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1 Lecture 2 Advanced Pharmaceutical Analysis IR spectroscopy Dr. Baraa Ramzi
2 Infrared Spectroscopy It is a powerful tool for identifying pure organic and inorganic compounds. Every molecular compound has a unique infrared absorption spectrum (fingerprint). Infrared spectroscopy is a less satisfactory for quantitative analyses than UV, because it has lower sensitivity, deviation from Beer s law and less precision.
3 Infrared Spectroscopy The unit used in IR spectra is called wavenumber (cm-1). The range of IR that is usually used (mid infrared) is cm-1. The area between cm-1 contains functional groups. The area between cm-1 is called fingerprint region.
4 Infrared spectrum for butyraldehyde
5 Fingerprint Region The region to the right-hand side of the spectra (from about 1500 to 500 cm-1) usually contains a very complicated series of absorptions. These are mainly due to all manner of bending vibrations within the molecule. This is called the fingerprint region.
6 Regions of IR (Mid IR)
7 Fingerprint Region
8 Infrared Spectroscopy The energy of infrared radiation can excite vibrational and rotational transitions, but it is insufficient to excite electronic transitions. IR can affect bonds in molecules, changing their length or angles. Each bond can absorb IR at a certain energy giving unique vibrational patterns.
9 The Spectrum
10 Vibration Modes Vibration in the IR region can be either stretching or bending. Stretching can be symmetric or asymmetric. This involves changing the bond length. Bending can be : scissoring, rocking, wagging, and twisting. This involved changing the angle between the bonds.
11 Vibration Modes Symmetric Symmetric stretching Antisymmetric Antisymmetric stretching
12 Vibration Modes Symmetric Antisymmetric Scissoring Rocking Wagging Twisting
13
14 Absorption Location of Different Bonds
15
16 Infrared spectrum for butyraldehyde
17 IR Spectrum These basic vibrations are called fundamental absorption. In reality, a spectrum is complicated due to overtones and combinations of bands. Overtone results from a multiplication of a fundamental absorption (e.g. c=o overtone in last spectrum).
18 Instrumentation IR spectrometer can be either Dispersive Infrared Spectrometers or Fourier Transform Spectrometers. The Dispersive one Is not used anymore as it takes longer time for analysis.
19 Fourier Transform Spectrometer FTIR
20 Fourier Transform Spectrometers Fourier transform infrared (FTIR) spectrometers offer the advantages of high sensitivity, resolution, and speed of data acquisition. All wavelengths are detected and measured simultaneously, the theory behind that is beyond the scope of this lecture.
21
22 Background Spectra To obtain a spectrum of a compound, the analyst first obtains an interferogram of the background spectra. IT consists of the infrared-active atmospheric gases, carbon dioxide and water vapor (oxygen and nitrogen are not infrared active).
23 Qualitative Applications of IR IR spectra is useful for the identification of functional groups. Finger print region helps in identification of samples using a pre-saved library.
24 Sample Preparation Glass and plastic absorb strongly throughout the infrared region of the spectrum. Sodium chloride and potassium bromide are used as sample holders (disks). Sodium chloride begins to absorb at 650 cm-1, but has low cost.
25 Sample Preparation Liquids :A drop of a liquid organic compound is placed between a pair of polished sodium chloride or potassium bromide plates, referred to as liquid film forms between them. Salt plates break easily and are water soluble.
26 Salt plates
27 Salt plates
28 Sample Preparation Solids (KBr Disk or pellet) : This method involves mixing the finely ground solid sample with powdered potassium bromide and pressing the mixture under high pressure.
29
30 KBr Disk The main disadvantage of this method is that potassium bromide absorbs water, which may interfere with the spectrum that is obtained, price is also an issue. Advantage : the spectrum obtained will have no interfering bands since potassium bromide is transparent down to 400 cm 1.
31 Sample Preparation Solids (NaCl Disk) :Nujol mull, involves grinding the compound with mineral oil (Nujol) to create a suspension of the finely ground sample dispersed in the mineral oil. The thick suspension is placed between salt plates.
32
33 Nujol mull The main disadvantage of this method is that the mineral oil obscures bands that may be present in the analyzed compound. Nujol bands appear at 2924, 1462, and 1377 cm-1. However, it is a cheaper method.
34 Attenuated total reflection (ATR) It is a technique that can be used to study solid IR samples directly without sample preparation. An IR beam is directed onto a crystal (diamond) with a high refractive index at a certain angle. The beam reflects inside the crystal creating evanescent wave.
35 Attenuated total reflection (ATR) In regions of the IR spectrum where the sample absorbs energy, the evanescent wave will be attenuated. The attenuated beam returns to the crystal, then exits the opposite end of the crystal and is directed to the detector in the IR spectrometer.
36 Principle of ATR
37 ATR
38 ATR
39 Advantages include : Easy method which requires no training. Saves time. Low costs (no chemicals required). Better spectra (no interference from nujol). Does not require pressure or physical force that can destroy the sample.
40 Triple, Double and Single Bonds In general, triple bonds are stronger than double or single bonds between the same two atoms and have higher frequencies of vibration (higher wavenumbers):
41 Effect of Mass The C-H stretch occurs at about 3000 cm 1. As the atom bonded to carbon increases in mass, the frequency of vibration decreases (wavenumbers get smaller):
42 Hybridization Bonds are stronger in the order sp > sp2 > sp3 and the observed frequencies of C-H vibration illustrate this.
43 Interpreting IR spectra To some extent C-O and C-C bonds absorb in the same region of the infrared spectrum, but C-O has a much sharper peak than C-C.
44 A comparison of the intensities of the C-O and C-C absorption bands.
45 Example
46 Interpreting IR spectra Although the N-H and O-H regions overlap,the shape and fine structure of a peak often give clues to its identity as well.
47 A comparison of the shapes of the absorption bands for the O-H and N-H groups.
48 Example
49 Example
50 ALKANES, ALKENES, AND ALKYNES
51
52
53 Example
54 Aromatic Compounds Aromatic compounds show a number of absorption bands in the infrared spectrum. The C=C-H stretching peaks for appear at values greater than 3000 cm-1. The C=C stretching bands for aromatic rings usually appear cm-1 where the C=C appears for alkenes 1650 cm-1. Why? Also has C-H bending peaks that appear in the range cm 1.
55 Example
56 Example
57 Alcohols and phenols Hydroxyl group (O-H) that has hydrogen bonding will show a broad peak at cm-1. Hydroxyl group that does not have hydrogen bonding will show sharp stretch peak at cm -1, this band appears in combination with the hydrogen-bonded O-H peak when the alcohol is dissolved in a solvent.
58 Examples The O-H stretch region. (a) Hydrogen-bonded O-H only (neat liquid). (b) Free and hydrogen-bonded O-H (dilute solution). (c) Free and hydrogen-bonded O-H (very dilute solution)
59 Example of Phenols
60 Carbonyl compounds The carbonyl group is present in aldehydes, ketones, acids, esters and amides. This group absorbs strongly in the range from 1850 to 1650 cm 1. Hydrogen bonding if present can lower the absorption frequency. Presence of electron withdrawing group and resonance effect can also change the absorption frequency.
61 Aldehydes Aldehydes show a very strong band for the carbonyl group (C=O) that appears in the range of cm-1. Conjugation can lower this value. A very important doublet can be observed in the C-H stretch region for the aldehyde C-H near 2850 and 2750 cm-1. (helps to distinguish from ketones).
62 Example
63 Example (conj. Ald.)
64 Example (conj. Ald.)
65 Ketones Ketones show a very strong band for the C=O group that appears in the range of cm-1. Conjugation can lower this value.
66 Example
67 Carboxylic acid Carboxylic acid exists in monomeric form only in very dilute solution, and it absorbs at about 1760 cm 1. However, acids in concentrated solution can give a lower wavenumber, why?
68 Carboxylic acid It tend to dimerize via hydrogen bonding. This dimerization weakens the C-O bond and lowers the stretching force constant, resulting in a lowering of the carbonyl frequency of saturated acids to about 1710 cm-1.
69 Carboxylic acids Carboxylic acids show a very strong band for the C=O group that appears in the range of cm-1 for simple aliphatic carboxylic acids in the dimeric form. The O-H stretch appears in the spectrum as a very broad band extending from 3400 to 2400 cm-1.
70 Example
71 Amides
72 Example
73 Amines Primary amines, R-NH2, show two N-H stretching bands in the range cm-1, whereas secondary amines, R-N-H, show only one band in that region. Tertiary amines will not show an N-H stretch. Bending in primary amines results in a broad band in the range cm-1. Secondary amines absorb near 1500 cm 1.
74 Example
75 Example
76 Step By Step Approach for Beginners
77 Step By Step Approach for Beginners
78 Step By Step Approach for Beginners
79 Step By Step Approach for Beginners
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