Spectrophotometry

Photometric techniques are among the most important instrumental

techniques available to the pharmaceutical analyst. Instrumental ranges

from the simple flame photometers to much more expensive

spectrometers such as ultraviolet-visible, infrared, mass spectrometry and

nuclear magnetic resonance spectrometers which are used in structural

and quantities analysis of molecules.

The basis of all these instrumental techniques is that they measure the

interaction of electromagnetic radiation with matter in specific energy

levels.

Electromagnetic radiation:

The white light from an electric lamp is made up of a large number of

individual waves of varying wavelengths. This is readily shown by

passing a beam of light through a prism, the so called continuous spectra

is formed, in which each color correspond to a particular wavelength.

Wavelength (λ):- Is the distance between any two consecutive parts of the

wave e.g. from A to B or B to C. its symbol λ(lambda). The units in

which it is expressed A◦ or nm or μm

A◦ = 10-10 m visible and ultra violet

nm = 10-9 m

μm = 10-6 m Infrared

Wave number defined as the reciprocal of the wave length expressed in

cm-1 i.e. the number of waves per cm.

1

Frequency (υ):- the number of the waves passing a point in one

second i.e. the number of cycles per second (υ) the units is S-1 on Hz

(hertz).

ounds are heated, light of characteristic colors in emitted. For example

sodium salts emit yellow light and potassium salts emit lilac light. This is

due to the emission of light at wave lengths characteristic of the metallic

elements in the sample. This is the simplest quantitative analytical

procedures.

When a sample containing metallic atoms is heated above 2000◦C, it first

decomposes into free atoms then volatilize to free gaseous atoms. The

electrons of the gaseous atoms exist in discrete quantized energy levels

i.e. they are in orbital which have specific energy levels that are

characteristic of the elements. The electron in the outer orbital of the

atom may absorb thermal energy and be promoted to one or higher

energy states. The gain in energy of each electron during a transition is a

specific quantity corresponding to the difference between the energy

levels before and after excitation. Deactivation of the thermally excited

atoms to lower energy states occur very rapidly and photons of light are

emitted which have energy ΔE equal to the difference between the upper

(Eu) and the lower energy (El) states. The energy of the emitted light is

2

directly proportional to the frequency (υ) and inversely proportional to

the wave length (λ) hence

ΔE = Eu – El = hc / λ

h = planck's constant = 4.132 × 10-15 eVs

C = speed of light = 3 × 108 ms-1 in vacuo

The wave lengths of light emitted from the sample as a result of thermal

excitation may be viewed through a spectroscope which is a simple

instrument containing an entrance slit and a dispersing device e.g. a

prism. Alternatively they may be recorded on a photographic film using

spectrograph.

3

The intense line at 589.3 nm is yellow light imparted when sodium salts

are introduced. The lilac coloration of a flam in presence of potassium

salts is due to the emission of two line in the red region of the spectrum at

767nm (4p 4s) and 694nm (4d 4p) and one light in the violet region

at404nm (5p 4s).

The atomic emission techniques are based on the measurement of the

light emitted from thermally excited atoms.

Atomic absorption Spectrophotometery is a technique for the quantitative

determination of metallic elements which is based on the measurement of

the absorption of monochromatic light by ground stat atoms.

Molecular spectra:-

Molecular spectra are characterized by the absorption or emission of light

over a much wider range of wave length called spectral bands than atomic

spectra. This is due to the very large number of transitions which

molecules can undergo in comparison to relatively few electronic

transitions of atoms.

Molecular absorption:-

The absorption of electromagnetic radiation of a suitable wavelength by

molecules can promote:

a) Increase in the energy of electrons to one or more high energy states.

b) An increase in the intermolecular vibrational energies of the constituent atoms.

c) An increase in the energy of rotation of the atoms round the bonds joining the atoms.

The total energy of a molecule is the sum of electronic, vibrational and

rotational energies. The relative energy required to induce electronic

4

vibrational and rotational transitions between quantized energy levels are

approximately

10,000: 100:1 respectively

Molecular emission

After absorption of light the excited molecules are short-lived and

deactivation occurs due to:-

1) Internal collisions (internal conversion).

2) Cleavage of chemical-bonds, initiating photochemical reactions.

3) Re-emission as light (luminescence)

Molecules on excitation normally possess higher vibrational energy than

they had in the ground state. This extra vibrational energy is lost by

collision after which the molecules return to the ground electronic state

with the emission of light as fluorescence. Deactivation as fluorescence is

a rapid process occurring with 10-6 - 10-9seconds of excitation.

Phosphorescence is a slower process than fluorescence at after 10-8 sec

and may be showered even several minutes or hours after source of

excitation is removed.

The difference in the energy levels (ΔE)

ΔE (absorption) >ΔE (fluorescence) > ΔE (phosphors)

5

6

7

Instrumentation

The essential components of spectrometers are

The light source:

A source of electromagnetic radiation, these are different from one

spectrometer to another depending on the amount of the energy required.

- Infrared radiation: Globar and Nernst glowers are common source of

infrared radiation. The globar in an electrically heated rod of silicon

carbide and the Nernst glower is a small rod of refractory oxides.

- Visible radiation: The tungsten filament lamp is a satisfactory light

source for the region of 350 to about 2000nm.

-Ultraviolet radiation: The most convenient light source for measurement

in the ultraviolet region is the deuterium discharge lamp; it gives five

times greater energy than the hydrogen lamp. The ultraviolet-visible

spectrophotometers normally have both a deuterium lamp and a tungsten

lamp.

Monochromators:-

Filters: Glass filters are pieces of colored glass which transmit limited

wavelength ranges of the spectrum. The color is produced by

incorporating oxides of such metals as vanadium, chromium, manganese,

iron, nickel, and copper in the glass. The color absorbed is the

8

complement of the color of the filter. A filter absorbing yellow light

(575-625 nm) appears blue. Other filters, gelatin filters, interferometric

filters may be used.

Prisms:

Light passing through a prism, lead to formation of spectrum from

which the required wavelength may be selected for spectrometric

measurements. Prisms are made of quartz for use in ultraviolet region

while glass prisms are preferable for the visible region of spectrum. For

infrared region, the transparent substances usually used for prisms are

sodium chloride (2-15 μm), and potassium bromide (12-25 μm), lithium

fluoride (0.2-6 μm) and cesium bromide (15-38 μm).

Cells:

Samples presented for spectrometric analysis may be in the solid, liquid

or gaseous state. The material that contains the sample ideally should be

transparent at the wavelength (s) of measurement. Cells or cuvettes of

silica for ultraviolet - visible region and of sodium chloride or potassium

bromide for infrared region.

Detectors:-

Barrier layer cell: One of the simplest detectors which has the advantage

that it requires no power supply but gives a current which under suitable

condition is directly proportional to the light intensity.

9

Ordinarily, selenium and metallic oxides and sulphides have small

electrical conductivity. Light of suitable frequency will impart sufficient

energy to the electrons so that they leave the selenium and enter the

transparent layer a current will flow as will be shown by the deflection of

the galvanometer needle. The current produced is proportional to the

intensity of light.

Other detectors like thermocouples, photomultipliers, and photo iodides

are present also.

Recorder (Readout systems):-

The signal from the detector is proportional to the intensity of the

incident light on the detector and after amplification may be displayed as

% transmittance (%T) or as absorbance (log 1/T).

10

Ultraviolet-visible absorption spectrophotometery:-

The technique ultraviolet-visible spectphotometry is one of the most

frequently employed in pharmaceutical analysis. It involves the

11

measurement of the amount of ultraviolet (190-380 nm) or visible (380-

800 nm) radiation absorbed by a substance in solution.

When a beam of light is passed through a transparent cell containing a

solution of an absorbing substance, reduction of the light may occur, this

is due to:-

1- Reflection at the inner and outer surfaces of the cell.

2- Scatter by particles in the solution.

3- Absorption of light by molecules in the solution.

The reflection of the cell surfaces can be compensated by a reference cell

containing the solvent only and scatter may be eliminated by filtrations of

the solution. The intensity of the absorbed light is

I = Io - IT

Io = intensity of the incident light.

IT = intensity of transmitted light from the cell.

The transmittance is the ratio IT/ Io and the % transmittance (%T) is

%T = 100 IT/ Io

12

Lambert (1760) found that there is a relation between the thickness of the

media and the incident light

ln Io / IT = K' b

On conversion to a common logarithm the expression becomes

log Io / IT = K' b/ 2.303

The quantity of log Io / IT is called absorbance (A)

A= log 1/T = log Io / IT =2 – log (%T)

The older terms for absorbance such as extinction, optical density and

absorbency are now obsolete.

The Lambert's law is defined as:-

The absorbance is proportional to the thickness of the solution.

Beer (in 1852) showed that there is a similar relationship between

absorbance and the concentration.

ln Io / IT = K''c

K'' = proportionality constant.

c = is the concentration.

log Io / IT = K''c / 2.303

Beer's law is defined as the absorbance is proportional to the

concentration.

A combination of the two law yields the Beer-Lambert law.

A = log Io / IT = a b c

The proportionality constants K'/2.303 and K''/2.303 are combined as a

single constant called absorptivity (formerly molar extinction coefficient)

and has the symbol Є (Greek letter epsilon)

13

The equation takes the form

A = Є b c Є = A1% x molecular weight/10

The molar absorptivity Є at a specified wavelength of a substance in

solution is the absorbance at that wave length of a 1 mol l-1 solution in a 1

cm cell. The units of Є are 1 mol -1 cm-1.

Another form of the Beer-Lambert proportionality constant is the specific

absorbance, which is the absorbance of a specified concentration in a cell

of specified path length. The most common form in pharmaceutical

analysis is the A (1%, 1cm) which is the absorbance of 1g/100 ml (1%,

w/v) solution in a 1cm cell. The Beer-Lambert equation therefore takes

the form.

A = A1% 1cm b c

Where c is in g/100 ml and b is in cm.

Quantitative spectrophotometric assay of medicinal substances:-

This may be quickly carried out by preparing a solution in a transparent

solvent and measuring its absorbance at a suitable wavelength. The

wavelength normally selected is a wavelength of maximum absorption

e.g. 1

Calculate the concentration of methyl testosterone in an ethanol solution

of which the absorbance in a 1 cm cell at its λmax was found to le 0.890.

14

The A1%1cm in the B.p. monograph of methyl testosterone is give as 540 at

241 nm.

A = A1%1cm bc

o.890 = 540 × 1× c

c = 0.00165 g /100 ml

E.g. 2

The absorbance values at 250 nm of five standard solutions, a blank

solution, and a sample solution of a drug are given in the table below;

calculate the concentration of the sample.

Concentration (µg ml -1 ) (x) A 250nm (y)

0 0.002

10 0.168

20 0.329

30 0.508

40 0.660

50 0.846

Sample 0.611

Assay of substances in multi component samples:-

15

The spectrophotometric assay of drugs rarely involves the measurement

of absorbance of samples containing only one absorbing component.

If the recipe of the sample formulation is available to the analyst, the

identity and the concentration of the interferents are known and the extent

of interference in the assay may be determined. Alternatively interference

which is difficult to quantify may arise in the analysis of the formulation

form manufacturing impurities, decomposition products and formulation

excipients. Unwanted absorption from these sources is termed irrelevant

absorption and if not removed, it will impart a systematic error to the

assay of the drug in the sample.

The basis of all the spectrophotometric techniques for multi component

sample is the property that at all wavelengths:-

1.The absorbance of a solution is the sum of the absorbance of the

individual components.

2.The measured absorbance is the difference between the total

absorbance of the solution in the sample cell and that of the solution in

the reference blank cell.

The concentration of a component in a sample which contains other

absorbing substances may be determined by a simple spectrophotometric

measurement of absorbance, provided that the other components have a

sufficiently small absorbance at the wavelength of the measurement of

the interfering substance, their absorptivity or the pathlength of the

solution are sufficiently small i.e. the absorbance can be ignored. A

systematic error of less than 1% would be acceptable (0.01).

An example of this approach is the assay of paracetamol in pediatric

paracetamol elixir. At the large over all dilution (approximately 3250

time) of the sample, the absorbance of interferents is negligible.

16

If the identity, concentration and absorptivity of the absorbing interferents

are known, it is possible to calculate their contribution to the total

absorbance of a mixture.

E.g. 3

The λmax of the ephedrine hydrochloride and chlorocresol are 257 nm and

279 nm and the A1%1cm values in 0.1M hydrochloric acid solution are:-

Ephedrine hydrochloride at 257 nm = 9.0

Ephedrine hydrochloride at 279 nm = 0

Chlorocresol at 257 nm = 20.0

Chlorocresol at 279 nm = 105.0

Calculate the concentrations of ephedrine hydrochloride and chlorocresol

in a batch of ephedrine hydrochloride injection, diluted 1 to 25 with

water, giving the following absorbance values in 1 cm cells.

A279 = 0.424 A257 = 0.972

At 279 nm ephedrine absorbance = 0

0.424 = 105 × 1 × c

C = 0.00404 g/100ml

.:. The concentration of chlorocresol in the injection

= 0.00404 × 25

= 0.1010 g/100 ml

= 1.01 mg/ml

The absorbance of chlorocreseol at 257 nm in the diluted injection:-

A = 20 × 1 × 0.00404

= 0.081

17

The concentration of ephedrine:-

Corrected absorbance of 257 = 0.972 – 0.081= 0.891

0.891 = 9 ×1× c

C = 0.099 g/100 ml

In the injection

=0.099 × 25 = 2.475 g/100ml = 24.75 g/ml

Another way of separating the interferents from the analyte is by solvent

extraction procedures. These are particularly appropriate for acidic or

basic drugs whose state of ionization determines their solvent partitioning

behavior. These substances may completely separated by the right choice

of the pH and the immiscible solvents. An example of this type of assay

is the B.p. assay of caffeine in Aspirin and caffeine tablets.

Chemical derivatization :-

Indirect spectrophotometric assays are based on the conversion of the

analyte by a chemical reaction to a derivative that has different spectral

properties. The majority of indirect spectrophotometric procedures

involve the conversion of the analyte to a derivative that has a longer λmax

and or higher absorptivity. Chemical derivatization could be for any of

the following reasons:-

a) If the analyte absorbs weakly in the ultraviolet region, a more sensitive

method of assay is obtained by converting the substance to a derivative

with a more intensely absorbing chromophore.

b) The interference from irrelevant absorbance may be avoided by

converting the analyte to a derivative which absorbs in the visible region,

where irrelevant absorption is negligible.

Diazotization and coupling of primary aromatic amines:-

18

Ar – NH2 + HNO2 + H Ar – N+ N + 2H2O

Ar - N+ N + Ar' – M Ar – N = N – Ar' + M+

e.g.

chloramphenicol can be reduced and then assayed by diazotization and

coupling.

Condensation reactions:

A colorimetric procedure based on the rapid reaction between amines and

carbonyl compounds.

Oxidation reactions:-

Oxidation of the side chain of weakly absorbing compounds containing a

simple phenyl group produce a carbonyl derivative that has much greater

absorptivity than the parent compounds.

19

Commonly used oxidation reagents are alkaline potassium permanganate

solution, acidified potassium dichromate, or potassium periodate e.g.

oxidation of ephedrine or propanolamine:-

Metal ligand complexation:

Many organic reagents form complexes with metal atoms by the

formation of coordinate bonds. Characteristic colors which vary with the

change in pH are given by the reaction of iron (ferrous) ions with phenols

that contain two adjacent hydroxyl groups. E.g. mixing a solution of

adrenaline with a buffued solution containing iron (ferrous) sulphate a

purple color (λmax = 540 nm) is formed that has maximum intensity at

pH= 8-8.5 (assay of adrenaline in procaine and adrenaline injection).

Body fluids

The determination of drugs and metabolits in body fluids form part of

investigation in bioavailability, biochemistry drug metabolism and

toxicology. The examination of body fluids is more difficult than that of

pharmaceutical preparations in:

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a) A small quantity of drug or metabolites is usually present in a large volume of blood, urine and tissue.

b) Solvent extraction of body fluids gives rise to an extract that may contain in addition to the drug, endogenous pigments or compounds which make analytical methods subject to error unless great care is taken e.g. choice of solvent, pH of extraction and subsequent purification methods.

c) The drug may occur in free form and combined as conjugates e.g. glucuromide or ethereal sulfate, both of which are polar and water soluble.

d) Protein-binding of the drug may occur and this lead to poor recovery unless the protein in denatured during the extraction procedure.

e) A control should be run in which the drug to be determined is added to the normal body fluids to compensate for any loss and error during the process of extraction and separation.

Structural analysis:-

The electronic transitions that take place in the visible and ultra violet

regions of spectrum are due to absorption of radiation by specific types of

groups, bonds and functional groups within the molecule. The

wavelength of absorption and intensity are dependent on the type of the

molecule structure. Electronic transition involves the promotion of an

electron from one orbital to another orbital of higher energy in a

molecule.

The wavelength of absorption is a measure of the energy required for

transition. The intensity is dependent on the probability of the transition

occurring when the electronic system and the radiation intensity and on

the polarity of the excited state.

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Multiple bonds separated by just one single bond each are said to be

conjugated, the orbital overlap which decrease the energy between

adjacent orbital, results in bathochromic shift.

Aromatic ring exhibit conjugation, the spectra are somewhat different

than other conjugation systems, being more complex absorbs strongly at

200 nm and a weaker band at 230 – 270 nm due to influence of vibration

transitions.

The electronic transition taking place in the ultraviolet and visible regions

is of the following types:-

a) A б electron, an electron present in a б molecular orbital.

b) A π electron, an electron present in a π molecular orbital.

c) An n electron, an electron present in a nonbonding orbital

a) б б* Transition

Electrons occur in fully saturated systems such as the bonds of alkanes

and they require so much energy that absorption bands lie in the vacuum

ultraviolet region of the spectrum. Such compounds do not show

absorption in the ordinary ultraviolet region.

b) n б* Transition

These are excitation from a nonbonding atomic orbital to antibonding б*

orbital. Compounds having nonbonding electrons on oxygen, nitrogen,

sulpher or halogen can show n б* transitions. These transitions are at

lower energy and hence occur at a longer wavelength б б* transitions,

but they are also of not much importance.

c) n π* Transition

22

These are transitions in which an electron in a nonbonding atomic orbital

is promoted to an antibonding π* orbital. Compounds having double bond

between heteroatom, e.g. C=O, C=S, N=O, show these transitions which

appear as weak absorption bands.

d) π π* Transition

π electrons are those of multiple bonds, the so called mobile electrons.

The basic absorption due to the transition of π π*, lies in the 180-200

nm region. The absorption of two or more chromophres which are

separated by more than one bond or usually additive but when such

chromophores are conjugated pronounced effects are produced. The

maximum absorption is shifted to a longer wavelength such an effect is

called a bathochromic shift and the increase in λmax as the number of

conjugation increases, result in that the absorption moves to longer wave

length. If there are enough double bonds in conjugation , absorption will

ultimately move to the visible region and the compound will be colored

e.g. β-carotene, a naturally occurring yellow pigment containing eleven

double bonds in conjugation and it owes its color to absorption in the

visible part of light (λmax = 451nm).

Transition probability

It is not necessary that exposure of a compound to ultraviolet or visible

light must always lead to electronic transition. On he other hand,

probability of a particular electronic transition depends upon the value of

the molecular extinction coefficient (Єmax) and certain other factors

accordingly transition has been divided into:-

1- Allowed transition

2- Forbidden transition

23

1- Allowed transition are those that have Єmax = 104 or more. These are

generally due to π π* transition e.g. butadiene which show absorption

at 217nm and has an Єmax 2100 represents an allowed transition.

2- Forbidden transitions are transitions for which Єmax is less than 104.

These are usually n π* transition e.g. saturated aldehyde or ketones

show weak absorption near 290nm and having an Єmax less than 100 is a

forbidden transition.

Chromophore and related terms

Chromophore: The term chromophore was originally used to describe an

unsaturated group of atoms which were thought to be essential for color

e.g. NO2 is a chromphore since its presence imparts yellow color to the

compound.

24

Table : Some simple chromophoric groups

SolventЄmaxλmax (nm)TransitionChromosphere

Vapor15,000175

Hexane10,000175

Hexane

10,000180

18,000160

15285

Methanol60205

Ethanol5338

Methanol15274

25

It may be seen that chromophores are of two types:-

a) Chrmophores such as , and which contain both π

and n electrons. They can undergo π π* as well as n π* transitions.

b) Chromophores such as

which contain π electrons and undergo π π* transitions.

For isolated chromophores i.e. multiple bonds that are not conjugated

absorption occur in far ultraviolet region which cannot easily studied but

the maximum of absorption and intensity can be modified by change of

the solvent or by derivatization.

Types of shift in absorption:-

a) Bathochromic shift or red shift:- it involves the shift of absorption

maximum towards longer wavelength due to the presence of certain

groups such as OH or NH2 called auxochromes or by change of the

solvent, e.g. decreasing the polarity of the solvent cause red shift in the

n π* transitions of carbonyl compounds.

Bathochromic shift is also produced if two or more chromophores are

present in conjugation in a molecule e.g. ethylene shows transition at

170nm, while 1,3-butadiene ( two double bonds are in conjugation shows

λmax at 217nm.

b) Hypsochromic shift or blue shift. It involves the shift of absorption

maximum towards shorter wavelength. It may be brought about by

removal of conjugation in a system or by change of solvent. For example,

protanation of aniline causes a blue shift from 280 nm to 203 nm since

26

Ph – NH3 has no lone pair of electrons on nitrogen which can conjugate

with beuzene ring.

c) Hyperchromic effect. This effect involves an increase in the inteusity

of absorption. It is usually brought about by introduction of an

auxochrome. For example, introduction of methyl group in position 2 of

pyridine increases Єmax (λmax 257 nm) from 2750 to 3560 (λmax 262 nm) for

π π* fransition.

d) Hypochromic effect. It involves a decrease in the intensity of

absorption. It is brought about by groups which distort the geometry of

the molecule. For example, when a methyl group is introduced in position

2 of biphenyl group hypochromic effect takes place due to distortion

caused by methyl group.

e) Auxochromes.

An auxochrome is a group which itself does not act as a chromophore but

when attached to a chromophore it shifts the absorption maximum

towards longer wavelength along with an increase in the intensity of

absorption, some commonly known auxochromic groups are –OH, -NH2,

-OR, -NHR and –NR2. For example, when the auxochrome –NH2 group is

attached to benzene ring, its absorption changes from λmax 255 nm

(Єmax = 203) to λmax 280 nm (Єmax = 1430).

All auxochromes contain one or more non-bonding pair of electrons.

When an auxochrome is attached to a chromophore it helps in extending

the conjugation by sharing of non-bonded pair of electrons as shown

below:

CH2 = CH –NR2 ↔:CH2-CH = +NR2

The extended conjugation is responsible for bathochromic effect of

auxochrome.

27

8. Solvent Effects and Choice of Solvent.

A. Solvent effects:-

The solvent used for dissolving a compound has a profound effect

on the UV spectrum of the compound as given below:

1) As a result of interactions of solute with solvent molecules, the

fine structure of the absorption bands is blurred out and a

smooth curve is obtained. The smoothing out effect is more in

polar solvents than in non-polar solvents. The fine structure of

UV spectrum is best observed in vapour phase.

2) Changing the polarity of the solvent may also lead to changes in

the position and intensity of absorption maximum. Generally,

changing the polarity of the solvent does not cause any

significant shift in the spectra of nonpolar compounds such as

dienes and conjugated hydrocarbons. For polar compounds like

α,β-unsaturated carbonyl compounds, the absorption maximum

(n π*) can shift to a shorter wavelength e.g. acetone absorbs

at 279nm in hexane, while the absorption maximum is 264nm in

aqueous solution. On the other hand increase in the polarity of

the solvent shifts the π to π* transition to longer wavelength (red

shift).

A suitable solvent for ultraviolet spectroscopy should meet the

following requirements:-

1- It should not absorb radiations in the region under

investigation.

2- It should be less polar so that it has minimum interaction

with the solute molecules.

28

The most commonly used solvent is 95% ethanol. It is

transparent above 210nm. Some other solvents which are

transparent above 210 are n-hexane, cyclohexane, methanol,

water, and ether. Benzene, chloroform, and carbon tetrachloride

cannot be used because they absorb in the range of about 240-

280nm. Hexane and other hydrocarbons are preferred to polar

solvents because they have minimum interactions with the

solute molecules.

29

Table: Some typical ultraviolet absorption bands due to the presence of different groups.

SolventЄmaxλmax (nm)

Example of Compound

Group

Vapor15,530171CH2 = CH2

Cyclohexane450187CH3 = C CH

VaporVapor

Hexane

20,000160180290

VaporHexaneHexane

16,00090017

166189279

Ethanol32208

Hexane20,900217CH2 = CH-H=CH2

Ethanol76007800

219228

CH2=CH-C CH

Hexane18,00030

218320

CH3-CH=CH-CH=O

Ethanol975038

224314

Water7400204

203.5254

Benzenoid ring

Water600217269

30

Water9800245.5

Woodward - Feiser rules for calculating absorption maxima:-

These rules enable us to calculate the λmax for conjugated dienes. It is

useful to consider some terms involved before discussing the rules.

1- Homoannular diene means a cyclic diene which contains conjugated

double bonds in the same ring e.g.

2- Heteroannular diene is acyclic diene in which double bonds in

conjugation are in separated rings e.g.

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3- Endocyclic double bond, it is the double bonds present as shown

below

4- Exocyclic double bond, it is a double bond in which one of the doubly

bonded atoms is part of a ring system as shown above.

Woodward-Fieser rules:-

According to these rules, each type of diene or triene system has a certain

fixed values at which absorption takes place. This constitute the basic

value or parent value added to it contributions made by various alkyl

substituents, ring residues, double bonds extending conjugation and polar

groups as -Cl, -Br, and –OR are added to the basic value to get λmax for a

particular compound.

Basic values1- Acyclic conjugated diene and heteroannular conjugated diene 215nm2- Homoannullar conjugated diene 253nm3- Acyclic triene 245nm IncrementsEach alkyl substituent or ring residue 5nmExocyclic double bond 5nmDouble bond extending conjugation 30nm Auxochromes -OR 6nm -SR 30nm -Cl, -Br 5nm

32

-NR2 60nm -OCOCH3 0nm

E.g.1

Calculate λmax for 2,4-Hexadiene CH3-CH=CH-CH=CH-CH3

Parent value 215nm Two methyl substituents 10nm Calculated λmax 225nm Observed λmax 227nm

E.g.2

Calculate λmax for 1,4-Dimethylcyclo1,3-hexadiene

parent value 253nm 4 alkyl sustituents 20nm Calculated λmax 273 Observed λmax 265nm

E.g.3

Compute λmax for the compound:

Parent value for heteroannular diene = 215 nm Four ring residues 4 x 5 = 20 nm

Calculated = 235 nm Observed = 236 nm

E.g.4

Calculate λmax for the compound having the structure:

33

Parent value for heteroannular diene = 215 nmThree ring residues x 5 = 15 nmOne exocyclic double bond = 5 nm

Calculated = 235 nm Observed = 235 nm

Woodward – Fieser Rules for α, β- Unsaturated Carbony Compounds.

These rules were modified by Scott and may be summed up as given in the following table:

Table : Parent values and increments for different substituents / groupsa) Parent values i- α ,β- unsaturated acyclic or six membered ring ketone 215 nm ii- α ,β- unsaturated five membered ring ketone 202 nm iii- α ,β- unsaturated oldehyde 207 nmb) Increments i- Each alkyl substituent or ring residue:

at α position 10 nm at β position 12 nm at γ and higher positions 18 nm

ii- Each exocyclic double bond 5 nm iii- Double bond extending conjugation 30 nm iv- Homoannular conjugated diene 39 nm v- Auxochromes Position

α β γ -OH 35 30 50

-OR 35 30 17 -SR - 85 -

OCOCH3 6 6 6 -Cl 15 12 - -Br 25 30 -

-NR2 - 95 -

34

E.g.Calculate the λmax for the compound

The compound is α,ß unsaturated acyclic ketone having an alkyl

substituent in α position

Parent value for α,ß unsaturated acyclic ketone = 215nm One alkyl substituent in α position = 10nm Calculated value = 225nm Observed value = 220nm

Applications of ultraviolet spectroscopy:

Ultraviolet spectroscopy can be used to reveal the presences of

conjugation and the type of that conjugation for example, comparison of

the absorption of calciferol with that of its stereoisomers iso-vitamin D2

(all-trans) and percalciferol illustrates the effects produced by cis-trans

isomerism and also by steric factors.

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The intense absorption of many keto-steroids is entirely due to the

conjugated system and even when a third double bond

present in conjugation as in prednisolone and prednisone very little

change is absorbed in position of maximum absorption. These are

examples of crossed conjugation and under these conditions the

absorption is always characteristic of the main chromophoric system.

Benzene gives ultraviolet absorption spectrum at 230 nm. Substitution of

one of the hydrogen by an alkyl group shifts the absorption to 250 nm e.g.

When a phenolic hydroxyl groups are present a marked bathochromic

shift of the absorption to about 280nm occurs, and the effect of solvent on

the absorption is considerable in neutral or acid media, the auxochrome is

–OH, in alkaline media OH O- .

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E.g. The absorption of chlorocresol in acid solution occur at 279nm but in

alkaline media the peak is found at 296nm.

The amino group is a powerful auxochrome when attached directly to a

benzene system. Thus in procainamide the 200nm high intensity

absorption of benzene sufferes a bathochromic displacement at 280nm.

Similarly, the sulphonamide drugs of general formula.

Show high intensity absorption at about 251nm. The effect of change in

PH used for ultraviolet measurements is very striking. In alkaline

solution, the absorbing system is at about 251nm but in acidic solution

the amino group will be protonated (–NH3 +) which is considerably less

efficient as an auxochrome.

The fusion of two or more benzene rings causes bathochromic shift from

200 to 255nm, while pyridine and its derivatives are completely analogue

to those of benzene and its derivatives.

Optimum conditions for spectrophotometric measurements.

Sample conditions:-

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Solvent:- the choice of the solvent is governed by the solubility of the

absorbing substance and the absorption of the solvent at the analytical

wavelength. The solubility of the substance in polar or non polar solvents

can often be predicted from its chemical structure e.g. vitamin A is

lipophilic so it is soluble in non polar solvent such as cyclohexane.

Substances with several functional groups are normally hydrophilic and

soluble in polar solvents such as water or ethanol. Water is the ideal

solvent, it is transparent at all wavelengths in the visible and ultraviolet

regions, it is also cheap and may be readily purified in the laboratory. If

the analyte is affected by PH then the PH of the solution should be

adjusted.

Organic solvents show either end absorption below 220nm due to

n π* as in ethanol or π π* as in toluene so their use are restricted

to measurements at wavelength where the solvents are reasonably

transparent. Compensation for weak absorption by the solvent in the

sample cell may be made by using a reference cell containing solvent

only.

Concentration and path length

In practice absorbance in the range of 0.3-1.5 are sufficiently reliable so

the concentration and the path length of analyte should be adjusted to

give an absorbance within this range.

Instrumental parameters:-

Slit width: too narrow slit width result in low signal noise ratio this lead

decrease in the precision of the measurement. The optimum slit width is

therefore the widest setting that provides spectral fidelity.

Scanning speed:

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The optimum scan speed is normally the fastest speed that maintains per

fidelity.

Problems in UV-Visible Spepectrometry

Question 1 Which of the following relationships between absorbance and %Transmittance is incorrect?

a) A = log10 100 / %Tb) A = 2 - log10 %Tc) A = log10 1 / %T

Question 2 In the equation, A = bc, what quantity is represented by "" ?

a) Absorbtivityb) Molar absorbtivityc) Path length

Question 3 Why is it generally preferable to use absorbance as a measure of absorption rather than % Transmittance?

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a) Because %T cannot be measured as accurately as absorbanceb) Because %T is dependant on the power of the incident radiationc) Because absorbance is proportional to the concentration of the analyte, whereas %T is not.

Question 4 Does a compound with high molar absorbtivity have a higher or lower limit of detection than a compound with low molar absorbtivity?

Question 5

Ergosterol, a precursor of vitamine D, has λmax =282 nm and molar absorptivity ε =11,900. What is the concentration of ergosterol in a solution whose absorbance A = 0.065 with a sample path length L =1.00cm.

Question 6

The absorbance of a certain dye is 0.42. Calculate the percent transmission?

Question 7

A cell of 2-cm thickness contains a dye having a molar concentration of 2.5 x 10–4 moles/liter and a molar absorptivity of 8.4 x 102 liter/mole cm. Find the absorbance of the cell.

Question 8

A compound shows λ max 235 nm in ethanol, and a 0.0001 M solution in a 1.0-cm quartz cell shows absorbance (A) equal to 1.03. Calculate εmax for the compound.

Question 9

The percentage transmittance of a solution of sodium fumarate in water at 250 nm and 25 C is 19.2% for a 5 x 10–4 molar solution in a 1-cm cell. Calculate the absorbance and the molar absorptivity. What is the percentage transmittance of a 1.75 x 10–4 molar solution?

Question 10

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When a 1.9 -cm absorption cell was used, the transmittance of the light at a wavelength of 435 nm by bromine in carbon tetrachloride solution was found to be as follows:

Calculate the molar absorptivity. What percentage of the incident light would be transmitted by 2 cm of solution containing 0.00155 mole of bromine per liter of carbon tetrachloride?

Question 11

A solution of a dye containing 1 g per 100 milliliters transmits 80% of the blue light at 435.6 nm in a glass cell 1 cm thick. What percent of light will be absorbed by a solution containing 2 g per 100 milliliters in a cell 1 cm thick? What concentration is required to absorb 50% of the light?

Question 12

A police officer swabs a suspect’s hands, and sends the swabs to the crime lab for analysis. Barium (Ba) and antimony (Sb) are usually present in gunpowder. An atomic absorption spectrophotometer with furnace atomizer detects the presence of 0.43? g of Ba and 0.09 ? g of Sb.

Question 13

The absorption of UV and Visible radiation by drug molecules is contributed to the presence of

1- A chromophoric group in the form of conjugated double bonds only2- A chromophoric group in the form of auxochromic groups3- Both groups in acidic media

Conc (moles/1) 0.00546 0.00350 0.00210 0.00125 0.00066

Transmittance T 0.010    0.050    0.160    0.343    0.570   

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Question 14

The measured absorbance, is proportional to

1- The path length and the concentration2- The wave length, path length and the concentration.3- The specific absorbance, the wave length, path length and the concentration.

Question 16

The cells intended to contain the solution being examined by spectrophotometers

1- Should be clean and handled with care 2- Should have a path length of 1cm3- Should be transparent to UV-Visible and IR radiation4- 1 and3

Question 17

Haloperidol injection is diluted by adding 15ml HCl 1M to 5ml of the injection, this solution was extracted three times with ether and washing the ether extract with 2 x10 ml of water. The combined ether aqueous layers were diluted to 100ml with water. 10ml of the diluted solution give an absorbance of 0.843 at 245nm, the A1%

1cm value = 346. Calculate the concentration in the injection.

Additional problems

1. Calculate the absorbance (A) of a 0.00 1 M solution of acetone in heptane contained in a quartz cell of 0.5-cm path length at 279 nm.

2 A compound shows λmax 235 nm in ethanol, and a 0.0001 M solution in a 1.0-cm quartz cell shows absorbance (A) equal to 1.03. Calculate εmax for the compound.

3. List in order of increasing energy the types of electronic transitions that would be expected for the following compounds:

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(a) Formaldehyde,

(b) Methyl vinyl ketone,

(c)

4. The absorption spectrum of trans-azobenzene, Ph—N=N—Ph, shows a λmax at 430 nm (εmax ~ 500) and a second λmax at 320 nm (εmax 25,000). To what types of transitions do the absorption bands correspond?

5. For which of the following pairs of isomeric compounds will ultraviolet spectroscopy be a useful tool for distinguishing the members? Explain.

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6. Calculate the λmax for the compound

7. Calculate the λmax for the compound

Q8 Define the following:-

1- Chromophore

2- Spectroscopy

3- bathochromic shift

4- Chemical shift

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Q9 Explain fluorescence and phosphorescence. Use diagram

Q10 Procaine and adrenaline injection was assayed as follows

1 ml of the injection was diluted to 20ml, 1ml of the resulting solution was further diluted to 100ml. The final solution gives absorbance value of 0.650. Assuming the absorption caused by adrenaline is negligible at this dilution. Calculate the concentration of procaine in the injection. For procaine A1cm

1% = 680 at 290nm

Q11 How does conjugation in a molecule influence absorption in the IR region.

Q16 Using Diagram, explain the main parts of spectrometer?

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