Ir spectroscopy by dr. pramod r. padole

For B.Sc. & M.Sc. students

1. Infrared Spectroscopy By: Dr. P. R. Padole

2. Infrared Spectroscopy Infrared spectroscopy - a technique used to identify information about the functional groups present in the compound . The most common technique used is absorption spectroscopy.

3. Spectroscopy is an instrumentally aided study of the interactions between matter (sample being analyzed) and energy (any portion of the electromagnetic spectrum) EMR ANALYTE SPECTROPHOTOGRAPH 1 . UV-Visible radiations--------excitation of electrons---------UV-visible spectrum 2 . IR-radiations- -----------------vibration changes in electrons--------IR spectrum 3 . Radio frequency---------------spin rotational changes-------------N.M.R spectrum Conc. should be lower

4. Principle of IR spectroscopy: Q.1) Describe the principle or theory of IR spectra? Q.2) Explain, how the IR spectrum are presented (Diagram expected). Also name the solvents used to scan the IR spectra. (W-12, 2 Mark) Q.3) In which region of electromagnetic spectrum the vibrationals transitions are observed? (S-16, ½ Mark) (a) U.V. (b) Microwave (c) IR (d) Visible

5. Principle of IR spectroscopy: Defination : The spectra produced due to, ( i ) Absorption of IR radiation by molecule & (ii) Vibrational energy level transition, is called as Vibrational spectra or IR spectra . It is observed in IR region in spectral range 4000-667 cm -1 . When IR absorption occur? Rule-1) IR absorption only occurs when IR radiation interacts with a molecule undergoing a change in dipole moment as it vibrates or rotates. Rule-2) Infrared absorption only occurs when the incoming IR radiation has sufficient energy for the transition to the next allowed vibrational state Note: If the Rule-1 & 2, above are not met, no absorption can occur.

6. PRINCIPLE OF IR SPECTROSCOPY Molecules are made up of atoms linked by chemical bonds. The movement of atoms and the chemical bonds look like spring and balls (vibration). This characteristic vibration are called Natural frequency of vibration. Applied infrared frequency = Natural frequency of vibration Change in dipole moment is required

7. When a wave of infrared light encounters this oscillating EM field generated by the oscillating dipole of the same frequency, the two waves couple, and IR light is absorbed The coupled wave now vibrates with twice the amplitude IR beam from spectrometer EM oscillating wave from bond vibration “coupled” wave

8. Infrared Absorption When IR radiation is passed through molecule, the radiation of definite frequency (frequency equal to vibrational frequency of molecule) is absorbed by the molecule. The energy absorbed too small to cause rotational excitation. But the energy absorbed is sufficient to increase amplitude of vibration. So, the molecule is excited and undergoes vibrational transition from one vibrational energy level to higher vibrational energy level within same electronic energy level .

9. Principle of IR spectroscopy Covalent bond in molecule behave as tiny spring The atom will not remain in fix motion with respect to each other but the avg. distance remain same So the vibration motion is occurred When internal vibrational energy of molecule matches with energy of externally applied IR, quantized

10. Molecules absorb IR Molecule excited from lower to the higher vibrational level within same electronic energy level . i.e. Increases the amplitude of vibration P.S.Kalsi ; Spectroscopy Of Organic Compounds, Sixth Edition:2004, Page no. 65-68 .

11. Infrared Spectra The intensity of remaining transmitted light (I t ) is measure by IR spectrophotometer (detector). It is compared with intensity of incident light (I o ); to get absorbance (A) or percentage transmittance (%T) of the substance (molecule). Thus, the intensity of light absorbed by the molecule is measured interms of %T verses frequency or wave number (in cm -1 ) of IR region. The graph of %T verses frequency or wave number (in cm -1 ) of radiation passed is recorded on the charge. It is called as IR Spectra

12. The frequency of vibration producing IR spectra is given by cm -1 Where, ʋ is frequency of vibration or frequency of spectral line or position of spectral line in cm -1 . K is the force constant of bond. It is measured bond strength μ is reduced mass of molecule

13. Types of Vibrational modes: Only those vibrational modes (vibration) which produces change in dipole moment of molecule, absorbs IR radiation and show IR spectra. So, the dipole moment of molecule must change during vibration. Q.1) What are the essential conditions for a compound to absorb IR radiation? (S-11 & W-13, 2 Mark) Q.2) Give reasons: Chlorine molecule does not absorb IR radiation. (S-11, 2 Mark) Q.3) What are the conditions or absorption in IR region? (W-11, 2 Mark) Q.4) Give the selection and essential condition (requirement) for a molecule to show IR spectra? Q.5) Give the necessary condition for a molecule to absorb IR radiation? (or to become IR active). Q.6) What are the conditions (requirement) for a substance to absorb in the IR region? Which of the following compounds are IR active? (W-19, 4 Mark) ( i ) CO 2 (ii) H 2 O & (iii) NH 3

14. Dipole moment: For a molecule to show infrared absorpsions it must possess a specific feature: an electric dipole moment of the molecule must change during the vibration . A dipole moment , µ is defined as, ‘the product of magnitude of charge on any one of the atoms and the distance between them’. Mathematically, dipole moment is given by µ = q . r or µ = q .d (Unit: C.m ) Where, q = Charge on any one of the atoms r (or d) = distance between the atoms. When these charge atoms vibrates, they shows change in dipole moment, & molecule absorb IR radiation.

15. Essential conditions for a compound to absorb IR radiation: The vibration frequency of the molecule must match (equal) with frequency of IR radiation. The dipole moment of molecule must change during vibration. What is mean by IR Active Compounds?? If the vibration transition in molecule is capable of change in dipole moment so the molecule is said to be IR active. e.g. I- Cl , C=O,CHCl 3 , H 2 O,etc. What is mean by IR Inactive Compounds?? If the vibration transition in molecule is not producing change in dipole moment so the molecule is said to be IR inactive. e.g. H 2 , Cl 2 , N 2 , O 2 , C=C, etc.

16. Most of the analytical applications are confined to the middle (mid) IR region because absorption of organic molecules is high in the middle (mid) IR region. Approximate range of IR region: Q.1) What is approximate range of mid IR region? (W-09, 1 Mark) Q.2) Give the approximate range of IR region in cm -1 . (S-10, 1 Mark) Q.3) Give approximate range of IR (mid) region. (W-12, 1 Mark)

17. Name the radiation source in IR spectroscopy Radiation source Incandescent lamp Nernst Glower Mercury Arc Globar source Q.1) Name the common source of IR radiation. (W-11, 1 Mark) Q.2) Name the radiation source in IR spectroscopy. (S-12 & S-14, 1 Mark) Q.3) Write at least one source of Infra red radiation. (W-12, 1 Mark) )

18. SOURCE of RADIATION: COMMONLY USED GLOBAR FILAMENT & NERNST GLOWER Globar - resistance rod of silicon carbide Globar - for longer wavelengths Nernst glower- a spindle of rare earth oxide ( thorium,zirconium,etc ) Nernst glower- for shorter wavelengths

19. What is a vibration in a molecule? Any change in shape of the molecule- stretching of bonds, bending of bonds, or internal rotation around bonds called vibration.

20. Vibration or Vibrational motion: Defination : The periodic displacement of bonded atom about it’s equilibrium position in the molecule along bond axis or perpendicular to bond axis is called as vibration or vibrational motion . =

21. Fundamental or Normal modes of Vibration: Defination : The vibration motion in which all atom of the molecule vibrate in phase simultaneously and with same frequency is called as normal or fundamental mode of vibration .

22. Types of vibrational modes: Q.1) Explain the various types of fundamental (normal) modes of vibration. Q.2) Define and explain: ( i ) Stretching vibration & (ii) Bending vibration. Q.3) Name the various type of vibration? Q.4) What is bending vibration? (W-09, 1 Mark) Q.5) Identify type of vibrations indicated in each of the following figures. (S-11, 2 Mark) Q.6) Explain various types of stretching vibrations with suitable example. (W-11, 3 Mark) Q.7) What are bending vibrations? Indicate in-plane and out-plane bending vibrations by taking suitable example. (W-13, 2 Mark) Q.8) Identify type of vibrations indicated in each of the following figures. (S-14, 2 Mark) Q.9) Illustrate with diagrams the different types of bending vibrations. (S-13 & W-17, 4 Mark) Q.10) What are stretching vibrations? (W-15, 1 Mark) Q.11) Describe different types of bending vibrations in IR spectroscopy. (W-15, 4 Mark) Q.12) Discuss the types of mode of vibrations in IR spectroscopy. (W-16, 4 Mark) Q.13) Describe types of vibrational modes in IR spectroscopy. (W-18, 4 Mark) Q.14) Explain the following terms with diagram: (S-19, 4 Mark) ( i ) Scissoring & (ii) Twisting

23. MOLECULAR VIBRATIONS Bending vibrations Stretching vibrations Vibration or oscillation along the line of bond Change in bond length Occurs at higher energy: 4000-1250 cm -1 a) Symmetrical stretching b) Asymmetrical stretching Vibration not along the line of bond Bond angle is altered Occurs at low energy: 1400-667 cm -1 In plane bending Out plane bending

24. Molecular vibration divided into stretching bending back & forth movement involves change in bond angles symmetrical asymmetrical scissoring rocking twisting wagging in-plane vibration out of plane vibration IR Vibrational Modes : involves change in bond length

25. Number of Vibrational Modes in a Molecule: Symmetric Stretching Asymmetric Stretching Scissoring Rocking Wagging Twisting

26. Stretching vibration: (bond length changes) (2 bonds increase or decrease in length simultaneously) Defination : The fundamental mode of vibration in which, Bond length ( internuclear distance between two atoms) changes (increases or decreases) periodically. Bond angle remain constant , is called as stretching vibration. The molecule containing N number of total atom has (N-1) number of stretching vibration. There are two types of stretching vibration.

27. STRETCHING VIBRATIONS Asymmetrical Stretching vibrations Symmetrical Stretching vibrations All bond stretched and compressed alternately in symmetrical manner. having permanent dipole moment is IR active having zero dipole moment is IR inactive One bond stretched while other bond is compressed and vice - versa. In this, one bond length is increased and other is decreased Asymmetrical stretching vibration is IR active

28. Example: Symmetrical stretching vibration in CH 2 group Example: A symmetrical stretching vibration in CH 2 group

29. a) Symmetrical stretching: 2 bonds increase or decrease in bond length simultaneously. H H C STRECHING VIBRATIONS

30. b) Asymmetrical stretching In this, one bond length is increased and other is decreased. H H C

31. Bending vibrations : (bond angle changes) 1 Scissoring: In this type, two atoms approach each other with respect to central atom. 2 Rocking : In this type, the movement of atoms takes place in the same direction with respect to central atom. 3 Wagging: In this type, two atoms move “up and down” the plane with respect to central atom 4 Twisting: In this type, one of the atom moves up the plane while the other moves down the plane with respect to central atom Types of Bending Vibration

32. a) In plane bending Scissoring: 2 atoms approach each other Bond angles are decrease H H C C

33. a) In plane bending Rocking: Movement of atoms take place in the same direction with respect to central atom. (Bonds swing in one direction in a plane) . H H C C

34. b) Out plane bending i . Wagging: 2 atoms move to one side of the plane. They move up and down the plane. H H C C

35. b) Out plane bending ii. Twisting: One atom moves above the plane and another atom moves below the plane. H H C C

36. Fundamental modes of vibration PRP

37. Fundamental modes of vibration: Type of molecule containing Total “ N ” atom Total degrees of freedom Translational degrees of freedom Rotational degrees of freedom Vibrational degrees of freedom Linear molecule 3 N 3 2 3N - 5 Non-linear molecule 3 N 3 3 3N - 6 IR

38. Fundamental modes of vibration: The number of degrees of freedom is equal to the sum of the coordinates necessary to locate all the atoms of the molecule in space. Three coordinates (X, Y and Z) are required the position of the atom P in space. Degree of freedom for the single atom P is equal to Three . Because three coordinates are required to locate the atom in space. IR

39. Fundamental modes of vibration: If a single atom restricted to movement in a plane , then translational degrees of freedom is Two , because only two coordinates (X, Y) would be required to fix the position of the atom in the plane . If there are two atoms in space , total translational degrees of freedom = 6 (2x3). Each atom would have three translational degrees of freedom because three coordinates would be required to fix the position of each atom. If there are N atoms in space, total translational degrees of freedom = 3 N (N x 3). Each atom would have three translational degrees of freedom because three coordinates would be required to fix the position of each atom. IR

40. Fundamental modes of vibration: Atoms have three degrees of freedom, all of which are translational. When atoms combine to form molecules, no degrees of freedom are lost. That is, the total number of degrees of freedom of the molecule will be 3N W here, N is the number of atoms in the molecule. The “ 3N ” degrees of freedom of the molecule will be made up of rotational, vibrational and translational degrees of freedom. IR

41. Fundamental modes of vibration: Rotational degrees of freedom result from the rotation of a molecule about an axis through the center of gravity. These rotations result in a degree of freedom only if the positions of the atoms in space change during the rotation. IR

42. Fundamental modes of vibration: For examples: ( i ) For Linear molecule (Rotation about two axes (X and Y) will result in a change in the position of the atoms ); Rotational degrees of freedom = 2 (ii) For Non-linear molecule (Rotation about all ( three ) axes ( X, Y and Z) will result in a change in the position of the atoms); Rotational degrees of freedom = 3 IR

43. Fundamental modes of vibration: All degrees of freedom not accounted for by translational and rotational are vibrational degrees of freedom. 3 N degrees of freedom = = translational + rotational + vibrational Therefore , Vibrational degrees freedom = = 3 N - (Translational + Rotational) IR

44. Fundamental modes of vibration: Type of molecule containing Total N atom Total degrees of freedom Translational degrees of freedom Rotational degrees of freedom Vibrational degrees of freedom Linear molecule 3 N 3 2 3N - 5 Non-linear molecule 3 N 3 3 3N - 6 IR

45. Q.1) Calculate fundamental modes of vibrations in each case of following molecules:(W-12, 3 Mark) ( i ) Ammonia, (ii) Cyclohexanol & (iii) Benzene. Q.2) What are the fundamental modes of vibration in water molecule? (W-13, 2 Mark) Q.3) Calculate the number of vibrational modes in H 2 O. Explain it’s IR spectrum. (S-13, 4 Mark) Q.4) Calculate the number of vibrational degrees of freedom in the following compounds:(S-14, 4 ) ( i ) CO 2 , (ii) H 2 O, (iii) H-C≡C-H & (iv) NH 3 Q.5) Calculate the vibrational degrees of freedom (fundamental modes of vibration) for following molecules in IR spectroscopy. (S-15, 4 Mark) ( i ) CO 2 & (ii) NH 3 Q.6) Calculate fundamental modes of vibrations in each of the following molecules: (S-16, 4 Mark) ( i ) NO (ii) CO 2 (iii) CH 4 & (iv) Benzene. Q.7) Calculate the number of fundamental modes of vibrations for the following molecules: (S-17, 4 Mark) ( i ) Water (H 2 O) (ii) Ammonia (NH 3 ) (iii) Carbon dioxide (CO 2 ) & (iv) Benzene (C 6 H 6 ). Q.8) Calculate the vibrational degrees of freedom for the following molecules in IR spectroscopy: ( i ) CO 2 (ii) NH 3 (iii) Benzene (C 6 H 6 ) & (iv) CH 4 (S-19, 4 Mark) Q.9) Water molecule shows _____ different modes of vibrations. (W-19, ½ Mark) (a) two (b) three (c) four (d) five Q.10) Calculate the number of vibrational degrees of freedom in the following compounds: ( i ) NH 3 & (ii) H 2 O (W-19, 4 Mark)

46. Fundamental modes of vibration: IR Sr. No. Molecule Total no. of atoms (N) Geometry of the molecule Applicable formula Fundamental modes of vibration (FMV) 1. NO 2 Linear 3N – 5 1 2. CO 2 3 Linear 3N – 5 4 3. H 2 O 3 Non-linear 3N – 6 3 4. NH 3 4 Non-linear 3N – 6 6 5. CH 4 5 Non-linear 3N – 6 9 6. C 6 H 6 12 Non-linear 3N – 6 30 7. H-C≡C-H 4 Linear 3N – 5 7 8. Cyclohexanol 19 Non-linear 3N – 6 51 9. CS 2

47. Fundamental modes of vibration: IR Sr. No. Molecule Total no. of atoms (N) Geometry of the molecule Applicable formula Fundamental modes of vibration (FMV) 1. NO 2. CO 2 3. H 2 O 4. NH 3 5. CH 4 6. C 6 H 6 7. H-C≡C-H 8. Cyclohexanol 9. CS 2

48. Most of the analytical applications are confined to the middle (mid) IR region because absorption of organic molecules is high in the middle (mid) IR region. Approximate range of IR region: Q.1) What is approximate range of mid IR region? (W-09, 1 Mark) Q.2) Give the approximate range of IR region in cm -1 . (S-10, 1 Mark) Q.3) Give approximate range of IR (mid) region. (W-12, 1 Mark)

49. Mid IR region or Spectral Range: 1 Functional group region: ( 2.5 to 7.7 µm i.e., 4000-1300 cm -1 ) The common functional group show IR absorption band in this region due to O-H, N-H, C=O and C-H stretching vibration of functional group. So, it is called as functional group region. 2 Finger print region : (7.7-11 µm i.e., 1300 to 909 cm -1 ) In this region each compound shows its unique (characteristics) absorption band. Similar to finger prints. This region is very useful for sample comparison. 3 Aromatic region: (11-15 µm i.e , 909-667 cm -1 ) The aromatic compound show absorption band in this region due to aromatic character. This region also useful in determination of substitution patterns on aromatic compounds such as ortho , meta, para substitution. IR Q.1) What is meant by finger print region in IR spectroscopy? (S-15 & W-16, 1 Mark) Q.2) Define the term: Finger print region. (S-18, 2 Mark) Q.3) Give the range of fingerprint region in IR Spectroscopy. (W-18, 1 Mark) 4000 – 667 cm-1

50. Mid IR region or Spectral Range: 1 Functional group region: ( 2.5 to 7.7 µm i.e., 4000-1300 cm -1 ) The common functional group show IR absorption band in this region due to O-H, N-H, C=O and C-H stretching vibration of functional group . So, it is called as functional group region. 2 Finger print region : (7.7-11 µm i.e., 1300 to 909 cm -1 ) In this region each compound shows its unique (characteristics) absorption band. Similar to finger prints. This region is very useful for sample comparison. 3 Aromatic region: (11-15 µm i.e , 909-667 cm -1 ) The aromatic compound show absorption band in this region due to aromatic character . This region also useful in determination of substitution patterns on aromatic compounds such as ortho , meta, para substitutio n . IR C-C C-O C-N Q.1) What is meant by finger print region in IR spectroscopy? (S-15 & W-16, 1 Mark) Q.2) Define the term: Finger print region. (S-18, 2 Mark) Q.3) Explain: Aromatic region in IR spectroscopy. (S-18, 2 Mark) Q.4) Give the range of fingerprint region in IR Spectroscopy. (W-18, 1 Mark) Q.5) The range of finger print region is __________. (S-19, ½ Mark)

51. Hydrogen bonding: IR

52. Hydrogen Bonding These three bonds all have; A strong permanent dipole A hydrogen atom An atom with lone pair electrons The three types of bonds which give molecules significant hydrogen bonding are; (i) N – H (ii) O – H (iii) F – H

53. Hydrogen bonding Examples

54. Intermolecular Forces O +  - H H O +  - H H Inter molecular Intra molecular

55. Special Notes Inter molecular hydrogen bond: Hydrogen bond formed between two molecules Intra molecular hydrogen bond: Hydrogen bond formed between two different atoms in the same molecule Intermolecular hydrogen bond is stronger than van der Waals’ forces

56. Intramolecular Hydrogen Bonding When hydrogen bonding exists within the molecule it is called intramolecular hydrogen bonding. In such type of hydrogen bonding two groups of the same molecule link through hydrogen bond, forming a stable five or six membered ring structure e.g., salicylaldehyde , o - chlorophenol , acetylacetone , ethylacetoacetate etc.

57. This intramolecular hydrogen bonding was first called chelation (after the Greek word " Chela " meaning, claw) because in the same molecule the formation of a ring hydrogen bonding is a pincer like action resembling the closing of a Crab's claw. Some more examples of intramolecular hydrogen bonding are :

58. Mid IR region or Spectral Range: Functional Group Region 4000 – 1300 cm-1 Finger Print Region 1300 - 909 cm-1 Aromatic Region 909 - 667 cm-1

59. Eight spectral region to identify the compound: S.No . Wave number , cm -1 Band causing absorption 1 3750 - 3000 O-H, N-H stretching 2 3300 - 2900 -C≡C-H, >C=C<H, Ar -H (C-H stretching) 3 3000 - 2700 -CH 3 , -CH 2 -, >C-H, -CHO ( C-H stretching) 4 2400 - 2100 C≡C, C≡N stretching 5 1900 - 1650 C=O (acids, aldehydes , ketones , amides, esters , anhydride, acid halides) stretching 6 1675 - 1500 >C=C< ( aliphatic and aromatic ), >C=N- stretch 7 1475 - 1300 ≡ C-H bending 8 1000 - 650 >C=C<H, Ar -H bending (Out of plane)

60. Identification of characteristic absorption bands S.No . Wave number , cm -1 Band causing absorption 1 3750 - 3000 O-H, N-H stretching 1.1 3700 - 3500 Free (non hydrogen bonded) O-H stretching 1.2 3500 Free phenol O-H stretching 1.3 3450 - 3200 Hydrogen bonded O-H stretching 1.4 3500 - 3300 Non-bonded amines (Free) N-H stretching 1.5 3500 - 3100 Bonded amine N-H stretching Note that: Primary amines shows two bands, 2 o amines & imines show only one band, and 3 o amines show no bands. 1.6 3500 - 3300 Amides also show N-H stretching 1.7 3000 - 2500 Carboxylic acid (-COOH)

61. Identification of characteristic absorption bands S.No . Wave number , cm -1 Band causing absorption 2 3300 - 2900 -C≡C-H, > C=C-H , Ar -H (C-H stretching) 2.1 3030 Ar-H stretch 2.2 3300 C≡C-H stretch 2.3 3040 - 3010 C=C-H stretch 3 3000 - 2700 -CH 3 , -CH 2 - , Ξ C-H , -CHO ( C-H stretching) 3.1 2960 & 2870 -CH 3 (two bands) 3.2 2930 & 2850 -CH 2 - (two bands) 3.3 2890 >C-H 3.4 2720 -CHO (C-H stretching)

62. Identification of characteristic absorption bands S.No . Wave number , cm -1 Band causing absorption 4 2400 - 2100 C≡C, C≡N stretching 4.1 2140 - 2100 H-C≡C-H 4.2 2260 - 2190 H-C≡C-R’ 4.3 No absorption R-C≡C-R (Symmetrical vibration causes no change in the dipole moment) 4.4 2260 - 2240 R-C≡N

63. Identification of characteristic absorption bands S.No . Wave number , cm -1 Band causing absorption 5 1900 - 1650 C=O (acids, aldehydes , ketones , amides, esters , anhydride , acid halides) stretching 5.1 1740 - 1720 Aldehydes i.e. R-CHO 5.2 1725 - 1705 Acids i.e., R-COOH 5.3 1725 - 1705 Ketones i.e., R-CO-R 5.4 1740 - 1710 Esters i.e., R-COO-R 5.5 1815 -1720 Acid halides i.e., R-COX 5.6 1850 – 1800 & 1780 - 1740 Anhydrides (Two bands separated by approximately 60 cm -1 ) 5.7 1700 - 1640 Amides i.e., R-CONH 2

64. Identification of characteristic absorption bands S.No . Wave number , cm -1 Band causing absorption 6 1675 - 1500 >C=C< ( aliphatic and aromatic ), >C=N- stretch 6.1 1680 - 1620 >C=C< (aliphatic) 6.2 1500 - 1400 >C=C< (aromatic) 6.3 1690 - 1640 >C=N- 6.4 1630 - 1575 -N=N- 7 1475 - 1300 >C-H bending

65. Identification of characteristic absorption bands: S.No . Wave number , cm -1 Band causing absorption 8 1000 - 650 >C=C<H, Ar -H bending (Out of plane) 8.1 990 & 910 RCH=CH 2 (C=C-H bending) 8.2 690 RCH=CRH (cis) 8.3 970 RCH=CRH (trans) 8.4 890 R 2 C=CH 2 8.5 840 -790 R 2 C=CHR Substituted Benzene: Type of Substitution 8.6 750 & 700 Mono -substituted aromatic (5 adjacent H) 8.7 750 Ortho aromatic (4 adjacent H) 8.8 810 - 780 Meta aromatic (3 adjacent H) 8.9 850 - 800 Para aromatic (2 adjacent H)

66. Selfie

67. Selfie

68. Problems on Interpretation of Infrared Spectra: Q.1) In which IR spectral region, the following compound be expected to absorb light? What bond gives rise to each absorption? IR S.No . Wave number, cm -1 Band causing absorption 1. 3000 - 2700 CH 3 , CH 2 and aldehyde proton ( C-H stretch ) 2. 2960 & 2870 -CH 3 (two bands) 3. 2930 & 2850 -CH 2 - (two bands) 4. 2720 (->C-H stretching ) aldehyde proton 5. 1740 - 1720 Aldehydes 6. 1475 - 1300 >C-H bending

69. Problems on Interpretation of Infrared Spectra: Q.2) In which IR spectral region, the following compound be expected to absorb light? What bond gives rise to each absorption? IR S.No . Wave number, cm -1 Band causing absorption 1. 3500 – 3300 Or 3500 – 3100 Or 3500 - 3300 Non-bonded amines (Free ) N-H stretching Bonded amines N-H stretching Amides also show N-H stretching 2. 2930 & 2850 -CH 2 - (two bands) 3. 1725 – 1705 Or 1700 - 1640 Ketones i.e., R-CO-R or Amides i.e., R-CONH 2 4. 1475 - 1300 >C-H bending

70. Problems on Interpretation of Infrared Spectra: Q.3) An IR spectra shows a strong absorption band at 1715 cm -1 , the functional group likely to be present is >C=O . (W-16, ½ Mark) (a) -CH=CH- (b) >C=O (c) –NH 2 (d) -NO 2 IR S.No . Wave number, cm -1 Band causing absorption 1. 1725 - 1705 Ketones , >C=O 1900 - 1650 C=O (acids, aldehydes , ketones , amides, esters , anhydride, acid halides) stretching

71. Problems on Interpretation of Infrared Spectra: Q.4) In which region of IR, absorption bands of stretching vibration occur for the following functional groups? ( W-16, 4 Mark ) ( i ) C=O (ii) = C-H (iii) – N-H (iv) C≡C IR S.No . Wave number, cm -1 Band causing absorption 1. 1900-1650 >C=O 2. 3040 - 3010 C =C-H 3. 3500 – 3300 Non-bonded 3500 – 3100 Hydrogen-bonded -N-H 4. 2400 - 2100 C≡C

72. Problems on Interpretation of Infrared Spectra: Q.5) In which region of IR, absorption bonds of stretching vibrations occur for the following functional groups? (S-17, 4 Mark) ( i ) -C=O (ii) -N-H (iii) -C-H (iv) –C=C- IR S.No . Wave number, cm -1 Band causing absorption 1. >C=O 2. -N-H 3. -C-H 4. –C=C-

73. Problems on Interpretation of Infrared Spectra: Q.6) Write the appropriate frequency range in cm -1 for the following functional groups: (W-19, 4 Mark) ( i ) –N-H (ii) =CH (iii) -C≡N (iv) >C=O IR S.No . Wave number, cm -1 Band causing absorption 1. -N-H 2. =C-H 3. -C-≡H 4. >C=O

74. Structure of H 2 O molecule: Q.1) Explain different types of stretching vibrations in tri-atomic molecule. ( S-13, 3 Mark) Q.2) Calculate the number of vibrational modes in H 2 O. Explain it’s IR spectrum. ( S-13, 4 Mark) Q.3) Explain the structure of water molecule on the basis of IR spectroscopy. (S-15 & W-19 , 4 Mark) Q.4) Describe IR spectrum of H 2 O molecule. ( W-15, 4 Mark ) Q.5) What tupes of vibrational modes are expected in H 2 O molecule? Discuss its spectrum. (W-18, 4 Mark)

75. Structure of Organic Compounds: Structure of H 2 O molecule: The total fundamental mode of vibration of non-linear , H 2 O molecule are (3N-6) = = [(3 x 3) – 6] = 3 Out of three fundamental mode of vibration; The molecule containing N number of total atom has (N-1) number of stretching vibration. i.e., (3-1) = 2 vibration are stretching vibration and one vibration is bending vibration . Out of two stretching vibration; one is symmetrical stretching vibration and another is asymmetrical stretching vibration. IR

76. Structure of H 2 O molecule: The modes of vibrations of water molecule are represented as below: H 2 O molecule is polar having certain dipole moment . It ’ s dipole moment changes in each vibration modes . So, all vibrations are IR active .

77. Structure of H 2 O molecule: The modes of vibrations of water molecule are represented as below: So, IR spectrum of water molecule shows three absorption bands . One for bending vibration at 1600 cm -1 , second band due to symmetrical stretching at 3650 cm -1 and third band due to asymmetrical stretching vibration at 3760 cm -1 . 1600 cm -1 3650 cm -1 3760 cm -1

78. Structure of CO 2 molecule: Q.1) Discuss IR spectrum of carbon dioxide molecule. ( W-09, 3 Mark) Q.2) How many modes of vibrations the CO 2 molecule will have? Which of them will be IR active? (S-10 & W-14, 2-4 Mark) Q.3) Give reasons : Carbon dioxide molecule shows only two IR bands in it’s IR spectrum inspite of it’s four fundamental modes of vibrations.. (S-11, S-12 & W-13, 2 Mark) Q.4) Describe IR spectrum of CO 2 molecule. ( W-14, 4 Mark ) Q.5) Calculate the number of vibrational modes in CO 2 . Discuss it on the basis of IR spectrum. (S-18, 4 Mark) Q.6) Explain the structure of CO 2 molecule on the basis of IR spectroscopy. (W-19, 2 Mark)

79. Structure of CO 2 molecule: The modes of vibrations of CO 2 molecule are represented as below: Structure of CO 2 molecule: (Linear molecule) The Carbon dioxide (CO 2 ) is a linear molecule. CO 2 molecule has fundamental modes of vibration =( 3N-5 ) = [(3 x 3) - 5] = 4 . Out of four fundamental mode of vibration; The molecule containing N number of total atom has (N-1) number of stretching vibration. i.e., ( 3-1 ) = 2 vibration are stretching vibration and two are bending vibration . Out of two stretching vibration; one is symmetrical stretching vibration and another is asymmetrical stretching vibration.

80. Structure of CO 2 molecule: The modes of vibrations of CO 2 molecule are represented as below: The remaining three vibrations are IR active because that produce changes in dipole moment of CO 2 molecule. The two bending vibrations have same energy, because they are degenerate (doubly). So, IR spectrum of CO 2 molecule shows two absorption bands.

81. Structure of CO 2 molecule: The modes of vibrations of CO 2 molecule are represented as below: So, IR spectrum of CO 2 molecule shows two absorption bands . 2350 cm -1 Bending 667cm -1 Asymmetrical

82. Problems: Q.1) Which of the following vibrational modes are IR active or inactive? (S-18, 4 Mark) ( i ) Symmetric CO 2 stretching (ii) Antisymmetric CO 2 stretching (iii) Symmetric H 2 O stretching (iv) H 2 O bending

83. Choice of solvent in IR spectra: The essential condition or requirements (properties) for a good solvent in IR spectroscopy are as given below: The solvent should be transparent to IR radiation ( i.e., It should not be absorbed in IR region). It should not interact with the solute. It should be pure and the solute should be in it. It should be less polar or non polar. Examples: CS 2 and CCl 4 are good solvents in IR spectroscopy. IR

84. Distinguish following compounds on the basis of IR spectroscopy: Q.1) How would the infrared spectra of the following compounds differ: (W-09, 3 Mark) IR

85. Distinguish following compounds on the basis of IR spectroscopy: IR S.NO. Band causing absorption, in cm -1 Compound A Compound B 1. 3040 - 3010 C=C-H stretch 2. 2960 & 2870 -CH 3 (two bands) -CH 3 (two bands) 3. 2930 & 2850 -CH 2 - (two bands) 4. 1725 - 1705 Ketones i.e., R-CO-R Ketones i.e., R-CO-R 5. 1680 - 1620 >C=C< (aliphatic) 6. 1475 - 1300 >C-H bending >C-H bending

86. Distinguish following compounds on the basis of IR spectroscopy: IR S.NO. Band causing absorption, in cm -1 Compound A Compound B 1. 2. 3. 4. 5. 6.

87. Distinguish following compounds on the basis of IR spectroscopy: IR S.NO. Band causing absorption, in cm -1 Compound A Compound B 1. 2. 3. 4. 5. 6.

88. Distinguish following compounds on the basis of IR spectroscopy: IR Q.12) Differentiate the following pairs of compounds on the basis of IR spectroscopy. (S-17 & S-19, 4 Mark) ( i ) Acetone & Ethanol (ii) Acetamide & Acetic acid S.NO. Band causing absorption, in cm -1 Compound A Acetone Compound B Ethanol CH 3 CH 2 OH 1. 2. 3. 4. 5. 6.

89. Distinguish following compounds on the basis of IR spectroscopy: IR Q.12) Differentiate the following pairs of compounds on the basis of IR spectroscopy. (S-17 & S-19, 4 Mark) (ii) Acetamide & Acetic acid S.NO. Band causing absorption, in cm -1 Compound A Acetamide CH 3 CONH 2 Compound B Acetic acid CH 3 COOH 1. 2. 3. 4. 5. 6.

90. Distinguish following compounds on the basis of IR spectroscopy: IR Q.13) Differentiate the following pairs of compounds on the basis of IR spectroscopy. (W-17, 4 Mark) ( i ) Acetaldehyde & Acetone S.NO. Band causing absorption, in cm -1 Compound A Acetaldehyde CH 3 CHO Compound B Acetone CH 3 COCH 3 1. 2. 3. 4. 5. 6.

91. Distinguish following compounds on the basis of IR spectroscopy: IR Q.13) Differentiate the following pairs of compounds on the basis of IR spectroscopy. (W-17, 4 Mark) (ii) Acetamide & Acetic acid S.NO. Band causing absorption, in cm -1 Compound A Acetamide CH 3 CONH 2 Compound B Acetic Acid CH 3 COOH 1. 2. 3. 4. 5. 6.

92. Distinguish following compounds on the basis of IR spectroscopy: IR Q.14) Differentiate between following pairs on the basis of IR spectroscopy. (W-18, 4 Mark) S.NO. Band causing absorption, in cm -1 Compound A Ethyl bromide CH 3 CH 2 Br Compound B Ethanol CH 3 CH 2 OH 1. 2. 3. 4. 5. 6.

93. Distinguish following compounds on the basis of IR spectroscopy: IR Q.14) Differentiate between following pairs on the basis of IR spectroscopy. (W-18, 4 Mark) S.NO. Band causing absorption, in cm -1 Compound A Acetone CH 3 COCH 3 Compound B Acetamide CH 3 CONH 2 1. 2. 3. 4. 5. 6.

94. Distinguish following compounds on the basis of IR spectroscopy: IR Q.10) How will you distinguish following compounds on the basis of IR spectroscopy? (W-14, 4 Mark) S.NO. Band causing absorption, in cm -1 Compound A Ethanol CH 3 CH 2 OH Compound B Dimethyl ether CH 3 OCH 3 1. 2. 3. 4. 5. 6.

95. Distinguish following compounds on the basis of IR spectroscopy: IR Q.10) How will you distinguish following compounds on the basis of IR spectroscopy? (W-14, 4 Mark) S.NO. Band causing absorption, in cm -1 Compound A Acetone CH 3 COCH 3 Compound B Acetic acid CH 3 COOH 1. 2. 3. 4. 5. 6.

96. Distinguish following compounds on the basis of IR spectroscopy: IR Q.9) Distinguish the following pairs of compounds on the basis of their IR spectra? (W-14, 4 Mark) S.NO. Band causing absorption, in cm -1 Compound A Compound B 1. 2. 3. 4. 5. 6.

97. Distinguish following compounds on the basis of IR spectroscopy: IR Q.8) How would you differentiate the following compounds on the basis of IR spectra? (W-13, 3 Mark) Q.7) How will you distinguish following compounds on the basis of IR spectroscopy? (W-13, 3 Mark )

98. Distinguish following compounds on the basis of IR spectroscopy: IR Q.6) Using IR spectrum of individual compound, how will you distinguish following pair of compounds? (W-12, 3 Mark) ( i ) Acetaldehyde and acetone (ii) Methyl amine and dimethyl amine Q.5) How the I.R. spectra of the following compounds differ? (S-12, 3 Mark)

99. Distinguish following compounds on the basis of IR spectroscopy: IR Q.4) On the basis of infrared spectra, how will you differentiate the following compounds? (W-11 & S-14, 2 Mark) Q.3) How the following compounds can be differentiated on the basis of infrared spectroscopy? (S-10 & S-14, 3 Mark)

100. Distinguish following compounds on the basis of IR spectroscopy: IR Q.2) How the infrared spectra of the following compounds differ? (S-10 & W-14, 3 Mark)

101. THE END... NEXT TOPIC University Examination Summer-2019

102. I R R I

103. A small truth to make life 100%

104. If A B C D E F G H I J K L M N O P Q R S T U V W X Y Z is equal to 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

105. Hard Work H+A+R+D+W+O+R+K 8+1+18+4+23+15+18+11 = 98% Knowledge K+N+O+W+L+E+D+G+E 11+14+15+23+12+5+4+7+5 = 96%

106. Love L+O+V+E 12+15+22+5 = 54% Luck L+U+C+K 12+21+3+11 = 47% (why is it that most of us think this is the most important ???)

107. Then what makes 100%? Is it Money ? ... NO ! ! ! M+O+N+E+Y 13+15+14+5+25 = 72% Is it Leadership ? ... NO ! ! ! L+E+A+D+E+R+S+H+I+P 12+5+1+4+5+18+19+9+16 = 89%

108. Every problem has a solution, but only if we change our attitude. To go to the top and achieve 100% We really need to go further...

109. ATTITUDE A+T+T+I+T+U+D+E 1+20+20+9+20+21+4+5 = 100% It is OUR ATTITUDE towards Life and Work that makes OUR Life = 100%

110. ATTITUDE IS EVERYTHING Change Your Attitude … … and You Change Your Life ! ! !

111. Thank You !

112. THE END... NEXT TOPIC University Examination Summer-2019

113. I R R I