21/23 54
Infrared Spectroscopy Sample Preparation
Liquid Samples 1 to 2 drops of liquid sample are placed between
two single crystals of sodium chloride (Plates) Note: NaCL plates are water soluble – keep dry
Solid Samples soluble in Acetone
Dissolve sample in acetone Evaporate on Salt Plate
Solid Samples not soluble in acetone Make Potassium Bromide (KBR) pellet
Put plates in plate holder Place holder in IR Spectrometer Obtain IR Spectrum Clean Plates with Acetone
04/21/23 55
Infrared Spectroscopy Fourier Transform (FT) Single Beam IR
Set background (air) Press “Scan” button Press “Background” button Verify No. of Scans is “4”; if not, push soft key
to set “4” Press “Execute”
Obtain Sample Spectra Insert Cell Holder into beam slot Press “SCAN” button Select Memory location ( X, Y, or Z) Press “Execute”
04/21/23 56
Infrared Spectroscopy Fourier Transform (FT) Single Beam IR (Con’t)
If spectrum bottoms out (might have to check with instructor), then remove Cell Holder; remove top of Salt Plate; wipe lightly with tissue; reassemble; and insert cell holder into beam slot.
Rerun Scan again
Push “Plot” to produce chart
Remove Cell Holder and disassemble
Clean Salt Plate; dry; return to instructor; place in desiccator
04/21/23 57
Infrared Spectroscopy The Infrared Spectrum
A plot of absorption intensity (% Transmittance) on the y-axis vs. frequency on the x-axis.
Transmittance (T) - the ratio of the radiant power transmitted by a sample to the radiant power incident on the sample.
Absorbance (A) - the logarithm to base 10 of the reciprocal of the Transmittance.
A = log10 (1 / T)
Frequency - The x-axis is represented by two scales:
Wavelength(2.5 to 25 ) (Bottom) Wavenumber (4000 cm-1 to 400 cm-1)(Upper)
04/21/23 58
Methyl IsopropylKetone
C5H10O CAS – 563-80-4
C=OCarbonyl
AliphaticC-H Stretch
CH3
CH2
C=OCarbonylOvertone
Infrared SpectroscopyIR SpectrumKetone
04/21/23 59
Infrared Spectroscopy IR Spectrum Peak Characteristics
Primary Examination Regions of the Spectrum
High Frequency Region - 4000 to 1300 cm-1
Intermediate (Fingerprint Region) - 1300 to 900 cm-1
High Frequency Region (Functional Group Region)
Characteristic Stretching frequencies of such groups as:
=CH, OH, NH, C=O, CO, C≡N, C≡C, C=C
The Fingerprint Region - 1300 to 900 cm-1
Absorption patterns frequently complex
Bands originate from interacting vibrational modes
Valuable when used in reference to other regions
Absorption unique for every molecular species
Effective use comes from experience04/21/23 60
Infrared Spectroscopy IR Spectrum Peak Characteristics (con’t)
Shape
Sharp (narrow)
Broad
Intensity
Weak (w)
Medium(m)
Strong (s)
Note: Peak intensity is dependent on amount of sample and sensitivity of instrument; therefore, the actual intensity can vary from spectrum to spectrum
04/21/23 61
Infrared Spectroscopy Principal Frequency Bands
O-H 3600 cm-1 (Acids, Alcohols)
N-H 3300 - 3500 cm-1 (Amino)
(1o - 2 peaks, 2o - 1 peak, 3o – 0 peaks) NO2 1450 – 1650 cm-1 (2 absorptions)
C≡N 2250 cm-1 (Nitrile)
C≡C 2150 cm-1 (Acetylene)
-C≡C-H 3300 cm-1 (Terminal Acetylene)
C=O 1685 - 1725 cm-1 (Carbonyl)
C=C 1650 cm-1 (Alkene) 2 absorptions
C=C 1450 – 1600 cm-1 (Aromatic) 4 absorptions
04/21/23 62
Infrared Spectroscopy Principal Frequency Bands (Con’t)
CH2 1450 cm-1 (Methylene)
CH3 1375 & 1450 cm-1 (Methyl)
C-O 900 - 1100 cm-1 (Alcohol, Acid, Ester, Ether, Anhydride)
−C-H Right side of 3000 cm-1 (Saturated Alkane)
=C-H Left side of 3000 cm-1 (Unsaturated Alkene)
=C-H 1667 – 2000 cm-1 (Aromatic Overtones)
≡C-H 2150 cm-1 (Stretch)
04/21/23 63
Functional Type of FrequencyGroup Vibration cm-1Intensity
Alkanes (C-H) (stretch) 3000-2850 s -CH3 (bend) 1450 & 1375 m
-CH2 (bend) 1465 m
Alkenes (C=C) (stretch) 3100-3000 m
(bend) 1000-650 s
Aromatics (stretch) 3150-3050 s
(OOP bend) 1000-650 s
Alkyne (C) (stretch) 3300 s
Aldehyde (CHO) (stretch) 2900-2800 w
(stretch) 2800-2700 w
Infrared Spectroscopy
04/21/23 64
Infrared Spectroscopy Correlation Table
Functional Group Frequency (cm-1) Intensity
CC Alkane Not UsefulC=C Alkene 1680-1600 m-w
Aromatic 1600-1400 m-wC≡C Alkyne 2250-2100 m-wC≡C-H Alkyne (terminal) 3300 sC=O Anhydride ~1810 s
~1760 sEster 1750-1730 sAldehyde 1740-1720 sKetone (acyclic) 1725-1705 sCarboxylic Acid 1725-1700 sAmide 1700-1640 s
04/21/23 65
Infrared SpectroscopyCorrelation Table Functional Group Frequency(cm-1) Intensity
C-O Alcohols, Ethers 1300-1000 s Esters, Acids
O-H Alcohols, Phenols Free 3650-3600 m H-Bonded 3400-3200 m
Carboxylic Acids 3300-2500 mN-H Primary & Sec Amines ~3500
mC≡N Nitriles 2260-2240 mN=O Nitro (R-NO2) 1600-1500 s
1400-1300 sC-X Fluoride 1400-1000 s
Chloride 800-600 sBromide, Iodide <600 s
04/21/23 66
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 1. Check for the presence of Carbonyl group (C=O) in the range 1660 – 1820 cm-1 (~1700 cm-1)
If the Carbonyl Group is present, one of the following types of compounds is present:
Carboxylic Acid Ester Amide Anhydride Aldehyde Ketone Acid Halide
If the molecule is conjugated (alternating double & single bonds), the strong (C=O) absorption will be shifted to the right by ~30 cm-1
04/21/23 67
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 2. Check for the presence of Saturated Alkane structures Compounds containing just Methyl (CH3) & Methylene (CH2)
groups produce generally simple IR spectra
C–H sp3 absorption is a stretch in the range 3000 – 2840 cm-1
Note: It is important to remember that the Alkane sp3 stretch occurs on the right side of the 3000 cm-1 mark in the
IR spectrum and that Alkene and Aromatic sp2 stretches occur on the left side of the 3000 cm-1 mark (see next slide).
CH3 Methyl groups (CH3) have a characteristic bending at 1375 cm-1 and a smaller absorption at 1450 cm-1.
CH2 Methylene groups (CH2) have characteristic bending at approximately 1465 cm-1
04/21/23 68
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 3. Check for the presence of unsaturated (=C–H) sp2 structures.
=C–H sp2 absorption is a stretch in the range 3000 – 3100 cm-1, i.e., on the left side of the 3000 cm-1
mark on the x-axis scale.
Step 4. Determine whether the =C–H bond is Aliphatic Alkene, Aromatic, or both.
For Alkene =C–H bonds, look for the C=C stretch at 1600 – 1650 cm-1, usually an unequal pair of absorptions.
Out-of-Plan (OOP) bending at 650 – 1000 cm-1
Note: See next slide or the table on page 895 of Pavia text for guide to substitution patterns on substituted alkenes.
04/21/23 69
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Out of Plane (OOP) substitution patterns (substituted alkenes)
04/21/23 70
1-Hexene
C6H12 CAS 592-41-6
Unsat=C-H Stretch
Sat’d-C-H Stretch
AliphaticC=C
Stretch
CH3
CH2
OOP BendingMonosubstitution
Infrared SpectroscopyIR SpectrumAliphatic Alkene
04/21/23 71
Infrared Spectroscopy
04/21/23 72
IR SpectrumCyclic Alkene
Cyclohexene
Unsat=C-H Stretch Sat’d
-C-H Stretch
AliphaticC=C
Stretch
CH2
OOP BendingCIS
Disubstitution
C6H10 CAS 110-83-8
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 4 (Con’t)
Aromatic =C-H bonds. Look for C=C stretch - (pair of absorptions at 1450 cm-1
and a pair of absorptions at 1650 cm-1
Overtone/Combination bands appear between1667 & 2000 cm-1
Out-of-Plain (OOP) bending between 650 – 1000 cm-1
Note: See next slide or the table on page 897 of Pavia text for guide to substitution patterns on Benzene ring.
Note: The substitution pattern information in the “Overtone” area and the OOP area is duplicative. Use both tables to confirm substitution pattern
04/21/23 73
Infrared Spectroscopy
04/21/23 74
OOP – Substitution Patterns (Aromatic)
Overtone Area Substitution Patterns (Aromatic)
Toluene (Methyl Benzene)
C7H8CAS 108-88-3
CH3
AromaticC=C
StretchOOP Bending
Mono-Substitution
AromaticOvertones
Mono-Substitution
Sat’nUnsat’d
Infrared SpectroscopyIR Spectrum(Aromatic)
04/21/23 75
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 5. Carbonyl Compounds (Carboxylic Acids)
Strong band of C=O group appears in range 1700-1725 cm-1.
Very broad absorption band of the OH group in the range2400-3400 cm-1.
This broad band will usually obscure the Alkane C-H stretch bands from 2849-3000 cm-1.
Medium intensity C-O stretch (as in C-OH) occurs in the range 1210-1320 cm-1
04/21/23 76
Isobutyric Acid
C4H8O2CAS 79-31-2
C=OCarbonyl
OH Stretch
C-O
CH3
sp3 C-HStretch
Infrared SpectroscopyIR SpectrumCarboxylic Acids
04/21/23 77
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 6. Carbonyl Compounds (Esters)
C=O stretch appears in the range 1730-1750 cm-1
Check for 2 or more C-O stretch bands, one stronger and broader than the other, in the range 1100-1300 cm-1
04/21/23 78
Methyl Benzoate
C8H8O2CAS 93-58-3
Infrared SpectroscopyIR SpectrumEsters
04/21/23 79
C=OCarbonyl
AliphaticC-H Stretch
Unsat’d=C-H Stretch
AromaticOvertones
AromaticOOP
C-O
Aromatic RingC=C Absorptions
C-O
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 7. Carbonyl Compounds (Anhydrides)
2 C=O stretch bands (1740-1775 cm-1 & 1800-1830 cm-1)
Conjugation will move these bands to lower frequency
Multiple C-O stretch bands in the range 900 – 1300 cm-1
04/21/23 80
Propionic Anhydride
C6H10O3 CAS 123-62-6
Pair ofC=O
Stretch bands
C-HAliphatic Stretch
C-O Stretch
CH3CH2
C=OOvertone
Infrared SpectroscopyIR SpectrumAnhydrides
04/21/23 81
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 8. Carbonyl Compounds (Amides)
C=O stretch at approximately 1640-1700 cm-1
N-H stretch (medium absorptions) near 3500 cm-1
Primary Amino (-NH2) - 2 Peaks (3180 & 3350 cm-1)
Secondary Amino (-NH) - 1 Peak (3300 cm-1)
N-H Scissoring - 1550 - 1640 cm-1
N-H Bend - 800 cm-1
04/21/23 82
IR SpectrumAmides Benzamide
C7H7NO CAS 55-21-0
NH2 Stretch2 peaks
Primary Amino
C=OCarbonyl
N-HScissoring
AromaticOvertones
C=CAromatic
Unsat’d=C-H Stretch
{
Infrared Spectroscopy
04/21/23 83
-C-N str
Acetanilide(N-Phenylacetamide)
C8H9NO CAS 103-84-4
OOP BendAromatic
MonosubstitutionNH Stretch
1 PeakSec Amino
C=CAromatic
N-HBend
C=OCarbonyl
{
AromaticOvertonesUnsat’d
=C-H Stretch
CH3
Infrared SpectroscopyIR SpectrumAmides
04/21/23 84
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 9. Carbonyl Compounds (Aldehydes)
C=O stretch appears in the range 1720 - 1740 cm-1
2 weak Aldehyde C-H stretch absorptions near 2850 and 2750 cm-1)
04/21/23 85
Nonanal
C9H18O CAS 124-19-6
Infrared SpectroscopyIR SpectrumAldehydes
04/21/23 86
AliphaticC-H Stretch
AldehydeHydrogen
Stretch2 Peaks
C=OCarbonyl
CH2
CH3
C=OOvertone
The Ketone structure produces a medium to strong absorption in the 1100 – 1300 cm-1 range due to coupled Stretching and Bending vibrations
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 10. Carbonyl Compounds (Ketones)
C=O stretch occurs at approximately 1705 – 1725 cm-1
Ketones are confirmed when the other five compound types containing a Carbonyl group have been eliminated.
Ketone IR Spectra can sometimes be confused with Ester spectra because of an absorption in the 1100 -1300 cm-1 range similar to the location of the C-O stretch in esters. Usually, however, the ester will have 2 or more of the C-O stretch absorptions.
04/21/23 87
Ethyl Isopropyl Ketone(2-Methyl-3-Pentanone)
C6H12O CAS – 565-69-5
C=OCarbonyl
CH2
CH3
AliphaticC-H Stretch
C=OOvertone
Infrared SpectroscopyIR SpectrumKetones
04/21/23 88
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 11. Triple Bonds
Alkynes
R – C ≡ C – R weak, sharp stretch near 2150 cm-1
R – C ≡ C – H (Terminal Acetylene)
Weak, sharp stretch near 2150 cm-1
and a second stretch at 3300 cm-1
Nitriles
C ≡ N Medium, sharp stretch near 2250 cm-1
04/21/23 89
IR SpectrumAlkynes (CC)
Propargyl Alcohol (2-Propyn-1-ol)
C3H4O CAS 107-19-7
OHH - Bonded
≡C-H Terminal AlkyneStretch
C-O
C≡CStretch
CH2
AliphaticC-H Stretch
Infrared Spectroscopy
04/21/23 90
Benzonitrile
IR SpectrumNitriles
C7H5N CAS 100-47-0
-C≡NStretch
Unsat=C-H Stretch
Aromatic ringC=C Absorptions
AromaticOOP Bending
Monosubstitution
AromaticOvertones
Infrared Spectroscopy
04/21/23 91
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 12. - Alcohols & Phenols
Broad absorption near 3600 - 3300 cm-1
Confirm presence of C–O (C–OH) near 1000 - 1300 cm-1
04/21/23 92
2-Naphthol (Nujol Mull)
IR SpectrumAlcohols & Phenols
C10H9O CAS 135-19-3
Infrared Spectroscopy
04/21/23 93
OHH - Bonded
Aromatic ringC=C Absorptions
Unsaturation=C-H Stretch
Saturation-C-H Stretch
2-Naphthol (CCl4 Soln)IR SpectrumAlcohols & Phenols
C10H9O CAS 135-19-3
OHH - Bonded
Aromatic ringC=C Absorptions
Unsat=C-H Stretch
C-O
Infrared Spectroscopy
04/21/23 94
2-Naphthol (KBr Disc)IR SpectrumAlcohols & Phenols
C10H9O CAS 135-19-3
Infrared Spectroscopy
04/21/23 95
OHH - Bonded Aromatic ring
C=C Absorptions
Unsat=C-H Stretch
C-O
IR SpectrumAlcohols & Phenols
C4H10O CAS 78-92-2
2-Butanol
OH
C-OCH2 CH3
AliphaticC-H Stretch
Infrared Spectroscopy
04/21/23 96
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 13. Ethers
C–O absorptions near 1000 - 1300 cm-1
Absence of OH
Absence of C=O group
Aliphatic Ethers give a single strong C-O band at1120 cm-1
Unbalanced Ethers will show 2 C–O groups
Phenyl Alkyl Ethers give two (2) strong bands at about 1040 & 1250 cm-1
04/21/23 97
IR SpectrumEthers
Butyl Ether(Balanced Ether)
C8H18O CH3(CH2)3 – O – (CH2)3CH3 CAS 142-96-1
C-O
CH2
CH3
AliphaticC-H Stretch
Infrared Spectroscopy
04/21/23 98
IR SpectrumEthers
C8H10O CAS 103-73-1
Phenetole(Unbalanced Phenyl Alkyl Ether)
Infrared Spectroscopy
04/21/23 99
AliphaticC-H Stretch
Unsat=C-H Stretch
C-O
C-O
CH3
Aromatic ringC=C Absorptions
AromaticOvertones
OOP BendingAromatic
Monosubstitution
CH2
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 14. Amines
N-H stretch (Medium absorptions) near 3500 cm-1
Primary Amino - 2 Peaks
Secondary Amino - 1 Peak
Tertiary Amino - No peaks
N-H Scissoring at 1560 - 1640 cm-1
N-H Bend at 800 cm-1
04/21/23 100
n-Butylamine(Primary Amine)
IR SpectrumAmines
C4H11N CAS 109-73-9
H-N-H Stretch2 Peaks
Primary Amine
N-HScissoring
CH2
CH3
-C-NStretch
-N-HOOP BendingAliphatic
(sat’n)C-H Stretch
Infrared Spectroscopy
04/21/23 101
N-Methylbenzylamine(Sec Amine)
IR SpectrumAmines
C6H11N CAS 103-67-3
Infrared Spectroscopy
04/21/23 102
N-HScissoring
CH2
CH3 -N-HOOP Bending
AliphaticC-H Stretch
AromaticOvertones
OOP BendingAromatic
Monosubstitution
Unsat =C-H Stretch
Aromatic ringC=C Absorptions
Sec-Amino
N-HScissoring
CH2
CH3Sat
– C-H Stretch
C-N Str
Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach
Step 15. Nitro Compounds
Two strong absorptions
Aliphatic Nitro Compounds
Asymmetric strong stretch 1530 -1600 cm-1
Symmetric medium stretch 1300 -1390 cm-1
Aromatic Nitro Compounds
Asymmetric strong stretch 1490 -1550 cm-1
Symmetric strong stretch 1315 - 1355 cm-1
04/21/23 103
Nitro BenzeneIR Spectrum
Nitro Compounds
C6H5NO2CAS 98-95-3
Infrared Spectroscopy
04/21/23 104
Unsat=C-H Stretch
NO2 (-N=O) Stretch2 Absorptions
C=CAromatic ringAbsorptions
AromaticOvertones
Mono-Substitution
1-Nitro PropaneIR SpectrumNitro Compounds
C3H5NO2CAS 108-03-2
NO2 (-N=O) Stretch2 Absorptions
AliphaticC-H Stretch
Infrared Spectroscopy
04/21/23 105
Infrared Spectroscopy Step 16. If none of the above apply then the compound
is most likely a:
Hydrocarbon
Alkyl Halide (see slides 105 - 109).
Hydrocarbons
Generally, very simple spectrum
–C-H Sat’d Alkanes – 2900 - 3000 cm-1
Methyl (CH3) – 1370 cm-1
Methylene (CH2) – 1450 cm-1
t-Butyl Group – 525 cm-1
Long Alkane (CH2) Chain – 720 cm-1
04/21/23 106
Decane
CH3(CH2)8CH3
IR SpectrumAlkane
C10H22CAS 124-18-5
CH2
CH3
AliphaticC-H Stretch
Long AlkaneChain (CH2)
Bending
Infrared Spectroscopy
04/21/23 107
Infrared Spectroscopy Step 17. Halogens
The Halogens as CH2 - X absorptions show up in the region (1000 – 1300 cm-1).
Halogens (Cl, Br, I) show in the Fingerprint region (485 – 800 cm-1) as one or two absorptions – see next slide.
Using IR to identify Halogens in this region can be difficult, especially if OOP Bending absorptions (used for “Substitution Pattern information) from Alkene and Aromatic unsaturated Pi () bond structures are present.
Halogen identification should be restricted to Aliphatic Alkane structures containing mainly CH2 & CH3 groups.
Iodide and Bromide absorptions in the range 485 – 650 cm-1 are generally out range on NaCL Salt Plates, however, if other substrates, e.g.,KBr pellets, are used, the absorptions are extended to this range.
04/21/23 108
Infrared Spectroscopy Step 17. Halogens (Con’t)
Fluoride 1000 – 1400 cm-1
Monofluorides 1000 – 1200 cm-1
Polyfluorides 1100 – 1300 cm-1
Aryl Fluorides 1100 – 1250 cm-1
Chloride (2 or more bands) 540– 785 cm-1
CH2-CL (Bend Wagging) 1230 – 1300 cm-1
t-Butyl Group – 525 cm-1
Bromine (KBr Pellets) 510 –650 cm-1
CH2-Br (Bend Wagging) 1190 – 1250 cm-1
Aryl Bromides 1030 – 1075 cm-1
Iodide (KBr Pellets) 485 –600 cm-1
CH2-I (Bend Wagging) 1150 – 1200 cm-1
04/21/23 109
IR SpectrumHalogens 2-Bromobutane
Br
CH2-Br
CH3
CH2
-C-HSat’n
C4H9Br CAS 78-76-2
Infrared Spectroscopy
04/21/23 110
IR SpectrumHalogens 1-Chloropropane
C3H7Cl CAS 540-54-5
CH3
CH2
-C-HSat’n Cl
CH2-Cl
Infrared Spectroscopy
04/21/23 111
IR SpectrumHalogens
C7H7Cl
o-Chlorotoluene
CAS 95-49-8
-C-HSat’n
=C-HUnsat’n
AromaticOvertones
O-Disubstitution
-C=C-Aromatic
CH3
CH2-Cl
Cl
OOPo-disubstitution
(750 cm-1)(missing)
{
Infrared Spectroscopy
04/21/23 112
Infrared Spectroscopy
04/21/23 113
IR SpectrumHalogens
C5H14CL CAS 594-36-5
T-Pentyl525 cm-1CH2
CH3
CH2-ClSaturatedAliphatic
C-H Stretch
T-Pentyl Chloride(2-Chloro-2-MethylButane
Carbonyl (C=O) @ 1715-1685(Conjugation moves absorption to right ~30 cm-1
Acid
Ester
Amide
Anhydride
Aldehyde
Ketone
Alcohol
Amine
Ether
Alkanes -C-HMethylene -CH2
Methyl -CH3
Alkenes (Vinyl) -C=CAlkynes (Acetylenes) -C≡CAromatic -C=C
Nitriles Nitro
Saturation< 3000 cm-1
Yes No
Unsaturation> 3000 cm-1
Hydrocarbons
IR Analysis Scheme
04/21/23 114
Carbonyl (C=O) is Present
Acid - Broad OH Absorption @ 3300-2500 cm-1
Ester - C-O Absorption @ 1300-1000 cm-1
Amide - NH Absorption @ 3500 cm-1 (1 or 2 peaks)
Anhydride - 2 C=O Absorptions 1810 & 1760 cm-1
Aldehyde - Aldehyde C-H Absorptions @ 2850 & 2750 cm-1
Ketone - None of the above except C=O
Carbonyl is Absent
Alcohol - Broad OH absorption @ 3300 - 3000 cm-1
Also C-O absorption @ 1300 - 1000 cm-1
Amine - 1 to 2 equal NH absorptions @ 3500 cm-1
Ether - C-O absorption @ 1300 - 1000 cm-1
IR Analysis Scheme
04/21/23 115
Saturation
Unsaturation
Alkanes -C-H Stretch – several absorptions to “right” of 3000 cm-1
Methylene -CH2 1450 cm-1
Methyl -CH3 1375 cm-1
Double Bonds =C-H Stretch – several absorptions to “left” of 3000 cm-1
OOP bending at 1000 – 650 cm-1
Alkenes (Vinyl) -C=C- Stretch (weak) @ 1675 – 1600 cm-1
Conjugation moves absorption to the rightAlkynes -C≡C-H Terminal Acetylene Stretch at 3300 cm-1
Alkynes (Acetylenes) -C≡C Stretch @ 2150 cm-1 Conjugation moves absorption to the right
Aromatic (Benzene) =C-H Stretch absorptions also to left of 3000 cm-1
OOP bending at 900 – 690 cm-1
OOP absorption patterns allow determination of ring substitution (p. 897 Pavia text)
-C=C 4 Sharp absorptions (2 pairs) @ 1600 & 1450 cm-1
Overtone absorptions @ 2000 – 1667 cm-1
Relative shapes and numbers of peaks permit determination of ring substitution pattern (p. 897 Pavia text).
IR Analysis Scheme
04/21/23 116