INJIBARA UNIVERSITY
Instrumental Analysis II(Chem2052)
Prepared By Alemu Talema.
March 2 / 03 /2019
Injibara, Ethiopia
1
CHAPTER ONE: INTRODUCTION TO
SPECTROSCOPY
Introductions
Analytical chemistry is the study of the separation, identification,
and quantification of the chemical components of natural and
artificial materials.
Qualitative analysis gives an indication of the identity of the
chemical species in the sample,
And quantitative analysis determines the amount of certain
components in the substance.
Analytical methods can be separated into classical and instrumental.
2
CONT…
 Classical methods use separations such as precipitation, extraction,
and distillation.
 And classical qualitative analysis by color, odor, or melting point.
 Classical quantitative analysis is achieved by measurement of direct
weight or volume.
 Instrumental methods use an apparatus to measure physical quantities
of the analyte such as light absorption, fluorescence, or conductivity.
 The separation of materials is accomplished using chromatography,
electrophoresis or field flow fractionation methods.
3
INSTRUMENTAL METHODS
This methods involve the use of modern analytical instruments for chemical
analysis
Instruments serve as communication device between chemical species and the
chemist.
Instrumental Methods for the Analysis of Environmental Samples
1. Spectroscopic Methods
 Molecular UV/Vis Spectroscopy
 Atomic Absorption (AA) Spectroscopy
 Inductively Coupled Plasma (ICP) Spectroscopy
2. Chromatographic Methods = separations techniques based on differential
migration of solutes or analytes in a column
 Gas Chromatography (GC)
 High Performance Liquid Chromatography (HPLC)
4
WHAT IS SPECTROSCOPY?
 Spectroscopy is a general term for methods that investigate
interactions between electromagnetic radiation and matter.
 Spectroscopy uses electromagnetic radiation to investigate
properties of substances and for quantitative analysis of the
substance.
 Deals with the interactions of various types of radiation (mainly
electromagnetic radiation or light) with matter.
 Spectroscopic techniques are widely used to detect molecules, to
measure the concentration of a species in solution, and to
determine molecular structure.
5
EMR
 Electromagnetic radiation and its interaction with matter
It is kind of energy with wave character that can be
characterized by using wavelength (), frequency (), velocity
and amplitude.
 Electromagnetic radiation requires no supporting medium for its
transmission and passes readily through a vacuum.
 The electromagnetic wave consists of two fluctuating fields—one
electric field component and other a magnetic field component.
 The two vectors are at right angles (orthogonal) to one another,
and both are perpendicular to the direction of travel.
6
CONT…
 EMR can be either ionizing(UV, X-ray) or non-ionizing (Visible
light, Radio-wave), depending on its energy and ability to penetrate
matter.
Non-ionizing radiation, such as visible light, is not harmful.
7
CONT…
8
CONT…
 Radiation can travel up to 300,000 km/s (light speed) and slows
down when it passes through matter (s, l, g) but still travels
much faster than sound or water waves.Light
 The stream of discrete particles or wave packets of energy of
Electromagnetic radiation is a called photons where energy is
proportional to the frequency of the radiation. is el
In a vacuum
 All EM waves travel at the same speed
 Frequency and wavelength vary
9
CONT…
 All EMR / light behave as packets of energy called photons
 A photon is a particle of EMR
 It is the smallest quantity of any type of EM radiation
 Recall that the energy of a photon is determined by its frequency
(f) or wavelength (λ)
 And is given by E = hf = hc/ λ ,
where h is Planck's constant = 6.6254·10-34 J·s.
and c is the speed of light in vacuum.
10
CONT…
EX: Photons in a pale blue light have a wavelength of 500 nm. (The symbol nm is
defined as a nanometer = 10–9 m).What is the energy of this photon?
[ 6.625× 10−34 J.s × 2.998× 108 m / s ]
Solution: E = hc/λ =
=
500 × 10−9 m
6.625× 10−34 J × 2.998× 108
500 × 10−9
= 3.97 × 10-19 J
EX: Sodium vapor lamps are sometimes used for public lighting. They give off a
yellowish light with wavelength of 589 nm. What is the frequency of this
radiation?

c

11
CONT…
Common spectroscopic methods based on EMR
12
CONT…
When EMR strikes an object it is:
1)
2)
3)
Reflected/scattered
Transmitted
Absorbed
Absorbed
PO
Reflected
Transmitted
13
P
ABSORPTION OF RADIATION
 Atoms and molecules can absorb electromagnetic radiation.
 This can be rationalized as absorption of photons by molecules:
M + hν → M∗
 In above equation M stands for the molecule of substance, hν energy (photon) absorbed by the molecule. M* represents excited
molecule.
Planck's constant h = 6.6254·10-34 J·s.
 Excitation of a molecule means giving it more energy (molecule
passes to higher energy state). Lifetime of the excited state is
usually very short: about
10-12
to
10-9
seconds. Different
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CONT…
 Amount of light absorbed gives information about concentration of the
substance.
 When radiation passes through a sample certain frequency may be
selectively removed by absorption, a process in which electromagnetic
energy is transferred to the atoms, ions or molecules.
 Absorption promotes these particles from ground state to one or more
higher energy states to excited.
 If all the light passes through a solution without any absorption, then
absorbance is zero, and percent transmittance is 100%.
 If all the light is absorbed, then percent transmittance is zero, and
absorption is infinite.
15
CONT…
16
CONT…
17
CONT…
EXAMPLE:
(a) A sample has a percent transmittance of 50.0%. What is its
absorbance?
(b) A sample has a percent transmittance of 34.0%. What is its
absorbance?
SOLUTION:
(a) With a percent transmittance of 50.0%, the transmittance of the sample
is 0.50.
A = -log (0.5) = 0.301
(b) With a percent transmittance of 34.0%, the transmittance of the sample
is 0.34.
A = -log (0.34) = 0.469
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TYPES OF ABSORPTION BASED ON THE SAMPLES
1. Atomic Absorption: The passage of polychromatic ultraviolet or visible
radiation through a medium that consists of monoatomic particles results
the absorption of a few well-defined frequency.
 Such spectra is very simple due to the small number of possible energy
states for the absorbing particles.
2. Molecular Absorption: Absorption spectra for polyatomic molecules are
considerably more complex than atomic spectra because the number of
energy states of molecules is generally enormous when compared with the
number of energy states for isolated atoms.
 The energy E of a molecule is made up of three components (Er << Ev
<< Ee ), E = Eelectronic + Evibrational + Erotational
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EMISSION OF RADIATION:
 Emission of Radiation: Electromagnetic radiation is produced when excited particle (atoms, ions,
or molecules) relax to lower energy levels by giving up their excess energy as photons.
 For every emission process, there is an absorption process, radiation from an excited source is
characterized by means of an emission spectrum.
X  X*  X + h
Excitation can be done by:
1. Bombardment with electrons
2. Electric current ac spark
3. Heat of a flame
4. A furnace (produce uv, vis or ir radiation)
5. Beam of electromagnetic radiation
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CONT…
21
CONT…
Two types of emission:
1.Atomic emission
2.Molecular emission
22
THE ELECTROMAGNETIC SPECTRUM
Electromagnetic Spectrum—name for the range of electromagnetic
waves when placed in order of increasing frequency or it is the
distribution of photon energies coming from a light source.
Spectra are observed by passing light through a spectrograph:
- Breaks the light into its component wavelengths and spreads them
apart (dispersion).
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24
ABSORPTION LAWS
The Beer-Lambert Law: For monochromatic radiation, absorbance
is directly proportional to the path length “b” through the
medium and the concentration “c” of the absorbing species.
According to Beer’s law, absorbance is directly proportional to:
 concentration, c , of the absorbing species.
 path length, b , of the absorbing medium.
 proportionality constant, ε , called the absorptivity.
 These relationships are given by,
A = 𝒍𝒐𝒈𝑷𝟎/ P = ε 𝒃𝒄
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CONT…
Because absorbance is a unit less quantity, the absorptivity unit
depends on the units of b and c (must have units that cancel the
units of b and c). When we express the concentration in moles per
liter (molar, M) and b in cm, the proportionality constant is called
the molar absorptivity (or molar absorption coefficient) and is
given the symbol ε. Thus, where ε has the units of L mol-1 cm-1.
EXAMPLE:
A 5.00 x 10-5M solution of an analyte is placed in a sample cell that
has a path length of 1.00 cm. When measured at a wavelength of
490 nm, the absorbance of the solution is found to be 0.338. What
is the analyte’s molar absorptivity at this wavelength?
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CONT…
SOLUTION:
ε = A /𝒃𝒄 =
0.338
(1.00 cm) × (5.00 ×10−4M)
= 676 cm-1 M-1
EXAMPLE:
A 7.25x10-5 M solution of potassium permanganate has a transmittance of
44.1% when measured in a 2.10 cm cell at a wavelength of 525 nm.
Calculate (a) the absorbance of this solution and (b) the molar absorptivity
of KMnO4.
SOLUTION:
A = -logT = -log 0.441 = -(-0.356) = 0.356
ε =𝐀 /𝐛𝐜 0.356/(2.10 cm x 7.25 x 10-5mol L-1) = 2.34 x 103 L mol-1 cm-1
27
CONT…
The Law says that the fraction of the light absorbed by each layer
of solution is the same. For our illustration, we will suppose that this
fraction is 0.5 for each 0.2 cm "layer" and calculate the following data:
Path length /cm
%T
0
100
0.2
50
0.4
25
0.6
12.5
0.8
6.25
1
3.125
Absorbance
0
0.3
0.6
0.9
1.2
1.5
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CONT…
A = εbc tells us that absorbance depends on the total quantity of the
absorbing compound in the light path through the cuvette. If we
plot absorbance against concentration, we get a straight line
passing through the origin (0,0).
Note that the Law is not obeyed at high concentrations. This
deviation from the Law is not deal with here.
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LIMITATIONS TO BEER’S LAW
According to Beer’s law, a calibration curve of absorbance versus
the concentration of analyte in a series of standard solutions
should be a straight line with an intercept of 0 and a slope of ab
or εb. In many cases, however, calibration curves are found to be
nonlinear.
Deviations from linearity are divided
into three categories:
A. Fundamental or real,
B. Chemical,
C. Instrumental.
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CONT…
A. Fundamental or Real Limitations to Beer’s Law:
 Beer’s law is a limiting law that is valid only for low concentrations of analyte.
At high concentration:
- particles too close
- average distance between ions and molecules are diminished
- affect the charge distribution and extent of absorption.
- cause deviations from linear relationship.
 A second contribution is that the molar absorptivity depend on the sample’s
refractive index. Since the refractive index varies with the analyte’s
concentration, the values of molar absorptivity will change.
 For sufficiently low concentrations of analyte, the refractive index remains
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essentially constant.
CONT…
B. Chemical Limitations to Beer’s Law:
Occur when the absorbing species undergoes association,
dissociation, or reaction with the solvent to give products that
absorb differently from the analyte.
Consider, as an example, an analysis for the weak acid, HA. Since
HA is a weak acid, it exists in equilibrium with its conjugate weak
base, A–.
HA + H2O ⇄ H3O+ + AIf both HA and A– absorb at the selected wavelength, then Beer’s
law is written as: A = εHA 𝒃cHA + εA 𝒃cA
–
where CHA and CA are the equilibrium concentrations of HA and A
32 .
CONT…
C. Instrumental Limitations to Beer’s Law:
Polychromatic Radiation
Beer’s law is strictly valid for purely monochromatic radiation.
The value of ε is wavelength dependent. However, even the best
wavelength selector passes radiation with a small, but finite
effective bandwidth.
Using polychromatic radiation always gives a negative deviation
from Beer’s law, but is minimized if the value of ε is essentially
constant over the wavelength range passed by the wavelength
selector.
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CHAPTER TWO: INSTRUMENTS FOR
OPTICAL SPECTROSCOPY
The instruments for measuring each differ in configuration, most of
their basic components are remarkably similar.
Typical spectroscopic instruments contain five components:
(1) a stable source of radiant energy,
(2) a transparent container for holding the sample,
(3) a device that isolates a restricted region of the spectrum for
measurement,
(4) a radiation detector that converts radiant energy into a signal
detector,
(5) a signal processor and readout.
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CONT…
(1) The arrangement for absorption measurement
Light
Source
Wavelength
selector
Sample
Detector
Signal processor
and readout
(2) The arrangement for emission measurement
Sample
Wavelength
selector
Detector
Signal processor
and readout
Thermal
(3) The arrangement for fluorescence measurement
Light
Source
Wavelength
selector
Sample
Wavelength
selector
Detector
Signal processor
and readout35
CONT…
(4) The arrangement for chemiluminescence measurement
Sample +
Reagent
Wavelength
selector
Detector
Signal processor
and readout
Emission spectroscopy and Chemiluminescence spectroscopy differ
from the others in that no external radiation source is required. The
sample itself is the emitter
36
CONT…
Emission: sample container is a plasma, or flame that contains the
analyte, and also causes it to emit radiant light .
Chemiluminescence: radiation source is a solution of the analyte plus
reagents held in a transparent sample holder
(1) In the arrangement for absorption measurements, note that source
radiation of the selected wavelength is sent through the sample, and the
transmitted radiation is measured by the detector-signal processingreadout unit.
37
CONT…
(2) In the configuration for emission spectroscopy, here, a source
(no need of external sources) of thermal energy, such as a flame or
plasma, produces an analyte vapor that emits radiation isolated by
the wavelength selector and converted to an electrical signal by the
detector.
(3) In the configuration for fluorescence measurements, here, two
wavelength selectors are needed to select the excitation and emission
wavelengths.
The selected source radiation is incident on the sample and the
radiation emitted is measured, usually at right angles to avoid
scattering. Light source perpendicular to detector.
38
CONT…
39
1) SOURCES OF ENERGY
In order to be suitable for spectroscopic studies, a source:
- must generate a beam of radiation with sufficient power for easy
detection and measurement.
- All forms of spectroscopy require a source of energy. In
absorption and scattering spectroscopy this energy is supplied by
photons.
- Emission and luminescence spectroscopy use thermal, or
chemical energy to promote the analyte to a less stable, higher
energy state.
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CONT…
1. Sources of Electromagnetic Radiation:
A source of electromagnetic radiation must provide an output that is both intense and
stable in the desired region of the electromagnetic spectrum.
Sources of electromagnetic radiation are classified as either continuum or line sources.
A continuum source emits radiation over a wide range of wavelengths, with a relatively
smooth variation in intensity as a function of wavelength.
Line sources, on the other hand, emit radiation at a few selected, narrow wavelength
ranges.
2. Sources of Thermal Energy:
The most common sources of thermal energy are flames and plasmas. Flame sources
use the combustion of a fuel (acetylene, hydrogen…) and an oxidant(air, oxygen…) to
achieve temperatures of 2000–3400 K.
Plasmas, which are hot, ionized gases, provide temperatures of 6000–10,000 K.
41
CONT…
3. Chemical Sources of Energy
Exothermic reactions also may serve as a source of energy.
In chemiluminescence the analyte is raised to a higher-energy state
by means of a chemical reaction, emitting characteristic radiation
when it returns to a lower-energy state.
 When the chemical reaction results from a biological or enzymatic
reaction, the emission of radiation is called bioluminescence.
 Commercially available “light sticks” and the flash of light from a
firefly are examples of chemiluminescence and bioluminescence,
respectively.
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CONT…
- Output power should be stable for reasonable periods.
Sources are of two types.
A. Continuum sources
Continuous Sources
Visible and near IR
Ultraviolet
radiation
radiation
Tungsten Lamp
Deuterium Lamp
320-2500 nm
200-400 nm
B. Line Sources (HCL, LASERS)
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A. CONTINUUM SOURCES
Continuum Sources: - Continuum sources emit radiation that
changes in intensity only as a function of wavelength.
It is widely used in absorption and fluorescence spectroscopy.
UV region
Hydrogen and Deuterium Lamp,
High pressure gas‐filled lamp contains argon, xenon or mercury for
high intensity source
Visible region
Tungsten filament lamp
IR region
Inert solid ceramics heat to 1500‐2000k
44
CONT…
UV Region sources are:
Hydrogen Lamp ‐most common source for UV absorption measurements H2
emission is from 160 nm to 375 nm of wave length.
Deuterium Lamp: a truly continuous spectrum in the ultraviolet region is
produced by electrical excitation of deuterium at low pressure, (160nm~375nm
of wave length).
 Same λ distribution as H2 but with higher intensity (3 to 5 times),
 D2 is a heavier molecule & moves slower so there is less loss of energy
by collisions High pressure D2 . The reactions for deuterium are:
D2 + Ee  D2*  D’ + D" + hv,
 Deuterium gives a somewhat larger and brighter ball than hydrogen, which
45
accounts for the widespread use of deuterium.
CONT…
Mercury and Xenon Arc Lamps
- Produce radiation at wavelengths from 200 to 800 nm
- UV and Visible regions
Xenon Lamp–Xe at high pressure (10‐20 atm) ‐high pressure needed
to get lots of collisions for broadening leading to continuum..
Visible Region sources are : a)
Tungsten (W) filament‐normally operated at ~3000K
with inert atmosphere to prevent oxidation.
Useful from 350 nm to 2000 nm, below 350 nm glass envelope absorbs &
emission weak
b)Tungsten‐Halogen lamps‐can be operated as high as 3500K
. More intense (high flux).
46
CONT…
Function of halogen is to form volatile tungsten halide which
redeposite W on filament, i.e., keeps filament from burning out.
Requires quartz envelope to withstand high temps (which also tran
smits down to shorter wavelengths).
IR Region thermal sources (Black Body) are:
a) Nernst Glower–fused mixture of ZrO2, Y2O3, and ThO2
normally operated at 1900 oC –better for shorter IR λ’s (near IR)
47
CONT…
b) Globar
–silicon carbide normally operated at 1200 to 1400 oC –
better at longer IR λ’s .
- Produces radiation at wavelengths from 1200 to 40000 nm
c) Nichrome wire
–cheapest way.
NiChrome wire (750 nm to 20000 nm)
ZrO2 (400 nm to 20000 nm)
All operated at relatively low temperature.
Good for IR and give some visible emission.
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B. LINE SOURCES
Line Sources: Sources that emit a few discrete lines find wide use in
atomic absorption spectroscopy, atomic and molecular
fluorescence spectroscopy.
Examples
- Hollow cathode lamp (UV-VIS region)
- Electrodeless discharge lamp (UV-VIS region)
- Sodium vapor lamp (UV-VIS region)
- Mercury vapor lamp (UV-VIS region)
- Lasers (UV-VIS and IR regions)
49
HOLLOW CATHODE LAMP (HCL)
 The most common source for atomic absorption measurements is
the hollow-cathode lamp
50
CONT…
Ar ions bombard cathode and sputter cathode atoms
Repeated bombardment of the metal atom by the gas causes it to be excited. It
ultimately relaxes, producing specific atomic to be excited. It ultimately relaxes,
producing specific atomic emission lines. The metal atoms eventually diffuse back to
the emission lines. The metal atoms eventually diffuse back to the cathode surface.
51
CONT…
 This type of lamp consists of a tungsten (W) anode (positive-charged)
and a cylindrical cathode (negative charged) sealed in a glass tube filled
with neon or argon at a pressure of 1 to 5 torr. The cathode is
constructed of the metal (Li, Cu, Ni, Pb, etc) whose spectrum is desired
or serves to support a layer of that metal.
 Ionization of the inert gas (Ar) occurs when a potential difference on the
order of 300 V is applied across electrodes, which generates a current of
about 5 to 15 mA as ions and electrons migrate to the electrodes.
 If the voltage is sufficiently large, the gaseous cations acquire enough
kinetic energy to dislodge some of metal atoms from the cathode surface
and produce an atomic cloud in a process called sputtering.
52
CONT…
A portion of the sputtered (vaporized) metal atoms are in excited states and thus emit
their characteristic radiation as they return to the ground state. Eventually, the
metal atoms diffuse back to the cathode surface or to the glass walls the tube and
are re-deposited.
The cylindrical configuration of the cathode tends to concentrate the radiation in a
limited region of the metal tube; this design also enhances the probability that redeposition will occur at the cathode rather than on the glass walls. An hollow
cathode lamp will only produce the emission lines for the cathode element.
Disadvantage of hollow cathode lamps:
- A separate lamp source is needed for each element.
Example: Sodium vapor lamp for Na element.
Lithium vapor lamp for Li element.
Copper vapor lamp for Cu element.
53
ELECTRODELESS DISCHARGE LAMPS (EDL)
- Light intensity is about 100 times greater than that of HCL. A salt
of the metal of interest is sealed in a quartz tube along with an inert
gas.
- No electrode – argon is energized by microwaves or radio
frequency (RF) .
- Ionized argon is accelerated to excite the atoms and less stable
than HC
54
LASER SOURCES
* The term ‘LASER’ is an acronym for Light Amplification by
Stimulated Emission of Radiation.
* This is a device to produce a beam of monochromatic light in
which all the waves are in phase or are coherent.
* Laser are highly useful because of their
 - very high intensities,
 - narrow bandwidths,
 - single wavelength, and
 - coherent radiation.
* Laser are widely used in high-resolution spectroscopy.
55
COMPONENTS OF LASERS
Lasers contain four primary components regardless of style, size or
application.
- lasing medium,
- pumping source, and
- Partially Transmissive and High Reflectance Mirror.
56
CONT…
1. Active Medium: The heart of the device is the lasing medium. It may
be a solid crystal such as ruby, a semiconductor such as gallium
arsenide, a solution of an organic dye or a gas such as argon or
krypton.
 Active mediums contain atoms whose electrons may be excited to a
metastable energy level by an energy source.
2. Excitation Mechanism: Excitation mechanisms pump energy into the
active medium by one or more of methods; optical, electrical or chemical.
3. High Reflectance Mirror: A mirror which reflects essentially 100% of
the laser light.
4. Partially Transmissive Mirror: A mirror which reflects less than 100%
of the laser light and transmits the remainder.
57
CONT…
58
PROPERTIES OF LASER LIGHT
 The light emitted from a laser is monochromatic, that is , it is of
one or the same color/wavelength throughout their journey. In
contrast , ordinary white light is a combination of many colors (or
wavelengths) of light.
 Lasers emit light that is highly directional, that is, laser light is
emitted as a relatively narrow beam in a specific direction.
Ordinary light, such as from a light bulb, is emitted in many
directions away from the source.
 The light from a laser is said to be coherent, which means that the
wavelengths of the laser light are in phase in space and time.
Ordinary light can be a mixture of many wavelengths.
59
CONT…
Incoherent light waves
Coherent light waves
60
TYPES OF LASERS
1) Solid State Lasers
a) Ruby laser:
Al 2O3 + Cr(III) ‐694.3 nm pumped with Xe arc flashlamp –
pulsed (can be continuous)
b) Nd/YAG laser:
yttrium aluminum garnet + Nd (Neodymium) ‐1064 nm
2) Gas Lasers:
a) Neutral atom–He ‐Ne –632.8 nm continuous
b) Ion lasers–Ar + or Kr+ – 514.5 nm
61
CONT…
c) Molecular lasers –CO2 (10,000 nm) or N2 (337.1 nm) pulsed
d) Excimer lasers:
inert gas + fluorine creates excimers ArF+ (193 nm), KrF+ (248 nm)
,
XeF+ (351nm) pulsed
3) Dye Lasers:
tunable over 20 - 50 nm many dyes available for wide range of λ’s
4) Semiconductor Diode Lasers:
wide range of λ’s available, continuous
62
WAVELENGTH SELECTOR
 Wavelength selectors are important instrumental components that are
used to obtain a certain wavelength or a narrow band of wavelengths.
 Three main approaches:
 1) Block off unwanted radiation –optical filters
 2) Disperse radiation & select desired band – monochromator (prism)
 3) Modulate wavelengths at different frequencies ‐interferometer
The quality of a wavelength selector is measured by the inverse of the
effective bandwidth.
• Effective bandwidth is defined as the peak width at half height of a plot of
the output of a wavelength selector (% transmittance) as a function of
wavelength.
63
CONT…
64
TWO TYPES OF WAVELENGTH SELECTOR
1. Filters: (a) interference filter (UV , VIS , IR)
(b) absorption filter (VIS)
2. Monochromators －include 5 parts : slit , lense , mirror , window ,
and grating or prism.
(a)prism
(b) grating
Wavelength
selectors
Filters
Monochromators
65
CONT…
1.
Filters: Filters are wavelength selectors that usually allow the passage of a band of
wavelengths and can be divided into three main categories.
A) Absorption Filters: This type of filters absorbs most incident wavelengths and
transmits a band of wavelengths.
Sometimes, they are called transmission filters.
It is colored glass, colored film, colored solutions –cheapest way.
Properties:
 Cheap and can be as simple as colored glasses or plastics.
 They transmit a band of wavelengths with an effective bandwidth: (The effective
band width is the width of the band at half height) in the range from 30-250 nm).
 Their transmittance is usually low where only about 10-20% of incident beam is
transmitted.
66
CONT…
Absorption filters are also known as bandpass filters
Usually exhibit low peak transmittance
Can use two or more absorption filters together to produce desired
transmittance characteristics
Generic filters are 2 x 2 inch glass or quartz
Relatively inexpensive
Disadvantage
 Range of wavelengths transmitted is very broad (50 – 300 nm)
67
INTERFERENCE FILTERS
Made up of multiple layers of materials
The thickness and the refractive index of the center layer of the material control
the wavelengths transmitted
Range of wavelengths transmitted are much smaller (1 – 10 nm)
Amount of light transmitted is generally higher
Transmits light in the IR, VIS, and UV regions
Interference-operate by internal reflections and constructive/destructive
interference. Set up a cavity (metal film/CaF or MgF dielectric/metal film)
which has dimensions which is a multiple of the desired wavelength, all other
wavelengths will then be rejected by destructive interference.
68
2. MONOCHROMATOR
Monochromator (Monochrome = “one colour”) to spread out or
disperse light into its component wavelengths and select the
required wavelength for analysis.
One limitation of an absorption or interference filter is that they do
not allow for a continuous selection of wavelength.
A further limitation is that filters are available for only selected
nominal ranges of wavelengths.
An alternative approach to wavelength selection, which provides for
a continuous variation of wavelength, is the monochromator.
69
COMPONENTS OF A MONOCHROMATOR:
 Entrance slit (restrict unwanted radiation)
 Dispersing element (separate the wavelengths of the
polychromatic radiation)
 prism and reflection grating
 Exit slit – adjustable (control the width of the band of
wavelengths)
1. Entrance slit: provides a rectangular optical image of the
incoming polychromatic radiation.
 Polychromatic radiation (radiation of more than one wavelength)
enters the monochromator through the entrance slit.
70
CONT…
2) Colliminating Lens or Mirror: The beam is collimated, and then
strikes the dispersing element at an angle or provides a parallel beam
of radiation that impinges upon the dispersive element.
The radiation is collected by a collimating mirror, which reflects a
parallel beam of radiation to a diffraction grating.
71
CONT…
3.Prism or Grating: The beam is split into its component wavelengths by
the grating or prism.
It (dispersive element) disperses the polychromatic radiation by the
process of diffraction. The diffraction grating is an optically reflecting
surface with a large number of parallel grooves.
4) Focusing Lens or Mirror - Focuses the dispersed radiation on the exit
slit. By moving the dispersing element or the exit slit, radiation of only a
particular wavelength leaves the monochromator through the exit slit.
Diffraction by the grating disperses the radiation in space, where a second
mirror focuses the radiation onto a planar surface containing an exit slit.
72
CONT…
5) Exit Slit: Isolate the wavelength band of interest or radiation of only a
particular wavelength leaves the monochromator through the exit slit. Radiation
exits the monochromator and passes to the detector.
A narrower exit slit provides a smaller effective bandwidth and better resolution,
but allows a smaller throughput of radiation.
Monochromator Slits:
Two pieces of carefully machined metal to give sharp
edges that are exactly parallel to one another.
Commonly the slits are connected to a micrometer mechanism so the slit width
can be adjusted. The effective bandwidth of the radiation exiting a
monochromator is directly proportional to the slit width.
Throughput
Resolution
Quant
Qual
Wide slits
High
Low
Good
Poor
Narrow slits
Low
High
Poor
Good
73
PERFORMANCE CHARACTERISTICS OF
MONOCHROMATORS
1. Spectral Purity - The purity of the exit beam from the Monochromator.
Exit beams are usually contaminated with some quantity of scattered or
stray radiation of a different wavelength than the preferred wavelength.
Sources of impurities:
Reflections of the radiation beam from various optical components due to
mechanical imperfections.
Scattering due to dust particles in the optical path.
2. Resolution - The ability of a monochromator to resolve different
wavelengths.
74
CONT…
3. Light Gathering Power -A measure of the amount of radiation
that reaches the detector.
The more radiant energy that reaches the detector the greater the
signal-to-noise ration of the resulting measurement.
4. Spectral Bandwidth - The bandwidth of radiation that is output
by the monochromator.
• Usually expressed as the effective bandwidth of the sharpest peak
in a spectrum.
Bandwidth of a monochromator is most affected by the width of the
entrance and exit slits of a monochromator.
75
SAMPLE CONTAINERS
Sample containers are required for all spectroscopic studies except emission
spectroscopy. In common with the optical elements of monochromators, the
cells or cuvettes that hold the samples must be made of material that passes
radiation in the spectral region of interest. Quartz or fused silica is required
for work in the ultraviolet region (below '150 nm); both of these substances are
transparent in the visible region. Silicate glasses can be employed in the region
between 350 and
2000 nm. Plastic containers have also found application in the visible region.
Crystalline sodium chloride is the most common substance employed for cell
windows in the infrared region.
Spectral Region
Material
UV
Fused silica
Must be made of material that is transparent to the VIS
spectral region of interest over range of wavelengths.IR
Plastic, glass
NaCl
76
RADIATION TRANSDUCERS (DETECTORS)
Modern instruments contain devices that convert the radiation to an
electrical signal.
A device that converts a chemical or physical property, such as pH
or photon intensity, to an easily measured electrical signal, such as
a voltage or current.
A. Types of Radiation Transducers and Ideal Properties
1.Two general types of radiation transducers
A. Photon detectors
B. Thermal detectors
77
A. PHOTON DETECTORS
Photoelectric (or quantum) detectors have an active surface, which
is capable of absorbing EM radiation.
 Phototubes and photomultipliers contain a photosensitive surface
that absorbs radiation in the ultraviolet, visible, and near infrared
(IR), producing an electric current proportional to the number of
photons reaching the transducer.
 Other photon detectors use a semiconductor as the photosensitive
surface. When the semiconductor absorbs photons, valence
electrons move to the semiconductor’s conduction band,
producing a measurable current.
78
CONT…
Several types of photon detectors are available:
1.
Vacuum phototubes
2.
Photomultiplier tubes
3.
Photovoltaic cells
4.
Silicon photodiodes
5.
Diode array transducers
6. Photoconductivity transducers
Types 1 & 2 relay on the release of electrons from a photosensitive
surface
79
1. VACUUM PHOTOTUBES
 Composed of a semicylindrical cathode and a wire anode inside of an
evacuated, transparent tube.
 The cathode is coated with a photoemissive
material (e.g. Ga/As) that emits
electrons when struck by EM radiation.
 By applying a potential across the cathode and anode, the electrons will flow
from the cathode to the anode resulting in a photocurrent.
 The photocurrent is amplified and sent to a readout device.
80
2.
PHOTOMULTIPLIER TUBE (PMT)
 Works on the same principle as the vacuum phototube.
 Useful for the measurement of low radiant powers.
 Contains several additional electrodes known as dynodes, each capable of
releasing electrons.
It is a commonly used detector in UV-Vis spectroscopy and visible radiation. It
consists of a photoemissive cathode (a cathode which emits electrons when struck by
photons of radiation), several dynodes (which emit several electrons for each electron
81
striking them) and an anode.
CONT…
82
CONT…
A photon of radiation entering the tube strikes the cathode, causing the emission of
several electrons. These electrons are accelerated towards the first dynode (which
is 90V more positive than the cathode). The electrons strike the first dynode,
causing the emission of several electrons for each incident electron. These electrons
are then accelerated towards the second dynode, to produce more electrons which
are accelerated towards dynode three and so on. Eventually, the electrons are
collected at the anode. By this time, each original photon has produced 106 - 107
electrons. The resulting current is amplified and measured.
Photomultipliers are very sensitive to UV and visible radiation. They have fast
response times. Intense light damages photomultipliers; they are limited to
measuring low power radiation.
83
3. PHOTOVOLTAIC CELL
Constructed of a semiconducting layer (Se) deposited on an iron or
copper cathode and the semiconductor is coated with a thin
metallic layer (Au or Ag), which serves as the anode.
• When radiation reaches the semiconductor, covalent bonds are
broken resulting in conduction electrons and holes.
• The electrons flow towards the metallic layer (anode) and the holes
flow towards the base of the semiconductor (cathode). The electrons
then flow through the circuit resulting in a current that is
proportional to the power of the radiation.
• Maximum sensitivity at 550 nm and falls of at 350 nm and 750 nm,
so they are most useful for visible radiation.
84
B. THERMAL DETECTORS
Used for infrared spectroscopy because photons in the IR region lack
the energy to cause photoemission of electrons. IR photons don't
have enough energy to dislodge photoelectrons so detect by the
heat they generate. Have a blackened surface which absorbs heat
photons, raising temperature and resulting in an increase an
electrical signal. Must measure temperature differences of a few
milli °C.
Three types of thermal detectors
1. Thermocouples
2. Bolometers
3. Pyroelectric transducers
85
CONT…
1. Thermocouples: Consist of a pair of bimetal junctions whose
potential (voltage) varies as a function of temperature.
Usually constructed of two copper leads fused to a dissimilar metal.
Thermocouples have a slow response time. They are more common
in older scanning IR instrument.
2. Bolometers: Constructed of strips of metal (platinum or nickel)
whose resistance (W) changes as a function of temperature.
3. Pyroelectric Transducers: Constructed of crystalline wafers
(triglycine sulfate) that have a strong, sensitive temperature
dependent polarization. TGS crystals have a fast response time and
are suitable and commonly used as detectors for IR.
86
CONT…
3. Generally, detectors have the following Properties:
a. High sensitivity
b. High signal-to-noise ratio
c. Constant response over a large range of wavelengths
d. Fast response time
87
CHAPTER THREE
Atomic Absorption and
emission spectroscopy
88
1. ATOMIC ABSORPTION SPECROSCOPY
ATOMIC ABSORPTION SPECROSCOPY : Atomic absorption
spectroscopy the most widely used method for the determination of
single elements in analytical samples. „
AAS is an elemental analysis
technique capable of providing quantitative information on
approximately 70 elements in almost any type of sample. A
quantitative method of analysis is based on the absorption of light by
atoms in the free atomic state. „
„
The method relies on the Beer-Lambert relationship. Lambert
relationship calculations are the same as with absorption
methods.„
89
BASIC PRINCIPLE
Atomic absorption spectroscopy (AAS) is an analytical technique that
measures the concentrations of elements in the atomic level.
 Applicable to many metals and few non metals.
 It makes use of the absorption of light by these elements in order to
measure their concentration
 Atomic-absorption spectroscopy quantifies the absorption of ground
state atoms in the gaseous state .
 The atoms absorb ultraviolet or visible light and make transitions to
higher electronic energy levels.
 The analyte concentration is determined from the amount of absorption.
90
GENERALLY…
- Liquid sample is sucked
- Sample passes through a plastic tube into a flame
- Flame breaks molecules into atoms (atomization)
- Monochromator selects wavelength that reaches the detector
The concentration of elements is measured by
emission or absorption radiation
- Concentrations are measured at the ppm level
In most cases our analyte is in solution form. If our sample is a solid, then we must
bring it into solution before the analysis.
When analyzing a lake sediment for Cu, Zn, and Fe, for example, we bring the analytes
into solution as Cu2+, Zn2+, and Fe3+ by extracting them with a suitable reagent.
91
THEORY OF OPERATION
• When atoms are subjected to
heat or some form of EMR, one
or more electrons jump to a
higher energy level, leaving a
vacancy in the inner shell
Atoms
• We say the electron is excited
• As this happens, energy is
absorbed
92
THEORY OF OPERATION
• The
excited electron in the
outer orbital returns to the
lower energy level of the inner,
vacant orbital, energy is
released in the form of a
photon
The solvent of the solution is evaporated and all materials present in the
sample are vaporised and dissociated to atoms at the very high
temperature.
93
GENERAL SET UP OF AAS
Light
source
Po
Atomizer
P
monochromator
detector
readout
(λ selector)
Sample
• Atomizer converts liquid sample into
free atoms
which absorb energy from the lamp
• Monochromator selects wavelength
used for measurement
• Detector measures light absorbed by
free atoms
94
CONT…
If operating a single beam AAS always allow sample warm-up time
for radiation sources because the intensity drifts with time.
95
CONT…
If operating a Double beam AAS, drift is minimized by the use of a
reference beam, and little to no warm-up time is required.
96
RADIATION SOURCE
The light source is usually a hollow cathode lamp of the element that
is being measured .
It contains a tungsten anode and a hollow cylindrical cathode made
of the element to be determined.
These are sealed in a glass tube filled with an inert gas (neon or
argon ) . Each element has its own unique lamp which must be
used for that analysis .
97
CONT…
Cathode material made of the element of interest, e.g. Na HCL for
the analysis of Na.
An individual lamp is needed for each element. So AAS is a oneelement-at-a-time measurement!
98
SAMPLE ATOMIZATION
Elements to be analyzed needs to be in atomic state.
The process of converting an analyte in solid, liquid, or solution
form to a free gaseous atom is called atomization.
Converting an aqueous analyte into a free atom requires that we
strip away the solvent, volatilize the analytes, and, if necessary,
dissociate the analyte into free atoms.
Desolvating an aqueous solution of CuCl2, for example, leaves us
with solid particulates of CuCl2.
Converting the particulate CuCl2 to gas phases atoms of Cu and Cl
requires thermal energy.
CuCl2(aq)  CuCl2(s)  CuCl2(g )  Cu( g) + 2Cl(g )
99
CONT…
100
PROCESSES OCCURRING DURING ATOMIZATION
Basic steps:
a) Nebulization – solution sample, get into fine droplets by
spraying through thin nozzle or passing over vibrating crystal.
b) Desolvation - heat droplets to evaporate off solvent just leaving
analyte and other matrix compounds
c) Volatilization – convert solid analyte/matrix particles into gas
phase
d) Dissociation – break-up molecules in gas phase into atoms.
e) Ionization – cause the atoms to become charged
f)
Excitation – with light, heat, etc. for spectra measurement.
101
PROCESSES OCCURRING DURING ATOMIZATION
102
CONT…
Some of the atoms in the gas ionize to form cations and electrons.
Other molecules and atoms are produced in the flame as a result of
interactions of the fuel with the oxidant and with the various
species in the sample.
Types of atomization
There are two common atomization methods: flame atomization and
electrothermal atomization, although a few elements are atomized using
other methods.
103
FLAME ATOMIZERS
Flame AA can only analyze solutions , where it uses a slot type
burner to increase the path length, and therefore to increase the
total path length, and therefore to increase the total absorbance .
Sample solutions are usually introduced into a nebuliser, by being
sucked a capillary tube.
In the nebuliser the sample is dispersed into tiny droplets, which can
be readily broken down in the flame.
104
CONT….
105
TYPES OF FLAMES
Vary flame temperature by fuel/oxidant mixture.
Fuel
Oxidant Temperature (K)
Acetylene
Air
2,400 - 2,700
Acetylene Nitrous Oxide 2,900 - 3,100
Acetylene
Oxygen
3,300 - 3,400
Hydrogen
Air
2,300 - 2,400
Hydrogen
Oxygen
2,800 - 3,000
Cyanogen
Oxygen
4,800
Selection of flame type depends on the volatilization temperature of the
atom of interest.
106
BURNING VELOCITY
The burning velocities are important because flames are stable only in
certain ranges of gas flow rates.
If the gas flow rate does not exceed the burning velocity, the flame
propagates itself back in to the burner, giving flashback.
As the flow rate increases, the flame rises until it reaches a point above the
burner where the flow velocity and the burning velocity are equal.
This region is where the flame is stable.
At higher flow rates, the flame rises and eventually reaches a point where
it blows off of the burner.
With these facts in mind, it is easy to see why it is so important to control
the flow rate of the fuel-oxidant mixture.
107
FLAME STRUCTURE
Flame profile: depends on type of fuel and oxidant and mixture ration
108
CONT…
Flame Structure – selection of correct flame region is important for
optimal performance
a) primary combustion zone – is recognizable by its blue luminescence
arising from the band emission of C2, CH and other radicals, in a
hydrocarbon flame . Thermal equilibrium is usually not achieved in this
region, and it is therefore, rarely used for flame spectroscopy
b) Interzonal region
- region of highest temperature (rich in free atoms)
- often used in spectroscopy
- can be narrower in some flames (hydrocarbon) tall in others
(acetylene), primary region for spectroscopy.
c) Outer zone / Secondary zone : oxygen present so stable molecular oxides
are formed for some metals and are dispersed into the surroundings.
- cooler region
- rich in O2 (due to surrounding air)
- gives metal oxide formation
109
PERFORMANCE CHARACTERISTICS OF FLAME
ATOMIZERS
In terms of reproducible behavior, flame atomization appears to be superior to all
other methods for liquid sample introduction.
ADVANTAGES:
1. Uniform drope size
2. Homogeneous flame
3. Quiet flame and a long path length
DISADVANTAGES:
1. A large portion of the sample flows down the drain , ~90% of sample is lost
2. The residence time of individual atoms in the optical path in the flame is brief
(10-4s).
3. Flash back, if Vburning > Vflow
110
ELECTROTHERMAL ATOMIZATION
A significant improvement in sensitivity is achieved by using the resistive
heating of a graphite tube in place of a flame. A typical electrothermal
atomizer, also known as a graphite furnace, consists of a cylindrical
graphite tube approximately 1–3 cm in length and 3–8 mm in diameter.
The graphite furnace has several advantages over a flame.
 First it accept solutions, slurries, or solid samples.
 Second it is a much more efficient atomizer than a flame and it can
directly accept very small absolute quantities of sample.
Samples are placed directly in the graphite furnace and the furnace is
electrically heated in several steps to dry the sample, ash organic
matter, and vaporize the analyte atoms.
111
CONT…
A, water-cooled electrical graphite
contact cylinders; B, graphite
tube; C, sample injection port; D,
light path of the spectrometer
112
CONT…
During electrothermal atomization, a sample goes through three phases to
achieve atomization.
First, the sample is dried at a low temperature.
Then the sample is ashed in a graphite furnace ,
Followed by a rapid temperature increase within the furnace where the
sample becomes a vapor containing atoms from the sample. Absorption
is measured above the heated surface where the sample was atomized.
A graphite furnace is made up of a graphite tube open at both ends with a
hole in the center for sample introduction. The tube is encased within
graphite electrical contacts at both ends that serve to heat the sample. A
supply of water is used to keep the graphite furnace cool.
113
CONT…
An external stream of inert gas flows around the tube to prevent outside
air from entering the atomization environment, Outside air can
consume and destroy the tube.
An internal stream of inert gas flows through the tube, carrying away
vapors from the sample matrix.
Electrothermal atomizers provide enhanced sensitivity because samples
are atomized quickly and have a longer residence time compared to
flame AAS systems, which means more of the sample is analyzed at
once.
Electrothermal atomization also offers the advantage of smaller sample
size and reduced spectral interferences because of the high temperature
of the graphite furnace.
114
CONT…
However, electrothermal atomizers have disadvantages including
slow measurement time because of the heating and cooling required
of the system.
Additionally, analyte and matrix diffuse into the graphite tube, and
over time, the tube needs replacing, increasing maintenance and cost
associated with electrothermal atomization
115
GRAPHITE FURNACE AND FLAME AA
METHODS
Graphite Furnace
Flame AA
Use small sample size
0.05 ul-100ul or 0.005 ml-0.1ml
Use larger sample size 3-5 ml
Analyse sample in pbb range
Analyse sample in ppm range
Longer analysis time 5 mins.
Fast. 10 secs.
Single measurement technique.
Sample is continuously aspirated.
A fixed vol of sample is analysed at one Mamy measurements can be taken during
time
an aspiration
Sample prep is minimised.
Involve sample prep. Extraction, digestion
etc.
Sample type can be solid, slurry, powder
Sample must be in solution only.
or soln.
Longer residence time for atoms in
furnace
Shorter residence time for atoms in flame.
116
CONT…
Graphite Furnace
Flame AA
Electrothermal analyser - Graphite
tube.
Flame atomiser-Burner and nebulizer.
Temperature programming.
Depends on temperature of flame.
Can analyse about 40 elements from Can analyse about 70 of the elements.
periodic table..
High cost
Lower cost
Measure ground state atoms.
Also measure ground state atoms.
Base on Principles of AA
spectroscopy.
Also, base on Principles of AA spectroscopy.
117
APPLICATION OF AAS
There are many applications for atomic absorption:
- Clinical analysis : Analyzing metals in biological fluids such as blood and urine.
- Environmental analysis : Monitoring our
environment – e g finding out the levels of various elements in rivers, seawater,
drinking water, air, and petrol.
- Pharmaceuticals. In some pharmaceutical
manufacturing processes, minute quantities of a
catalyst used in the process (usually a metal) are
sometimes present in the final product. By using
AAS the amount of catalyst present can be
determined.
118
CONT…
- Industry : Many raw materials are examined and
AAS is widely used to check that the major elements
are present and that toxic impurities are lower than
specified – e g in concrete, where calcium is a major
constituent, the lead level should be low because it is
toxic.
-
Mining: By using AAS the amount of metals such as gold in rocks can be
determined to see whether it is worth mining the rocks to extract the gold .
- Trace elements in food analysis
- Trace element analysis of cosmetics
- Trace element analysis of hair
119
2. ATOMIC EMISSION SPECTROSCOPY
Atomic emission spectroscopy is also an analytical technique that is
used to measure the concentrations of elements in the samples.
It uses quantitative measurement of the emission from excited
atoms to determine analyte concentration.
Schematic Diagram of an Atomic Emission spectrometer
120
CONT…
The instrumentations of AES:
 Similar to AA, but no need for external light source (HCL)
 look at light from flame
 flame acts as sample cell & light source
Uses the intensity of light emitted from a flame, plasma, determine the quantity
(concentration) of an element in a sample.
A simple flame photometer (AES) consists of burner, nebulizer, monochromator,
detector and recorder.
Principles: The thermal energy of the source excites the atoms in to higher
energy levels (i.e., excited states) that subsequently emit light when they return to
the ground state.
The emitted light does not have a continuum therefore is expressed as an atomic
spectral lines specific for given element.
121
CONT…
 This is simply called as ‘Flame Photometry’, and measures the
atoms excited by a flame (temperature range: 2000 – 31000 K).
 The analyte atoms are promoted to a higher energy level by the
sufficient energy that is provided by the high temperature of the
atomization sources .
 The excited atoms decay back to lower levels by emitting light .
 The wavelengths of the emitted light will almost be similar as
those that were absorbed in the atomic absorption, since exactly
the same energy transitions occur, except in the order of reverse!
122
CONT…
 The source of energy in Atomic Emission could be a flame like the
one used in atomic absorption or an inductively coupled plasma
(ICP )
 The flame (1700–3150oC) is most useful for elements with
relatively low excitation energies like sodium potassium and
calcium .
 The ICP (6000–8000oC) has a very high temperature and is useful
for elements of high excitation energies .
123
A) FLAME SOURCE
- Used mostly for alkali metals
> easily excited even at low temperatures
- Na, K
- need internal standard (Cs usually) to correct for variations
flame
Advantages
- cheap
Disadvantage
- not high enough temperature to extend to many other elements
124
INDUCTIVELY COUPLED PLASMA (ICP)
 Plasma is a type of electrical discharge
 Plasma is any type of matter that contains electrons and +ve ions.
 Plasma has 2 characteristics:
i- can conduct electricity
II- affected by magnetic fields
 Plasma is highly energetic ionized gases usually inert, recently reactive oxygen
is used.
 Other plasmas include direct current plasma (DCP) and microwave induced
plasma (MIP)
125
CONT…
Magnetic field
Temperature Regions
in Plasma Torch
126
CONT…
The ICP torch consists of three concentric quartz tubes, with the inner tubes
containing sample aerosol and Ar support gas and outer tube containing an Ar gas
flow to cool the tubes. A radio frequency (RF) generator produces an oscillating
current in an induction coil that wraps around the tubes. The induction coil creates
an oscillating magnetic field.
The magnetic field in turn set up in an oscillating current in the ion and electrons of
the support gas. The ions and electrons transfer energy to the other atoms in the
support gas by the collisions to create a very high temperature plasma.
The sample is mixed with a stream of Ar using a nebulizer, and is carried to the plasma
through the torch’s central capillary tube.
Plasma formation is initiated by a spark from a Tesla coil. An ICP is a very high
temperature (8000 – 10,000K) excitation source that efficiently desolvates,
vaporizes, excites and ionizes atoms. Molecular interferences are greatly reduced
with this excitations sources but are not eliminated completely.
127
CONT…
Generally, plasma – electrically conducting gaseous mixture (cations & electrons)
- temperature much higher than flame
- possibility of doing multiple element analysis
> 40-50 elements in 5 minutes
Advantages
- uniform response
- multi-element analysis, rapid
- few inter-element interferences
- can use with gas, liquid or solids sample
128
Disadvantages: many spectral and non spectral interferences, the lowest detection
CHAPTER FOUR
ULTRAVIOLET AND
VISIBLE (UV-VIS)
SPECTROSCOPY
129
CONT…
UV-Visible: This absorption spectroscopy uses electromagnetic
radiations between 190 nm to 800 nm and is divided into the
ultraviolet (UV, 190-400 nm) and
visible (VIS, 400-800 nm) regions. Since the absorption of
ultraviolet or visible radiation by a molecule leads transition among
electronic energy levels of the molecule, it is also often called as
electronic spectroscopy.
130
Basic Principles
 Ultraviolet and visible (UV-Vis) absorption spectroscopy is the
measurement of the attenuation (intensity) of a beam of light after
it passes through a sample.
 A beam of light from a visible and/or UV light source is separated
into its component wavelengths by a prism or diffraction grating.
 Each monochromatic (single wavelength) beam in turn is split
into two equal intensity beams by a half-mirrored device.
131
CONT…
132
CONT…
 One beam, the sample beam , passes through a small transparent
container (cuvette) containing a solution of the compound being
studied in a transparent solvent.
 The other beam, the reference, passes through an identical
cuvette containing only the solvent.
 The intensities of these light beams are then measured by
electronic detectors and compared.
 The intensity of the reference beam, which should have no light
absorption, is defined as I0.
133
CONT…
The intensity of the sample beam is defined as I.
Over a short period of time, the spectrometer automatically scans all
the component wavelengths in the manner described.
The ultraviolet (UV) region scanned is normally from 190 to 400 nm,
and the visible portion is from 400 to 800 nm.
134
INSTRUMENTATION
Instrumentation: It consists of a dual light source, tungsten lamp for
visible range and deuterium lamp for ultraviolet region, grating
monochromator, photo-detector, mirrors and glass or quartz cells.
NOTE: For measurements to be made under visible region both
glass and quartz cells can be used.
For the measurements under ultraviolet region, only quartz cell
should be used, since, glass cells absorb ultraviolet rays.
Light Source
Tungsten filament lamps and Hydrogen-Deuterium lamps are most
widely used and suitable light source as they cover the whole UVVis region.
135
CONT…
Tungsten filament lamps are rich in red radiations; more specifically they
emit the radiations of 375 nm,
while the intensity of Hydrogen-Deuterium lamps falls below 375 nm.
Monochromator
Monochromators generally is composed of prisms and slits.
Most of the spectrophotometers are double beam spectrophotometers.
The radiation emitted from the primary source is dispersed with the help
of rotating prisms.
The various wavelengths of the light source which are separated by the
prism are then selected by the slits.
136
SAMPLE AND REFERENCE CELLS
 One of the two divided beams is passed through the sample solution
and second beam is passé through the reference solution.
 Both sample and reference solution are contained in the cells.
 These cells are made of either silica or quartz.
 Glass can’t be used for the cells as it also absorbs light in the UV region.

Detector
 Generally two photocells serve the purpose of detector in UV spectroscopy.
 One of the photocell receives the beam from sample cell and second detector
receives the beam from the reference.
 The intensity of the radiation from the reference cell is stronger than the beam
of sample cell.
 This results in the generation of alternating currents in the photocells.
137
AMPLIFIER
The alternating current generated in the photocells is transferred to
the amplifier.
Generally current generated in the photocells is of very low intensity,
the main purpose of amplifier is to amplify the signals many times
so we can get clear and recordable signals.
138
NATURE OF ELECTRONIC TRANSITIONS
The total energy of a molecule is the sum of its electronic, its
vibrational energy and its rotational energy.
Energy absorbed in the UV region produces changes in the
electronic energy of the molecule.
As a molecule absorbs energy, an electron is promoted from an
occupied molecular orbital (usually a non-bonding n or bonding π
orbital) to an unoccupied molecular orbital (an antibonding π∗ or
σ* orbital)
139
CONT…
~ 150-250 nm
~ 115 nm
~ 400 - 700 nm
~ 200 – 400 nm
140
Absorption: Physical Basis
Absorption occurs when the energy contained in a photon is
absorbed by an electron resulting in a transition to an
excited state
Since photon and electron energy levels are quantized, we
can only get specific allowed transitions
E=h
~ 400 - 700 nm
(h = 6.626*10-34 Js)
~ 115 nm
~ 200 – 400 nm
~ 150-250 nm
http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/uvvisab1.htm
CONT…
Depending on the functional groups the organic molecules may
undergo several possible transitions which can be placed in the
increasing order of their energies:
n → π* < n → σ* < π → π* < σ → π* < σ → σ*.
142
CONT…
Alkanes can only undergo σ → σ* transitions.
 These are high-energy transitions and involve very short
wavelength ultraviolet light (< 150 nm).
 These transitions usually fall out-side the generally available
measurable range of UV-visible spectrophotometers (200-1000
nm).
 The σ → σ* transitions of methane and ethane are at 122 and
135 nm, respectively.
In alkenes amongst the available σ → σ* and π → π* transitions,
the π → π* transitions are of lowest energy and absorb radiations
between 170-190 nm.
143
CONT…
In saturated aliphatic ketones the lowest energy transition involves
the transfer of one electron of the nonbonding electrons of oxygen
to the relatively low-lying π* anti-bonding orbital.
This n→π* transition is of lowest energy (~280 nm) but is of low
intensity as it is symmetry forbidden.
Two other available transitions are n → π* and π → π*.
The most intense band for these compounds is always due to π → π*
transition.
144
CONT…
In conjugated dienes the π → π* orbitals of the two alkene groups
combine to form new orbitals – two bonding orbitals named as π1
and π2 and two antibonding orbitals named as π3* and π4*.
It is apparent that a new π → π* transition of low energy is
available as a result of conjugation. Conjugated dienes as a result
absorb at relatively longer wavelength than do isolated alkenes
O
O
O
max = 238, 305 nm
max = 240, 311 nm
max = 173, 192 nm
145
CHROMOPHORE
A chromophore is that part of a molecule that absorbs UV or visible light.
Alkanes: molecules contain single bonds and the only possible transitions
are σ to σ* .
Absorb ultraviolet energy at very short wavelengths below 200 nm.
 Alcohols, Ethers, Amines, and Sulfur Compounds: In saturated
molecules that contain atoms bearing nonbonding pairs of electrons,
possible transitions of the n to σ*. They are also high-energy transitions.
 Alcohols and amines absorb in the range from 175 to 200 nm;
Organic thiols (RSH) and sulfides absorb between
200 and 220 nm.
146
CONT…
Alkenes and Alkynes: Possible transitions are π to π*. These
transitions are of rather high energy (170 nm) as well, but their
positions are sensitive to the presence of substitution
Carbonyl Compounds: Unsaturated molecules that contain atoms
such as oxygen or nitrogen may also undergo n to π*transitions
(280 to 290 nm).
Carbonyl compounds also have a π to π* transition at about 188
nm
147
SUBSTITUENT EFFECTS
The attachment of substituent groups in place of hydrogen on a basic
chromophore structure changes the position and intensity of an absorption
band of the chromophore. Substituents that increase the intensity of the
absorption and the wavelength, are called auxochromes.
The substituents like methyl, hydroxyl, alkoxy, halogen, amino group etc. are some
examples of auxochromes.
148
CONT…
 Generaly – Substituents may have any of four effects on a chromophore
i. Bathochromic shift (red shift): a shift (an absorption maximum) to
longer  ; lower energy.
ii. Hypsochromic shift (blue shift): shift (an absorption maximum) to
shorter ; higher energy.
iii. Hyperchromic effect – an increase in absorption intensity
iv. Hypochromic effect – decrease in absorption intensity
Solvent Effects
Highly pure, non-polar solvents such as saturated hydrocarbons do not
interact with solute molecules either in the ground or excited state and the
absorption spectrum of a compound in these solvents is similar to the one
in a pure gaseous state.
149
CONT…
Polar solvents such as water, alcohols etc. may stabilize or destabilize the
molecular orbitals of a molecule either in the ground state or in excited
state and the spectrum of a compound in these solvents may
significantly vary from the one recorded in a hydrocarbon solvent.
π→ π* Transitions
In case of π→ π* transitions, the excited states are more polar than the
ground state
The dipole-dipole interactions with solvent molecules lower the energy of
the excited state more than that of the ground state.
Therefore a polar solvent decreases the energy of π → π* transition and
absorption maximum appears ~10-20 nm red shifted in going from
hexane to ethanol solvent.
150
n→ π* transitions
In case of n → π* transitions, the polar solvents form hydrogen
bonds with the ground state of polar molecules more readily than
with their excited states.
Therefore, in polar solvents the energies of electronic transitions are
increased.
For example, the following figure shows that the absorption
maximum of acetone in hexane appears at 279 nm which in water
is shifted to 264 nm, with a blue shift of 15 nm.
151
CONT…
UV-spectra of acetone in hexane and in water
152
CONT…
153
APPLICATION
1) Qualitative
2) Quantitative
1) Qualitative
A) detection of impurities: to detect the presence of impurities we can use UV
spectrophotometric measurements.
Additional peaks can be due to impurities in the sample and can be compared
with that of standard raw material. also by absorbance measurements at specific
wavelength
B) structure elucidation of organic compounds: UV spectroscopy helps in
structure elucidation of organic molecules
It is useful in the structure elucidation of organic molecules, such as in detecting
154
the presence or absence of unsaturation, the presence of hetero atoms.
CONT…
2) Quantitative
The determination of an analyte’s concentration based on its
absorption of ultraviolet or visible radiation is one of the most
frequently encountered quantitative analytical methods.
One reason for its popularity is that many organic and inorganic
compounds have strong absorption bands in the UV/Vis region of
the electromagnetic spectrum.
A. Environmental Applications
The analysis of waters and wastewaters often relies on the
absorption of ultraviolet and visible radiation.
155
CONT…
B. Clinical Applications
The analysis of clinical samples is often complicated by the
complexity of the sample matrix, which may contribute a
significant background absorption at the desired wavelength.
The determination of serum barbiturates provides one example of
how this problem is overcome.
C. Industrial Analysis
UV/Vis molecular absorption is used for the analysis of a diverse
array of industrial samples including pharmaceuticals, food,
paint, glass, and metals
156
CONT…
D. Forensic Applications
UV/Vis molecular absorption is routinely used for the analysis of
narcotics and for drug testing. One interesting forensic
application is the determination of blood alcohol using the
Breathalyzer test.
157
CHAPTER FIVE
INFRARED (VIBRATIONAL)
SPECTROSCOPY
158
INTRODUCTION
• IR deals with the interaction of infrared radiation with matter. The IR
spectrum of a compound can provide important information about its
chemical nature and molecular structure.
• Most commonly, the spectrum is obtained by measuring the
absorption of IR radiation.
•
IR radiation does not have enough energy to induce electronic
transitions as seen with UV.
• IR Spectroscopy is an extremely effective method for determining
a wide variety of functional groups in a molecule.
159
CONT…
IR spectroscopy:

Fast and cheapest
 It is the measurement of the absorption of IR frequencies by organic
compounds placed in the path of the beam of light
 The transitions responsible for IR bands are due to molecular
vibrations, i.e. to periodic motions involving stretching or bending
of bonds. Gas, liquid or solid sample can be measured...
 Used to elucidate molecular structure, particularly the recognition
of functional group and their environment.
160
IR REGION OF THE EM SPECTRUM
 Pure rotation gives rise to absorption in the microwave region or in
the far infrared
 Molecular vibrations give rise to absorption bands throughout most of
the IR region of the spectrum.
 Infrared radiation lies between the visible and microwave portions of
the electromagnetic spectrum.
 Infrared waves have wavelengths longer than visible and shorter than
microwaves, and have frequencies which are lower than visible
and higher than microwaves.
161
CONT…
 The Infrared region is divided into: near, mid and far-infrared.
 Near-infrared refers to the part of the infrared spectrum that is
closest to visible light and far-infrared refers to the part that is
closer to the microwave region.
 Mid-infrared is the region between these two. For chemical
analysis, we are interested in mid IR region (2.5 m-15 m).
162
CONT…
Infrared (IR) spectroscopy: based on IR absorption by molecules as
undergo vibrational and rotational transitions
Potential Energy (E)
rotational transitions
Vibrational transitions
Interatomic Distance (r)
163
MOLECULAR VIBRATIONS
Radiation in the IR region will cause stretching and bending vibrations of the bonds in
most covalent molecules.
Modes of Vibration
There are 2 types of vibrations.
1)
Stretching vibrations
2)
Bending vibrations
1) Stretching - the rhythmic movement along a bond axis with a subsequent
increase and decrease in bond length or in this bond length is altered.
They are of 2 types
a) symmetrical stretching: 2 bonds
increase or decrease in length.
164
CONT…
b) Asymmetrical stretching: in this one bond length is increased and
other is decreased.
2) Bending - a change in bond angle or movement of a group of atoms
with respect to the rest of the molecule. These are also called as
deformations.
• In this bond angle is altered.
These are of 2 types
a) In plane bending→ scissoring, rocking
b) Out plane bending→ wagging, twisting
165
CONT…
Scissoring:
This is an in plane bending.
In this bond angles are decreased.
2 atoms approach each other.
Rocking:
In this movement of atoms takes place in same direction.
166
CONT…
Wagging:
It is an out of plane bending.
In this 2 atoms move to one side of the plane.
They move up and down the plane.
Twisting:
In this one atom moves above the plane and the other atom moves below
the plane.
167
NUMBER OF VIBRATIONAL MODES
A molecule can vibrate in many ways, and each way is called a
vibrational mode.
If a molecule contains ‘N’ atoms, total number of vibrational modes:
 for non-linear molecules, number of types of vibrations: 3N-6
 for linear molecules, number of types of vibrations: 3N-5
Eg: H2O, a non-linear molecule, will have 3 × 3 – 6 = 3 degrees of
vibrational freedom, or modes.
1) HCl : 3(2)-5
2) CO2 :
= 1 mode
3(3)-5 = 4 modes
168
CONT….
169
IR ACTIVE SPECIES
Molecular species with small energy differences between various vibrational and
rotational states ( most organic species) are IR active species.
Only bonds which have significant dipole moments will absorb infrared radiation.
Bonds which do not absorb infrared include
• Symmetrically substituted alkenes and alkynes.
R
R
R
R
C
C
R
R
• Symmetric diatomic molecules. H2 , Cl2 , O2 , N2
• Ionic salts NaCl, KBr absorb only in the far IR region ( < 700 cm-1) so are suitable
as sample holders for most Mid-IR measurements.
170
WHAT ABOUT FOR CO2 ?
The CO2 molecule on the left is undergoing a symmetric stretch, the one
in the middle an asymmetric stretch and the one on the right an inplane bend.
μ > 0; IR active
μ = 0; IR inactive
O
C
O
O
C
μ > 0; IR active
O
O
C
O
Stretching involves a change in bond lengths
Bending involves a change in bond angle
The symmetric stretch is an easier deformation than the asymmetric stretch occurs at
lower wavenumbers. The bending vibration is much easier than stretching so it
occurs at lower wavenumber
171
CONT…
Not all covalent bonds display bands in the IR spectrum. Only polar bonds do so. These
are referred to as IR active.
The intensity of the bands depends on the magnitude of the dipole moment associated
with the bond in question:
Strongly polar bonds such as carbonyl groups (C=O) produce strong bands.
Medium polarity bonds and asymmetric bonds produce medium bands.
Weakly polar bond and symmetric bonds produce weak or non observable bands.
17
2
VIBRATIONAL FREQUENCY
The covalent bond between two atoms acts like a spring, allowing the atoms to
vibrate (stretch and bend) relative to each other.
The stretching frequency of a bond can be approximated by Hooke’s Law. This
approximation, two atoms and the connecting bond are treated as a simple harmonic
oscillator composed of 2 masses (atoms) joined by a spring:
173
CONT…
According to Hooke’s law, the frequency of the vibration of the spring is related to
 the mass and the force constant of the spring, f , by the following formula:
174
CONT…
How does the mass influence the vibration?
H2
I2
MM =2 g/mole
MM =254 g/mole
The greater the mass - the lower the wavenumber ( vibrational
frequency
17
5
GENERAL TRENDS:
i) Stretching frequencies are higher than corresponding bending frequencies.
(It is easier to bend a bond than to stretch or compress it)
ii) Bonds to hydrogen have higher stretching frequencies than those to
heavier atoms.
iii) Triple bonds have higher stretching frequencies than corresponding
double bonds, which in turn have higher frequencies than single bonds.
NB: No net change in dipole moment occurs during the vibration or
rotation of homonuclear species such as O2, N2, or Cl2.
As a result, such compounds cannot absorb IR radiation.
176
CONT…
177
CONT…
 The vibrational frequency of a bond would increase with the
decrease in reduced mass of the system. It implies that C-H and
O-H stretching absorptions should appear at higher frequencies
than C-C and C-O stretching frequencies
 As the force constant increases, the vibrational frequency
(wavenumber) also increases.
Example: calculate the approximate wave number and wave length
of the fundamental absorption due to the stretching vibration of a
carbonyl(
) group
178
CONT…
Solution: The mass of the carbon atom in kg is given by
m1= 12 × 10-3 kg /mol
× 1 atom = 2.0 × 10-26 kg
6.022 × 1023 atom /mol
Similarly, for oxygen,
M2 =
12 × 10-3 kg /mol
× 1 atom = 2.7 × 10-26 kg
6.022 × 1023 atom /mol
And the reduced mass μ is given by
μ = 2.0
× 10-26 kg × 2.7 × 10-26 kg
(2.0
= 1.1 × 10-26 kg
+ 2.7 ) × 10-26 kg
The force constant for the double bond is 1
× 103 N/m
ṽ =5.3 × 10-12 s/cm √[(1 × 103 N/m) / (1.1 × 10-26 kg) ] = 1.6 × 103 1/cm
The carbonyl stretching band is found experimentally to be in the region of 1600 to
1800 cm-1.
179
IR SPECTRUM
IR radiation is passed through a sample. Some of the infrared radiation is
absorbed , the rest is transmitted.
No two unique molecular structures produce the same infrared spectrum.
This makes infrared spectroscopy useful for several types of analysis.
Each molecule has a unique IR spectrum.
 The IR spectrum is a “fingerprint” for the molecule.
IR spectrum results from a combination of all possible stretching
and/or bending vibrations of the individual bonds and the whole
molecule.
180
CONT…
 Simple stretching: ~1600-4000 cm-1.
 Complex vibrations: 600-1400 cm-1, called the “fingerprint region.”
 Stretching absorption of a bond appears at higher frequency in the IR spectrum than
bending absorption frequency of the same bond.
C-H < 3000 cm-1
C= O 1715 cm1
181
CONT…
An IR spectrum is used to identify functional groups that are present
(or absent).
 Carbon-Carbon Bonds
Increasing bond order leads to higher frequencies:
 C-C
 C=C
 CC
1200 cm-1
(fingerprint region)
1600 - 1680 cm-1
2200 cm-1
182
CARBON-HYDROGEN BONDS
Bonds with more s character absorb at a higher frequency.
 sp3 (alkane) C-H
 just below 3000 cm-1 (to the right)
 sp2 (alkene or aromatic hydrocarbon) =C-H
 just above 3000 cm-1 (to the left)
 sp (alkyne) =C-H
 at 3300 cm-1
183
CONT…
184
O-H and N-H bonds
Both O-H and N-H stretches appear around 3300 cm-1, but they look
different.
 Alcohol O-H
 broad with rounded tip when hydrogen bonding is present (sharp in the absence of
hydrogen bonding)
O-H str at 3600-3200 cm-1
C-O str at 1050-1150 cm-1
 Secondary amine (R2NH) : Broad (usually) with one sharp peak
 Primary amine (RNH2): Broad (usually) with two sharp peaks.
 No signal for a tertiary amine (R3N)
185
CONT…
186
CYCLOHEXANOL
187
1O AMINES
188
2O AMINE
189
CARBONYLS
Carbonyl stretches are generally strong:
 Aldehyde
 Ketone
 Carboxylic acid
 Ester
 Amide
~1710 cm-1
~1710 cm-1
~1710 cm-1
~1730 - 1740 cm-1
~1640-1680 cm-1
Conjugation shifts all carbonyls to lower frequencies.
Ring strain shifts carbonyls to higher frequencies.
-1
1745 cm
O
H3C
190
ALDEHYDES
191
KETONE and ALDEHYDE
192
CONT…
193
CONT…
Stretching absorption of a bond appears at higher frequency in the IR
spectrum than bending absorption frequency of the same bond.
194
INSTRUMENTATION
Basic Design
- normal IR instrument similar to UV-VIS
- main differences are light source & detector
195
Light Source:
i.) Light Source: An inert solid is electrically heated to a temperature
in the range 1500-2000 K. The heated material will then emit infra
red radiation.
- must produce IR radiation
- can’t use glass since absorbs IR radiation
- Several possible types
A) The Nernst glower:
Zr, Ce, Th
V
196
CONT…
- rare earth metal oxides (Zr, Ce, Th) heated electrically
- apply current to cylinder, has resistance to current flow
generates heat (1200o – 2200o C).
- causes light production similar to blackbody radiation
- range of use ~ 670 – 10,000 cm-1 (wave number)
B) Globar
- similar to Nernst Glower but uses silicon carbide rod instead of rare
earth oxides
- similar usable range
197
CONT…
C) Incandescent Wire Source
- tightly wound nichrome or rodium wire that is electrically heated
- same principal as Nernst Glower
- lower intensity than Nernst Glower or Globar, but longer lifetime
ii) Detectors
Two main types in common IR instruments
A) Thermal Detectors
1) Thermocouple
- two pieces of dissimilar metals fused together at the ends
- when heated, metals heat at different rates
198
CONT…
- potential difference is created between two metals that varies with their
difference in temperature
- usually made with blackened surface (to improve heat absorption)
- placed in evacuated tube with window transparent to IR (not glass or
quartz)
- IR “hits” and heats one of the two wires.
- can use several thermocouples to increase sensitivity.
h
metal1
metal2
- +
V
IR transparent
material (NaCl)
199
CONT…
2) Bolometer
- strips of metal (Pt, Ni) or semiconductor that has a large change in
resistance to current with temperature.
- as light is absorbed by blackened surface, resistance increases and
current decreases
- very sensitive
h
A
200
CONT…
B) Photoconducting Detectors
 thin film of semiconductor (ex. PbS) on a nonconducting glass surface
and sealed in a vacuum.
- absorption of light by semiconductor moves from non-conducting
to conducting state
- decrease in resistance  increase in current
- range: 10,000 -333 cm-1 at room temperature
h
vacuum
Semiconductor
glass
201
Transparent
to IR
CONT…
C) Pyroelectric Detectors
- pyroelectric (ceramic, lithium tantalate) material get polarized
(separation of (+) and (-) charges) in presence of electric field.
- fast response, good for FTIR
202
CONT…
iii) Other Components
a.) Sample Cell
- must be made of IR transparent material (KBr pellets or NaCl)
b.) monochromator
- reflective grating is common
- can’t use glass prism, since absorbs IR
203
INFRARED INTERPRETATION
Step 1
 Look first for the carbonyl C=O band.
 Look for a strong band at 1820-1660 cm-1.
This band is usually the most intense absorption band in a
spectrum. It will have a medium width. If you see the carbonyl
band, look for other bands associated with functional groups that
contain the carbonyl by going to step 2.
 If no C=O band is present, check for alcohols and go to step 3.
Step 2
 If a C=O is present you want to determine if it is part of an acid,
an ester, or an aldehyde or ketone. At this time you may not be
able to distinguish aldehyde from ketone.
204
ACID
 Look for indications that an O-H is also present.
 It has a broad absorption near 3300-2500 cm-1.
 This actually will overlap the C-H stretch. There will also be a CO single bond band near 1100-1300 cm-1.
 Look for the carbonyl band near 1725-1700 cm-1.
 ESTER
 Look for C-O absorption of medium intensity near 1300-1000 cm1. There will be no O-H band.
205
ALDEHYDE
 Look for aldehyde type C-H absorption bands. These are two weak
absorptions to the right of the C-H stretch near 2850 cm-1 and 2750
cm-1 and are caused by the C-H bond that is part of the CHO aldehyde
functional group.
 Look for the carbonyl band around 1740-1720 cm-1.
KETONE
 The weak aldehyde CH absorption bands will be absent. Look for the
carbonyl CO band around 1725-1705 cm-1.
Step 3
 If no carbonyl band appears in the spectrum, look for an alcohol O-H
band.
ALCOHOL
 Look for the broad OH band near 3600-3300 cm-1 and a C-O
absorption band near 1300-1000 cm-1.
206
Step 4
 If no carbonyl bands and no O-H bands are in the spectrum,
check for double bonds, C=C, from an aromatic or an alkene.
ALKENE
 Look for weak absorption near 1650 cm-1 for a double bond.
There will be a CH stretch band near 3000 cm-1.
AROMATIC
 Look for the benzene, C=C, double bonds which appear as
medium to strong absorptions in the region 1650-1450 cm-1. The
CH stretch band is much weaker than in alkenes.
Also check the region 3030 cm-1 ( the –CH stretching)
207
C-H stretching region
• Alkanes C-H sp3 stretch < 3000 cm-1
• Alkenes C-H sp2 stretch > 3000 cm-1
• Alkynes C-H sp stretch ~ 3300 cm-1
• C-H Bending region
• CH2 bending ~ 1460 cm-1
• CH3 bending (asym) appears near the same value
• CH3 bending (sym) ~ 1380 cm-1
208
Step 5
 If none of the previous groups can be identified, you may have an
alkane.
ALKANE
 The main absorption will be the C-H stretch near 3000 cm-1. The
spectrum will be simple with another band near 1450 cm-1.
209
CONT…
210
CARBONYL COMPOUNDS
211
EXAMPLE:
C6H12O
CH3
C-H stretch
O
CH3 CH CH2 C
CH3
C=O stretch
21
2
C 8H8O
C-H stretch
O
C
CH3
aromatic C=C
conj C=O
213
C6H12O
O-H
stretch
sp3 C-H stretch
OH
bending
C-O
stretch
214
Carboxylic Acid
215
APPLICATION OF IR SPECTROSCOPY
1.Identification of drug substance: IR spectrum of sample and standard
can be compared to identify a substance, if the spectra same then the
identify of the sample can be confirmed.
2. Structure determination: This technique helps to establish the structure
of a unknown compound. all major functional groups absorb at their
characteristic wave number
216
CONT…
3. Identification of functional groups: The presence or absence of absorption bands help
in predicting the presence of certain functional groups in the compounds.
Functional group region 4000 -1600 cm -1 (streching vibrations occur). Finger print
region 1550-600 cm -1 (bending vibrations occur)
4. Identifying the impurities in a drug sample:
Impurities having different chemical nature when compared to the pure drug.
Hence these impurities give rise to additional peaks than that of the pure drug.
By comparing these ,we can identify the presence of impurities.
217
PROBLEMS
Problem #1: Unknown molecule with molecular formula C5H10O. Which of these five
molecules is it most likely to be?
218
CONT…
Problem #2: Unknown molecule with molecular formula C6H12O.
219
CONT…
Problem #3: Unknown molecule with molecular formula C6H14O .
220
CONT…
Problem #4: Unknown molecule with formula C4H8O2 (Also, smells like vomit)
221
ANSWERS
Problem 1:
You’re given the molecular formula, which is C5H10O. This corresponds
to an index of hydrogen deficiency (IHD) of 1, so either a double
bond or ring is present in the molecule. This immediately rules
out d) whose IHD is zero and thus has a molecular formula of
C5H12O.
Looking at the spectrum we see a broad peak at 3300 cm-1 and no
dominant peak around 1700 cm-1 (That peak halfway down around
1700 cm-1? It’s too weak to be a C=O. )
That broad peak at 3300 tells us that we have an alcohol (OH group).
The only option that makes sense is e) (cyclopentanol) since it has
both an OH group and an IHD of 1. It can’t be b) since that molecule
lacks OH.
a) and c) are further ruled out by the absence of C=O ; B is ruled out by
222
the presence of the OH at 3300
CONT…
Problem 2:
A molecular formula of C6H12O corresponds to an IHD of 1 so either a double
bond or ring is present in the molecule.
There is no strong OH peak around 3200-3400 cm-1 (that little blip around 3400
cm-1 is too weak to be an OH). We can immediately rule out a) and e) .
However, we do see a peak a little above 1700 cm-1 that is one of the strongest
peaks in the spectrum. This is a textbook C=O peak. We can safely rule
out b) which lacks a carbonyl.
The only option that makes sense is d) (2-hexanone) since c) doesn’t match the
molecular formula (two oxygens, five carbons).
Note also that the C-H region shows all peaks below 3000 cm-1 which is what
we would expect for a saturated (“aliphatic”) ketone.
223
CONT…
Problem 3:
A molecular formula of C6H14O corresponds to an IHD of zero. No double bonds or rings
are present in the molecule.
Using this we can immediately rule out d) and e) since their structures cannot correspond
to molecular formula (they are both C6H12O)
There is no OH peak visible around 3200-3400 cm-1. We can rule out a) and b) .
This leaves us with c) . It’s an ether.
Useful tip: ethers are “silent” in the prominent parts of the IR spectrum; this functional
group is best identified through a process of deduction. Seeing an O in the formula but
no OH or C=O peaks, the only logical selection is c) .
Final note: e) is a cyclic ether called an “epoxide”. The important clue to distinguish c)
and e) was the fact that we were given the molecular formula. In the absence of that
information it would have been difficult to tell the difference without a close
consultation of an IR peak table.
224
CONT…
Problem 4:
The immediate giveaway is the smell of puke. That’s butyric acid for sure!
More seriously: the formula of C4H8O2 corresponds to an IHD of 1. We can
immediately rule out c) .
Looking at the IR spectrum we see a huge peak in the 3300-2600 cm-1 region that blots
out everything else. This seems like a textbook “hairy beard” typical of a carboxylic
acid, but let’s look for more information before confirming it. We can at least rule
out a) , which has no OH peaks. We also see a strong peak a little above 1700 cm1 which is typical of a C=O.
We can safely rule out e) which lacks carbonyl groups entirely. This leaves us with two
reasonable choices: b) (the carboxylic acid) and d) the ketone / alcohol. How to
choose between the two? The “hairy beard” is diagnostic. Alcohol OH peaks don’t
fill up 600 wavenumber units the way that carboxylic acid peaks do.
A more subtle way to distinguish the two might be the position of the carbonyl
peak, but carboxylic acids (1700-1725 cm-1) show up largely in the same range
as do ketones (1705-1725 cm-1).
225
QUIZ 1
Which compound is this?
a) 2-pentanone
b) 1-pentanol
c) 1-bromopentane
d) 2-methylpentane
1-pentanol
226
CONT…
What is the compound?
a) 1-bromopentane
b) 1-pentanol
c) 2-pentanone
d) 2-methylpentane
2-pentanone
227
CHAPTER SIX
NUCLEAR MAGNETIC RESONANCE
SPECTROSCOPY (NMR)
228
INTRODUCTION
Nuclear magnetic resonance spectroscopy measures the absorption of electromagnetic
radiation in the radio-frequency region and it is a powerful analytical technique used to
characterize organic molecules by identifying carbon-hydrogen frameworks within
molecules or that gives us information about the number and types of atoms in a
molecule.
Nuclei (instead of outer electrons) are involved in absorption process.
Two common types of NMR spectroscopy are used to characterize organic structure:
 1H NMR is used to determine the type and number of H atoms in a molecule;

13C
NMR is used to determine the type of carbon atoms in the molecule.
The source of energy in NMR is radio waves which have long wavelengths, and thus
low energy and frequency. When low-energy radio waves interact with a molecule, they
can change the nuclear spins of some elements, including 1H and 13C ,15N, 19F, 31P .
229
• When a charged particle such as a proton spins on its axis, it creates a
magnetic field. Thus, the nucleus can be considered to be a tiny bar magnet.
A spinning charged nucleus generates a
magnetic dipole
Such nuclear magnetic dipoles are characterized
by nuclear magnetic spin quantum numbers which are designated by the
letter I and can take up values equal to 0, ½ , 1, 3/2, ... etc
A nucleus with an odd atomic number or an odd mass number has a nuclear
spin that create a magnetic field as they spin..
230
TYPE I
Nuclei with I = O. These nuclei do not interact with the applied magnetic field and
are not NMR chromophores.
Nuclei with I = 0 have an even number of protons and even number of neutrons
and have no net spin. The most prominent examples of nuclei with I = 0 are 12C
and 160.
The rules for determining the net spin of a nucleus are as follows;
A. If the number of neutrons and the number of protons are both even, then the
nucleus has NO spin.
B. If the number of neutrons plus the number of protons is odd, then the nucleus
has a half-integer spin (i.e. 1/2, 3/2, 5/2)
C. If the number of neutrons and the number of protons are both odd, then the
nucleus has an integer spin (i.e. 1, 2, 3)
231
CONT…
Type 2: Nuclei with I = ½. These nuclei have a non-zero magnetic
moment and are NMR visible and have no nuclear electric quadrupole
(Q). The two most important nuclei for NMR spectroscopy belong to this
category: IH and 13C.
Also,two other commonly observed nuclei 19F and 31P have I = 1/2
Type 3:
Nuclei with I >1/2, These nuclei have both a magnetic
moment and an electric quadrupole . This group includes some common
isotops ( e.g 2H and 14N) but they are more difficult to observe.
232
The most important consequence of nuclear spin is that in a uniform magnetic
field, a nucleus of spin I may assume 2I + 1 orientations.
For nuclei with I = 1/2 there are 2 permissible orientations.
These two orientations will be of unequal energy (by analogy with the parallel
and antiparallel orientations) and it is possible to induce a spectroscopic
transition (spin-flip) by the absorption of a quantum of electromagnetic energy
(ΔE) of the appropriate frequency (v):
-1/2 antiparallel
E
+1/2 parallel
no field
applied field
Bo
233
• Normally, these tiny bar magnets are randomly oriented in space.
However, in the presence of a magnetic field B0, they are oriented
with or against this applied field. More nuclei are oriented with
the applied field because this arrangement is lower in energy.
• The energy difference between these two states is very small
234
ENERGY
• When an external energy source (h) that matches the energy
difference (E) between these two states is applied, energy is
absorbed, causing the nucleus to “spin flip” from one orientation to
another.
The energy difference between these two nuclear spin states
corresponds to the low frequency RF region of the electromagnetic
spectrum.
aligned against the field
b
Eh
a
aligned with the field
235
Fig. Change in spin state energy separation with increase by applied magnetic
field, B0
• Thus, two variables characterize NMR: an applied magnetic field
B0, the strength of which is measured in tesla (T),
and the frequency  of radiation used for resonance, measured in
hertz (Hz), or megahertz (MHz)—(1 MHz = 106 Hz).
236
CONT...
A line diagram of NMR spectrophotometer along with its
components are as follows;
237
CONT…
The principle behind NMR is that many nuclei have spin and all nuclei are electrically
charged. If an external magnetic field is applied, an energy transfer is possible
between the base energy to a higher energy level (generally a single energy gap).
238
CONT...
• Protons in different environments absorb different frequencies,
so they are distinguishable by NMR.
• The size of the magnetic field generated by the electrons around
a proton determines where it absorbs.
• Modern NMR spectrometers use a constant magnetic field
strength B0, and then a narrow range of frequencies is applied
to achieve the resonance of all protons.
239
SOLVENTS
 The solvent used for dissolving sample should have following
properties;
 Should not contain proton,
 Inexpensive
 Low boiling point and non polar in nature.
 Generally deuterated chloroform CDCl3 is used as solvent.
 If sample is soluble in polar solvent, then deuterium oxide (D2O),
DMSO, CCl4, CS2, are used as solvent.
240
NMR SPECTROSCOPY
INTERPRETATION OF 1H NMR SPECTRA
NMR spectra provide information about the structure of organic molecules from the:

number of different signals in the spectrum

position of the signals (chemical shift)

intensity of the signals
OBTAINING SPECTRA
• A liquid sample is placed in a tube which spins in a magnetic field
• Solids are dissolved in solvents which won’t affect the spectrum - CCl4, CDCl3
•TMS, tetramethylsilane, (CH3)4Si, is added to provide a reference signal
• When the spectrum has been run, it can be integrated to find the relative peak
241
TETRAMETHYLSILANE - TMS
PROVIDES THE REFERENCE SIGNAL

non-toxic liquid - safe to use

inert - doesn’t react with compound being analysed

has a low boiling point -can be distilled off and used again

all the hydrogen atoms are chemically equivalent - produces a single peak

twelve hydrogens so it produces an intense peak - don’t need to use much

signal is outside the range shown by most protons

given the chemical shift of d = 0

the position of all other signals is measured relative to TMS
The molecule contains four methyl groups
attached to a silicon atom in a tetrahedral
arrangement. All the hydrogen atoms are
chemically equivalent.
242
CHEMICAL SHIFT
•The chemical shift is the difference between the field strength at which it absorbs
and the field strength at which TMS protons absorb
• Each proton type is said to be chemically shifted relative to a standard (usually
TMS)
• The TMS peak is assigned a value of zero (d = 0.00)
• All peaks of a sample under study are related to it and reported in ppm.
• H’s near to an electronegative species are shifted “downfield” to higher d values
243
CONT…
Approximate chemical shifts
The actual values depend on the environment
H
- C-X
ROH
-CHO
- C-H
-COOH
13
12
-C=CH-
11
10
9
8
7
6
TMS
5
4
3
2
1
0
d
Chemical Shift Scale (ppm)
More deshielded
More shielded
244
CONT…
Shielding:
The higher the electron density around the nucleus, the higher the
opposing magnetic field to B0 from the electrons, the greater the
shielding. Because the proton experiences lower external magnetic
field, it needs a lower frequency to achieve resonance, and therefore,
the chemical shift shifts upfield (lower ppms) .
245
CONT….
Deshielding:
If the electron density around a nucleus decreases, the
opposing magnetic field becomes small and therefore, the
nucleus feels more the external magnetic field B0, and
therefore it is said to be deshielded. Because the proton
experiences higher external magnetic field, it needs a higher
frequency to achieve resonance, and therefore, the chemical
shift shifts downfield (higher ppms) .
246
247
248
CONT...
Chemical shift depends upon following parameters:
 Electro negativity of nearby atoms
 Hybridization of adjacent atoms
 Diamagnetic effects from adjacent pi bonds
249
CONT...
Electro negativity of nearby atoms
Electron egativ ity of X
Chem ical
Shif t (d)
CH3 OH
4.0
3.5
4.26
3.47
CH3 Cl
3.1
3.05
CH3 Br
CH3 I
2.8
2.5
2.68
( CH3 ) 4 C
( CH3 ) 4 Si
2.1
0.86
1.8
0.00
CH3 - X
CH3 F
2.16
250
CONT...
Diamagnetic effects from adjacent pi bonds
A carbon-carbon triple bond shields an acetylenic hydrogen and
shifts its signal to lower frequency (to the right) to a smaller
d value. A carbon-carbon double bond deshields vinylic hydrogens
and shifts their signal to higher frequency (to the left) to a larger d
value.
Type of H
N ame
RCH3
Alk yl
RC CH
R2 C=CH2
Acetylenic
Vin ylic
Chemical
Shift (d)
0.8- 1.0
2.0 - 3.0
4.6 - 5.7
251
CONT...
252
MULTIPLICITY(SPIN-SPIN SPLITTING) (N+1 RULE)
• Often a group of hydrogens will appear as a multiplet rather than as a
single peak. This happens because of interaction with neighboring
hydrogen and is called SPIN-SPIN SPLITTING.
Nonequivalent protons on adjacent carbons have magnetic fields that
may align with or oppose the external field.
If you have N number of magnetically equivalent hydrogens causing the
splitting then you have N+1 peaks in the spectrum. A hydrogen does
not cause splitting with itself, but only with neighboring hydrogens.
253
SIGNAL SPLITTING (N + 1 RULE)
 1H-NMR spectrum of 1,1-dichloroethane.
For these hydrogens, n = 1;
their signal is split into
(1 + 1) = 2 peaks; a doublet
For this hydrogen, n = 3;
CH3 - CH- Cl its signal is split into
(3 + 1) = 4 peaks; a quartet
Cl
254
RULES FOR SPLITTING OF PROTON SIGNALS
 Equivalent protons do not split each other.
 Protons bonded to the same carbon will split each other if they
are nonequivalent.
 Nonequivalent protons on adjacent carbons normally will split
each other.
 Protons separated by four or more bonds will not split each
other.
255
CONT…
Number of peaks = number of chemically different H’s on adjacent atoms + 1
1 neighbouring H
2 peaks
“doublet”
1:1
2 neighbouring H’s
3 peaks
“triplet”
1:2:1
3 neighbouring H’s
4 peaks
“quartet”
1:3:3:1
4 neighbouring H’s
5 peaks
“quintet”
1:4:6:4:1
Signals for the H in an O-H bond are unaffected by hydrogens on adjacent
atoms - get a singlet
0 neighbouring H’s signal isn’t split
1 peak “singlet”
256
EQUIVALENT & NONEQUIVALENT PROTONS
Equivalent protons are like this;
These equivalent protons do not split each other because equivalent
protons have the same chemical shift.
257
CONT…
For alkanes normally observe splitting only for hydrogens attached to adjacent carbons.
2 H's
3 adjacent H'
therefore
quartet
C
A
O
B
H3C
O
C
3 H's
no adjacent H's
therefore singlet
A
CH3
B
3 H's
2 adjacent H's
therefore triplet
258
CONT.…
In normal organic compounds, splitting is only observed with hydrogens attached to
adjacent atoms
A
2 H's
5 adjacent H's
therefore hextet
O
A
B
C
CH3
H3C
singlet
D
triplet
D
B
C
triplet
259
CONT...
 Non equivalent protons have different chemical shift with
splitting occurs
260
CONT…
1H
NMR Spectrum of 4-Methylbezaldehyde
CH3
H
H
O
H
261
PROPYL ETHER:
262
cyclohexane
a
singlet 12H
263
benzene
a singlet 6H
264
p-xylene
H3C
CH3
a
a
b
a singlet 6H
b singlet 4H
265
1H
NMR Peak Integration or Peak Area
The relative peak intensity or peak area is proportional to the number of protons
associated with the observed peak.
266
267
1H
NMR—Structure Determination
268
1H
NMR—Structure Determination
269
1H
NMR—Structure Determination
270
1H
NMR—Structure Determination
271
13C
NMR SPECTROSCOPY
• The most abundant isotope of carbon (12C) cannot be observed by NMR.
• The 13C nucleus is present in only 1.08% natural abundance.
• Unlike 1H NMR, the area of a peak is not proportional to the number of
carbons giving rise to the signal. Therefore, integrations are usually not
done.
• When running a spectrum, the protons are usually decoupled from their
respective carbons to give a singlet for each carbon atom. This is called a
proton-decoupled spectrum.
• Notice that carbon-carbon splitting is not detected in NMR
272
CONT…
The 13C NMR is directly about the carbon skeleton not just the proton
attached to it.
a. The number of signals tell us how many different carbons or set of
equivalent carbons
b. The splitting of a signal tells us how many hydrogens are attached to
each carbon. (N+1 rule)
c. The chemical shift tells us the hybridization (sp3, sp2, sp) of each carbon.
d. Integration: Not useful for 13C NMR
 For each carbon the multiplicity of the signal depends upon how many
protons are attached to it.
273
CONT…
Note: Due to low natural abundance, 13C NMR spectra do not ordinarily
show carbon-carbon splitting two 13C being next to other is 1.1 %x
1.1% = 0.012 % (because 12C does not have a magnetic moment, it
cannot split the signal of an adjacent 13C ).
Chemical Shift in 13C NMR spectrum arises in the same way as in the
proton NMR spectrum. Each carbon nucleus has its own electronic
environment, different from the environment of other, non-equivalent
nuclei; it feels a different magnetic field, and absorbs at different
applied fields strength.
Electronegative atoms and pi bonds cause downfield shifts
13C
chemical shift range 0-250 ppm
274
CONT…
In 13C NMR spectrum, the more electronegative group bonded to carbon
atom
deshielding increases.
How many signals are in the 13C NMR spectrum?
O
f
g
a
e
b
b
c
c
g. 14.2 ppm
e. 41.4 ppm
f. 60.6 ppm
a. 135.ppm
b.129 ppm
c. 127 ppm
d. 126 pm
d
Note: there are two methyl groups and one corresponding to –
CH2 downfield (60.6 ppm) is attached to O cause deshielded
Benzyl CH2 (41.1 ppm) . Aromatic ring carbons have resonance
over range from 126 ppm to 135 ppm.
27
5
CONT…
EX: Determine the structure from this formula C4H8O2 in 13C spectrum.
179.9 ppm (triplet), 51.5 ppm (quartet), 27.5 ppm (triplet) and 9.2
ppm (quartet). 179.9 ppm corresponding to ester or ketone carbonyl
group; 51.5 ppm is downfield must be close to carbonyl or oxygen
(OCH3) ; 27.5 ppm (CH2) and 9.2 ppm (quartet) –CH3 group further
away from carbonyl group in upfield region.
O
O
O
O
d
b
O
O
a
c
Note: b. this carbon must downfield and
probably over 200 ppm.
276
CONT…
277
POSITION OF FUNCTIONAL GROUPS IN 13C NMR
The relative placement is similar to 1H NMR, but the
scale is much larger
278
CONT…
Example: Two alcohols (-OH compound) with formula C3H8O. DoU= 0.
Possible structures:
O
H
O
H
t h e
s a m
e
279
CONT…
The C-NMR spectrum of isopropanol only shows two different carbons!
O
H
These two carbons are identical by symmetry:
280
CONT…
13C
peaks are in reality split by bonded protons.
C
H
H
C
H
Appears as quartet.
H
The general rule is: The number of peaks observed is equal to the
number of attached protons, (N), plus one (N+1 ).
281
CONT…
In a typical 13C NMR, the peak areas are dependent upon how many
hydrogens are attached to carbon, not the relative number of carbons
causing the signal. CH3 groups are the biggest followed by CH2, CH
and quaternary carbons are the smallest
A
A
A
B
A
E
C
D
E
E
C
B
D
282
NMR PROBLEM
A compound has molecular formula C8H8O.
The proton NMR has three peaks;
singlet at d 2.2 (3H),
singlet at d 10.0 (1H)
two doublets centered around d 7.6. Assign the structure.
SOLUTION:
.The doublets centered at d 7.6 are in the aromatic region; the fact that
two doublets are observed (2H each) suggests a 1,4-disubstituted
aromatic compound. The peak at d 2.2 is in the region for a methyl
group adjacent a mildly electronegative group. The singlet at d 10 is in
the region observed for aldehydic protons.
283
HENCE STRUCTURE IS
The presence of two doublets in the aromatic region is highly
characteristic of 1,4-disubstitution.
O
C
H
CH3
Structure:
IUPAC Name: 4-methylbenzaldehyde
284
CHAPTER SEVEN
Mass spectroscopy (MS)
285
INTRODUCTION
 Uses the interaction of electric and/or magnetic fields (eletromagnetic
radiation) with matter to determine weight or mass of the molecule.
 Mass to charge ratios (m/z) is measured (not absorption or emission of
EMR)
 Sample vaporized and subjected to bombardment by electrons that remove
an electron
 Neutral molecules that contain even number of electrons but ejection of
one electron gives an odd number of electrons to form M+.
 When the electron beam ionizes the molecule, the species that is formed is
called a cation radical, and symbolized as M+•.
286
CONT…
The radical cation M+• is called the molecular ion or parent ion.
Atom or molecule is hit by high-energy electron from an electron beam
at 10ev forming a positively charged, odd-electron species called the
-
molecular ion
-
e beam
e
+
•
287
CONT…
 Basic Principle. A mass spectrometer generates multiple ions from
the sample under investigation, it then separates them according to
their specific mass-to-charge ratio (m/z), and then records the relative
abundance of each ion type. A mass spectrum of the molecule is thus
produced.
 The substance is bombarded with a beam of electrons so the atoms or
molecules it contains are turned into ions. A computerized, electrical
detector records a spectrum pattern showing how many ions arrive
for each mass/charge ratio.
 This can be used to identify the atoms or molecules in the original
sample.
288
CONT…
 The first step in the mass spectrometric analysis of compounds is the production of
gas phase ions of the compound, basically by electron ionization. This molecular ion
undergoes fragmentation.
 Each primary product ion derived from the molecular ion, in turn, undergoes
fragmentation, and so on.
 The ions are separated in the mass spectrometer according to their mass-to-charge
ratio, and are detected in proportion to their abundance.
 It displays the result in the form of a plot of ion abundance versus mass-to-charge
ratio.
Ions provide information concerning the nature and the structure of their
molecule. In the spectrum of a pure compound, the molecular ion, if present,
appears at the highest value of m/z (followed by ions containing heavier isotopes)
and gives the molecular mass of the compound.
289
MS spectrometers
Mass spectrometer is an instrument that measures the mass-to charge ratio (m/z) values
and their relative abundances of ions
Basic components of mass spectrometry instrument are:
1. Inlet: Introduce sample to the instrument (HPLC, GC, Syringe, Plate , Capillary)
2. Ionizer -Generates ions in the gas phase. find a way to “charge” an atom or a
molecule
3. Mass analyzer –Separate charged atoms or molecules in a magnetic field (separate
ions on the basis of differences in m/z with a mass analyzer)
4. Detector - Detect ions. detector detects the ions using micro-channel plate or
photomultiplier tube
290
CONT…
291
CONT…
How does a mass spectrometer work?
 Ionization methods
– Electron Impact (EI)
– Chemical Ionization (CI)
– Electrospray (ESI)
– Atmospheric Pressure Chemical Ionization (APCI)
– Photo-ionization (APPI)
Why Ionize? Difficult to manipulate neutral particles on molecular scale. If they
are charged, then we can use electric fields to move them around.
292
CONT…
As a general, a mass spectrometer should always perform the following
processes:
 Produce ions from the sample in the ionization source.
 Separate these ions according to their mass-to-charge ratio in the mass
analyzer.
 Eventually, fragment the selected ions and analyze the fragments in a second
analyzer.
 Detect the ions emerging from the last analyzer and measure their abundance
with the detector that converts the ions into electrical signals.
 Process the signals from the detector that are transmitted to the
computer.
293
CONT…
Acquiring a Mass Spectrum
294
IONIZATION METHODS
1. Electron Ionization (EI)

Most widely used method
 Sample introduced into instrument by heating it until it evaporates
 Analytes are bombared with high-energy electrons (usually 70eV)- from rhenium or
tungsten filament.

As a result of collision, an electron is removed from the analytes (M), generating a
molecular ion M+ (radical cation) (M + e- → M+ + 2e- )
 Radical species are generated initially
 Due to excess internal energy, fragmentation of the molecular ion will occur.
 The fragmentation is reproducible and characteristic of the compound.
295
CONT…
o Electron impact on the analyte results in either loss of electron (to produce cation) or
gain of electron (to produce anion). Chemical bonds in organic molecules are
formed by pairing of electrons. Electron impact may knock out one of the electron.
o This leaves the bond with a single unpaired electron. This is radical as well as being
cation written as M+., where (+) indicates ionic state while (.) indicates radical.
o Electron impact may result in electron capture (extra unpaired electron).
296
CONT…
Advantages
 inexpensive, versatile and reproducible
 fragmentation gives structural information
 large databases if EI spectra exist and are searchable
 most common ionization technique
Disadvantages
• Sample must be relatively volatile
• “Hard” ionization method leads to significant fragmentation
• Ionization is efficient but non-selective
• limited to relatively low MW compounds
297
2. CHEMICAL IONIZATION (CI)
 It is soft ionization technique , a technique that produces ions with little energy.
 Vaporized sample reacts with pre-ionized reagent gas via proton transfer, charge
exchange, electron capture etc.
 This technique presents the advantage of yielding a spectrum with less
fragmentation.
 Generally, chemical ionization is complementary to electron ionization.
 Reagent species is ionized by high-pressure EI
 Indirect ionization of sample
 Collisions between sample & gas ions cause proton transfers à produces [M+H]+
ions, not M+ ions (so parent is M+1).
 Provides less information about structure
 Common CI reagents are methane, ammonia, iso-butane, hydrogen, methanol
 Interaction between reagent species and electron beam results in electron ejection.
298
CONT…
• Choice of reagent allows tuning/process of ionization
Softer ionization technique
Less fragmentation
Easier to
find molecular ions.
Ionization of the analyte molecule, M, is achieved through reaction
with a reagent ion, R+
Radical species are generated initially
M(reagent) + e-
M+• + 2e-
CH4+. + CH4
CH3. + CH5+
CH3+ + CH4
H2 + C2H5+
299
CONT…
A. Form Reagent Ions First
For Example : Methane as a reagent gas in CI
1. First CH4 undergoes electron ionization:
CH4 + e-
CH4+. + 2e-
– Fragmentation forms of methane : CH3+, CH2+. , CH+
2. Ion-molecule reactions create stable reagent ions:
CH4+. + CH4
CH3 + CH5+
CH3+ + CH4
H2 + C 2 H5 +
CH5+ and C2H5+ are the dominant methane CI reagent ions .Thus, the peaks at
m/z 15, 14, 13 and 12 are due to these lower molecular weight fragments.
300
CONT…
B. Reagent ions react with analytes
• Several types of reactions may occur
– Form pseudo-molecular ions (M+1)
CH5+ + M
CH4 + MH+ (H+ transfer)
Form molecular ion by charge transfer
CH4+ + M
M+ + CH4 (charge transfer)
301
CONT…
Advantages of CI
 Gives little fragmentation
 “Soft” ionization
 Parent Ion
Disadvantages of CI
• Need Volatile Sample
• Need Thermal Stability
• Quantification Difficult
• Low Mass Compounds (<1000
 Interface to GC
amu)
 Insoluble Samples
• Solids Probe Requires Skilled
Operator
302
ELECTROSPRAY IONIZATION (ESI)
 Electrospray Ionization (ESI) is a preferred method of ionization
when the sample is in liquid form. This is also a soft method of
ionization and results in less fragmentation.
 ESI is a very valuable method for analysis of biological samples.
 The analyte is introduced either from a syringe pump or as the eluent
flow from liquid chromatograph.
 The analyte solution passes through the electrospray needle (Stainless
steel capillary with 75-150 µm internal diameters) that has a high
potential difference applied to it.
303
CONT…
This forces the spraying of charged droplets from the needle with a
surface charge of the same polarity to the charge on the needle.
As droplet moves towards counter electrode cone (which passes it to
analyzer), solvent evaporation occurs and droplet shrinks until it
reaches the point that the surface tension can no longer sustain the
charge (the Rayleigh limit) and at that point droplets break.
This produces smaller droplets and the process is repeated.
Finally after all solvent evaporated, charge is passed on to analyte.
These charged analyte molecules can have single or multiple
charges.
304
ION SEPARATION- MASS FILTER
Once molecules are ionized, they immediately feel the forces
• Electric field steer the ions
• Molecular ion passes between poles of a magnet and is deflected by magnetic
field
• Only cations are detected. Radicals are “invisible” in MS.
• The amount of deflection observed depends on the mass to charge ratio (m/z).
– Most cations formed have a charge of +1 so the amount of deflection
observed is usually dependent on the mass of the ion.
– Highest m/z deflected least
305
M/Z RATIO ANALYSIS
Different elements or compounds can be uniquely identified by their
mass
Types of analyzers:
Magnetic Sector Analyzer (MSA) • High resolution, exact mass, original
MA Quadrupole Analyzer (Q) • Low resolution, fast, cheap
Time-of-Flight Analyzer (TOF) • No upper m/z limit, high throughput
Ion Trap Mass Analyzer (QSTAR) • Good resolution, all-in-one mass
analyzer
Ion Cyclotron Resonance (FT-ICR) • Highest resolution, exact
mass, costly
306
CONT…
Magnetic Sector Analyzer
307
Ions of non-selected
mass/charge ratio
are not detected
Ions of selected
mass/charge ratio
are detected
Ionization
chamber
308
RESOLUTION
A measure of how well a mass spectrometer separates ions of different mass.
– low resolution: Refers to instruments capable of separating only ions that differ in
nominal mass; that is ions that differ by at least 1 or more atomic mass units.
– high resolution: Refers to instruments capable of separating ions that differ in mass by as
little as 0.0001 atomic mass unit.
– C3H8O and C2H4O2 both have nominal masses of 60 and can not be distinguished by lowresolution MS.
– A molecules with nominal mass of 44 could be C3H8, C2H4O, CO2, or CN2H4.
– However exact masses for C3H8 , C2H4O, CO2, CN2H4 are 44.06260, 44.02620,
43.98983 , 44.03740 respectively
– Can be distinguished between them by high-resolution MS.
309
RESOLVING POWER
Width of peak indicates the resolution of the MS instrument
• The better the resolution or resolving power, the better the instrument
and the better the mass accuracy
• Resolving power is defined as: M/ΔM
Where M is the mass number of the observed mass and ΔM is the
difference between two masses that can be separated
310
RESOLUTION AND RESOLUTION POWER
 Resolution is the ability to separate ions of nearly equal mass/charge
 e.g. C6H5Cl and C6H5OF, 112 m/z
 C6H5Cl = 112.00798 amu (all 12C, 35Cl, 1H)
 C6H5OF = 112.03244 amu (all 12C, 16O, 1H, 19F)
Two definitions
– Resolution
= Δm/m (0.024/112.03 = 0.00022 or 2.2x10-4)
– Resolving power = m/Δm (112.03/0.024 = 4668)
311
FRAGMENTATION PATTERNS
Alkanes
 Hydrocarbon chains characterized by successive losses of m/z
 Fragmentation often splits off simple alkyl groups:
Loss of methyl
M+ - 15
Loss of ethyl
M+ - 29
Loss of propyl
M+ - 43
Loss of butyl
M+ - 57
 Branched alkanes tend to fragment forming the most stable carbocations.
 More stable carbocations will be more abundant.
 For simple linear alkanes fragmentation will occur towards the middle of the chain.
312
CONT…
Mass spectrum of 2-methylpentane
313
ALKENES
Fragmentation typically forms resonance stabilized allylic
carbocations
Alkenes can yield allylic stabilized carbocations by fragmentation,
splitting out a radical since resonance-stabilized cations favored
314
AROMATICS
Fragment at the benzylic carbon, forming a resonance stabilized
benzylic carbocation (which rearranges to the tropylium ion)
315
CONT…
Aromatics may also have a peak at m/z = 77 for the benzene ring
316
ALCOHOLS
 Alcohols have several characteristic fragmentation patterns.
 Fragment easily resulting in very small groups.
 alpha (α) cleavage (at the bond next to the C-OH) and
 dehydration (loss of H-OH) to give C=C
 An alcohol radical/cation can undergo α fragmentation to produce a
radical and a resonance stabilized carbocation.
 May lose hydroxyl radical or water
M+ - 17 or M+ - 18
317
CONT…
Commonly lose an alkyl group attached to the carbinol carbon forming an oxonium ion.
1o alcohol usually has prominent peak at m/z = 31 corresponding to H2C=OH+
318
MS FOR 1-PROPANOL
319
MS FOR HYDROCINNAMALDEHYDE
320
CONT…
321
CARBOXYLIC ACID
Carboxylic Acids can also undergo α cleavage
322
MS INTERPRETATION
Important Regions of the MS Spectrum, 3600-2700 cm-1, X-H stretch region
3600-3300 cm-1
3300-2500 cm-1
3200-3000 cm-1
3000-2800 cm-1
2850 and 2750 cm-1
The alcohol OH stretch is usually a broad and
Alcohol O-H
strong absorption near 3400. The NH stretch is
Amine or Amide
typically not as broad or strong as the OH, and in
the case of an NH2 it may appear as two peaks. The
N-H
terminal alkyne C-H may be confirmed by a weak
Alkyne C-H
CC triple bond stretch near 2150 cm-1
This is normally a very broad signal centered near
Acid O-H
3000 cm-1.
The aromatic CH's usually appear as a number of
Aromatic (sp2) = C-H
weak absorptions, while the alkene C-H is one or a
2
Alkene (sp ) =C-H
couple stronger absorptions.
Almost all organic compounds have alkyl CH's so this
is not usually too informative. However, the intensity
Alkyl (sp3) C-H
of these peaks relative to other peaks gives a hint as to
the size of the alkyl group.
Two medium intensity peaks on the right hand
Aldehyde C-H
shoulder of the alkyl C-H's. Look for confirming
carbonyl C=O peak.
323
CONT…..
2300-2100 cm-1,
C
X
2260-2210 cm-1
A sharp, medium intensity peak. Carbon
Dioxide in the atmosphere may also result in
an absorption in this area if not subtracted
N
C
out.
2260-2100 cm-1
This peak's intensity varies from medium to
A
l
k
y
n
e
C
C
nothing. Since the intensity is related to the
change in dipole moment, symmetrical alkynes
will show little or no absorption here!
324
1850-1500 CM-1, C=X STRETCH REGION
Anhydrides have two absorptions, one near 1830-1800 and one
near 1775-1740. The absorption frequency increases as the ring
1850-1750 cmAnhydride C=O
size decreases. For example: cyclohexanone=1715,
1
3-4 membered ring C=O
cyclopentanone=1745, cylobutanone=1780,
cyclopropanone=1850.
Aldehyde C=O
1750-1700 cm- Ketone C=O
1
Ester C=O
Acid C=O
1700-1640 cm1
1680-1620 cm1
1600-1400 cm1
This is usually the most intense absorption in the spectrum.
Amide C=O
Conjugated C=O
Because of the weakening of the C=O due to resonance, amides and
conjugated carbonyl's come slightly lower than "normal" C=O. In
general, conjugation lowers the absorption by 20-50 cm-1.
Alkene C=C
This absorption is not as intense as that seen for C=O. It is variable
and may be fairly small in symmetrical, or nearly symmetrical
cases. Look for confirming alkene C-H peaks above 3000.
Aromatic C=C
Multiple sharp, medium peaks. The pattern of peaks varies
depending upon the substitution pattern. Usually there is one peak
around 1600 and several others at lower wavenumbers. Look for
confirming aromatic C-H peaks slightly above 3000.
325
1500-400 CM-1, FINGERPRINT REGION
1300-1000 cm-1 C-O
1500-400 cm-1
Various
A strong absorption.
Interpretation of peaks in the fingerprint
region is complicated by the large number
of different vibrations that occur here.
These include single bond stretches and a
wide variety of bending vibrations. This
region gets its name since nearly all
molecules(even very similar ones) have a
unique pattern of absorptions in this
region.
326
TYPICAL MASS SPECTRAL DATA
The most abundant peak (largest) in the mass spectrum is called the base peak.
It is assigned a value of 100% and all other detectable masses are indicated
as a percent of the base peak.
The molecular weight peak is called the mass peak or molecular ion peak or
parent peak and symbolized with an M. Since this peak is a radical cation, it
often also has a + or +. .
Base peak = largest peak in spectrum = 100% peak or The most abundant
ion formed in the ionization chamber gives to rise the tallest peak in the
mass spectrum, called the base peak.
Molecular ion = M = M+ = M+. = parent peak or By using one of the
many ionization methods, the simple removal of an electron from a
molecule yields a positively charged radical cation, known as the
molecular ion and symbolized as [M]+.
.
327
CONT…
328
Mass Spectrum of Isobutyrophenone
105
(base peak)
C6H5CO+
O
molecular
weight = 148
77
C6H5+
molecular ion, M
(148)
M+1
Example: 3-pentanone, C5H10O
330
The radical cation (M+•) is unstable and will fragment into smaller ions
m/z=15
Relative abundance (%)
m/z=16 (M+)
-e
+
H
_
H C H
H C H
H
H
H
H C+
H C H
+
H
charge neutral
not detected
m/z = 14
H H H
H C C C H
m/z=29
-e
_
-e
m/z=44
(M)
H C C C
_
+
H C C
H H
+
H
charge
neutral not
detected
H
+
H H
H H
charge neutral
not detected
m/z = 43
H H
H C C C H
+
H H H
m/z = 44
H H H
H C C
m/z
H C C C
H
m/z = 29
m/z=45
(M+1)
H H H
H H H
H H H
m/z=15
+
H H H
H H H
m/z=43
H
+
m/z=17
(M+1)
m/z=14
+
H
charge neutral
m/z = 15 not detected
m/z = 16
m/z
Relative abundance (%)
H
+
C H
H
charge neutral
not detected
H
+
+C H
H
m/z = 15
331
35Cl
m/z=112
(M+)
Cl
37Cl
34.96885
36.96590
75.77 %
24.23 %
m/z=114
(M+ +2)
m/z=113
(M+ +1)
m/z=77
m/z=115
(M+ +3)
m/z
Br
m/z=77
79Br
m/z=156
(M+)
m/z=157
(M+ +1)
m/z
m/z=158
(M+ +2)
81Br
78.91839
80.91642
50.69 %
49.31 %
m/z=159
(M+ +3)
332
CONT….
A.T
333
CHAPTER -EIGHT
Structure Elucidations
By Joint Application of Different
Spectroscopic Methods: UV, IR, NMR and MS
334
Introduction
 In this chapter you will employ jointly all of the spectroscopic
methods.
 It involves analysis of the mass spectrum (MS), the infrared (IR)
spectrum, and proton and carbon (1H and 13C) NMR.
 In general, you should first try to gain an overall impression by
looking at the gross features of the spectra provided in the problem.
 As you do so, you will observe evidence for pieces of the structure.
 Once you have identified pieces, you can assemble them and test
against each of the spectra the validity of the structure you have
assembled.
335
1. Mass spectrum
 You should be able to use the mass spectrum to obtain a molecular
formula by performing the Rule of Thirteen calculations on the molecular
ion peak (M) labeled on the spectrum.
 In most cases, you will need to convert the hydrocarbon formula to one
containing a functional group. For example, you may see a carbonyl group
in the IR spectrum or 13C spectrum.
 Make appropriate adjustments to the hydrocarbon formula so that it fits
the spectroscopic evidence.
 When the mass spectrum is not provided in the problem, you will be
given the molecular formula.
 Some of the labeled fragment peaks may provide excellent evidence for
the presence of a particular feature in the cpd being analyzed.
336
2. Infrared spectrum.
The IR spectrum provides some idea of the functional group or
groups that are present or absent.
 Look first at the left-hand side of the spectrum to identify functional
groups such as O-H, N-H, C ≡ N, C≡ C, C= C.C= O, NO2, and
aromatic rings.
 Ignore C-H stretching bands during this first’’glance’’ at the
spectrum as well as the right-hand side of the spectrum.
 Determine the type of C=O group you have and also check to see if
there is conjugation with a double bond or aromatic ring.
A complete analysis of the infrared spectrum is seldom necessary.
3. Proton NMR spectrum.
The proton (1H) NMR spectrum gives information on the numbers
and types of H-atoms attached to the C-skeleton.
You will need to determine the integral ratios for the protons by
using the integral traces shown.
337
Cont…
In most cases, it is not easy to see the splitting patterns of multiplets
in the full 300MHz spectrum.
We have, therefore, indicated the multiplicities of peaks as doublet,
triplet, quartet, quintet, and sextet on the full spectrum.
4. Carbon NMR spectra
The carbon (13C) NMR spectrum indicates the total number of nonequivalent carbon atoms in the molecule.
In some cases, because of symmetry, carbon atoms may have
identical chemical shifts.
Commonly, sp3 carbon atoms appear to the up field (right) side of
the CDCl3 solvent peak, while the sp2 carbon atoms in an alkene or in
an aromatic ring appear to the left of the solvent peak.
Carbon atoms in a C= O group appear furthermost to the left in a
carbon spectrum.
338
Cont…
5. DEPT-135 and DEPT-90 spectra.
 In some cases, the problems list information that can provide
valuable information on the types of C-atoms present in the unknown
cpd.
6. Ultraviolet/visible spectrum.
 The UV spectrum becomes useful when unsaturation is present in a
molecule.
7. Determining a final structure.
A complete analysis of the information provided in the problems
should lead to a unique structure for the unknown compound.
Note that more than one approach may be taken to the solution of
these example problems.
339
The Rule Of Thirteen
 High-resolution MS provides molecular mass information from
which the user can determine the exact molecular formula directly.
 When such molar mass information is not available, however, it is
often useful to be able to generate all the possible molecular formulas
for a given mass.
 By applying other types of spectroscopic information, it may then be
possible to distinguish among these possible formulas.
 A useful method of generating possible molecular formulas for a
given mass is the Rule of Thirteen.
 As a first step in the Rule of Thirteen, we generate a base formula,
which contains only C& H.
 The base formula is found by dividing the molecular mass, M by
13( the mass of 1 C plus 1H). This calculation provides a numerator, n
& a remainder, r.
M/13 = n + r/13
The base formula thus becomes
CnHn+r
340
Index of H-deficiency(unsaturation index)
 The number of π bonds and/or rings a molecule contains. That
corresponds to the preceding formula is calculated easily by applying
the relationship
U = (n-r+2)/2
 Of course, you can also calculate the index of H-deficiency using
the ff method for a molecular formula, CcHhNnOoXx,
U = (2c+2-h-x+n)/2
 If we wish to derive a formula that includes other atoms besides C
& H,
 we must subtract the mass of a combination of Cs & Hs that equals
the masses of the other atoms being included in the formula.
 For example, if we wish to convert the base formula to a new
formula containing one O-atom, then we subtract 1 C & 4 Hs at the
same time that we add one O-atom.
 Both changes involve a molecular mass equivalent of 16
O = CH4 = 16
341
Cont…
 The table shown below includes a number of C/H equivalents for
replacement of C & H in the base formula by the most common
elements likely to occur in an organic cpd.
 To comprehend how the Rule of Thirteen might be applied,
consider an unknown substance with a molecular mass of 94 amu.
 Application of the formula provides
94/13 = 7 +3/13
 According to the formula, n = 7 & r = 3. the base formula must be
C7H10 The index of H-deficiency is U =(7-3+2)/2 = 3
342
Cont…
A substance that fits this formula must contain some combination of 3
rings or multiple bonds. a possible structure might be
C7H10
U=3
If we were interested in a substance that had the same molecular mass
but that contained 1 O-atom, the molecular formula would become
C6H6O.
This formula is determined according to the ff scheme.
1. Base formula = C7H10
U=3
2. Add: +O
3. Subtract: -CH4
4. Change the value of U:
∆U = 1
5. New formula = C6H6O
343
Cont…
6. New index of H-deficiency:
A possible substance that fits these data is
U=4
C6H6O
U=4
 There are additional possible molecular formulas that confirm to a
molecular mass of 94 amu:
C5H2O2
U=5
C5H2S
U=5
C6H8N
U = 3.5
CH3Br
U=0
 As the formula C6H8N shows, any formula that contains an even
number of H-atoms but an odd number of N-atoms leads to a
functional value of U, an unlikely choice.
 Any cpd with a value of U less than zero(i.e., negative) is an
impossible combination.
 Such a value is often an indicator that an O or N-atom must be
present in the molecular formula.
344
Cont…
 When we calculate formulas using this method, if there are not
enough Hs, we can subtract 1 C & add12 Hs ( & make the
appropriate correction in U).
 This procedure works only if we obtain a positive value of U.
 Alternatively, if the value of U is greater than 7,we can obtain
another potential molecular formula by adding 1 C & subtracting
12 Hs ( & correcting U)
345
Cont…
Example-1:The UV spectrum of this cpd shows only end absorption.
determine the structure of the cpd using the ff spectroscopic data.
346
Cont…
347
Cont…
 Note that this problem does not provide a molecular formula.
 We need to obtain it from the spectral evidence.
 The molecular ion peak appears at m/e=102.using the Rule of
Thirteen, we can calculate a formula of C7H18 for the peak at 102.
 The IR spectrum shows a strong absorption at 1740cm-1, suggesting
that a simple unconjugated ester is present in the cpd.
 The presence of a C-O(strong and broad) at 1200cm-1 confirms the
ester.
 We now know that there are 2 oxygen atoms in the formula.
 Returning to the mass spectra evidence, the formula calculated via
the Rule of Thirteen was C7H18.
we can modify this formula by converting carbons and hydrogens
(1C & 4 Hs per O atom) to the 2-O atoms, yielding the formula
C5H10O2.
 This is the molecular formula for the cpd.
348
Cont…
 We can now calculate the index of H-deficiency for this cpd, which
equals 1, & that corresponds to the unsaturation in the C=O group.
 The IR spectrum is also shows sp3(aliphatic) C-H absorption at less
than 3000cm-1.
 we conclude that the cpd is an aliphatic ester with formula C5H10O2.
 Notice that the 13C NMR shows a total of 5 peaks, corresponding
exactly to the number of carbons in the molecular formula.
 This is a nice check on our calculation formula via the Rule of
Thirteen (5-C atoms).
 The peak at 174ppm corresponds to the ester C=O carbon.
 The peak at 60 ppm is a deshielded C-atom caused by a neighboring
single-bonded O-atom.
 The rest of the C-atoms are relatively shielded.
 These 3 peaks correspond to the remaining part of the C-chain in
the ester.
349
Cont…
We could probably derive a couple of possible structures at this point.
The 1H NMR spectrum should provide confirmation.
Using the integral traces on the spectrum , we should conclude that
the peaks shown have the ratio 2:2:3:3(downfield to up field).
These numbers add up to the 10 total H-atoms in the formula.
Now, using the splitting patterns on the peaks, we can determine the
structure of the cpd. It is ethyl propanoate.
The downfield quartet at 4.1ppm (d protons) results from the
neighboring protons on C-b,
 While the other quartet at 2.4 ppm (C protons) results from the
protons on C-a.
 Thus, the proton NMR is consistent with the final structure.
350
Cont…
The UV spectrum is uninteresting but supports the identification of
structure. Simple esters have weak n Π* transitions(205 nm) near the
solvent cutoff point.
Returning to the mass spectrum, the strong peak at 57 mass units
results from an α cleavage of an alkoxy group to yield the acylium ion
(CH3-CH2-C+=O), which has a mass of 57.
Example-2;-determine the structure of a cpd with the formula
C10H12O2 using the following spectroscopic data.
351
Cont…
Solution:
 We calculate an index of hydrogen deficiency of 5.
 The 1H NMR and 13C NMR spectra, as well as the IR spectrum,
suggest an aromatic ring(index =4).
352
Cont…
The remaining index of 1 is attributed to a C=O group found in the
IR spectrum at 1711cm-1.
 This value for the C=O is close to what you might expect for an
unconjugated carbonyl group in a ketone and is too low for an ester.
 When inspecting the 1H NMR spectrum, notice the nice para
substitution pattern between 6.8 & 7.2 ppm, which appears as a
nominal pair of doublets, integrating for 2 Hs in each pair.
Also notice in the 1H NMR that the up field portion of the spectrum
has Hs that integrate for 3:2:3 for a CH3, a CH2, & a CH3 respectively.
Also notice that these peaks are unsplit indicating that there are no
neighboring Hs.
The down field methyl at 3.8 ppm is next to an O-atom, suggesting a
methoxy group.
Keeping in mind the para disubstituted pattern & the singlet peaks in
the proton NMR, we derive the ff structure for 4-methoxy phenyl
353
acetone.
Cont…
 Further confirmation of the para disubstituted ring is obtained from
the C-spectral results.
 Notice the presence of 4 peaks in the aromatic region of the CNMR spectrum.
 Two of these peaks(126 & 159 ppm) are ipso C-atoms(no attached
protons) that do not show in the DEPT-135 or DEPT-90 spectra.
The remaining 2 peaks at 114 & 130 ppm are assigned to the
remaining 4 Cs (2 each equivalent by symmetry).
The 2 C atoms d show peaks in both of the DEPT experiments,
which confirms that they have attached protons(C-H).
Likewise the 2 C atoms e have peaks in both DEPT experiments
confirming the presence of C-H.
354
Cont…
 The IR spectrum has a para substitution pattern in the out-of-plane
region(835cm-1),which helps to confirm the 1,4-disubstitution on the
aromatic ring.
Example-3:- This cpd has the molecular formula C9H11NO2. include
in this problem are the IR spectrum, proton NMR with expansions,
and C-NMR spectra data.
355
Cont…
356
oWe
Cont…
Solution:o we calculate an index of H-deficiency of 5.
o All of the spectra shown in this problem suggest an aromatic ring
(index = 4).
o The remaining index of 1 is assigned to the C=O group found at
1708cm-1.
o This value for the carbonyl group is too high for an amide.
o It is in a reasonable place for a conjugated ester.
357
Cont…
While the NO2 present in the formula suggests a possible nitro
group, this can't be the case, because we need the 2-Os for the ester
functional group.
The doublet at about 3400 cm-1 in the IR spectrum is perfect for a
primary amine.
The C-NMR spectrum has 9 peaks, which correspond to the 9 Catoms in the molecular formula.
The ester C=O C-atom appears at 167 ppm.
The remaining down field Cs are attributed to the 6 unique aromatic
ring Cs.
From this we know that the ring is not symmetrically substituted.
The DEPT results confirms the presence of 2 C-atoms with no
attached protons(131 & 147 ppm) & 4 C-atoms with 1 attached proton
(116,119.120, & 129).
From this information we now know that the ring is disubstituted.
358
Cont…
We must look carefully at the aromatic region in the 1H-NMR
spectrum to determine the substitution pattern on the disubstituted
ring.
Notice in the full 300 MHz spectrum that there are 4 protons
between 6.8 & 7.5 ppm, & that each set of peaks represents 1 proton
(integrals).
Although it is difficult to see the patterns on the full spectrum, if you
look closely you will observe a doublet at 7.42 ppm, a singlet at 7.35
ppm, a triplet at 7.19 ppm, & a doublet at about 6.84 ppm.
A pattern of this type suggests a 1,3-disubstituted benzene ring
(meta).
The singlet proton is between the 2 attached groups on the ring.
Anisotropy causes the protons next to the C=O group to shift
downfield.
The other 2 protons are upfield in the aromatic region.
359
Cont…
Because of their splitting patterns & chemical shifts, we should be
able to assign the protons in the aromatic region.
Although not as reliable as 1H-NMR evidence, the aromatic out-ofplane bending bands in the IR spectrum suggests meta disubstitution:
680, 760, & 880 cm-1.
The proton NMR spectrum shows an ethyl group because of the
quartet & triplet found upfield in the spectrum(4.3 & 1.4 ppm,
respectively, for the CH2 & CH3 groups).
 Finally, a broad NH2 peak, integrating for 2 protons, appears in the
proton NMR spectrum at 3.8 ppm.
The cpd is ethyl 3-amino benzoate.
360
Cont…
Example-4:- This cpd has the molecular formula C5H7NO2. ff are the
IR, 1H-NMR, & 13C-NMR spectra.
singlet
361
Cont…
Solution: we calculate an index of H-deficiency of 3.
A quick glance at the IR spectrum reveals the source of unsaturation
implied by index of 3.
A nitrile group at 2260cm-1(index = 2) & a carbonyl group at 1747 cm1 ( index = 1).
The frequency of the carbonyl absorption indicates an unconjugated
ester. The appearance of several strong C-O bands near 1200 cm-1
confirms the presence of an ester functional group.
We can rule out a C≡C bond because they usually absorb at a lower
value (2150 cm-1) & have a weaker intensity than cpds that contain
C≡N.
362
Cont…
o The C-NMR spectrum shows 5 peaks & thus is consistent with the
molecular formula, which contains 5 C-atoms.
o Notice that the C-atom in the C≡N group has a characteristic value
of 113 ppm.
o In addition, the C-atom in the ester C=O appears at 163 ppm.
o One of the remaining C-atoms (63 ppm) probably lies next to an
electronegative O-atom.
o The remaining 2 C-atoms, which absorb at 25 & 14 ppm, are
attributed to the remaining methylene & methyl carbons.
oThe 1H-NMR spectrum shows a classical ethyl pattern: a quartet (2
H) at 4.3 ppm & a triplet (3 H) at 1.3 ppm.
o The quartet is strongly influenced by the electronegative O-atom,
which shifts it downfield.
oThere is also a 2-proton singlet at 3.5 ppm.
oThe structure is:
363