AYUSHI (IR SPECTROSCOPY)2221940018new.pptx

IR spectroscopy

1. INFRARED ABSORPTION SPECTROSCOPY SHRI GURU GOBIND SINGH JI GOVERNMENT COLLEGE PAONTA SAHIB AYUSHI CHOUDHARY

2. Introduction Basic Theory of IR Spectroscopy: Molecular Vibrations Principle Number of fundamental Vibrations Factors influencing vibrational Frequency Characteristic Absorption regions of Various Bonds Spectral features of classes of Compounds Applications Numerical Problems Contents:

3. Introduction IR Spectroscopy is the study of absorption of IR radiation which results in vibrational transistions Mainly used in structural elucidation to determine the functional group IR region in present between visible and microwave region in EM spectrum IR or Em Spectrum is divided into 3 regions

4. Basic Theory of IR Spectroscopy: Molecular Vibrations STRETCHING VIBRATION BENDING VIBRATION

5. In simple diatomic molecule A-B the only vibrations which occur in a rhythmic compression and extension along the A-B bond. This type of periodic back and forth movement of the atoms along the covalent bond axis which can be compared to that of a coiled spring is known as stretching vibration or bond stretching. STRETCHING VIBRATIONS OR BOND STRETCHING

6. TYPES OF STRETCHING VIBRATION Symmetric stretching : It is the stretching vibration in which the movement of the atoms with respect to a central atom in the molecule is in the same direction. Asymmetric stretching : It is the stretching vibration in which one atom moves towards the central atom while the other moves away from it.

7. Bending vibrations can be defined as periodic changes in bond angles between bonds formed by two atoms. BENDING VIBRATION

8. Scissoring . This is an in plane bending vibrations, in which the two atoms move towards each other. Rocking. This is an in plane bending vibration in which the two atoms move in the same directions. Wagging . This is an out of plane bending vibration in which the two atoms more simultaneously either above or below the plane with respect to the central atoms, Twisting . This is an out of plane bending vibration in which one atoms move above the plane while the other moves down the plane with respect to the central atom . NOTE: SCISSORING AND ROCKING REPRESENT IN PLANE BENDING WHILE WAGGING AND TWISTING REPRESENT OUT OG PLANE BENDING T ypes of bending vibrations.

9. Principle

10. Number of fundamental Vibrations Absorption of infrared radiation having energy equal to the difference between Two vibrational energy level will lead to a vibrational transistion . In the Transition from ground to the first excited state light is absorbed strongly and as a result Intense bands called fundamental absorption bands are produced. The number of fundamental absorption bands exhibited by a molecule is related to the Fundamental ways of vibrational modes availab le to the molecule Number of fundamental Number of fundamental Absorption bands = vibrational modes

11. Since a diatomic molecule has only one fundamental vibrational mode,it give rise to one fundamental absorption bands . However , polyatomic moldecule can have more than one kind of vibrational movement , they exhibit more than one fundamental absorption band NON- LINEAR POLYATOMIC MOLECULE Non linear polyatomic molecule has 3n -6 modes of fundamental vibrations where n is the number of atom present in the molecule. For example water (n=3) has three vibrational modes.each of which can have absorption band.

12. Linear polyatomic molecule: A linear polyatomic molecule having n atoms can have 3n-5 fundamental vibrations. For example (n=3) has four vibrational modes. In case of large molecule the number of fundamental vibration goes on increasing very rapidly with the increase in the value of n. But it has been observed that the number of fundamental bonds actually obtained is usually less than that expected from the theorectical number of fundamental vibrations. This may be due to the following reason.

13. Some of the fundamental vibration may be very weak and therefore may not be recorded as bands. Some of the fundamental vibration may be very close and as a result overlapping of bands may take place. Some of the fundamental vibrations may involve energy changes which fall outside of usual IR range. It is also interesting to note that sometimes additional bands such as combination bands , difference bands and overtones may also be exhibited .

14. Depending upon the types of bonds present in a molecule various wavelength are found to be absorbed in the infrared spectrum of the molecule giving rise to absorption band . The position of absorption band depend upon the nature of the bond or group. Absorption band depends upon characteristics of a particular bonds or group indicates the presence of the bond or group in the molecule. Infrared Spectrum can be usually divided into two regions of the purpose of it’s interpretions . Functional Group Regions: absorption bands from 5.5 to 8 um are associated with the change in the vibrational states of the various bonds. Accordingly this region is the characteristic of the types of the bonds present and may be termed as functional group region of the infrared spectrum. POSITION AND INTENSITY OF BANDS

15. Finger print region : Absorption band in the region 8um are generally associated with complex vibrational and rotational energy changes of the molecule as a whole. This region is very useful in establishing whether changes of the molecule as a whole. This region is very useful for establishing whether two compounds are identical or different from each other. It is therefore, referred to as finger print region . There are so many absorption bands in the finger print region that no compound, however closely related can have identical IR spectrum in all details including intensity

16. The intensity of an absorption band depends on the change in the dipole moment of the bond associated with the vibration. The greater the change in dipole moment, the more intense is the absorption. The dipole moment of a bond is equal to the magnitude of the charge on one of the bonded atoms, multiplied by the distance between the two charges. If we have two different atoms, there will be an electro-negativity difference and larger the electro-negativity difference, the more intense the absorption will be. The stretching vibration of an O-H bond will be associated with a greater change in dipole moment than that of an N-H bond because O-H bond is more polar. Consequently, the stretching vibration of the O-H bond will be more intense. Likewise, the stretching vibration of an N-H bond is more intense than that of a C-H bond because the N-H bond is more polar. The Intensity of Absorption bands

17. The intensity of an absorption band also depends on the number of bonds responsible for the absorption. For example, if you compare the IR spectra of methane to that of octane, the octane molecule will have a much more intense C-H peak because it has many more C-H bonds than methane. The concentration of the sample used to obtain an IR spectrum also affects the intensity of the absorption bands. Concentrated samples have greater numbers of absorbing molecules and, therefore, more intense absorption bands

18. Factors influencing vibrational Frequency

19. Characteristic Absorption regions of Various Bonds

20. Spectral features of classes of Compounds HYDROCARBON IR Spectra of Alkanes

21. Infrared Spectra of Alkenes

22. IR SPECTRA OF ALCOHOLS

23. IR SPECTRA OF ALDEHYDES AND KETONES

24. Applications Infrared spectroscopy is widely used in industry as well as in research. It is a simple and reliable technique for measurement, quality control and dynamic measurement. It is also employed in forensic analysis in civil and criminal analysis. 1. Identification of functional group and structure elucidation Entire IR region is divided into group frequency region and fingerprint region. Range of group frequency is 4000-1500 cm -1  while that of finger print region is 1500-400 cm -1 . In group frequency region, the peaks corresponding to different functional groups can be observed. According to corresponding peaks, functional group can be determined. Each atom of the molecule is connected by bond and each bond requires different IR region so characteristic peaks are observed. This region of IR spectrum is called as finger print region of the molecule. It can be determined by characteristic peaks.

25. 2. Identification of substances IR spectroscopy is used to establish whether a given sample of an organic substance is identical with another or not. This is because large number of absorption bands is observed in the IR spectra of organic molecules and the probability that any two compounds will produce identical spectra is almost zero. So if two compounds have identical IR spectra then both of them must be samples of the same substances. IR spectra of two enatiomeric compound are identical. So IR spectroscopy fails to distinguish between enantiomers . For example, an IR spectrum of benzaldehyde is observed as follows. C-H stretching of aromatic ring-                        3080 cm -1 C-H stretching of aldehyde -                              2860 cm -1  and 2775 cm -1 C=O stretching of an aromatic aldehyde -         1700 cm -1 C=C stretching of an aromatic ring-                  1595 cm -1 C-H bending-                                                     745 cm -1  and 685 cm -1

26. 3. Studying the progress of the reaction Progress of chemical reaction can be determined by examining the small portion of the reaction mixure withdrawn from time to time. The rate of disappearance of a characteristic absorption band of the reactant group and/or the rate of appearance of the characteristic absorption band of the product group due to formation of product is observed. 4. Detection of impurities IR spectrum of the test sample to be determined is compared with the standard compound. If any additional peaks are observed in the IR spectrum, then it is due to impurities present in the compound.

27. Numerical Problems

28. The approximate position of the characteristic absorption bands of the carbonyl group in the given compound are as follows. CH3CHO 1740cm-1 CH3CH2COCH3 1715cm-1 CH3COCH2=CH2 1675cm-1 C6H5CHO 1700cm-1 Q. Give the approximate position of characteristic absorption bands of the carbonyl group in the IR spectra of the following compounds . CH3CHO, CH3COCH2CH3, CH3COCH=CH2,C6H5-CHO

29. The structural features responsible for each of the given peaks in the IR spectra of ethyl acetate, CH3COOCH2-CH3 are as follows 3002 cm-1 C-H stretching of CH3/CH2-CH3 1742 cm-1 C=O stretcting 1240 cm-1 C-O stretching Q The IR spectra of ethyl acetate shows three important peaks at 3002cm-1 , 1742cm-1 , and 1240cm-1 . Attribute these peaks to the following structural features of the molecules. CH3/CH2-CH3, C=O, C-O

30. CH3-CH2-CO-CH3 and CH2=CH-O-CH3 can be distinguish from each other on the basis of following characteristic absorption bands in their respective IR spectra. 1) A strong absorption band at about 1715 cm-1 due to c=o stretching in the IR spectrum of CH3-CH2-CO-CH3, No such band will be present in the IR spectrum of CH2=CH-O-CH2 There will be band at about 3015 cm-1 due to C-H stretching of CH2=CH-group and another band at about 1100 cm-1 due to C-O strecting in the IR spectrum of CH2=CH-O-CH3. There will be no such band in the spectrum of CH3-CH2-CO-CH3. 2) Both CH3-C-CO-H and CH3-CO-CH3 will be show a C=O strecting band in their respective IR spectrum at about 1700cm-1 -1750cm-1 . However the IR spectrum of CH3COOH will have s characteristic broad band in the range if 2500-3000 cm-1 due to hydrogen bonding O-H group: there being no such band in the spectrum of CH3COCH3 . Thus this band can help in distinguish between the two compounds. How will you distinguish between the two compounds in each of the following pairs on the basis of IR spectroscopy CH3CH2COCH3, CH2=CH-O-CH3, CH3COOH, CH3COCH3

31. C6H5CH3CH3 Absorption band at 1602, 1580 and 1460 cm-1 are due to C=C stretching of aromatic ring. The band at 3016 cm-1 is due to C-H stretching of Ar -H bond Bands at 705 cm-1 and 790 cm-1 are due to C-H bending of m- disubstitued benzene. . Q.1 A hydrogen having the molecular formula C8H10 exhibits the following absorption bands in its infrared spectrum 3016 cm-1, 1602 cm-1, 1580 cm-1 , 1460 cm-1, 705 cm-1, 790 cm-1. What is the possible structure of the hydrocarbon.

32. Only CH3-CO-OH show absorption band due to O-H stretching at 3000-2500 cm-1 Only (CH3)2CHNH2 Show absorption due to N-H strecting at about 3500cm-1. Q Which peak in the IR spectrum could be used to distinguish between the two compounds? CH3COOH, CH3COOC2H5 (CH3)3N and (CH3)2CHNH2

33. Q.How will you distinguish between CH3CH2CH2NH2 and (CH3)3N show characterstics . CH3CH2CH2NH2 show characterstics N-H stretching At 3400 cm-1 and N-H bending bond at 1600 cm-1 These bands are absent in the spectrum of (CH3)3N

34. Percentage of hydrogen= 14.3 percentage of carbon= 100-14.3= 85.7 According to this data the empirical formula of the compound work out to be CH2 Since the compound contain 4 carbon atom , it s molecular formula should be C4H8 Because the compound is a straight chain hydrocarbon it can be the either 1-butene (CH3-CH2-CH=CH2) or 2-butene (CH3-CH=CH-CH3) The absorption band at 960 cm-1 can be only due to C-H out of plane bending for a disubstituted alkene having trans configuration . Therefore the given hydrocarbon is trans-2 – butene . That is CH3 Q. A 4-carbon straight chain hydrocarbon having 14.3 per cent hydrogen exhibits the following absorption bands. 1) 3015 cm-1, 960 cm-1, 1680 cm-1 C=C H CH3 H

35. The absorption band at 3015 cm-1 Is due to C-H stretching while the Band at 1680 cm-1 is for C=C Stretching of non conjugated alkanes .

36. Q The infrared spectrum of methyl salicylate, shows the following peaks. 3300, 1700, 1540, 2590, 2990 cm-1 Which of these peaks represents which of the following structure.? CH3, C=O, OH group on the ring, aromatic ring. 3300 cm-1 (hydrogen bonded O-H group 1700 cm-1 (C=O) stretching 1590,1540 cm-1 (aromatic ring) 2990 cm-1 ( for C-H for methyl group )

37. A compund having the molecular formula C8H8O showing peakat 1685cm-1 Which of the following is likely structure of the compound C6H5COCH3, C6H5CH2CHO, C6H5OCH=CH2. C6H5COCH3 REASON : In C6H5COCH3 carbonyl group is in Conjugation with benzene ring ,Therefore it show the Absorption peak due to C=O stretching at lower Frequency.

38. There can be five isomers having the molecular formula. C3H6O: CH3CO CH3CH2CHO CH2=CH2-CH2OH CH2=CH-OCH3 1) Absorption peak at 1710 cm-1 can be due to carbonyl group of CH3COCH3( for CH3-CH2-CHO, the carbonyl absorption peak is near 1740 cm-1 . The remaining three isomers do not show any absorption in the region )Hence X is CH3COCH3. 2) Absorption peak at 3300 cm-1 is due to hydrogen bonded O-H group which rules out the possibility of CH2=CH-OCH3. The peak at 1640 cm-1 is due to C=C stretching and cannot be accounted for by Q. Two isomers X and Y having the moleculat formula C3H6O exhibit the following peaks in the infrared spectrum (X) 1710 cm-1, (Y) 3300 cm-1, , 1640 cm-1. Write structures of X and Y on the basis. CHOH CH2

39. H2C Only CH2=CH=CH2OH can account for both the peaks at 3300 cm-1 and 1640 cm-1 hence Y is CH2=CH-CH2OH CH2 CH2OH

40. ethyl acetate 1680 cm-1 acetic anhydride 1750 cm-1 acetamide 1800 cm-1 Q Match the carbonyl absorption bands for the following compounds.

41. For methyl acetylene 3300 cm-1 ( C-H stretching ) and 2100 cm-1 ( C For ethyl alcohol 3200 cm-1 ( hydrogen bond O-H stretching ), 2850-2960 cm-1 (C-H stretching ) 1000-1300 cm-1 for ( C-C stretching) Q Indicate the expected IR peak for the following compounds Methyl acetylene, CH3-CCH Ethyl alcohol C2H5OH C )

42. Ans Q Deduce the structure of an aromatic compound C8H8O which has an infrared absorption maximum at 1690 cm-1 but no break near 3300 cm-1 c o CH3

43. CH3CH2OCH2CH3 CH3OCH2CH2CH3 CH3O-CH-CH3 Q, A compound having the molecular formula C4H10O show no infrared bands above 1500 cm-1 except the sharp bend at about 2900 cm-1 . What possible strecture can it have. CH3

44. o o CH3-CO-CH3 1725 cm-1 1670 cm-1 1710 cm-1

45. . In an infrared (IR) spectrum, which of the following functional groups has the highest frequency? a.) Ketone b.) Aldehyde c.) Ester d.) Alcohol Correct Answer– (d.) Alcohol.

46. A carbonyl group will cause a sharp dip at about ____ cm–1 a.) 1700 b.) 2800 c.) 3400 d.) 1200 Correct Answer– (a.) 1700

47. https://youtu.be/d8eKl7I8XI0?si=Tnc6y3cp_szCQvrc https://youtu.be/nkLCXDcmFLA?si=Xdx8qziTEYzlbDWO https://youtu.be/pkYChfNDMz8?si=9b5qUCDX5M7jF_Dy https://youtu.be/Bxo_DQpFh1c?si=hFYyh_tJO4qBvTSi https://youtu.be/6xSSG330Awc?si=BPAGRrinwe_bv5CA Bibliography