chp1 UV- Visible last edited -2025 (1).pdf

Pharamacuetical Analaysis II
(Phar 3122)
Deneke A. 11/11/2024
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Introduction
 Analysis is the qualitative and quantitative determination of an
analyte (i.e target substance to be determined) in a given
sample.
 Pharmaceutical analysis is the qualitative and quantitative
determination of active constituents or impurities in
formulated pharmaceutical products.
 Based on Interaction there areTwo types
 Classical method (or so-called wet chemical methods) and
 Instrumental Method (physicochemical method)
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Introduction…
Instrumental methods involve studying the physical properties of analytes.
 i,e Conductivity, electrode potential, light absorption or emission
 Instrumental method commonly classified as:
Optical method : measure physical properties of analyte interaction with EMR (light)
e.g Absorbance or transmittance in Molecular or Atomic Spectrophotometry
Electrochemical method: measure physical properties of Analytes :
e.g. conductance in Conductometry, current in polarography potential in Potentiometery
Chromatography: measure Properties of analytes i,e retention time (Rt) in GC and HPLC
 Qualitative - identification by measuring physical property
Quantitative - measuring property and determining relationship to conc.
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Types of Instrumental Methods
Property Example
 Radiation emission Emission spectroscopy: fluorescence,
 Radiation absorption Absorption spectroscopy -,
Spectrophotometry, photometry
 Radiation scattering Turbidity, Raman
 Radiation refraction Refractometry, interferometry
 Radiation diffraction X-ray
 Radiation rotation Polarimetry, circular dichroism
 Electrical potential Potentiometry
 Electrical charge Coulometry
 Electrical current Voltammetry - polarography
 Electrical resistance Conductometry
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Intro…
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Spectroscopy
• Is the study of interaction between electromagnetic radiation and
matter.
Spectrophotometry
 It is more specific than the general term Electromagnetic spectroscopy
in that spectrophotometry deals with visible light, near-ultraviolet,
and near-infrared.
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Introduction
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 UV-Visible spectrophotometry is the method of choice in most
laboratories in:
 pharmaceuticals
 nucleic acids
 proteins, foodstuffs and fertilizers
 in mineral oils and in paint.
• Modern spectrophotometers are:
quick
 accurate
reliable and
make only small demands on the time and skills of the
operator.
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Electromagnetic radiation
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 EMR is a form of energy whose behavior is described by the properties of
both waves and particles.
 The optical properties of EMR, such as diffraction, are explained best by
describing light as a wave.
 Many of the interactions between EMR and matter, such as absorption
and emission, however, are better described by treating light as a
particle,or photon.
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1.Wave Properties of EMR
• EMR consists of oscillating electric and magnetic fields that propagate
through space along a linear path and with a constant velocity.
• In a vacuum, EMR travels at the speed of light, c,which is 3 x 108 m/s
EMR……
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 An electromagnetic wave is characterized by several fundamental
properties, such as:
1-Wavelength (λ,lambda):which is the linear distance measured along the line of
propagation, between crest of one wave to that of the next wave.
 It can be expressed in: Angstrom (o), nanometer (nm) , milimicrons (mμ )
centimeter (cm) or micrometer (μm).
 1 m = 102 cm = 103 mm = 106  = 109 nm = 1010 o.
 the unit of λ widely used in Uv-visible spectrometry is the nanometer (nm)
2-Amplitude:which is the vertical distance from midline of a wave to the
peak or trough.
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EMR……
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3- Frequency (v,nu) is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s)
4- Wave number ( ,nu par):number of waves per centimeter
and which is expressed in cm-1.
It is a reciprocal of wave length
 (  ) = 1/ , cm-1 .
EMR……
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Example:
if we have a visible radiation of 500 nm, then:
 in cm = 500 x 10-7 = 5 x 10-5 cm.
 = 1/ = 1/5 x 10-5 = 0.2 x 105 = 2 x 104 cm-1.
and  = C X  = 3 X 1010 . 2 X 104 = 6 X 1014 Hz
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EMR……
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Relations between ,  and  : are given by the following equations:
C =  x , Since  = 1/
Then  = 1/ = /C Or C =  / 
Where C is the velocity of light in vacuum = 3 x 1010 cm/Sec.

EMR…
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2.Light as energy
 Light like any other matter consists of energy packets called photons.
 The absorption and emission of light by compounds occur in these
packets (photons).
 The energy (E) of a photon is directly proportional to the frequency and
inversely proportional to the wavelength.
 It can be related to C,  and  by the following equation:
E = h = h C/
Where h is a constant called Planck’s constant , which equal to 6.625 x 10-27 erg. sec.
or 6.625 X 10-34 J.sec
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EMR…..
Remark:
 The higher the frequency, the higher the energy of radiation
 i.e. a photon of high frequency or shorter wavelength has higher energy
content than photons of lower frequency (longer wavelength).
 The Intensity EMR is proportional to the total number of photons
 It is independent of energy of each photon
 since energy per unit time is power:
 Intensity is often referred as the power emitted by the source.
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Example: What is the energy of a 500 nm photon?
 = c/  = (3 x 1010 cm s-1)/(5.0 x 10-5 cm)
 = 6 x 1014 s-1
E = h =(6.626 x 10-27 erg.s)(6 x 1014 s-1) = 4.0 x 10-12 erg
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Electromagnetic spectrum
 Arrangement of all types of EMR in order of their increasing wavelength or
decreasing frequency is known as electromagnetic spectrum
 For convenience, EMR is divided into different regions based their energy
 It varies from the highly energetic gamma rays to very low energy radio waves.
 The EM spectral region are based on the methods required to generate and detect
various types of radiation.
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Electromagnetic spectrum….
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Diagrammatic representation of electromagnetic radiation spectrum
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Electromagnetic spectrum……………
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Electromagnetic spectrum …..
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Electromagnetic spectrum
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The various regions of electromagnetic spectrum
according to the wavelength/ wave no.
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How Light InteractsWith Matter
As radiation passes from a
vacuum through the surface of a
portion of matter, the electrical
vector of the radiation interacts
with the atoms and molecules
of the medium.
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How Light Interacts….
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 The nature of the interaction depends upon the properties of the
matter.
 Each interaction can disclose certain properties of the matter.
refraction
transmission
absorption
reflection scattering
 scattering :Turbidity, Raman
 refraction :Refractometry, interferometry
 diffraction : X-ray
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Absorption : the interaction results in transfer of energy from EMR to the
matter atom/molecules)
Emission is the reverse process in which a portion of the internal energy of
matter converted into radiant energy
 In emission, e`s in exited state emit photons of energy and returning to the
lower energy state.
 In Absorption, e`s at lower energy state absorb photons of energy and transit
to the higher energy states .
There are two types of absorption
 Atomic Absorption
 Molecular Absorpition
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Atomic absorption
 promotes valance e` of an atom from their ground state to one or
more higher energy exited states
 Under normal situation an e- stays at the lowest possible shell- the e- is
said to be at its ground state.
 Upon absorbing energy (excited), an e- can change its orbital to a
higher one - we say the e- is at its excited state.
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 The excitation can occur at different degrees
 Low E tends to excite the outmost e-’s first
 An e- at its excited state is not stable and tends to return its ground state.
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ΔE = S2 - S0 = 380 ΔE = S1 - S0 , =590nm (Absorption) ΔE of emission= 590 nm
Absorption and emission for the sodium atom in the gas phase, Illustrates discrete energy transfer

Molecular Absorption
 More complex than atomic absorption because many more potential transitions exist.
 A molecule may absorb light energy in three ways:
 By raising an electron to a higher energy level (electronic).
 By increasing the vibration of constituent nuclei (vibrational).
 By increasing the rotation of molecule about its axis (rotational)
E = Es - Eg = h = h C/
 The energy E associated with the absorption bands of a molecule
E = E electronic + E vibrational + E rotational
E electronic > E vibrational > E rotational
 The number of possible energy levels for a molecule is much greater than for
an atomic particles

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Molecular spectrum
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A plot of absorbance as a function wavelength or wave number is called the absorption spectrum.
 The nature of Absorption spectrum is influenced by differences b/n
 absorption spectra for atoms and
 absorption spectra for molecules.

Spectrum
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A


Line
spec.(atoms)
max
Spectrum: is the display of the energy level of
EMR as a function of wave number or wavelength.
The energy level of EMR expressed in terms of
Absorbance A (Intensity,I) or transmittance(T)
 Spectrum can be:
a) line spectrum/ Atomic Spectra : occur with
atoms such as sodium metal which has a sharp line of 
at 590 nm.
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Spectrum…
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b) band spectrum: occurs with molecules due to the presence of different vibrational and
rotational sub-levels which the molecules may occupy on transition to excited state.
 What an spectrum tells
 There are two parameters which define an absorption band :
1. Its position (max) on wavelength scale
2. Its intensity on the absorbance scale.
 The height of a peak (A) at max corresponds the amount absorption.
 thus can be used as a quantitative information (e.g. conc).
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UV-Visible spectrophotometry
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Chapter 2

UV–Vis Spectroscopy
 A type of Spectroscopy which utilizes the UV-Visible region of the EMR
 It is Molecular spectroscopy that involves study of the interaction of UV-
Visible radiation with molecules
 UV region: 200 nm- 400 nm
 Visible region: 400-800 nm
 NB : < 200nm Known as Vacuum UV
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A visible Region spectrum
 constitutes a small part of the total electromagnetic spectrum
 The human eye is only sensitive to a tiny proportion of the total EM spectrum in the
region (400-800nm )
In this region (400-800nm):we perceive the colors of the rainbow from
 Violet (small wavelength) through to red (large wave length).
 Common colors of the spectrum with increasing wavelength
 represented as: VIBGYOR
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UV–Vis Spectroscopy
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 InVisible region a portion of the light is absorbed by the substance and
 the balance is reflected
 the color of the sample is determined by the reflected light.
Example:
 if violet is absorbed, the sample appears
yellow-green
 if yellow is absorbed, the sample appears blue.
 The colors are described as complementary.
UV–Vis ……
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A visible Region spectrum…
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 Substances which appear colorless do have absorption spectra.
 In this instance, the absorption will take place in the ultraviolet
or infra red region and not in the visible region
UV–Vis ……
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Principles of UV-Visible spectroscopy
 The absorption of UV/visible radiation occurs through Transition of e`s form
 lower energy orbital to higher energy orbitals
 Absorption of light in these region mainly causes electronic transition
UV-Vis Radiation can transit e`s (bonding or non-bonding e`s) from
 lower energy Filled orbitals (bonding/ nonbonding orbitals) in to
 higher energy Unfilled orbitals (antibonding orbitals) - the so called
electronic transition
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The difference in energy between molecular bonding, non-bonding and anti-
bonding orbitals ranges from 125kJ/mol - 650 kJ/mol
 This energy corresponds to EMR in the UV region, 200-400 nm, and
VIS regions 400-800nm of the spectrum
Principles of UV-Visible…………
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Types of electronic transitions
 Absorption of radiation in the UV-VIS region depends upon
 the number and arrangement of electrons in absorbing molecules.
 The outer electrons in an organic molecule may occupy
 one of three different energy levels in ground state : - , - or n- energy level.
Accordingly,there are three types of electrons;
a)  -electrons: bonding e`s (single bond): possess the lowest energy level (most
stable)
b)  -electrons; forming the  -bond (double/triple bond)and possess higher energy
than δ-electrons.
c) n-electrons; non bonding e`s present in atomic orbitals of hetero atoms
(N, O, S or halogens).
 They usually occupy the highest energy level of the ground state.
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Types of electronic…
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 In excited state: these e`s occupy high energy level anti bonding orbitals
The  -electrons occupy an anti-bonding energy level ( *) and the
transition is termed  -  * transition.
 -electrons occupy an anti-bonding energy level (П *) and the
transition is termed -- * transition
the n-electrons may occupy  * or * levels to give
n-  * or n - * - transition.
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UV-Visble….
Types of electronic transition….
 Energy level of the four possible transition:
 -  * > -- * > n -  * > n - *
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 Example electronic transitions of formaldhyde
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1. -  *
 Requires large energy
 saturated hydrocarbons are  -  *
 E.g. Methane (C-H bonds) shows an absorbance maximum at125 nm.
 occur well in the vacuum UV(< 200 nm)
 Not used much in UV/VIS
 They are transparent in the near UV region (200 - 400 nm) and this make them
ideal solvents for other compounds studied in this range.
 Absorption maxima due to  -  * transitions are not seen in typical UV-VIS
spectra (200 - 700 nm)
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2.n -  * Transition
Requires less energy than  -  * transitions.
 Occurs in the range 150 - 250 nm.
 Few organic functional groups show n -  *peaks in the UV region(200-
400)
 E.g. Saturated alcohols, amines, halides, aldehydes, ketones, ethers
 most of them are useful as common solvents in UV region.
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Solvent , nm Solvent  , nm
Water 190 Chloroform 247
Ether 205 Carbon
tetrachloride
257
Ethanol 207 Benzene 280
Methanol 210 Acetone 331
Cut-off wavelengths of some common solvents:
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2. n -  *Transition………..
 However, their intense absorption usually extends to the edge of near UV
producing what is called end absorption (cut off wavelength)
 mostly in the 200 - 250 nm region.
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3.-* and n- *
 Unsaturated compounds containing no hetero atoms are characterized by
 -* transitions,
 compounds, such as ethylene (CH2= CH2) OR CH ≡CH
-* transitions occurs in 180- 200nm
 When these compounds containing hetero atoms,
 they can undergo n-* transitions,
 example acetone (CH3-COCH3).
 n-  * transition occurs in 275-300 nm
 In general in these cpds - *and n-* transitions occurs in the range 200 -700 nm
regions
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 From all type of transition the most important transitions are the n-π* and π - π *
 because they involve functional groups that are characteristic of the analyte in the
UV-Vis ( 200nm - 800nm)
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UV-Visble….
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Characteristics of UV-Vis spectrum
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• The amount of the radiation absorbed (A) at each wavelength is measured and plotted
against the wavelength (λ)
 UV-vis spectrum is a plot of Absorbance versus wavelength (λ)
 UV-Vis spectrum is band spectrum
 The UV-Vis spectrum is characterized by two major parameters,
 maximum Absorbance wave length (λmax), and
 the intensity of the bands (ε).
UV-Visble….
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The λmax = The wavelength at which the absorbance (A) is highest
The intensity (ε) : indicate the probability that light of a given wavelength will be
absorbed by the chromospheres ( Absorbing groups)
Chromophores -functional groups each of which absorbs a characteristic UV or visible radiation
 λmax (‘lambda max’) is a characteristic of a particular chromophore
 The λmax of a compound is sometimes used in the BP
 for identification of drugs and unknown compounds
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UV-Visible….
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Characteristics of UV-Vis spectrum…………

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A simple UV/visible absorption spectrum
• Y axis is Absorbance and the x- axis is the wave length(nm)
UV-Visble….
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Characteristics of UV-Vis spectrum…………

Some important terms
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Chromophores: (Chrome = color, phore = carrier).
 They are functional groups, which confer color on substances capable of absorbing UV
and/or visible light (200 - 780 nm).
 functional groups which exhibits a characteristic absorption in the UV-Vis region.
Consists unsaturated group (double or triple bonds),benzene ring and
unsaturated group with hetroatom
Some of the most important Chromophores are:
 E.g. N = N, N=O, C=O, C=N, C≡N, C=C, C=S
 The most common Chromophores found in drug molecules is a benzene ring
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UV-Visble….

Some important terms………….
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Auxochromes:
 They are functional groups which can not confer colors on substances
but have the ability to increase the coloring power of Chromophores.
 They does not absorb radiations longer than 200(absorbed far) nm,
 but when attached to a given chromophore, causes a shift to a longer
wavelength with increase in absorption intensity.
These include. -OH , OCH3 -NH2 , -SH, Cl, Br and I
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UV-Visble….

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UV-Visble….
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 Bathochromic (Red) shift: shift of absorption to longer wavelength
 Hypsochromic(Blue) shift: it is shift of absorption to shorter wavelength.
 Hyperchromic effect it increases in absorption intensity and
 Hypochromic effects:it decrease in absorption intensity
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Red shift
Blue shift
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UV-Visble….

Factors influencing absorption of EMR in UV-Vis
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 There are various factors that govern measurement of absorption of EMR
 Absorption band can be changed in its intensity or Position, or both effect by
various factors
These factors are
– Absorbing group - Chromophores
– Presence of Auxochromes ,ring residue or Alkyl substitution
– Solvents
– PH of the solution
• The absorbance readings should preferably be at the wavelength where the
analytes have their absorption maxima λmax
UV-Visble….
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1.Absorption characteristics of Chromophores
A- Ethylenic chromophores:(-CH2=CH2-)
Their bands are difficult to observe in near UV region (<180 nm), are not that much useful
 However, substitution, and certain structural features may cause red shift
rendering the band observable in the near UV region.
 Auxochromes, alkyl substitution, ring residue,
exocyclic double bond or extra double bond
Examples:
 Alkyl substitution: cause red shift due to hyper-conjugation and stabilization of excited
state.
 Attachment to auxochromes: cause red shift and Hyperchromic effect due to conjugation.
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UV-Visble….
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B- Carbon-hetero atom chromophores:
These are: -C=O, -C=N, -C=S, -N=O, ….etc.
They contain common transition n- π * with Absorption band in the range of 275-300 nm.
But some factors Such as :
Auxochromes, alkyl substitution, ring residue,
exocyclic double bond or extra double bond
Cause red shift due to hyper-conjugation.
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UV-Visble….
1.Absorption characteristics of Chromophores…………
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 When Separated Chromophores (by two or more single
bonds)
eg CH2 = CH – CH2 – CH = CH2 :
• have additive effect only : Hyperchromic shift only
• because there is little or no electronic interaction
between separated chromophores.
• Due to -orbitals overlap decreases the energy gap
b/n adjacent orbitals
eg CH2 = CH – CH = CH2 or CH2 = CH – CH = O
CH2 = CH2
CH2 = CH – CH2 – CH = CH2
170-180
nm
170-180 205-215 nm
CH2 = CH – CH = CH2
C- Conjugated chromophores
UV-Visble….
CH2 = CH – CH2 – CH = CH2
when two chromophoric groups are conjugated:
 the * transition is red shifted by 15 - 45 nm
1.Absorption characteristics of Chromophores……………
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Effect of Conjugation Chromophores
 If any of the simple chromophores is conjugated
Conjugation: - raises the energy of the HOMO and
-lowers the energy of the LUMO
These the less energy is required for transition of the e`s
• Therefore The λmax shifts to longer (Red shift)
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UV-Visble….
Absorption characteristics…………
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Effect of Conjugation Chromophores…..
Example : molecules given below undergo π – π* Transition (conjugation of π e`s)
“The λmax moves to a longer wavelength (Red Shift) with high intensity (Hyperchromic
effect)
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UV-Visble….
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Effect of Conjugation Chromophores…..
 The same effect occurs when groups containing n e`s are conjugated with a π e`s group;
 e.g., the number of conjugated double bonds increases.”
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UV-Visble….
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D. Aromatic Systems :
I. Benzene ring
Benzene has three maxima at 184 nm ( the most intense), 204 nm and at 254 nm.
The first two bands have their origin in the ethylenic π-π* transition,
while the longest B-band (254nm) is a specific feature of benzenoid compounds.
 B-band, is characterized by vibrational fine structures.
Both the B-band and the 204-nm ethylenic band ( E-band) are useful
while the far UV band (184 nm) is unsuitable for analytical purposes.
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UV-Visble….
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Benzene ring ...
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UV-Visble….
184
nm
204
nm
254
nm
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II. Monosubstituted benzenes :
When the benzene ring is substituted with a single functional group
 a Red shift occurs for both the E- and B-bands with increase in the absorption
intensity.
This occurs whether the substituent is an e` donating or withdrawing group.
 In addition the B band loses most of its fine structure.
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UV-Visble….
D D
W X W X
h
h
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 Which compound in each of the following pairs is likely to absorb
radiation at longer wavelength (Give reasons) :
CH3-CH2-CH3 or CH3-CH=CH2
CH3-CH2-CH=CH2 or CH3-CH2-CH=O
CH3-CH2COOH or CH3-CH2CH=O
CH2=CH-CH=CH2 or CH2=CH-CH2-CH=CH2
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UV-Vis…
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2.Effect of pH on absorption spectra
 The spectra of compounds containing acidic or basic groups are dependent on
the pH of the medium (e.g.) phenols and amines.
 UV-spectrum of phenol in acid medium is completely different from its
spectrum in alkaline medium
 Spectrum in alkaline medium exhibits bathochromic shift with
hyperchromic effect.
 The red shift is due to the participation of the pair of electrons in
resonance with the  electrons of the benzene ring, thus increasing the
delocalization of the  electrons.
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Effect of pH on….
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-
+
H
in acid medium in alkaline medium
O
O
OH
OH
(Phenol)max = 270 nm (phenate anion) max= 290 nm
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UV-Vis…

 On the other hand, UV spectrum of aniline in acid medium shows
hypsochromic (blue) shift with hypochromic effect (decrease in
absorption intensity).
This blue shift is due to the protonation of the amino group, hence the
pair of electrons is no longer available and the spectrum in this case is
similar to that of benzene (thus called benzenoid spectrum).
NH2 NH3
In alkaline medium in acid medium
Aniline, max= 280 nm Anilinium ion max= 254 nm
+
+ H+
- H+
Effect Of pH On….
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3.Effect of Solvents on absorption spectra
 The solvents may have a strong effect on the position of max due to its effect on
the energy of transition.
Two cases arise:
I Non polar cpds (dienes and conjugated Dienes):
 -*Transition :position of max not shifted by any change of solvent polarity
due to absence of charge separation in either ground or excited states.
II. Polar cpds (enones):
 position of max shifted with the change in the polarity of the solvents.
There are two transitions
-* and n-*Transition bands of enones:
.
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UV-Vis…

a.-* Transition bands of enones
max shifted to a longer wave lengths (red shift) with increasing solvent polarity.
 Due to stabilization of excited state by dipole-dipole solvent interaction
Dipole interaction is more strong with the excited state(π* orbital)than with the ground
state (π orbital)
 Thus results Lowering the energy of π * orbital and
 max is shifted to longer wavelength (red Shift.)
Effect of Solvents…
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b.n-* Transition bands of enones
max shifted to shorter wavelengths (blue shift) with increasing solvent polarity.
 Due to stabilization of excited state by Hydrogen bonding with the solvent
since Hydrogen bonding is more strong with the ground state (n orbital) than excited
state (π*orbital) HB: R-C=O…….HOR
 thus results lower the energy of the ground state (n-orbital)
 max shifts to shorter wavelengths (blue shift)
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UV-Visble….
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For example, the figure below shows that acetone CH3-CO-CH3
(n → π* ) Transition
max of acetone in hexane (non polar) appears at 279 nm
 But max of acetone in water( polar)is shifted to 264 nm,
a max shifted to shorter wave length of 15 nm. (blue shift) Why?
Generally For polar cpds ( enones)
 Increase in polarity of solvents η→ π* blue shift
 Increase in polarity of solvents shifts π → π* Red shift
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UV-Visble….
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Effect of Solvent in Benzene
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UV-Visble….
• In benzene Polar solvents form solute solvent complex through H bonding,
• hence fine structure may disappear
• Non polar solvents do not form H-bond with solute so fine structure often observed
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I. Woodward's rules: Named after Robert Burns Woodward,
He attempt to predict ( λmax) in an UV-Visible spectrum of a given compound.
A. Woodward's Rules for conjugated dienes
 These rules specify a base value for each type of conjugated dienes
 acyclic dien: open chain diens (1,3-butadiene) with base value 214 nm
 Heteroannular diene :presence of the two double bonds in two different rings base value214nm
 Homoannular diene : presence of the two double bonds within the same ring base value
253nm
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UV-Visble….
H2C=CH-CH=CH2
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Estimations of λmax of an organic compound
acyclic dien

A. Woodward's Rules for conjugated dienes…
 The base value each conjugated dienes is red shifted upon
 alkyl substitution or attachment of ring residues or olefin
Conjugated dienes base value also affected by
the presence of double bonds out side a ring (exocyclic),
Addition of extra double bonds in conjugation and
Attachment of auxochromes.
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UV-Visble….
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R
CH3
CH3
R
Counted Ring residues and alkyl substitutions
75
OCH3
SH
Auxochrom attachment
Extradouble bonds
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UV-Visble….

OCH3
OAc
Cl
SH
OH
Check that this chemical compound containing
2-Extra double bonds
5- Auxochromes attachements
5- Ring residues
No alkyl substitutions
3- Exocyclic double bounds
One homoannular nature
76 Deneke A. 11/11/2024
UV-Visble….

Woodward Rules for Conjugated Dienes can be summarized as :
Component nm
Base value for heteroannular or opened-chain dienes 214
Base value for homoannular dienes 253
Add the following Values to the base value:
(a) Each extra double bond in conjugation 30
(b) Each Alkyl Substituent or ring residue 5
(c) Each exocyclic nature 5
(d) Each auxochrome has its corresponding value:
- OAc 0
- OR (including OH) 6
- SR (including SH) 30
- Cl or Br 5
- NR2 (including NH2 & NHR) 60
(e) Solvent Correction 0
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UV-Visble….

The following examples illustrate the use of these rules:
Basic Value 214 253 253
Extra D.B. --- --- 30
Exocyclic D.B. 5 --- 5
Ring residue 15 10 15
Alkyl Substituent 5 5 10
Auxochromes
OR 6 6 6
SR --- --- 30
Cl & Br --- 5 5
NR2 60 60 ---
Calculated max 305 339 354
OR
NH2
OH
Cl
NH2
OH
Cl
Cl
SH
OCH3
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UV-Visble….

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UV-Visble….
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Calculations of λmax of an organic compound

UV-Visble….
Woodward's Rules for conjugated dienes… Examples
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Calculations of λmax of an organic compound

Exercise: Calculate the max of the following compounds :
Cl
OH
Cl
OH
Cl
OR
OCH3
NH2
OR
Cl SH
OH
Br
OR
OH
Cl
Cl
Br
OAc
SH
OH
OR
Cl Br
OR
NH2
Cl
NH2
CH2
Br
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UV-Visble….

B.Woodward's Rules for Conjugated enones
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UV-Visble….
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 Examples
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UV-Visble….
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 Examples
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UV-Visble….
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UV-Visble….
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α, β -unsaturated aldehydes, acids and esters follow the same
general trends as enones, but have different base values.
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UV-Visble….
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C. Woodward's Rules for Benzoyl Derivatives
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UV-Visble….
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 Example
The Woodward’s rules work well only for conjugated polyenes having four double
bonds or less.
For conjugated polyenes with more than four double bonds the Kuhn rules are used.
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UV-Visble….
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C. Woodward's Rules for Benzoyl Derivatives

 According to this rule
λmax = 134(n)1/2 +31
 Where n is the number of conjugated double bonds
Example
λmax =476 nm
λmax =476 nm
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UV-Visble….
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2. Simplified Kuhn and Hausser rule

Example : Calculate the max of the following compound :
max = 134 5 + 31 = 330.6 or 331 nm
This rule is also useful for calculating number of double bonds from the observed max as n
= (max - 31/134)2
Example : If max of a compound is 433 nm calculate the approximate number of double
bonds :
The number of double bonds (n) = [(433 –31) / 134]2
= 9
CH2OH
90
UV-Visible….
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Using the simplified Kuhn and Hausser rules, Calculate the approximate
λmax for the following compounds
Calculate the approximate number of double bonds present in each compound , if you
gave the following λmax for each:
1- 420 nm , 2- 530 nm , 3- 485 nm
4- 565 nm 5- 612 nm 6- 710 nm
CHOH
OH
C CH
CH
OH
91
UV-Visbile….
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Additional Examples of Dienes
UV-Visble….
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UV-Visble….
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UV-Visble….
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UV-Visble….
Additional Examples of Dienes….

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2. Additional Examples of Enones
UV-Visible….
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UV-Visble….
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UV-Visble….
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UV-Visble….
Additional examples of Benzoyl Derivatives
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UV-Visble….
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UV-Visble….

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UV-Visible….
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Additional Notes on: UV spectra of some representative drug molecules
a. Drugs containing steroid enones as chromophores: steroid enones.
All steroid enones have absorbance maxima of similar intensity, at around 240 nm.
 In The extra double bond in betamethasone as compared with hydrocortisone does not
make a great difference to the wavelength of maximum absorption since it does not extend
the original chromophoric linearly.
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UV-Visble….

Additional Notes…..
 However, the shape of the absorption band for betamethasone is quite different from
that for hydrocortisone.
 Such differences in the spectra can be employed in qualitative identity tests;
 these are used particularly in conjunction with high-pressure liquid chromatography
(HPLC) identification checks where the method of detection is by diode array UV
spectrophotometry
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UV-Visble….

104
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UV-Visble….

b. Drugs with benzoid chromophore: for instance ephedrine
UV spectrum of ephedrine
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UV-Visble….

c. Drugs with extended benzene chromophore:
ketoprofen, cyproheptadine, dimethindine, protripetyline and zimeldine.
UV spectrum of ketoprofen (λmax = 260 nm)
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UV-Visble….

d. Drugs with extended benzene ring chromophore and amino groups as auxochrome:
 Procaine, procainamide and proxymetacaine
In addition to the extended chromophore, procaine also contains an auxochrome in the
form of an amino group, which under basic conditions has a lone pair of electrons that can
interact with the chromophore producing a bathochromic shift.
Under acidic conditions, the amine group is protonated and does not function as an
auxochrome, but when the proton is removed from this group under basic conditions a
bathochromic shift is produced and an absorption with λmax at 270 nm.
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UV-Visble….
107

UV spectrum of procaine under acidic λmax 260 and basic conditions (λmax = 270
nm)
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UV-Visble….
108

e. Phenylephrine: hydroxyl group as auxochrome
 The chromophore of phenylephrine is not extended but its structure includes a phenolic
hydroxyl group.
 The phenolic group functions as an auxochrome under both acidic and alkaline conditions.
 Under acidic conditions it has two lone pairs of e`s which can interact with the benzene ring,
a
 under basic conditions it has three lone pair of e`s which interact with benzene ring
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UV-Visble….

UV spectrum of phenylephrine under acidic (λmax = 273 nm) and
basic (λmax = 292 nm)
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UV-Visble….
110

Instrumental design of UV-Visible
Spectroscopy
Components
 A spectrophotometer is an instrument for measuring theT orA
of a sample as a function of the wavelength of EMR.
The key components of a spectrophotometer are:
1. Source that generates a broad band of EMR
2. Wavelength selectors
3. Sample holder
4. One or more detectors to measure the intensity of
radiation
5. Signal Processor
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i- Light Sources.Two types:
1- Continuous Sources:which produce spectra over a broad range(e.g.):
 Tungsten lamp (provides visible spectrum; 400-1200 nm)
 Deuterium lamp (provides ultra-violet spectrum; 190-400 nm)
2- Discontinuous or Discrete Sources: which produce only specific (discrete) wavelengths .
 Hollow cathode lamp (HCL)
 Electrodeless discharge lamp (EDL)
Tungsten Lamp Deuterium Lamp Hollow cathode lamp
Instrum…
112
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Instrum…
 The ideal light source would yield a constant
intensity over all wavelengths with low noise
and long-term stability.
 Two sources are commonly used in UV-
visible spectrophotometers.
a) Deuterium arc lamp:yields a good
intensity continuum in the UV region
 Although modern deuterium arc lamps have
low noise.
 Over time, the intensity of light from a
deuterium arc lamp decreases steadily.
 Such a lamp typically has a half-life of
approximately 1,000 h.
deuterium arc lamp
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Instrum…
b) Tungsten-filament:consists of a tungsten
filament contained in a glass envelope.
 The life of the lamp is limited by the
evaporation of tungsten.
c)Tungsten-halogen lamp:
• The halogen gas prevents the evaporation of
tungsten and increases the lifetime of the lamp to
more than double that of the ordinary tungsten
lamp.
• yields good intensity over part of the UV
spectrum and over the entire visible range.
 This type of lamp has very low noise and low
drift and typically has a useful life of 10,000 h.
 Most spectrophotometers used to measure the
UV-visible range contain both types of lamps.
tungsten-halogen lamp
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Instrum…
 Either a source selector is used to switch between
the lamps as appropriate, or the light from the two
sources is mixed to yield a single broadband
source.
 An alternate light source is the xenon lamp which
yields a good continuum over the entire UV and
visible regions.
 The noise from currently available xenon lamps is
significantly worse than that from deuterium or
tungsten lamps
 Xenon lamps are used only for applications in
which high intensity is the primary concern.
115
xenon lamps
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ii.Wavelength selectors
 Narrower bandwidth tend to enhance the sensitivity and selectivity of the
absorbance measurements and give a more linear r/ship between the optical
signal and concentration of the substance to be determined
 i.e. narrower bandwidth representing better performance.
 Ideally, the output from a wavelength selector would be radiation of a single
wavelength.
 Two types of wavelength selectors are used:
 Filters and
 Monochromators.
Instrum…
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A. Filters:
 Either absorption or interference filters are used for wavelength selection:
1.Absorption filters:
 Usually function via selective absorption of unwanted wavelengths and
transmitting the complementary color.
 The most common type consists of colored glass or a dye suspended in gelatin
and sandwiched between two glass plates.
 They have effective bandwidths from 30 to 50 nm.
 They are inexpensive and widely used for band selection in the visible region.
Instrum…
117
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Instrum…
2. Interference filters:
 As the name implies, an interference filter relies on optical interference to
provide a relatively narrow band of radiation.
 It consists of a transparent material (calcium or magnesium fluoride)
sandwiched between two semitransparent metallic films coated on the
inside surface of two glass plates.
 The thickness of the dielectric layer is carefully controlled and determines
the wavelength of the transmitted radiation.
 When it is subjected to a perpendicular beam of light, a fraction passes
through the first metallic layer and the other is reflected.
118
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Instrum…
Figure 16; Interference filter
White
radiation
Narrow
band
radiation
Glass plates
Dielectric layer
Mealic films
Interference Filters
119
Fraction that is passed undergoes a similar partitioning upon passing
through the second metallic film, thus narrower bandwidths are
obtained.
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B) Monochromators:
 All monochromators contain
 an entrance slit,
 a collimating lens or mirror to produce a parallel beam of light
 a prism or grating to disperse the radiation into its component
wavelengths
 a focusing element and exit slit
Instrum…
120
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Dispersion devices
 It cause different wavelengths of light to be dispersed at different
angles.
 When combined with an appropriate exit slit, these devices can be used
to select a narrow waveband
 Two types of dispersion devices, prisms and holographic gratings,
are commonly used in UV-visible spectrophotometers.
 A prism generates a rainbow from sunlight.
 This same principle is used in spectrophotometers.
 Prisms are simple and inexpensive, but the resulting dispersion is
angularly nonlinear (see Figure).
 Moreover, the angle of dispersion is temperature sensitive.
Instrum…
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122

 For these reasons, most modern
spectrophotometers contain holographic
gratings instead of prisms.
 These devices are made from glass blanks, onto
which very narrow grooves are ruled.
 The dimensions of the grooves are of the same
order as the wavelength of light to be dispersed.
 Finally, an aluminum coating is applied to
create a reflecting source.
 Light falling on the grating is reflected at
different angles, depending on the wavelength.
 Holographic gratings yield a linear angular
dispersion with wavelength and are temperature
insensitive.
123
Instrum…
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Instrum…
 However, they reflect light in different orders, which overlap.
 As a result, filters must be used to ensure that only the light from the
desired reflection order reaches the detector.
 A concave grating disperses and focuses light simultaneously.
 A monochromator consists of an entrance slit, a dispersion device,
focusing mirror and an exit slit.
 Ideally, the output from a monochromator is monochromatic light.
 In practice, however, the output is always a band(group), optimally
symmetrical in shape.
 The width of the band at half its height is the instrumental bandwidth
(SBW).
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Instrum…
Optics
 Either lenses or concave mirrors are used to relay and focus light
through the instrument.
 Simple lenses are inexpensive but suffer from chromatic
aberration(devation from what is normal or desirable), that is,
light of different wavelengths is not focused at exactly the same
point in space.
 Achromatic lenses combine multiple lenses of different glass with
different refractive indices in a compound lens that is largely free
of chromatic aberration.
 Such lenses are used in cameras.
125 Deneke A. 11/11/2024

Instrum…
 They offer good performance but at relatively high cost.
 Concave mirrors- are less expensive to manufacture than achromatic
lenses and are completely free of chromatic aberration.
 However, the aluminum surface is easily corroded, resulting in a loss
of efficiency.
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Instrum…
iii- Sample cells (sample holders):
 For UV/Vis instrument, this is a light tight box in w/c the container
holding the sample so/n is placed.
The container is called cuvette.
 For the UV region sample compartment is made of quartz since quartz
will not absorbed in the UV region.
 For the Visible region, compartment composed of simple glass or
plastic cells since they absorb in the UV but not absorb in the visible region.
 In UV any solvent that does not have any Ñ-bonding can be used including water.
 In the visible region any solvent that is colourless can be used w/c also
including water.
127
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Instrum…
 The standard path-length of cells for
measurements of absorption in the
uv-visible range is 1 or ½ cm path-
length, although cells of path length
from 0.1 to 10 cm can also be used.
128
Deneke A. 11/11/2024

iV) Detectors:TwoTypes of detectors are used in this respect:
1- Heat Sensitive Detectors
2- Photoelectric Detectors
 Photoelectric detectors are the most frequently used for this purpose.
They give electrical signal, which is directly proportional to the intensity of
the transmitted light.
The following types of photoelectric detectors are used:
1- Photovoltaic cells 2- Phototubes
3- PhotomultiplierTubes (PMT’s) (The most widely used)
4- Photoconductivity tubes and Silicon photodiodes
The main three types are illustrated in the following part:
Instrum…
129 Deneke A. 11/11/2024

(A) Photocells (Phototubes):
 Light (radiant energy) falls on the
cathode surface which excites
electrons and generates an electric
current which is proportional to
light intensity
 (In other words) Converts the
energy of an incoming photon into a
current pulse. Conversion is done on
a photoemissive surface by the
“photoelectric effect”
Instrum…
130 Deneke A. 11/11/2024

B) PhotomultiplierTubes
 The PMT (see Figure below) combines signal conversion with several
stages of amplification within the body of the tube.
 The nature of the cathode material determines spectral sensitivity.
A single photomultiplier yields good sensitivity over the entire UV-
visible range.
Instrum…
131 Deneke A. 11/11/2024

This type of detector yields high
sensitivity at low light levels.
However, in analytical spectroscopic
applications, high sensitivity is
associated with low concentrations,
which result in low absorbances,
which in turn result in high intensity levels.
To detect accurately small differences
between blank and sample
measurements, the detector must have
low noise at high intensity levels.
Instrum…
132 Deneke A. 11/11/2024

Radiation enters over the grill and strikes
the cathode photo-emissive surface
 Radiation striking the cathode is converted
into photo-electrons
The photo-electrons are attracted to the
first (+) dynode which produces a cascade
of electrons which travel to dynode 2 due
to its higher potential.
Each electron strikes the second dynode
releases a cascade of new electrons which
travel on to the next dynode in the series
and so on to the last (main) anode.
The final photocurrent is thousand times
greater than the primary current (about
106 to 107 times greater than primary
current).
Instrum…
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Instrum…
C) Photodiodes
 Photodiode detectors have a wider dynamic range and, as solid-
state devices, are more robust (stronger) than photomultiplier
tube detectors
 In a photodiode, light falling on the semiconductor material allows
e- s to flow through it, thereby depleting the charge in a capacitor
connected across the material.
 The amount of charge needed to recharge the capacitor at regular
intervals is proportional to the intensity of the light.
 Earlier photodiodes had low sensitivity in the low UV range, but
this problem has been corrected in modern detectors.
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Instrum…
 Some modern spectrophotometers contain an array of
photodiode detectors instead of a single detector.
 A diode array consists of a series of photodiode detectors
positioned side by side on a silicon crystal.
 Each diode has a dedicated capacitor and is connected by a solid-
state switch to a common output line.
 The amount of charge needed to recharge the capacitors is
proportional to the number of photons detected by each diode,
which in turn is proportional to the light intensity.
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Instrum…
 The absorption spectrum is obtained by measuring the
variation in light intensity over the entire wavelength range.
 The array typically comprises between 200 and 1000
elements, depending on the instrument and its intended
application.
 Photodiode arrays are complex devices but, because they are
solid state, have high reliability.
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Instrum…
a)The conventional spectrophotometer
137 Deneke A. 11/11/2024

Instrum…
 The absorbance of a sample is determined by measuring the d/c b/n
intensity of light reaching the detector without the sample (the blank)
and with the sample.
 This design is well-suited for measuring absorbance at a single point in
the spectrum.
 It is less appropriate, however, for measuring different compounds at
different wavelengths or for obtaining spectra of samples.
 To perform such tasks with a conventional spectrophotometer, parts
of the monochromator must be rotated
 This introduces the problem of mechanical irreproducibility into the
measurements.
 Moreover, serial data acquisition is an inherently slow process.
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Instrum…
b)The diode array spectrophotometer
 Polychromatic light from a source is passed through the sample area
and focused on the entrance slit of the polychromator.
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Instrum…
 The bandwidth of light detected by a diode is related to the
size of the polychromator entrance slit and to the size of the
diode.
 Each diode in effect performs the same function as the exit
slit of a monochromator.
 The polychromator disperses the light onto a diode array, on
which each diode measures a narrow band of the spectrum.
 The polychromator and the diode array are contained in a
unit known as a spectrograph.
 This configuration often is referred to as reversed optics.
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Instrum…
 To minimize possible photochemical reactions, a shutter is used.
 When the measurement is initiated, the shutter is automatically
opened, and light passes through the sample to the array of
diodes.
 The difference in the intensities of the light reaching the detector
with and without the sample is measured.
 A diode array spectrophotometer :
 inherently very fast owing to its parallel data acquisition and electronic
scanning capabilities
 has excellent wavelength reproducibility, and is highly reliable.
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Instrum…
v) Signal Processors/Readout
 Signal Processing
 Amplifying the signal coming from the detector
 Converting the signal coming from detector into a form that is
easily displayed.
e.g. from electron current to (DC) voltage
 Many forms of readout can be used:
 Computer display
 Digital or analog readout
 Strip chart recorders
 Integrators
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Instrum…
Configuration
 Various configurations of spectrophotometers are available.
i) Single-beam design
 Both conventional and diode array spectrophotometers are single
beam.
 The reference spectrophotometers used by national standards
institutions such as the NIST in the US and NPL in the UK are
single beam.
 Diode array spectrophotometers in particular are well-suited to
single-beam configuration.
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Instrum…
 Figure below shows the optical system of a modern diode array
spectrophotometer.
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Instrum…
Dual-beam design
 In a conventional single-beam spectrophotometer, Lamp drift can
result in significant errors over long time intervals.
 The dual-beam spectrophotometer was developed to compensate
for these changes in lamp intensity between measurements on
blank and sample cuvettes.
 In this configuration, a chopper is placed in the optical path, near
the light source.
 The chopper switches the light path between a reference optical
path and a sample optical path to the detector.
 It rotates at a speed such that the alternate measurements of
blank and sample occur several times per second.
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Instrum…
 Figure below shows a schematic of a dual-beam spectrophotometer.
 Compared with single-beam designs, dual-beam instruments contain
more optical components, which reduces throughput and sensitivity.
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Instrum…
 In addition, the more complex mechanical design of the dual-beam
spectrophotometer may result in poorer reliability.
 Single-beam instruments offer higher sensitivity and greater ease of
use, with drift typically only a factor of two worse than that of
dual-beam instruments.
 The first commercially available diode array Spectrophotometer
was a multibeam design (see Figure below).
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Instrum…
 The beam director is used to shift the beam alternately through the
reference position and as many as four sample positions (for clarity
only one is shown in the figure).
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Instrum…
Split-beam design
 This configuration enables the blank and the sample to be measured at
the same time.
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Instrum…
 Although the split-beam design is mechanically simpler
than the true dual-beam instrument and requires fewer
optical elements, the use of two independent detectors
introduces another potential source of drift.
 This design provides high stability, although not as high as a
dual-beam instrument since two detectors can drift
independently, and good noise, although not as good as a
single-beam instrument since the light is split so that less
than 100 % passes through the sample.
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 When abeam of light is passed through a transparent cell containing a solution of an
absorbing substance, reduction of the intensity of the light may occur due to:
Absorption of light by molecules in the solution
 The intensity of light absorbed is then given by
 Pabsorbed = P0 - PT, Where,
Pabsorbed= intensity of light absorbed
P0 = is the original intensity of light
PT = is intensity of light transmitted from the cell
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Laws governing absorption of radiation
Beer- Lambert’s law
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Laws governing absorption …………..:
electromagnetic radiation passes through a sample is described quantitatively by
 two separate but related terms:
transmittance and absorbance.
 Transmittance is defined as the ratio of the original intensity of light (PT ) and intensity of
light transmitted from the cell (P0)
Multiplying the transmittance by 100 gives the percent transmittance (%T),
%T varies between 100% (no absorption) and 0% (complete absorption)
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absorbance, (A)
 Absorbance is defined as negative logarithm of Transmitance
A = -log T, = -log Pt/Po = log Po/pt
Absorbance is the more common unit used in UV
absorption of light by the sample
But some of the light may be
 reflected and absorbed by the sample container,
 absorbed by components of the sample matrix other than the analyte and scattered.
To compensate for this loss of light , we use a method blank.
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Laws governing absorption …………..:

 Beer-Lambert’s Law
The Law sates that: Amount of light absorbed by a sample is directly proportional to
the concentration of the analyte (C)
path length (b) of the sample holder
A=  bc, or A= abc, or A= (A1% 1cm )b c
Where
A is absorbance
a is absorptivity where the concentration is expressed in gm/100mL or gm/L
 ∈ is molar absorptivity where the concentration is expressed in mol/L(M)
C is concentration
 b is the path length of sample cell
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Assumptions of Bear`s law:
The incident beam is monochromatic
The individual particles of analytes absorb independently of each other.
Path length of the sample holder is uniform (b) over the cross section of
the beam.
Absorbing medium is homogenous and does not scatter the radiation.
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Limitations to Beer’s Law…
Deviations from linearity between the absorbance and concentration.
fundamental, chemical, and instrumental Limitations
I. Fundamental Limitations:
valid only for low conc/diluted solutions of analyte. valid for dilute (< 10-3 M)
At higher concentrations the individual particles of analyte no longer behave
independently of one another.
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II. Chemical Limitations
Deviations also arise when an analyte
associates,
dissociates or
reacts with a solvent
produce a product having a different absorption spectrum from the analytes.
III. Instrumental Limitations
 such as stray light, improper slit width, fluctuation in single beam
 Using polychromatic radiation always gives a negative deviation from Beer’s
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 Generally, the data over a wide range of concentrations will deviate from Beer's law.
 This indicates that Beer's law is only applicable up to a conc of c1
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Relations of specific Absorbanc and 
 A (1%, 1cm) is a constant known as specific absorbance
o It is an absorbance of a 1% w/v (1 g/100 ml) solution in a 1 cm cell;
 The Unit of a is L g-1 cm-1
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Example:
A 5.00x10–4 M solution of an analyte is placed in a sample cell that has a pathlength 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?
Ans(Molar A. = 676 cm-1 M-1
)
A sample has a percent transmittance of 50.0%. What is its absorbance? Ans (A= 0.301)
The molar absorptivity of a substance is 2.0 × 104 cm-1 mol-1 L. Calculate the
transmittance through a cuvette of path length 5.0 cm containing 2.0 × 10-6 mol L-1
solution of the substance. Ans (T= 0.63)
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Exercises
1. What are the concentrations of the following solutions of drugs in g/100 ml?
A. Carbimazole, A (1%, 1 cm) value=557 at 291 nm, measured absorbance=0.557 at 291 nm
B. Hydrocortisone sodium phosphate, A (1%, 1 cm) value=333 at 248 nm,
measured Absorbance =0.666 at 248 nm
C. Isoprenaline, A (1%, 1 cm) value=100 at 280 nm, measured absorbance= 0.5 at 280 nm
 Answers:
A. 0.001 g/100 ml
B. 0.002 g/100 ml
C. 0.005 g/100 m
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APPLICATIONS OF UV-VISIBLE SPECTROPHOTOMETER
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Application
Important advantages of spectrophotometric methods include:
1- Wide applicability; large number of organic and inorganic species absorb light in the UV-
Visible ranges.
2- High sensitivity; analysis for concentrations in the range from 10-4 to 10-6 M are ordinary in
the Spectrophotometric determinations.
3- Moderate to high selectivity; Due to selective reactions, selective measurements and different
mathematical treatments.
4- Good accuracy; Relative errors in concentration measurement lie in the range of 0.1 to 2 %.
5- Ease and convenience; Easily and rapidly performed with modern instruments.
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APPLICATIONS OF UV-VISIBLE…………

I. Qualitative Applications
In terms of qualitative analysis of the analyte, the UV-VIS spectrometry
is of a secondary importance for the identification and the determination
of structural details.
The information obtained from it needs to be supplemented by that from
IR, NMR and mass spectrometry.
Nonetheless, it can still provide information about the presence or
absence and the nature of the chromophore in the molecule.
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1- Identification of Chromophores
 Example,
 the presence of an absorbance band at a particular wavelength often is a good indicator of
the presence of a chromophore.
 Useful information about substance can be obtained via examination of its max and εmax,
 which could be correlated with the structural features (See the following table).
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APPLICATIONSOF UV-VISIBLE…………

2-Confirmation of identity
 Although UV-visible spectra do not enable absolute identification of an unknown,
they frequently are used to confirm the identity of a substance:
a. comparisonof themeasuredspectrumwitha referencespectrum.
Eg. An absorption band at 254 nm with characteristic vibrational fine structures may be
an evidence for existence of aromatic structure.
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b. IdentificationbyusingAbsorbanceratio
 Absorption ratio or molar absorptivity ratio determination
Q value is one of the characterstics of standard drugs
e.g. ASA λmax 265 &299, USP tolerance Q is ratio of absorbance
at 265/299 be 1.5-1.56
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2-Confirmation of identity…

3- Approximate determination of the number of double bonds:
By using Simplified Kuhn and Hausser rule :
max (nm) = 134 n + 31
where n is the number of conjugated double bonds.
4- Identification of the position and/or conformation of certain functional groups:
δ γ β α
C = C – C = C – C = O enones
 α-Alkyl cause red shift about 10 nm & α -OH about 35 nm
β-Alkyl cause red shift about 12 nm & β -OH about 30 nm
 γ/-Alkyl cause red shift about 18 nm & γ -OH about 50 nm
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4. Detection of impurities
Additional peaks can be observed due to impurities to the sample &
it can be compared with that of standard raw material.
E.g. UV spectra of paracetamol (PCM)
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II Quantitative Application
Scope
- Applications of spectrophotometric methods are so numerous and touch every
field in which quantitative chemical information are required.
- In general, about 90% of all the quantitative determinations are performed by
spectroscopic techniques.
- In the field of health alone, 95 % of all quantitative determinations are
performed by UV-Visible spectrophotometer and similar techniques.
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#1 Assay of single component
The assay of absorbing substance may be
quickly carried out by preparing a solution in a
transparent solvent and measuring its absorbance
at a suitable wavelength
Calculate by using beers law
 wavelength should be maximum
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There are three principal methods
A. Use of standard absorptivity value
Use of A1%
1cm or  values
Avoids preparation of standard solution
Reference std are expensive and difficult to obtain
E.g. calculate the concentration of methytestosterone in an ethanolic solution of
w/c the absorbance is a 1 cm cell at its max , 241nm was found to be 0.890. (
A1%
1cm =540 )
 Ans: 0.00165g/100 ml
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Problem
Calculate the concentration of in μg/ ml of a solution of
trypthophan (M.wt.=204.2) in 0.1 M HCl, giving an
absorbance at its max , 277nm of 0.613 in a 4 cm cell.
(=5432).
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Exercise…
Determine the concentration of the following injections:
a.Haloperidol injection:
Add 15 ml of 1 M HCl to 5 ml of injection.Extract three times with ether,washing the ether extracts with 10 ml of
water.Combine the aqueous layers and dilute to 100 ml.Take 10 ml of the diluted aqueous solution and dilute to
100 ml.
– Absorbance reading at 245 nm=0.873
– A (1%,1 cm) value at 245 nm=346
b.Isoxsuprine injection is diluted as follows:(i) 10 ml of injection is diluted to 100 ml and then 10 ml of the
dilution to 100 ml:
– Absorbance reading at 274 nm= 0.387
– A (1%,1 cm) value at 274 nm=73
Answers:
a.haloperidol injection=0.505 % w/v
b.Isoxsuprine injection= 0.530% w/v
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B. Use of a calibration graphs Y = ax + b
Example: the absorbance values at 250 nm of 5
standard solutions, and sample solution of a drug are
given below:
Conc. (μg/ml) A 250 nm
10 0.168
20 0.329
30 0.508
40 0.660
50 0.846
Sample 0.661
 Calculate the concentration of the sample.
(Y= 0.01679X-0.0008, C= 39.5 ug/ml)
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C. Singlepoint standardization
 Involves the measurement of the absorbance of a sample solution and of
a standard solution of the reference substance By proportionality
C test sample = (A sample * C std)/ A std
 Ctest and Cstd - concentrations of sample and standard solutions
 Atest and A std- absorbance's of the sample and standard solutions
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Example:
1. In a spectrophotometric determination of a drug in an aqueous
solution, the absorbance of 4.5 x 10-5 mg/ml of a standard solution of
drug at 273 nm was found to be 0.454 with a path length of 1 cm. The
absorbance reading of the sample solution was 0.367 at 273 nm, with
the path length of 1 cm.
Determine the concentration of a drug in mg/ml?
C test = [ Atest x Cstd] / A std
C =[ 0.367 x 4.5 x10-5 mg/ml]/0.454
C= 3.64 x10-5 mg/ml
C= 3.64 x10-5 mg/ml
C= 0.00364 mg/ml
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 Self exercise
An absorption spectrum of a drug at 400 nm gives absorbance of 0.456 when
32 μg/ml solution is taken. Unknown sample of a drug is treated in identical
fashion gives an absorbance of 0.501 assuming identical cell. Determine
unknown concentration?
 Answer : 35.1578 μg/ml
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#2. Simultaneous analysis of a two component mixture.
 When a solution of two light-absorbing substances is to be analyze
spectophotometrically,
If they do not interact or not affect to eachother
light absorption will be additive.
 The analysis of such components will wholly depend on the nature of their
individual absorption spectrum.
 A two-component mixture may be analyzed by
making absorbance measurements at two max max (one for each
component) and
solving the following pair of simultaneous equations:
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#2 Simultaneous analysis of a two component mixture…..
At max (1) : A1 = ax1 b Cx + ay1 b C1
 At max (2) : A2 = ax2 b Cx + ay2 b C2
A 1 and A 2are experimentally measured absorbances and
ax1 , ay2 , ax2 and ay2 can be evaluated from individual std solutions of cpds 1
and 2.
 from these equations C1 and C2 can be calculated.
Accuracy of this method could be increased by proper selection of max at
which d/ce in absorptivities are large.
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Simultaneous analysis of a two component mixture
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Example: a sample contains two absorbable samples X and y
Let λ1 and λ2 be their absorbing maxima
ax1 and ax2 be absorptivity of X at λ1 and λ2 respectively
ay1 and ay2be absorptivity of Y at λ1 and λ2 respectively
Absorbance of the sample at λ1 and λ2 be A1 and A2
Cx and Cy concentration of x and y
For measurements in 1 cm cells, b=1 Substituting the Cy value in the first equation
gives

Simultaneous analysis …
Binary mixtures cannot be analyzed unless:
Spectral data for the pure substances are available.
The absorptivity values for the components can be easily and accurately
determined
The absorptivity values for the components are sufficiently d/t at the chosen
wavelength to permit an accurate solution of the simultaneous equations.
The absorbance values for the mixture are accurately determined.
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Example
 The max of ephedrine HCl and Chlocresol are 257 nm and 279 nm respectively
and A1%1cm values in 0.1 M HCl solution are
 Ephedrine at 257=9
 Ephedrine at 279=0
 Chlorocresol at 257=20
 Chlorocresol at 279=105
Calculate the concentration of ephedrine HCl and Chlorocresol in a batch of
ephedrine HCl injection, diluted 1 to 25 with water, giving the following absorbance
values in 1 cm cell. (A279=0.424, and A 257=0.97)
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Simultaneous analysis …

Derivative spectroscopy
 Derivative spectroscopy uses first or higher derivatives of absorbance with
respect to wavelength for qualitative analysis and for quantification.
 If a spectrum is expressed asabsorbance, A, as a function of wavelength,,
the derivative spectra are:
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 A first-order derivative is the rate of change of
absorbance with respect to wavelength.
 It passes through zero at the same wavelength as λmax of
the absorbance band is characteristic of all odd-order
derivatives.
 A strong negative or positive band with minimum or
maximum at the same wavelength as λ max of the
absorbance band is characteristic of the even-order
derivatives.
The most characteristic feature of a second-order
derivative is a negative band with minimum at the same
wavelength as the maximum on the zero-order band.
 A fourth-order derivative shows a positive band.
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 Note that the number of bands observed is equal to the derivative order plus one.
Advantages
Derivative spectrum shows better resolution of overlapping bands
 The spectrum may permit the accurate determination of the λ max of the individual
bands.
It permits discrimination against broad band interferences, arising from turbidity or
non-specific matrix absorption.
It is a convenient solution to a number of analytical problems, such as
resolution of multi-component systems,
removal of sample turbidity, matrix background and
enhancement of spectral details.
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Background elimination Resolution
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Reading Ass.

Other Applications of UV Visible Spectroscopy
.
A.Spectrophotometric titration
 In photometric method of equivalent point detection of titrations, the appearance of an
absorbing species will give a linear or conc dependent change in absorbance, w/c will
yield two straight lines that intersect at the equivalent point.
 There are at least three components w/c may absorb light: the original sub, the titrant &
the resulting product (s).The usual procedure is to select some wave length at w/c only
one component absorb.
 In spectrophotometric titration, the absorbance of the so/n at a specified λ is measured
after each addition of the titrant.
 The results are plotted (A vs ml of titrant ), and the end point is determined graphically.
 The point at w/c the two straight lines intersects the end point.
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 graph is constructed on the basis of data obtained well before and well after the end pt.
 Some typical titration curves for the reaction X +Y → Z where X the component to be
determined,Y titrant & Z the product (s) of the reaction.
 If X absorbs radiation energy at a specified λ and it is contaminated with other absorbing sub,
a spectrophotometric titration can be carried out if a titrant can be found w/c react
selectively with X.
 If X & the contaminant are the only species that absorb, the absorbance of the so/n will
decrease as titrant is added.
 IfY is the only species in so/n w/c absorbs, the so/n will not absorb until the end pt is
reached.
 If Z is the only species in so/
 The n w/c absorbs, the absorbance will increase as product is formed.
189
Spectrophotometric titra…
Other Applications of UV Visibe………..
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The following are some of the spectrophotometric titration curves which can be
observed in the normal conditions:
Where:
S = Sample , P = titration Product ,T =Titrant ,
@ =Absorbing species , X = Non absorbing species.
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Examples:
1-Titration of potassium permanganate against ferrous sulfate
MnO4
- + Fe++ + H+ == Mn++ + Fe3+ + H2O
@ at 540 (X) at 540 (X) at 540 (X) at 540 nm
2-Titration of Bi3+ & Cu2+ mixture with EDTA;
 Bi3+ , Cu2+ & Bi-EDTA complex are non absorbing at 745 nm
Cu-EDTA complex is absorbing at 745 nm
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Advantages:
1. More accurate results than direct photometric analysis are obtained.
2. No interference from other absorbing substances because the end point
depends on the change in the absorbance curve and not on the absorbance
value
(affect only the curve shape and sharpness of end point).
3. Can be used for titration of very dilute solutions.
4. Not need favorable equilibrium constants as those required for titration that
depends upon observations near the end point.
5. Can be used for (applied to) all types of reactions (redox, acid-base,
complxometry, pptmetry ….etc.)
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B. Monitoring drug degradation kinetics
Can be simply done when the product has a different absorption spectrum than
that of un-degraded drug.
The rate of disappearance of the spectrum or appearance of other spectrum (as a
function of time ) may be used to determine rate constant for hydrolysis or
degradation.
Oxidation reactions and any other type of reactions that yield products whose
spectra are different from the reactants , may be followed and their rate constant
estimated.
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ReadingAss.

C. DETECTOR in Chromatography
 Mainly used in HPLC and HPTLC.
They are the most widely used detectors, because:
Most drugs absorb UV-Visible radiation.
More sensitive and more selective than the bulk property detectors (e.g. R.I.
detectors).
Some absorbance detectors have one or two fixed wavelengths (280 and/or 254 nm).
More modern HPLC instruments have variable wavelength detectors using the
photodiodes
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ReadingAss.

D. Determination of Equilibrium Constants (Pka)
Where a pH-dependent UV shift is produced, it is possible to use it to
determine the pKa of the ionisable group responsible for the shift. .
 A general equation for determination of pKa from absorbance
measurement at a particular wavelength is given below.
where A is the measured absorbance in a buffer of known pH at the
wavelength selected for analysis;
Ai is the absorbance of the fully ionised species; and Au is the absorbance of
the un-ionised species.
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ReadingAss.

Example:The absorbance of a fixed concentration of phenylephrine at 292 nm
is found to be 1.224 in 0.1 M NaOH and 0.02 in 0.1 M HCl. Its absorbance in
buffer at pH 8.5 is found to be 0.349. Calculate the pKa value of its acidic
phenolic hydroxyl group.
196
Determination of Equilibrium Constants (Pka)……….
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Reading Ass.

Determination of Equilibrium Constants (Pka)…………
.Exercise: Calculate the pKa value of the weakly basic aromatic amine in
procaine from the data given below. Absorbance of a fixed concentration of
procaine in 1 M HCl at 296 nm = 0.031; absorbance in 0.1 M NaOH=1.363;
absorbance in buffer at pH 2.6=0.837.
 Answer: 2.41
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Reading Ass.

E. Environmental applications
The analysis of water and wastewater often relies on the absorption of
ultraviolet and visible radiation.
 Although the quantitative analysis of metals in water and wastewater is
accomplished primarily by atomic absorption or atomic emission spectroscopy,
 many metals also can be analyzed following the formation of a coloured metal–
ligand complex.
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Reading Ass.

F. Forensic applications
UV/Vis molecular absorption is routinely used for the analysis of narcotics
and for drug testing in athlets.
One interesting forensic application is the determination of blood alcohol
using the Breathalyzer test.
In this test a 52.5-mL breath sample is bubbled through an acidified solution of
K2Cr2O7, which oxidizes ethanol to acetic acid.
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The concentration of ethanol in the breath sample is determined by the
decrease in absorbance at 440 nm where the dichromate ion absorbs.
A blood alcohol content of 0.10%, which is above the legal limit, corresponds
to 0.025 mg of ethanol in the breath sample.
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Forensic applications……

Colorimetry
Is a technique which involves measurement of absorbance in the visible region is
known as colorimetry.
Involves measurement of color intensity of compounds.
Requirements for colorimetry
the substance should be colored or
The substance should be able to be derivatized in to colored product.
While derivatizing
 The reagent should be specific
 The color produced should be stable enough until the analysis is completed
 Color intensity should be directly proportional to the concentration of the analyte.
Application- colored drugs and those drugs which can be derivatized.
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