Engineering Chemistry-II (BS-104)

Spectroscopic Techniques-1

Dr. Shruti Khanna Ahuja

Assistant Professor
USAR, GGSIPU
• Study of interaction of electromagnetic radiation with matter defines the science of spectroscopy.
• Spectroscopy is that science which attempts to determine what specific energies and amounts of incident light are
absorbed by specific substances, and what specific energies and amounts are later re-emitted.
• It includes different analytical techniques used for both quantitative and qualitative analyses of inorganic, organic and
biological compounds.
• Spectrometers are optical instruments that reveal the light energies absorbed and emitted in terms of specific
wavelengths or frequencies.
• Spectra are the records that provide us with important information pertaining to the atomic and molecular structure of the
substances on which the electromagnetic energy is focused.
• These spectra provide us with “fingerprints” that are characteristic of—and therefore uniquely connected to—different
elements and compounds.

A generic spectrometer
• Electromagnetic radiation is a type of energy that is transmitted through
space at enormous velocities. It is found to have dual nature - the particle
nature (photons) and wave nature.
• It is a propagating wave of electrical energy with an orthogonal magnetic
component oscillating with exactly the same frequency.
• Molecules absorb electromagnetic radiation in discrete packets of energy
called “quanta”, which are measurable by spectroscopy. Electromagnetic radiation
• The many different types of EM waves are categorized according to their
origins and their frequency/wavelength values.
c = λν
c = speed of light = 3 X 108 m/s
λ = wavelength of radiation
ν = frequency of electromagnetic radiation Electromagnetic
spectrum
E = hν = hc/ λ
E= energy of electromagnetic radiation
h = Planck’s constant = 6.62 X 10-34 Js

As the energy increases, wavelength decreases

• The field of spectroscopy is divided into two main classes:
i. Absorption ii. Emission

i. Absorption spectra are produced when white light is directed onto a particular
Higher energy level
substance and certain wavelengths in the spectrum of the white light are
absorbed via internal energy excitations.
• The wavelength absorbed characterizes some specific functional group of the h
compound or the compound itself.
• The regions of electromagnetic radiation of greatest interest to organic chemists
Ground state
are 200-400 nm (UV), 400-800 nm (visible) and 800 nm-1 mm (infra-red)
• The mechanisms of absorption of energies is different in the ultra-violet, infra-red
and nuclear magnetic resonance (NMR) regions, though the fundamental Absorption
phenomenon is the absorption of certain amount of energy.
• Absorption in UV-vis is due to electronic transitions, in infra-red region, it is due to
molecular vibrations, where as in NMR, nuclear spin transitions take place.
 Higher energy level
ii. Emission spectra for a substance occurs when the substance
receives energy from an external source, absorbs the energy to
h
raise its electrons to higher energy metastable states (electrons
having short life time here), then emits characteristic energy
amounts as the excited electrons return to their previous energy Ground state

levels.
Emission
• Fluorescent lights and colors obtained by heating salts of certain
elements in the flame are very common examples of emission spectra
• Fluorescence: The electron in the excited level return to its ground
state either directly or in steps with the emission of certain amount of
energy. When this emission of light is instantaneous the phenomenon
is known as fluorescence
• Phosphorescence: When the electron in the excited level return to
it’s ground state with the emission of light after some time lag, it is
known as phosphorescence.
1. UV-VISIBLE SPECTROSCOPY
• A region of electromagnetic radiation whose interaction with a molecule gives
rise to electronic transition exists at 200-800 nm
• It requires electromagnetic radiations of higher energy
• It is primarily used to measure the multiple bonds or aromatic conjugation Violet: 400 - 420 nm
Indigo: 420 - 440 nm
within the molecules. Blue: 440 - 490 nm
• When monochromatic light is passed through a medium containing particles, Green: 490 - 570 nm
Yellow: 570 - 585 nm
they interact with ions and molecules and electronic transitions occur by the Orange: 585 - 620 nm
excitation of electrons from the ground state to the excited state. Red: 620 - 780 nm

• Molecules that absorb in UV-vis region transfer the corresponding energy

from that photon to one or several of its outermost electrons.
• This transfer of energy results in a change on the total mechanical energy
associated with the molecule. The total energy will be equal to:
Etot = Eelec + Evib + Erot,
where, Eelec > Evib > Erot
• Instrumentation:

1. Light Source: Tungsten filament lamps and Hydrogen-Deuterium lamps are the most widely used and suitable light
sources as they cover the whole UV region. 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.
2. Monochromator: Monochromators generally are 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 such
the rotation of the prism results in a series of continuously increasing wavelengths to pass through the slits for recording
purposes. The beam selected by the slit is monochromatic and further divided into two beams with the help of another
prism.
3. Sample and reference cells: One of the two divided beams is passed through the sample solution and the second beam
is passed through the reference solution. Both sample and reference solution is 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.
4. Detector: Generally, two photocells serve the purpose of the detector in UV spectroscopy. One of the photocells receives
the beam from the sample cell and the second detector receives the beam from the reference. The intensity of the
radiation from the reference cell is stronger than the beam of the sample cell. This results in the generation of pulsating or
alternating currents in the photocells.
5. Amplifier: The alternating current generated in the photocells is transferred to the amplifier. Generally, the current
generated in the photocells is of very low intensity, the main purpose of the amplifier is to amplify the signals many times
so we can get clear and recordable signals.
6. Recording devices: Most of the time amplifier is coupled to a pen recorder which is connected to the computer. The
computer stores all the data generated and produces the spectrum of the desired compound.

Schematic Representation UV-Vis spectrophotometer

• Chromophores: It is a group of molecules, which is responsible for certain absorption of radiation. It is a minimum
structural requirement for the absorption of radiation in the UV region, e.g. molecules having π electrons such as C=C,
C=O, N=N
• Auxochromes: It is a saturated group containing unshared electrons which when attached to the chromophore changes
both intensity as well wavelength of the absorption maxima, e.g. OH, NH2, Cl
• λmax : It is a wavelength at which there is a maximum absorption. It is a physical constant and characteristic of a molecular
structure and is very useful for identification of compounds. It is independent of concentration.
• Bathochromic shift: It is the shifting of the absorption peak to the higher wavelength due to the effect of a substituent
group or solvent. It is also known as red shift. E.g., λmax of ascorbic acid is 243 nm, while that in alkali medium is 299 nm
• Hypsochromic shift: It is the shifting of the absorption peak to the lower wavelength. It is also known as blue shift. E.g.,
λmax of phenol in basic media is 297 nm, while that in acidic media is 277 nm
• Hyperchroism: Increase in absorption intensity due to substituent, solvent, pH or some other factors
• Hypochroism: Decrease in absorption intensity due to substituent, solvent, pH or some other factors
Shifts in absorption position and intensity
• Electronic energy levels and transitions:
• Types of electronic transitions:
i. σ–σ* transition: An electron is excited from bonding σ orbital to higher energy anti-bonding σ* orbital. Sigma bonds, in
general, are very strong and therefore the energy required for this transition is large and the band appear at short
wavelengths <150 nm. It is observed with saturated compounds such as alkanes e.g. λmax for methane is 122 nm. This
regions is usually less informative.
ii. n–σ* transition: This transition involves saturated compounds with one heteroatom having unshared pair of electrons (n
electrons) e.g. saturated alkyl halides, alcohols, ethers, aldehydes, ketones and amines etc. These transitions require
comparatively lesser energy than σ–σ* transitions. In saturated alkyl halides the energy required for this transition decreases
with increase in size of the halogen atom (less electronegativity => loosely bound outermost electron => less energy) e.g.
λmax for methyl chloride of 173 nm and that of methyl iodide is 258 nm. This transition is sensitive to H-bonding with the
solvent molecules as in alcohols, amines thereby shifting the absorption peaks to shorter wavelength.
iii. π–π* transition: It occurs in the compounds with unsaturated centers such as simple alkene, aromatics, carbonyl
compounds. This transition is lesser in energy than n–σ* transition. Unconjugated alkenes generally absorb in 170-205 nm.
iv. n–π* transition: An electron form non-bonding orbital of a heteroatom is promoted to anti-bonding π* orbital. It requires
least energy out of all transitions and the absorption band appear at higher wavelengths.
v. Conjugated systems: In conjugated dienes, the π orbitals of the separate alkene groups combine to give new orbitals, i.e.,
the new bonding orbitals designated as π1 and π2 and two anti-bonding orbitals, π3* and π4*. The relative energy of π2  π3* of
a conjugated diene is of very low energy than π  π* transition in an unconjugated alkene. e.g. λmax for ethene is 171 nm while
the peak is shifted bathochromically in butadiene with λmax at 217 nm.
• Types of absorption bands: They occur due to electronic transition of a molecule.
i. R-bands: These are observed in compounds containing double bonds with heteroatoms such as carbonyls (C=O), nitro (-
NO2) etc. An electron from non-bonding orbital is promoted to anti-bonding π* i.e. they undergo n–π* transition. This transition
require minimum energy of out of all transitions and shows absorption band at longer wavelength around 300 nm e.g. in
acetone absorption band appear at 279 nm. Saturated aldehyde show both type of transitions: lower energy n–π* and higher
energy π–π* thereby exhibiting absorption peaks at 290 and 180 nm, respectively. The εmax value is less than 100.
ii. K-bands: These arise from π–π* transitions in π–π conjugated systems like alkenes, alkynes, carbonyls, nitriles and
aromatic compounds. The show εmax greater than 10,000. e.g. 1,3,5-hexatriene exhibits λmax at 258 nm
iii. B-bands: These are due to aromatic and hetero-aromatic systems. The λmax values are between 230-270 nm and εmax less
than 2000. These bands are called benzenoid bands. In the presence of K-bands the position of the B-band is shifted to larger
wavelengths. The UV spectrum of benzaldehdye contains K, R and B bands.
iv. E-bands: These are also known as ethylenic bands and are characteristic of the aromatic systems as are the B-bands. They
are classified as E1 and E2 bands wherein E1 bands appear at smaller wavelengths and are usually more intense, while E2
bands for the same compound appear at longer wavelengths e.g. in naphthalene E1 band is at 221 nm and E2 band at 286 nm.
The εmax values of these bands vary from 2000- 14000. The presence of auxochromic group shifts an E-band to a higher
wavelength.
• Woodward-Fieser’s rules for calculation of λmax values in conjugated systems:
1) For α, β-unsaturated aldehydes and ketones: Compounds in which C=O is conjugated with a C=C (enones) exhibit
absorption band at longer wavelengths due to low energy n–π* (R-band) and π–π* transitions (K-band). e.g.

• Rules:
i. The base value of the chromophore (C=C and C=O) in an acyclic and a 6-membered ring system is assigned as 215 nm
ii. For each α-substituent (methyl) 10 nm is added to the base value
iii. For each β-substituent (methyl) 12 nm is added to the base value
iv. For each ring system (5 or 6 membered) to which the C=C is exocyclic (double bond outside ring), 5 nm is added to the base
value
v. If the C=C and the C=O are in 5-membered ring i.e. cyclopentenone, the base value is 202 nm
 When cross conjugation i.e. α, β –unsaturation on both sides, λmax is calculated by considering most highly substituted
conjugated system.
• Examples:
1. Methyl vinyl ketone: No α and β substituents present, Cal λmax: 215 nm (Obs. : 213 nm)

1
2. 2-Methyl-1-butene 3-one: one methyl at α-position, Cal. λmax: 225 nm (Obs. : 220 nm)

3. 3-pentene-2-one: one methyl at β-position, Cal. λmax: 227 nm (Obs. : 224 nm) 2

4 5 6
2) Conjugated dienes (acyclic): Conjugation shifts the absorption band to higher wavelength.
e.g. λmax for ethene, 1,3-butadiene and 1,3,5-hexatriene is 171, 217 and 258 nm, respectively
• Rules:
i. The base value for acyclic conjugated dienes and cyclic conjugated dienes containing a non-fused 6-membered ring system
is 217 nm
ii. For each acyclic alkyl substituent 5 nm is added to the base value
iii. If C=C is exocyclic (double bond outside ring), 5 nm is added to the base value
iv. Trans-isomer absorbs at a longer wavelength due to steric effects. For effective overlap of π-orbitals and low π–π* transition
energy, the molecule should be coplanar. Cis-isomer adopts a non-planar conformation due to steric effects
3) Conjugated dienes (cyclic):
• Rules:
i. The base value for heteroannular conjugated dienes (tansoid) is 214 nm
ii. The base value for homoannular conjugated dienes (cisoid) is 253 nm
iii. For each acyclic alkyl substituent 5 nm is added to the base value
iv. If C=C is exocyclic (double bond outside ring), 5 nm is added to the base value
• Examples:
• Examples:
• Applications of UV-Vis spectroscopy in analytical chemistry:

• Qualitative analysis:

 Detection of functional groups

 Extent of conjugation

 Identification of unknown compound

 Detection of impurities

• Quantitative analysis: Determination of unknown concentration of a given sample using Beer-Lambert’s Law
• Beer-Lambert’s Law:
 When the monochromatic radiation passes through a sample containing an
absorbing species, the radiant power of a beam is progressively decreased as
more energy is absorbed by the particles of that species. The decrease in
power is expressed by
I0
1) Lambert’s law
2) Beer’s law (or) Beer - Lambert’s law
 Iabsorbed = I0-It
It
where, I0 = intensity of incident light
It = intensity of transmitted light
The transmittance (T) is given by :
T= It / I0
and %T = 100 It / I0 Reduction of the intensity of light by
 Transmittance is defined as the amount of light that “successfully” passes reflection, absorption, scattering

through the substance and comes out from the other side
• Lambert’s Law:
 It states that “when a beam of monochromatic radiation is passed through a homogeneous absorbing medium, the rate of
decrease of intensity of the radiation ‘dI’ with thickness or path length of absorbing medium ‘dl’ is proportional to the
intensity of the incident radiation ‘I’ ”.
-dI/dl = kI ………. (1)
On integrating the equation (1), between the limits I = I0 at l = 0 and I = I at l = l, we get
I l
dI ……… (2)

0
I
   k dl
0

where, k = absorption coefficient

i.e., I or, log I0 = k1 l …………(3) This is called Lambert’s law
ln  k l
I0 It 2.303
where log I0 / It is called Absorbance or optical density

 Absorbance is reciprocal of transmittance

• Beer’s Law:
It states that “when the monochromatic radiation passes through a sample containing an absorbing species, the
decrease in the radiant power of a beam is directly proportional to the concentration (C) of the sample”.
log I0 = k1C …………(4) This is called Beer’s law
It 2.303
• Beer- Lambert’s Law: When a beam of monochromatic radiation is passed through a solution of an absorbing substance,
the rate of decrease of intensity of radiation ‘dI’ with thickness or path length of the absorbing solution ‘dl’ is proportional to
the intensity of the incident radiation ‘I’ as well as the concentration of the solution ‘C’.
On combining equations (3) and (4)
A = log I0 = kCl
It 2.303
or, A=εCl ………….(5) This is called Beer-Lambert's law
where ε = k/ 2.303 is known as molar absorptivity coefficient
 The larger the molar absorptivity, the more probable the electronic transition. In uv spectroscopy, the concentration of the
sample solution is measured in mol L-1 and the length of the light path in cm. Thus, given that absorbance is unitless, the
units of molar absorptivity are L mol-1 cm-1
• Applications of Beer- Lambert’s Law: It can be used to determine the unknown concentration of a solution.
 A standard solution of known concentration Cs is first taken and it’s absorbance As is measured. Then according to Beer-
Lambert’s law: As
  x ………(1)
Cs

Let Cu be the unknown concentration of the solution which is to be determined. The absorbance Au of the given solution is
measured.
Au
  x ……….(2)
Cu
From equation (1) and (2), we get
As Au

Cs Cu
Au
=> Cu   Cs ………...(3)
As

We already know the concentration of standard solution Cs and the absorbance Au and As were experimentally measured
and hence the unknown concentration Cu can be calculated from the equation (3).
• Limitations of Beer- Lambert’s Law:
• This law can be used only for
i. dilute solution.
ii. monochromatic radiation.
iii. the system maintained at constant temperature.

• Specific Absorbance: When absorbance is measured for 1% (w/v) solution in a 1 cm cell, then the equation takes the form:

A = A1%1cm . C. l
where, A1%1cm is known as specific absorbance

• Colorimetric analysis: It is also based on Beer-Lambert’s law. Here, the visible light is used to analyze colored samples
• Solved Examples:
Question 1: The solution of concentration of 2.5 X 10-4 M has the percentage transmittance of 25% in a 1 cm cell.
Determine i) the absorbance (A) and ii) the molar absorption coefficient (ε).
Solution : Given : % T = 25% (or) 0.25, C = 2.5 X 10-4 M ; l = 1 cm
i) Absorbance : A = – log T = – log 0.25 = 0.6021
ii) The molar absorption coefficient, ε = A / C l
= 0.6021 / 2.5 x 10-4
= 2408

Question 2: A solution of thickness 2 cm transmits 40% incident light. Calculate the concentration of the solution given that
ε= 6000 dm3 mol-1 cm-1.
Solution : Given : % T = 40% (or) 0.40, ε= 6000 dm3 mol-1 cm-1; l = 2 cm
i) Absorbance : A = – log T = – log 0.40 = 0.3980
ii) The molar absorption coefficient, C = A / ε l
= 0.3980 / 6000 x 2
= 3.316 10–5 mol dm–3