1. Spectroscopic Technique 1 Principles, Electronic Spectroscopy of Conjugated Molecules, Woodward-Fieser Rule, Applications (Topic For Engineering Chemistry) Prepared By Dr. Rupa Rani Dey RD’s Academy rdschemistry@gmail.com

2.  Fundamental of Molecular Spectroscopy By C B Banwell & E M McCash  Organic Spectroscopy By William Kemp  Spectroscopy Of Organic Compounds By P S Kalsi  Elementary Organic Spectroscopy By Y R Sharma Reference Books 2

3. 3 Electromagnetic Radiation

4. 4

5. 5 E2E2 E1E1 Energy

6. 6  Various forms of EM radiations are Ultra Violet (UV) Visible Infra-red (IR) Radio frequency etc.  The study of Spectroscopy deals with absorption as well as emission of EM of different wavelength

7. 7  When EM radiation is passed through the substance then radiations of certain wavelength are absorbed by the substance depending on the structure of the compound

8. UV-Visible Spectroscopy 8

9.  Far UV Range: Below 200 nm Far UV spectra is also known as Vacuum UV Spectroscopy since it is performed to in vacuum to avoid the absorption of gases e.g. O 2, N 2 9  Visible Range: 400-800 nm  UV Range: 200-400 nm

10. 10 When UV-Visible radiation (monochromatic * ) is passed through homogenous absorbing medium: A portion of radiation is absorbed by the compound and rest is transmitted. * Radiations of very narrow range of wavelength or frequency

11. 11 Function of monochromator is to separate desired set of wavelength from the band of spectrum

12. Plot of Electronic Spectra: The UV-Vis spectra is observed as a plot of intensity of absorbed(or absorbance) radiation against the wavelength of the absorbed radiation. 12 absorbance wavelength (nm) max = 217: maximum absorption at this wavelength

13. Lambert’s Law : Solid Sample Beer’s Law : Solution 13 The Absorption Law

14. 14 y x y > x Sample of same concentration but different path length

15. 15 CACA CBCB C A > C B Sample of same path length but different concentration

16. 16 The Beer-Lambert’s law l - A ∝ cl A = cl  The amount of light absorbed is directly proportional to the concentration (c) of the solution of the sample A ∝ c  The absorbance is directly proportional to the length of the light path (l), which is equal to the width of the cuvette. A ∝ l

17. Beer-Lambert’s Law The law states that “when a beam of monochromatic radiation on UV- visible range is passed through homogeneous absorbing medium, the rate of decrease of intensity of radiation with thickness of absorbing medium is proportional to the intensity of incident radiation as well as the concentration of the solution ”. 17

18. 18 Derivation

19. 19

20. 20 t

21. 21 absorbance wavelength (nm) Representative UV spectra for compound ‘X’ (X)

22. Absorption peaks broadens from atom to polyatomic molecules 22 vovo v1v1 v2v2 v3v3 vovo v1v1 v2v2 v3v3 rnrn rnrn rnrn rnrn rnrn rnrn Electronic Transition E2E2 E1E1

23.  Excitation of electrons are also accompanied by the constant vibratory and rotatory motion of the molecules. 23  Electronic levels are comprised of band of vibrational and rotational levels.

24. 24 E ele : Electronic Energy E vib : Vibrational Energy E rot : Rotational Energy E ele (10 5-6 cm -1 ) >>> E vib (10 3 cm -1 ) > E rot (1-10 cm -1 ) Born-Oppenheimer Approximation, vibrational and rotational energies are quantised and treated separately Total energy of a molecule E tot = E ele + E vib + E rot +..

25. 25 Excited state Ground state Transition can be represented as vertical line in the energy diagram Frank-Condon Principle assumed ‘an electronic transition takes place so rapidly that the nuclei of the vibrating molecule can be assumed to fixed during the transition’

26. 26  Explains the intensity of vibronic (vibrational and electronic) transition.  Classic Concept: An electronic transition takes place so rapidly that the nuclei of the vibrating molecule can be assumed to be fixed during the transition.  The vertical transition represents the most intense peak which is for (0,0) transition. Frank-Condon Principle

27. 27 Fig. Vibronic Transition with no change in Internuclear Distance

28. 28 Change of Internuclear distance due to vibration Relaxation of Electronic States: Typically when the electronical excitation of a molecule is considered, there is always displacement of charge and a new equilibrium Internuclear distance

29. 29 As the optical transition becomes less vertical, the intensity decreases due to the change in the Frank-Condon principle.

30. Broadening of Spectral Lines 30  The electronic excitation is superimposed upon rotational and vibrational levels.  There will be a large number of possible transitions corresponding to electronic, vibrational and rotational levels.  Large number of wave length which are close to each other will be absorbed, resulting in the formation of band.

31. Other Factors of Spectral Line Broadening  Collision broadening – Molecular interactions are more severe in liquids than in gas (Gas phase spectra are sharper than those of the corresponding liquids). I case of solids, the spectral lines are sharper but show evidence of interactions by splitting of lines. 31  Doppler effect – Atoms or molecules in gases and liquids are in continuous motions with different velocities in different directions from the detector which causes Doppler effect (More in gases than in liquids).

32. 32 Which depicts a transition to a dissociative state? Which depicts a transition in a molecule that has a larger bond length in the excited state? Which would show the largest intensity in the 0-0 transition? Which represents molecules that can dissociate after electronic excitation? Which represents the states of a molecule for which the v”=0 → v’=3 transition is strongest? Bond Length Energy

33.  Energy of Transitions 33  Number of Transitions  Intensity of Transitions

34. 34 HOMOLUMO Electronic Transitions

35. 35 HOMOLUMO Electronic Transitions

36. 36 Types of Electronic Transitions Four types of Electronic transitions are possible

37. 37 o High energy transition (<<200nm) o Colourless, since not absorbed light in visible region. o Saturated hydrocarbons shows this type of transitions e.g. Cyclopropane, λ max 190 nm. Cyclohexane, λ max 153 nm (vacuum UV)

38. 38 o Appear in the near UV or visible region o Compounds containing non-bonding electrons or lone-pair electrons show this type of transition

39. 39 Compounds MeCl169 nm MeI258 nm

40. 40 For compounds containing double, triple bonds or aromatic rings show this type of transition. This is usually a symmetry allowed and high intensity transition. e.g. Alkenes, Alkynes, Carbonyls, Cyanides, Azo compounds etc.

41. 41 Q. Why extended conjugation lowers the ΔE, and increase the λmax. CH 2 =CH 2 λmax 171 nm CH 2 =CH-CH=CH 2 λmax 217 nm CH 2 =CH-CH=CH-CH=CH 2 λmax 256 nm

42. 42 Q. Why introduction of an alkyl group to the double bond shift the λmax value to longer wavelength. CH 2 =CH 2 λmax 171 nm CH 3 -CH 2 -CH=CH-CH 2 CH 3 λmax 184 nm Q. Out of the given set of compounds, find out the one which absorbs at the longest wavelength?

43. 43 The excitation of an electron on a nonbonding (lone pairs) orbital, such as unshared pair electrons on O, N, S to an antibonding, π* orbital. Usually in an double bond with hetero atoms, such as C=O, C=S, N=O etc show this transition. A symmetry forbidden and low intensity transition.

44. 44 In carbonyl compounds, two types of transitions take place which are High energy transitions Low Energy Transition

45. 45 Q. Acetaldehyde is having absorption peak at 160 nm, 180 nm and 292 nm. What type of transition is responsible for each of these absorption? Q. Assign electronic transitions for the following compounds. Acetaldehyde, Benzene, Cyclohexane, Ethanol, Heptane, Aniline, Butadiene

46. 46 Q. UV spectra for the compound ‘m’ has given below assign electronic transitions and justify your answer m

47. Transition Probability i. Energy of the transition/incident light. ii. Orientation of the molecule/material. iii. Symmetry of the initial and final states. iv. Angular momentum (spin). 47 The probability of a particular transition taking place. As transition probability increases

48. Allowed transitions 48

49. e.g. C=O, C=C, NO 2, –COOH, -N=N-, C≡C etc. 49 Chromophore : Any functional group which absorb in UV-Visible range are Chromophores Can induce color in the molecule

50. e.g. Benzene absorbs at 255 nm but Aniline absorbs at 280 nm. 50 Auxochromes: Changes both intensity and λ max e.g. OH, NH 2, -SH, -Cl, OR, NHR 2 etc. Auxochromes has ability to extend conjugation of chromophores by sharing its nonbonding electrons

51. Absorption and intensity shift 51 absorbance wavelength (nm) Bathochromic (Red Shifted) Hypsochromic (Blue Shifted) Hyperchromic Hypochromic

52. Bathochromic shift: 52

53. 53 50,000 100,000 200 300 380 wavelength (nm) n = 5 n = 4 n = 3 R(CH=CH) n R 

54. 54 UV-spectra of acetone in hexane and in water absorbance wavelength (nm) 264 nm (water) 279 nm (hexane)

55. 55 Q. In alkaline medium, p-nitrophenol shows red shift Because negatively charged oxygen delocalizes more effectively than the unshared pair of electron

56. Hypsochromic Shift (Blue Shift ): 56

57. Hyperchromic and hypochromic shift 57 Double strand DNA absorbed less strongly than denatured DNA due to the stacking interactions between the bases. On denaturation, the bases are exposed thereby showing increased interaction. Hyperchromic (90 ºC) Hypochromic (40 ºC) absorbance wavelength (nm)

58. 58

59. 59 Influence of solvent on the UV λ max and  max of the n   and    excitation of 4-methylpent-3-en-2-one Solvent   * n  * n-hexane230(12,600)327(98) ether230(12,600)326(96) ethanol237(12,600)315(78) water245(10,000)305(60)

60. Solvent Effect Different types of transition are affected differently with change in polarity  Molecules showing only this type of transition are non-polar  Change in polarity of solvents has no effect on this type of transition

61. 61  Excited state is polar, but the ground state is neutral.  Polar solvent interacts with the excited state and stabilize it  Absorption shifts to longer wavelength Polar solvent Polar excited state Neutral ground state

62. 62 polar solvent neutral excited state polar ground state  Excited state is neutral and ground states are polar  Polar solvent will solvate the ground state  Ground state lowers its energy due to solvation.  Absorption shifts to shorter wavelength

63. 63

64. 64 Influence of solvent on the UV λ max and e max of the n   and    excitation of 4-methylpent-3-en-2-one Solvent   * n  * n-hexane230(12,600)327(98) ether230(12,600)326(96) ethanol237(12,600)315(78) water245(10,000)305(60)

65. 65 Auxochromes exert their bathochromic shifts by extension of the length of the conjugated system B: NH 2, OH, OR, X etc

66. 66 Woodward-Feiser rule

67. Parent values Heteroannular Conjugated Diene(transoid)214 nm Homoannular Conjugated Diene(cisoid)253 nm 67 Increments Each Alkyl Residue or ring residue5 nm Exocyclic double bond5 nm Double bond extending conjugation30 nm

68. 68 Heteroannular diene : 214 Alkyl substituents : (4x5) : 20 Exocyclic double bond : 5 Absorption maximum : 239 nm

69. 69 Homoannular diene : 253 Alkyl substituents : (4x5) : 20 Exocyclic double bond : (2x5) : 10 Absorption maximum : 283 nm

70. 70 Total : 234 nm Transoid (base) : 214 nm 3 ring residues : + 15 nm 1 exocyclic C=C : + 5 nm Observed λ max 235 nm

71. 71 Cisoid(base)253 3 ring residues+15 1 exocyclic C=C+5 Total273 Observed = 275 nm

72. 72 Cisoid(base)253 3 ring residues+15 1 exocyclic C=C+5 1 extra conjugation+30 Total : 303 nm Observed : 304 nm

73. 73 Transoid(base)214 2 ring residues+10 1 exocyclic C=C+5 Total229 Observed = 230 nm

74. 74 Transoid(base)214 2 ring residues+10 1 exocyclic C=C+5 Total229 Observed = 236 nm

75. Parent values Heteroannular Conjugated Diene214 nm Homoannular Conjugated Diene253 nm Increments Each Alkyl Residue or ring residue5 nm Exocyclic double bond5 nm Double bond extending conjugation30 nm Auxochrome -OR6 nm -SR30 nm -Cl, -Br5 nm -NR 2 60 nm -OCOCH 3 0 nm 75

76. 76 Transoid(base)214 3 ring residues+15 1 exocyclic C=C+5 -OR+6 Total : 240 nm Observed : 241 nm

77. 77 Transoid (base)214 5 ring residues+25 3 exocyclic C=C+15 1 extra conjugation+30 Total =284 nm Observed =283 nm Cisoid (base)253 3 ring residues+15 1 alkyl substituent+5 1 exocyclic C=C+5 Total =278 nm Observed =275 nm

78. 78 Cisoid (base)253 5 ring residues+25 3 exocyclic C=C+15 2 extra conjugation -OAc +60 +0 Total =353 nm Observed =355 nm

79. 79 Transoid (base)214 3 alkyl group+15 Total =229 nm Observed =232 nm Transoid (base)214 4 alkyl group+20 Total =234 nm Observed =235 nm

80. 80 Cisoid (base)253 5 ring residues+25 3 exocyclic C=C+15 1 extra conjugation+30 Total =323 nm Observed =325 nm Cisoid (base)253 4 ring residues+20 2 exocyclic C=C+10 Total =383 nm Observed =382 nm

81. Application of UV-Visible Spectroscopy 1.Detection of conjugation Addition of unsaturation shifts the absorption to longer wave length and pertains colour. 81 Phenolpthaline in aqueous methanol NaOH

82. 2. Distinction in conjugated & non conjugated isomers 82 Longer wave length Shorter wave length

83. 3. Detection of geometrical isomers 83 max = 234 nm  max = 12000-28000 max = 283 nm  max = 5000-15000 s-trans 214 nm s-cis 253 nm

84. ~Thank You For Your Attention~