Ultraviolet spectroscopy and its applications are discussed. UV spectroscopy involves promoting electrons from bonding to antibonding orbitals upon exposure to UV light, causing absorption. The wavelength of maximum absorption depends on factors like conjugation, substituents, and solvent polarity. Beer's law relates absorbance to concentration and path length. Woodward-Fieser rules allow predicting absorption maxima of conjugated dienes and α,β-unsaturated carbonyls. UV spectroscopy is useful for identification of functional groups, extent of conjugation, isomer differentiation, and detection of impurities in drugs.

1. Ultraviolet
Spectroscopy
&It’s Applications
Prof. A. Sreedevi
Institute of Pharmaceutical Technology,
SPMVV, Tirupati

2. Spectroscopy
• It is the branch of science that deals with
the study of interaction of electromagnetic
radiation with matter.
Electromagnetic radiation
• Electromagnetic radiations are produced
by the oscillation of electric charge and
magnetic field on the atom.

3. • There are various forms of electromagnetic
radiations such as Ultra-violet, Infra-red,
X-rays, Radio-waves, etc.,
• These are characterized by their
Wavelengths or Frequencies or Wave
numbers.

5. • The UV radiation region extends from 100 nm
to 400 nm.
• Far UV Region: below 200 nm
• Near UV Region: 200 nm to 400 nm

6. Beer Lambert Law
• Beer's law relates the absorption of light to
the concentration of absorbing solute and
Lambert’s law relates the total absorption
to the optical path length.
A=log(IO/I)=ԑcl
A=absorbance
IO= intensity of light incident upon sample cell
I= intensity of light leaving sample cell
c= molar concentration of solute
l=length of sample cell
ԑ= molar absorptivity (molar extinction coefficient)

7. • ԑmax is constant for a particular compound
at a given wave length. Commonly
expressed as ԑmax (absorption band
maximum).
• It ranges from 0-106. Values above 104 are
termed high- intensity absorptions ,while
values below 103 are low intensity
absorptions.
• Forbidden transitions have absorptivities in
the range from 0-100.

8. Principle of UV-Spectroscopy

9. • When ultraviolet light is passed through a
substance under examination, absorption of
energy results in the promotion of electron
from the ground state (bonding orbital )to
excited state (Anti-bonding orbital) or
highest occupied molecular orbital to
lowest available unfilled orbital.
• The anti bonding orbitals of σ, π are σ*, π*
respectively.

10. • Black - Bonding orbitals
• Red - Antibonding orbitals
• Dots - Atomic centers

11. UV spectrophotometer

12. 1,3-butadiene

13. Electronic Transitions

15. The possible electronic
transitions are
• σ → σ* transition
• π → π* transition
• n → σ* transition
• n → π* transition

16. σ→ σ* transition
• When σ electron is promoted to
corresponding anti-bonding orbital σ*,
σ→ σ* transition takes place.
• The energy required is large for this
transition. Eg: Saturated hydrocarbons.
• Methane has C-H bond only and can
undergo σ→ σ* transition and shows
absorption maxima at 125nm.

17. π- π* transition
• π- electron in a bonding orbital is excited
to corresponding anti- bonding orbital π*
• π- π* transitions require less energy than
σ→ σ* and n → σ* transitions.
• Compounds containing multiple bonds like
alkenes, alkynes, carbonyls, nitriles,
aromatic compounds, etc undergo π- π*
transition.
• Alkenes generally absorb in the region of
170-205nm.

18. n → σ* transition
• Saturated compounds containing
heteroatom with lone pair of electrons like
O, N, S and halogens are capable of
n → σ* transition.
• These transitions usually require less
energy than σ → σ* transitions.
• The n → σ* peaks in UV region range
from 150 – 250 nm.

19. n → σ* transition
• These transitions are very sensitive to
hydrogen bond.
• Alcohols, Amines form hydrogen bond
with solvent.
• Hydrogen bond shifts UV absorption to
shorter wavelengths.

20. n- π* transitions
• An electron from non-bonding orbital is
promoted to anti-bonding π* orbital.
• n → π* transitions require least amount of
energy out of all the transitions.
• Compounds containing double bond involving
hetero atoms (C=O, C≡N, N=O) undergo such
transitions.

21. •n → π* transitions require minimum
energy and show absorption at longer
wavelength (around 300 nm).
•Thus, n → π* & π → π* electronic
transitions show absorption in the region
above 200 nm which is accessible to UV-
visible spectrophotometer.
n- π* transitions

22. 4a-methyl-4,4a,5,6,7,8-hexahydronaphthalen-2(3H)-one

23. Terms used in
UV Spectroscopy

24. Chromophore
The part of a molecule responsible for
imparting color, is called as chromophore.
(or)
The functional groups containing multiple
bonds are capable of absorbing radiations
above 200 nm due to n → π* & π → π*
transitions.
e.g. NO2, N=O, C=O, C=N, C≡N, C=C

25. Auxochrome
• The functional group with non-bonding
electrons that does not absorb radiation in
near UV region but when attached to a
chromophore alters the wavelength &
intensity of absorption.
• Eg: OH, NH2, Cl
Benzene Aniline
λmax = 255 nm λmax = 280 nm
ԑmax = 203 ԑmax = 1430
NH2

26. Bathochromic Shift
(Red Shift)
• When absorption maxima (λmax) of a
compound shifts to longer wavelength, it is
known as bathochromic shift or red shift.
• The effect is due to substitution or by
extending conjugation or by the change of
solvent.
• e.g.-1,3 butadiene 1,3,5 hexatriene
217nm 258nm

27. Hypsochromic Shift
(Blue Shift)
• When absorption maxima (λmax) of a
compound shifts to shorter wavelength, it
is known as hypsochromic shift or blue
shift.
• The effect is due to removal of conjugation
or by the change of solvent.
λmax = 280 nm λmax = 254 nm
NH3
+
N
H2
..

28. Hyperchromic Effect
• When absorption intensity (ε) of a
compound is increased, it is known as
hyperchromic effect.
Pyridine 2-methyl pyridine
λmax = 257 nm λmax = 260 nm
ε = 2750 ε = 3560
N
N CH3

29. Hypochromic Effect
• When absorption intensity (ε) of a
compound is decreased, it is known as
hypochromic effect.
Biphenyl 2-methyl biphenyl
ε = 19000 ε = 10250
C
H3

30. Effects of solvent on transition

31. • Generally 95% ethanol, n- hexane, methanol,
cyclohexane, acetonitrile, diethyl ether, etc. are
used as solvents in UV region.
• The position and the intensity of absorption
maximum is shifted for a particular chromophore
by changing the polarity of the solvent.

32. • α, β unsaturated carbonyl compounds show
two different shifts.
• n → π* transition: The absorption band
moves to shorter wavelength by increasing
the polarity of the solvent.
• π → π* transition: The absorption band
moves to longer wavelength by increasing
the polarity of the solvent.

33. Woodward Fieser Rules

34. • UV absorption rules were developed
by Robert Burns Woodwards.
• Rules developed by Woodward in 1941
then expanded by Feiser in 1948.
• First (empirical) rule to predict the UV
absorption λmax for certain classes of
compounds.

35. Conjugated Diene
Correlations
Homoannular Diene
Heteroannular Diene

36. • Endocyclic conjugated
double bond
• Exocyclic conjugated
double bond
• Ring residue is a C-C bond, not a part of
conjugated system but attached to any one of
the carbon atoms of the conjugated polyene
system.

37. CONJUGATED DIENE CORRELATIONS:
i Acyclic diene 217nm
ii Basic value for homoannular diene 253 nm
iii Basic value for heteroannular diene 215 nm
iv Alkyl substituent or Ring residue attached
to the parent diene
5 nm
INCREMENTS FOR DIFFERENT SUBSTITUENTS/GROUPS:
Double bond extending conjugation + 30 nm
Exocyclic double bonds + 5 nm
-OR + 6 nm
-Cl,-Br + 5 nm
-NR3 +60 nm

38. Calculation of absorption
maxima

39. CH3-CH=CH-CH=CH-CH3
• Basic value (acyclic diene) = 217nm
• 2 alkyl substituents = 2x5 = 10nm
________
λmax 227nm
Observed value 227nm

40. • Base value (Hetero annular) = 215 nm
• Ring residue = 3 x 5 = 15 nm
• Exocyclic double bond = 1 x 5 = 5 nm
λmax = 234 nm

41. Homoannular or
heteroannular
Ring residues are present or
not
Exocyclic or endocyclic
Alkyl substituents
• Homoannular = 253nm
• 3 ring residues=3X5= 15nm
• 1 exocyclic = 5nm
• 1alkyl substituent = 5nm
__________
• λmax 278nm
• observed 275nm
CH3
CH3

42. • Homoannular or
heteroannular diene?
• ring residues ?
• Exocyclic or endocyclic?
• Any other?
• Heteroannular = 215nm
• 3 ring residues=3X5 = 15nm
• Exocyclic = 5nm
• Methoxy group = 6nm
________
• λmax 241nm
• Observed 241nm
O
CH3
CH3

43. α, β UNSATURATED CARBONYL COMPOUNDS
Acyclic ketones 215 nm
6 membered cyclic ketones 215 nm
5 membered cyclic ketones 202 nm
α, β unsaturated aldehydes 210 nm
α, β unsaturated carboxylic acids & esters 195 nm
Alkyl substituent or Ring residue in α position 10 nm
Alkyl substituent or Ring residue in β position 12 nm
Alkyl substituent or Ring residue in γ and higher positions 18 nm
Double bond extending conjugation 30 nm
Exocyclic double bonds 5 nm
Homodiene compound 39 nm

44. Chromophore increments in Carbonyl Compounds
Chromophore α β γ δ
-OH +35nm +30nm - +50nm
-OAC +6nm +6nm +6nm +6nm
-Cl +15nm +12nm - -
-Br +25nm +35nm - -
-OR +35nm +30nm 17nm 31nm
-SR - +85nm - -
-NR2 - +95nm - -

45. (CH3)2-C=CH-COCH3
• Base value = 215 nm
• β- Substituents = 2 x 12 = 24 nm
• ____________
• λmax = 239 nm
• observed value 237nm

46. Basic value = 215nm
2β ring residues =2X12 = 24nm
1 exocyclic double bond= 5nm
_______
λmax = 244nm
observed value- 241nm
CH3
CH3
O
C
H3

47. • Basic value = 215nm
• α ring residue = 10nm
• δ ring residue = 18nm
• 1 exocyclic double bond = 5nm
• Homoannular conjugated diene = 39nm
• 1 double bond extending conjugation= 30nm
___________
λmax = 317nm
observed value - 319nm
CH3
CH3
C
H3
O

48. • Basic value = 215nm
• One α ring residue = 10nm
• Two β ring residues = 2X12= 24nm
• Exocyclic double bond to two rings=10nm
• _________
• λmax = 259nm
O

49. • Basic value = 215nm
• Ring residue at β-position = 12nm
• Ring residue at δ-position = 18nm
• 1 exocyclic double bond = 5nm
• 1 double bond extending conjugation = 30nm
_________
• λmax = 280nm
• Observed value - 280nm
O
CH3
CH3 R

50. • Basic value?
• Any ring residue?
• Any other?
• Five membered = 202nm
• 2 ring residues
(at β-position) = 2X12 = 24nm
• 1 exocyclic double bond= 5nm
________
• λmax 231nm
observed = 226nm
O
C
H3

51. • Basic value ?
• Any ring residues?
• Any alkyl
substituents?
• Any other?
Six membered enone = 215nm
One ring residue at β-position
= 12nm
One alkyl substituent at
β-position = 12nm
OH substitution at α-position
= 35nm
__________
λmax = 274nm
• Observed value = 275nm
O
OH
C
H3
CH3

52. • Absorption maximum is solvent dependent.
• With the change in polarity of the solvent,
absorption maximum is shifted.
• More polar solvents will experience blue shift.
S.no Solvent correction
1. Hexane +11nm
2. Dioxane +7nm
3. Methanol 0nm
4. Water -8nm
5. Chloroform -1nm

53. Applications

54. UV spectroscopy is used in
Detection of functional groups.
Extent of conjugation.
Distinction in conjugated and non
conjugated compounds.
 Identification of an unknown compound.
 Determination of configuration of
geometrical isomers.
Detection of Impurities.

55. Drug Solvent used Wavelength
used (nm)
Acetazolamide 0.1NHcl 265
Chloramphenicol capsules Water 278
Diazepam tablets 0.5%H2SO4
In MeOH
284
Frusemide tablets 0.1N NaOH 271
Glibenclamide tablets Methanolic HCl 300
Indomethacin capsules Buffer pH 7.2 318
Paracetamol tablets 0.01N NaOH 257
The following is the list of some of the drugs
which show UV absorption:

56. Conclusion

57. • UV spectroscopy is an effective analytical
tool for identification of α, β unsaturated
carbonyl compounds and conjugated
dienes.
• UV spectroscopy is useful to differentiate
the conjugated dienes from non-conjugated
dienes.
• UV in combination with other analytical
techniques is helpful in the structural
elucidation of the compounds.

58. References
• Donald A. Pavia., Introduction to Spectroscopy,
Cengage Learning, 2008; 353-383.
• Y. R. Sharma., Elementary Organic
Spectroscopy, S.Chand, 2011; 8-60.
• William Kemp., Qualitative organic analysis:
Spectrochemical techniques, Mcgraw
Hill Book Co Ltd, 2008; 243-263.