Spectroscopy
Building A Toolset
For
The Identification of Organic Compounds
Physical
Chemical Tests
Properties
Hydrocarbons
Melting Point
Alkanes
Boiling Point
Alkenes
Density
Alkynes
Solubility
Halides
Refractive Index
Alcohols
Aldehydes
Ketones
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Spectroscopy
Mass
(Molecular Weight)
Ultraviolet/Visual
(Conjugation, Carbonyl)
Infrared
Functional Groups
NMR
(Number, Type, Location of
protons)
Gas Chromatography
(Identity, Mole %)
1
Spectroscopy
Spectroscopy Tools
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2
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 SPECTRUM
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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:
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
Electronic

Vibrational

Spin
4
Spectroscopy


Electromagnetic Spectrum
Cosmic
 (Gamma)
X-Ray
Ultraviolet
Visible
Infrared
Microwave
Radio
Energy States and the Electromagnetic Spectrum
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Electronic
–
Ultraviolet
Vibrational
–
Infrared
Spin
–
Radio
5
Spectroscopy
High
Frequency ()
Low
High
Energy (E)
Low
Short
Wavelength ()
Long
1.2 x 1014 Hz
Frequency
3 x 1019 Hz
3 x 1016 Hz
1.5 x 1015 Hz
3 x 108 Hz
2 x 1013 Hz
3 x 1011 Hz
1 x 109 Hz
6 x 107 Hz
4 x103cm-1
1.25 x104cm-1
Wave Number
1 x109cm-1
Cosmic
&
 Ray
1 x107cm-1
X-Ray
0.01 nm
Wavelength
2.5 x104cm-1
5 x104cm-1
667cm-1
Vacuum
UV
10 cm-1
Infrared
200 nm
400 nm
800 nm
Visible
Blue
Red
2.5 
3 cm-1
Microwave
1 mm
10 nm
Near
Ultraviolet
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0.002 cm-1
0.01 cm-1
Radio
30 cm
1m
Frequency
5m
15 
Vibrational
Infrared
Nuclear
Magnetic
Resonance
6
Spectroscopy
Quantization
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.
E = [E (excited) – E (ground)] = h
=c/
E = hc / 
Where:
 = Frequency (Hz)
 = Wavelength (cm)
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c = Velocity of Light (cm/sec)
h = Planck’s Constant
7
Spectroscopy
Spectroscopy 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
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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.”
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9
Mass Spectroscopy
M
Molecule
+
eHigh Energy
Electron

M+
+
2e-
Molecular Ion
(Radical Cation)
M - (H2O and CH2 – CH2)
Base Peak
1-Pentanol - MW 88
CH3(CH2)3 – CH2OH
M - (H2O and CH3)
Typical
Mass
Spectrum
M - H2O
CH2OH
Molecular Ion Peak
(M+ 88)
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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
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11
Mass Spectroscopy
Methyl Propyl Ketone (C5H10O) (CAS 107-87-9)
+
M – 43
(C2C2CH3) lost
Propyl Group
+
M – 28
(CH2CH2) lost
+
Molecular
Ion Peak
M – 15
(CH3) lost
+
M
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86
12
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.

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The actual presence of Nitrogen in the
compound is not explicitly indicated as it is with
an “Odd” value for the ratio.
13
Mass Spectroscopy

Halogens in Organic Compounds

Most elements exist in several isotopic forms:
Ex. 1H1, 2H1,

12C , 13C , 35Cl , 37Cl , 79Br , 81Br
6
6
17
17
35
35
“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.
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14
Mass Spectroscopy
 The Presence of Chlorine in a Compound
 The two (2) principal Chlorine Isotopes in nature are
Cl-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 the
Cl-35 peak will be 3 times the intensity of the Cl-37 peak
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15
Mass Spectroscopy

The Presence of Chlorine in a Compound (Con’t)
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 of
Cl-35 / Cl-37
Molecular Ion Peaks
M+ 78: M+2 80
very small
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16
Mass Spectroscopy
 The Presence of Bromine in a Compound
 The two (2) principal Bromine Isotopes in nature are
Br-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
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17
Mass Spectroscopy

The Presence of Bromine in a Compound (Con’t)
3-Bromo-1-Propene
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 of
Br-79 / Br-81
Molecular Ion Peaks
M+ 120; M+2 122
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18
Mass Spectroscopy
The Presence of Fluorine in a Compound

Fluorine exists in nature principally as a single isotope
19F
9

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A compound containing any number of Fluorine atoms
will have a single Molecular Ion peak (assuming no other
Halogens present)
19
Mass Spectroscopy

Multiple Halogens in a Compound

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Compounds containing two (2) Chlorine atoms
will produce three (3) Molecular Ion peaks
representing the 3 possible isotope combinations
of Chlorine:
35Cl
17
35Cl
17
(Rel Peak Intensity - 100.0)
35Cl
17
37Cl
17
(Rel Peak Intensity -
65.3)
37Cl
17
37Cl
17
(Rel Peak Intensity -
10.6)
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:
35Cl
35Cl
35Cl
17
35Cl
17
35Cl
17
37Cl
37Cl
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17
37Cl
17
35Cl
17
(Rel Peak Intensity
- 100.0)
17
37Cl
17
(Rel Peak Intensity
-
97.8)
17
37Cl
17
(Rel Peak Intensity
-
31.9)
17
37Cl
-
3.5)
17
(Rel Peak Intensity
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
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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
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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
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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 the
sigma () 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.
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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
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26
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:
 (sigma)
 (pi)
C:H
C::C
Sigma & pi bonds have
antibonding (unocuupied)
orbitals associated
with them
* & *
n (non-bonding)
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27
Ultraviolet/Visible (UV) Spectroscopy
Energy
Increasing
Energy
Increasing
Energy Transition Examples
n
n



 *
 *
 *
 *
 *
in carbonyl compounds
in oxygen, nitrogen, sulfur, and halogen compounds
in alkenes, alkynes, carbonyl and azo compounds
in carbonyl compounds
in alkanes
* Antibonding (single bonds)
* Antibonding (double bonds)
n


Nonbonding (as in lone electron
pairs or the propenyl (allyl) radical
Bonding (double bonds)
Bonding (single bonds)
Note:Electronic energy levels in aromatic molecules
are more complicated than are presented here.
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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).
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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

Log 
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– 15,000 cm-1
– 4.2
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)
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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

Carbonyl compounds  = 10 – 100

Nitro compounds
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 = 1,000 – 10,000 (Log  = 3.0 – 4.0)
 = 10
(Log  = ~ 2)
(Log  = ~ 1)
32
Ultraviolet/Visible (UV) Spectroscopy

Typical Transitions & Associated Wavelengths of Maximum
Absorption and Molar Absorptivities
max
(nm)
max
Log 
n  *
n  *
257
486
2.7
279
15
1.2
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
ca 200
255
8,000
215
3.9
2.3
Example
n-Butyl Iodide
Acetone
Acrolein
Benzene
(2 transitions)
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Transition
Ar   *
Ar   *
33
Ultraviolet/Visible (UV) Spectroscopy
Typical
Transitions and
Absorptions
of Simple
Isolated
Chromophores
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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-CN
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
34
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=cL
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/l
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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 =



cL
=A/cL
= 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 = 3.0
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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
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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
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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=CHCH2•)
 Vacant p orbital as in allyl cation
(CH2=CHCH2+)
 P orbital of another double bond
(CH2=CHCH=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: Good
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39
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 
2,5-Dimethyl-2,4-Hexadiene (in Methanol)

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)
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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
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41
Ultraviolet/Visible (UV) Spectroscopy

Carbonyl (C=O), Nitro Group (NO2) (Resonance effects on
substituted benzene)
n - * &
  *

Transitions

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.
1/17/2015
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:
- Conjugated , - Unsaturated
ketones, Dienes, Polyenes
 (3.0 – 4.0)
Log  (1.5 – 2.5)
Log  (1.0 – 1.5)
- Aromatic ring (Check IR, NMR)
Log

Log

1/17/2015
 (> 4.0)


- C=O
(Check IR)
- NO2
(Check IR)
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 
( )
1/17/2015
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/

Recall: E = h c / 

(cm /sec /cm = 1/sec)
Thus, wavenumbers are directly proportional to energy
1/17/2015
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.”
1/17/2015
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.
1/17/2015
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.
1/17/2015
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

Vibration - 10,000 cm-1 to 100 cm-1 (Spectrum is bands)
(Spectrum is lines)

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)
1/17/2015
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 polar covalent
bonds with dipole moments.

Stretching (requires more energy than bending)

1/17/2015

Symmetrical

Asymmetrical
Bending

Scissoring
(in-plane bending)

Rocking
(in-plane bending)

Wagging
(out-of-plane bending)

Twisting
(out of plane bending)
50
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):
H
C—H
C
H
Symmetric
Stretch
(2853 cm-1)
1/17/2015
H
C
H
Asymmetric
Stretch
(2926 cm-1)
51
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
H
Wagging
~1250 cm-1
(In Plane)
H
(Out of Plane)
Rocking
~750 cm-1
H
Twisting
~1250 cm-1
(Out of Plane)
Scissoring
~1450 cm-1
(In Plane)
1/17/2015
C
C
H
H
H
C
H
H
C
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.
1/17/2015
53
Infrared Spectroscopy

Instrumentation

Dispersive (Double Beam) IR Spectrophotometer
Split
Beams
Air
Detector
IR Source
Lenz
Slit
Recorder
Sample
Monochromator
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.
1/17/2015
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
1/17/2015
55
Infrared Spectroscopy

Fourier Transform (FT) Single Beam IR


1/17/2015
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”
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
1/17/2015
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)

1/17/2015
Wavenumber (4000 cm-1 to 400 cm-1) (Upper)
58
Infrared Spectroscopy
IR Spectrum
Ketone
Methyl Isopropyl Ketone
C=O
Carbonyl
Overtone
CH2
Aliphatic
C-H Stretch
C5H10O
1/17/2015
C=O
Carbonyl
CH3
CAS 563-80-4
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:
=CH, OH, NH, C=O, CO, C≡N, C≡C, C=C

1/17/2015
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 experience
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
1/17/2015
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)

C=C
1450 – 1600 cm-1 (Aromatic) 4 absorptions
1/17/2015
2 absorptions
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)
1/17/2015
63
Infrared Spectroscopy
Functional
Group
Frequency
cm-1
Intensity
(stretch)
3000-2850
s
-CH3
(bend)
1450 & 1375
m
-CH2
(bend)
1465
m
(stretch)
3100-3000
m
(bend)
1000-650
s
(stretch)
3150-3050
s
(OOP bend)
1000-650
s
Alkyne (C)
(stretch)
3300
s
Aldehyde (CHO)
(stretch)
2900-2800
w
(stretch)
2800-2700
w
Alkanes (C-H)
Alkenes (C=C)
Aromatics
1/17/2015
Type of
Vibration
64
Infrared Spectroscopy

Correlation Table
Functional Group
CC
C=C
Alkane
Alkene
Aromatic
C≡C
Alkyne
C≡C-H Alkyne (terminal)
C=O
Anhydride
Ester
Aldehyde
Ketone (acyclic)
Carboxylic Acid
Amide
1/17/2015
Frequency (cm-1)
Not Useful
1680-1600
1600-1400
2250-2100
3300
~1810
~1760
1750-1730
1740-1720
1725-1705
1725-1700
1700-1640
Intensity
m-w
m-w
m-w
s
s
s
s
s
s
s
s
65
Infrared Spectroscopy
Correlation Table
Functional Group
C-O
Frequency(cm-1)
Alcohols, Ethers
Esters, Acids
O-H Alcohols, Phenols
Free
H-Bonded
Carboxylic Acids
N-H Primary & Sec Amines
C≡N Nitriles
N=O Nitro (R-NO2)
C-X
1/17/2015
Fluoride
Chloride
Bromide, Iodide
Intensity
1300-1000
s
3650-3600
3400-3200
3300-2500
~3500
2260-2240
1600-1500
1400-1300
1400-1000
800-600
<600
m
m
m
m
m
s
s
s
s
s
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
1/17/2015
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 cm1
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
1/17/2015
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.
1/17/2015
69
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

1/17/2015
Out of Plane (OOP) substitution patterns (substituted alkenes)
70
Infrared Spectroscopy
IR Spectrum
Aliphatic Alkene
1-Hexene
CH3
Unsat
=C-H Stretch
CH2
Sat’d
-C-H Stretch
Aliphatic
C=C
Stretch
OOP Bending
Monosubstitution
1/17/2015
C6H12
CAS 592-41-6
71
Infrared Spectroscopy
IR Spectrum
Cyclic Alkene
Cyclohexene
Aliphatic
C=C
Stretch
CH2
Unsat
=C-H Stretch
Sat’d
-C-H Stretch
C6H10
1/17/2015
OOP Bending
CIS
Disubstitution
CAS 110-83-8
72
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 between
1667 & 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
1/17/2015
73
Infrared Spectroscopy
OOP – Substitution Patterns (Aromatic)
Overtone Area Substitution Patterns (Aromatic)
1/17/2015
74
Infrared Spectroscopy
IR Spectrum
(Aromatic)
Toluene (Methyl Benzene)
Aromatic
Overtones
Mono-Substitution
Unsat’d
CH3
Sat’n
Aromatic
C=C
Stretch
C7H8
1/17/2015
OOP Bending
Mono-Substitution
CAS 108-88-3
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 range
2400-3400 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
1/17/2015
76
Infrared Spectroscopy
IR Spectrum
Carboxylic Acids
Isobutyric Acid
CH3
OH Stretch
sp3 C-H
Stretch
C4H8O2
1/17/2015
C=O
Carbonyl
C-O
CAS 79-31-2
77
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

1/17/2015
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
78
Infrared Spectroscopy
IR Spectrum
Esters
Methyl Benzoate
Aromatic
Overtones
Unsat’d
=C-H Stretch
Aliphatic
C-H Stretch
Aromatic Ring
C=C Absorptions
C-O
C=O
Carbonyl
C8H8O2
1/17/2015
C-O
Aromatic
OOP
CAS 93-58-3
79
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

1/17/2015
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
80
Infrared Spectroscopy
IR Spectrum
Anhydrides
Propionic Anhydride
C=O
Overtone
C-H
Aliphatic
Stretch
CH2
Pair of
C=O
Stretch bands
1/17/2015
C6H10O3
CH3
C-O Stretch
CAS 123-62-6
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)
 Secondary
1/17/2015
Amino (-NH)
- 2 Peaks (3180 & 3350 cm-1)
- 1 Peak (3300 cm-1)

N-H Scissoring
- 1550 - 1640 cm-1

N-H Bend
- 800 cm-1
82
Infrared Spectroscopy
IR Spectrum
Amides
Benzamide
Aromatic
Overtones
Unsat’d
=C-H Stretch
NH2 Stretch
2 peaks
Primary Amino
N-H
Scissoring
C=O
Carbonyl
C7H7NO
1/17/2015
{
-C-N str
C=C
Aromatic
CAS 55-21-0
83
Infrared Spectroscopy
IR Spectrum
Amides
Acetanilide
(N-Phenylacetamide)
Unsat’d
=C-H Stretch
Aromatic
Overtones
NH Stretch
1 Peak
Sec Amino
N-H
Bend
C=O
Carbonyl
C8H9NO
1/17/2015
OOP Bend
Aromatic
Monosubstitution
{
CH3
C=C
Aromatic
CAS 103-84-4
84
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

1/17/2015
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)
85
Infrared Spectroscopy
IR Spectrum
Aldehydes
Nonanal
C=O
Overtone
CH3
Aldehyde
Hydrogen
Stretch
2 Peaks
Aliphatic
C-H Stretch
C9H18O
1/17/2015
CH2
C=O
Carbonyl
CAS 124-19-6
86
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.
The Ketone structure produces
a medium to strong absorption
in the 1100 – 1300 cm-1 range
due to coupled Stretching and
Bending vibrations
1/17/2015
87
Infrared Spectroscopy
IR Spectrum
Ketones
Ethyl Isopropyl Ketone
(2-Methyl-3-Pentanone)
C=O
Overtone
Aliphatic
C-H Stretch
C6H12O
1/17/2015
C=O
Carbonyl
CH3
CH2
CAS – 565-69-5
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
1/17/2015
Medium, sharp stretch near 2250 cm-1
89
Infrared Spectroscopy
IR Spectrum
Alkynes (CC)
Propargyl Alcohol
(2-Propyn-1-ol)
C≡C
Stretch
OH
H - Bonded
Aliphatic
C-H Stretch
≡C-H Terminal Alkyne
Stretch
C3H4O
1/17/2015
CH2
C-O
CAS 107-19-7
90
Infrared Spectroscopy
IR Spectrum
Nitriles
Benzonitrile
Aromatic
Overtones
Unsat
=C-H Stretch
-C≡N
Stretch
C7H5N
1/17/2015
Aromatic ring
C=C Absorptions
Aromatic
OOP Bending
Monosubstitution
CAS 100-47-0
91
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

1/17/2015
Step 12. - Alcohols & Phenols

Broad absorption near 3600 - 3300 cm-1

Confirm presence of C–O (C–OH) near 1000 - 1300 cm-1
92
Infrared Spectroscopy
IR Spectrum
Alcohols & Phenols
2-Naphthol (Nujol Mull)
Unsaturation
=C-H Stretch
OH
H - Bonded
Saturation
-C-H Stretch
Aromatic ring
C=C Absorptions
C10H9O
1/17/2015
CAS 135-19-3
93
Infrared Spectroscopy
IR Spectrum
Alcohols & Phenols
2-Naphthol (CCl4 Soln)
Unsat
=C-H Stretch
OH
H - Bonded
C10H9O
1/17/2015
Aromatic ring
C=C Absorptions
C-O
CAS 135-19-3
94
Infrared Spectroscopy
IR Spectrum
Alcohols & Phenols
2-Naphthol (KBr Disc)
Unsat
=C-H Stretch
OH
H - Bonded
C10H9O
1/17/2015
Aromatic ring
C=C Absorptions
C-O
CAS 135-19-3
95
Infrared Spectroscopy
IR Spectrum
Alcohols & Phenols
2-Butanol
CH2 CH3
OH
C4H10O
1/17/2015
C-O
Aliphatic
C-H Stretch
CAS 78-92-2
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 at
1120 cm-1

Unbalanced Ethers will show 2 C–O groups

1/17/2015
Phenyl Alkyl Ethers give two (2) strong bands at
about 1040 & 1250 cm-1
97
Infrared Spectroscopy
IR Spectrum
Ethers
Butyl Ether
(Balanced Ether)
CH3
CH2
Aliphatic
C-H Stretch
C8H18O
1/17/2015
CH3(CH2)3 – O – (CH2)3CH3
C-O
CAS 142-96-1
98
Infrared Spectroscopy
IR Spectrum
Ethers
Unsat
=C-H Stretch
Phenetole
(Unbalanced Phenyl Alkyl Ether)
Aromatic
Overtones
Aliphatic
C-H Stretch
CH3
C-O
CH2
Aromatic ring
C=C Absorptions
1/17/2015
C8H10O
C-O
OOP Bending
Aromatic
Monosubstitution
CAS 103-73-1
99
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

Step 14. Amines

1/17/2015
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
100
Infrared Spectroscopy
n-Butylamine
IR Spectrum
Amines
(Primary Amine)
-C-N
Stretch
N-H
Scissoring
CH3
H-N-H Stretch
2 Peaks
Primary Amine
CH2
Aliphatic
(sat’n)
C-H Stretch
C4H11N
1/17/2015
-N-H
OOP Bending
CAS 109-73-9
101
Infrared Spectroscopy
IR Spectrum
Amines
N-Methylbenzylamine
(Sec Amine)
Aromatic
Overtones
N-H
Scissoring
Sec-Amino
Unsat
=C-H Stretch
Sat
– C-H Stretch
Aromatic ring
C=C Absorptions
N-H
Scissoring
Aliphatic
C-H Stretch
C6H11N
1/17/2015
CH3
CH3
-N-H
OOP
C-N Str Bending
CH2
CH2
OOP Bending
Aromatic
Monosubstitution
CAS 103-67-3
102
Infrared Spectroscopy

Analyzing the Spectrum – A Suggested Approach

Step 15. Nitro Compounds

Two strong absorptions


1/17/2015
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
103
Infrared Spectroscopy
IR Spectrum
Nitro Compounds
Nitro Benzene
Aromatic
Overtones
Mono-Substitution
Unsat
=C-H Stretch
C=C
Aromatic ring
Absorptions
NO2 (-N=O) Stretch
2 Absorptions
C6H5NO2
1/17/2015
CAS 98-95-3
104
Infrared Spectroscopy
IR Spectrum
Nitro Compounds
1-Nitro Propane
Aliphatic
C-H Stretch
NO2 (-N=O) Stretch
2 Absorptions
C3H5NO2
1/17/2015
CAS 108-03-2
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).

1/17/2015
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
106
Infrared Spectroscopy
IR Spectrum
Alkane
Decane
CH3
Long Alkane
Chain (CH2)
Bending
CH2
Aliphatic
C-H Stretch
1/17/2015
C10H22
CH3(CH2)8CH3
CAS 124-18-5
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.
1/17/2015
108
Infrared Spectroscopy

Step 17. Halogens (Con’t)





Monofluorides
1000 – 1200 cm-1

Polyfluorides
1100 – 1300 cm-1

Aryl Fluorides
1100 – 1250 cm-1
Chloride
(2 or more bands)

CH2-CL

t-Butyl Group
Bromine
(Bend Wagging)
(KBr Pellets)

CH2-Br

Aryl Bromides
Iodide

1/17/2015
1000 – 1400 cm-1
Fluoride
CH2-I
(Bend Wagging)
(KBr Pellets)
(Bend Wagging)
540 – 785 cm-1
1230 – 1300 cm-1
– 525 cm-1
510 – 650 cm-1
1190 – 1250 cm-1
1030 – 1075 cm-1
485 – 600 cm-1
1150 – 1200 cm-1
109
Infrared Spectroscopy
IR Spectrum
Halogens
2-Bromobutane
Br
-C-H
Sat’n
1/17/2015
C4H9Br
CH3
CH2
CH2-Br
CAS 78-76-2
110
Infrared Spectroscopy
IR Spectrum
Halogens
1-Chloropropane
CH3
CH2
CH2-Cl
-C-H
Sat’n
C3H7Cl
1/17/2015
Cl
CAS 540-54-5
111
Infrared Spectroscopy
IR Spectrum
Halogens
o-Chlorotoluene
=C-H
Unsat’n
-C-H
Sat’n
Aromatic
Overtones
O-Disubstitution
CH3
{
Cl
-C=CAromatic
1/17/2015
C7H7Cl
OOP
o-disubstitution
(750 cm-1)
(missing)
CH2-Cl
CAS 95-49-8
112
Infrared Spectroscopy
IR Spectrum
Halogens
T-Pentyl Chloride
(2-Chloro-2-MethylButane
Saturated
Aliphatic
C-H Stretch
C5H14CL
1/17/2015
CH3
CH2
CH2-Cl
T-Pentyl
525 cm-1
CAS 594-36-5
113
IR Analysis Scheme
Carbonyl (C=O) @ 1715-1685
(Conjugation moves absorption to right ~30 cm-1
Yes
No
Acid
Alcohol
Ester
Amide
Amine
Ether
Saturation
< 3000 cm-1
Unsaturation
> 3000 cm-1
Alkanes
-C-H
Methylene -CH2
Methyl
-CH3
Alkenes (Vinyl)
-C=C
Alkynes (Acetylenes) -C≡C
Aromatic
-C=C
Anhydride
Aldehyde
Ketone
Nitriles
1/17/2015
Hydrocarbons
Nitro
114
IR Analysis Scheme
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
1/17/2015
Amine
- 1 to 2 equal NH absorptions @ 3500 cm-1
Ether
- C-O absorption @ 1300 - 1000 cm-1
115
IR Analysis Scheme
Saturation
Alkanes
-C-H Stretch – several absorptions to “right” of 3000 cm-1
Methylene -CH2 1450 cm-1
Methyl
-CH3 1375 cm-1
Unsaturation
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 right
Alkynes
-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).
Double Bonds
1/17/2015
=C-H
116