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IR Spectroscopy.pptx
- 1. IR SPECTROSCOPY Muhammad Naveed MSChemistry
- 2. Definitio n Infrared Spectroscopy isthe analysis of infrared light interacting with a molecule. This can be analyzed in three ways by measuring absorption, emission and reflection.
- 3. Main concern It's main functionfor the determination of Functional group. It give structural information. Wavenumber. It is the number of waves per unit distance. Typically, it is measured using cm-1 and is given by- ν ¯ = 1/ λ
- 4. Which molecules show response to IR Thathave dipole movement That are Asymmetric No response from symmetric. Every bond have different IR spectrum
- 5. Modes of vibrational motion Bending lowf • Scissoring Relative High f • Wagging low f • Rocking low f • Twisting low f Stretching high f Symmetric and asymmetric
- 6. Bending vibration A changein the angle occurring between two bonds is known as a bending vibration. Stretching vibration When there is a continuous change in the interatomic distance along the axis of the bond between two atoms,
- 7. In symmetric stretching,two or more bonds vibrate in and out together. In asymmetric stretching, some bonds are getting shorter at the same time as others are getting longer When there is a continuous change in the interatomic distance along the axis of the bond between two atoms, this process is known as a stretching vibration. A change in the angle occurring between two bonds is known as a bending vibration. Four bending vibrations exist namely, wagging, twisting, rocking and scissoring.
- 9. Combination band andDifference bands When two vibrational frequencies in a molecules couple to give rise to a vibration of a new frequency within the molecule and when such a vibration is IR active.. Not all possible combination occurs. When fundamental vibration couple with overtone or combination band is called Fermi resonance. • The observed frequency is from the difference between the two interacting bands • It's similar to combination band
- 11. IR SPECTRA COMPLICATIONS IRspectra become complicated because of Weak overtone Combination Difference band Any type of physical vibration cause overtone.
- 12. Why IR bandis broad The reason behind this is that infrared spectrum is rotational coupling Weak bond vibration at low f Strong bond vibration at high f Hybridization effect SP>SP²>SP³ Resonance effect. Infrared absorption at low frequency As ring size decrease absorption frequency increase because it remove the p character
- 13. IR spectrometer The instrumentthat determine the absorption spectrum for a compound Dispersive Fourier transform FT IR Spectrometer is faster than Dispersive instrument
- 14. Dispersive IR spectrometer Beamof IR produce from hot wire. Then decided two beams of equal intensity using mirrors. Sample placed in one beam other beam as reference beam. Then passed through monochromator.which disperses into a continuous spectrum. Monochromator has beam chopper that pass beam to a diffraction grating. It varies the frequencies of radiation reaching the thermocouple detector. The detector sence the ratio between the intensities of the reference and sample beam.Dispersive instrument record a spectrum in the frequency domain.
- 15. Preparation of sample Sampleplaced in cell. Cells made up of ionic substance. Sodium chloride plates mostly used due to low cast. Pottasium bromide plates are expensive. For liquid Drop of liquid OC is placed on plates, these plates squeezed gently a thin liquid film forms between them. The spectrum obtained is neat spectrum because no solvent used.. Organic compound must be free of water
- 16. For solid Make pelletswith pottasium bromide Or Nujol mull. Grinding the compound with mineral oil. Disadvantage is that mineral oil bands present in analyzing compound.
- 17. Because each functionalgroup absorb specific radiation. We can determine easily which functional group are present in given compound. For understanding that we need to remember all these values given in table.
- 19. Some key pointskeep in mind IR sort start from 4000-400cm Different functional groups are Aklane, Alkene, Alkyne, Alchol, Carboxlic acid, Aldehyde, Ketone, Amines, Amide, Imines, Oxime, Nitriles, Sulfoxide, Alleged, Ketenes, Isocyantes Aryl Halide Ester, Ether, and acid anhydried. IR spectra obtained for different groups .
- 20. Alkane C-C not useful C-H3000-2840 CH2 1465 CH3 1375 CH2 more than 4 720
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- 33. From the followinggiven spectra we see that If peaks above 3000 it's may be Alkene , Alkyne, Alchol, Anime, Amide,, carboxlic acid. If it's below 3000 it may be Alkane, Aldehyde If it's below 2500 it may be Thiols, Phosphine, Nitriles, Alkyne Ether don't have O-H peak
- 34. Carbonyl group detection Carbonylgroup peaks present between 1675-1810 Amide 1675 Carboxlic acid 1710 Ketone 1715 Aldehyde 1725 Ester 1735 Acid chloride 1800 Anhydride have two peaks At 1760 and 1810 Why Anhydride have two peaks?
- 35. Now we aremoving to solve some spectra Remember peak values is always important but it's not necessary to see all peaks.
- 37. What you understandfrom this given spectra Is it Alkane, Alkene, Ether Aldehyde , or carboxlic acid or amines?
- 39. We absorbed threepeaks One is below 3000 Other two are below 1500 Now recall the values It's not Carboxlic acid, Aldehyde or ketone because carbonyl peak is absent. It's also not Alkene or Alkyne. Nitriles peak is also absent So it's spectrum of Alkane
- 41. Is it Aldehyde, Ketone,Alken, Alkane, Alchole, Nitriles Ester or Ether
- 43. Peaks shown areat below 3000 Below 1500 Above 1000 Alkenes Alkynes peaks absent Carbonyl group peakis also absent Ester peak is also absent Nitriles peak is also absent So the given compound is ethers
- 45. What you sayabout this spectra. Is it Ketone, Ether, Amide, Nitriles acid Anhydried, or Sulphide .
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