![VIBRATIONAL FREQUENCY :
The value of the Stretching Vibrational frequency of a Bond can be
calculated by using Hooke’s Law :
V = 1 / 2 C [ K(m1+m2) / m1 . m2 ]1/2
V = 1 / 2 C [ K / m1 . m2 / (m1+m2) ]1/2
Where …. C = velocity of radiation (2.998×1010cm/sec)
K = force constant = strength of the bond
i.e.… for single bond = 5 × 105g/sec2
for double bond = 10 × 105g/sec2
for triple bond = 15 × 105g/sec2
m1 & m2 = Mol. Wt. of atoms m1 & m2
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IR introduction Introduction, Principle & Theory
- 1. INFRARED SPECTROSCOPY Introduction, Principle & Theory THE ELECTROMAGNETIC SPECTRUM INFRA RED GAMMA RAYS X RAYS UV VISIBLE Presented by : NIVEDITHA G 1st sem Mpharm Dept of pharmaceutics
- 2. INTRODUCTION IR-spectroscopy is also called as Vibrational spectroscopy. It is concerned with the study of Absorption of IR radiation, which results in Vibrational transitions. IR spectroscopy is mainly used to Structure elucidation to dtm the Functional groups. 2
- 3. DIFFERENT REGIONS OF IR Near IR /Overtone region (0.8 – 2.5µ) (12,500 - 4000/cm) Mid IR / Vib-Rot region (2.5 - 15µ) (4000 – 667/cm) Far IR / Rotation region (15 - 200µ) (667 – 50/cm) The absorption of IR radiation can be expressed either in terms of Wavelength (λ) / in Wave no. In IR spectra of an org-comp the graph is plotted as %Transmittance v/s Wave no. 3
- 4. The Relationship b/w Wavelength & Wave number w.k.t… Wave no = reciprocal of wavelength (1 / λ ) Wave no = 1 / λ = 1 / λ in cm (1μ = 10-4/cm) if λ = 2.5μ then ….. Wave no = 1 / 2.5 × 10-4cm = 4000/cm if λ = 15μ then….. Wave no = 1/ 15 × 10-4cm = 667/cm 4
- 5. CRITERIA FOR A MOLECULE TO SHOW IR ABSORPTION SPECTRUM 1. Dipole movement : A molecule can absorb IR radiation only when it’s absorption causes a change in it’s Dipole movement. 2. Correct Wavelength of Radiation : A molecule can absorb IR radiation only when the applied Infrared frequency is equal to the Natural frequency of vibration of that molecule 5
- 6. PRINCIPLE The technique is based on the simple fact that a chemical substance shows marked absorption in the IR region of the electromagnetic spectrum. After absorption of the IR radiation, the molecule of a chemical substance Vibrates at many rates of vibration giving rise to close packed Bands called the IR absorption spectra . Various bands will be present in the IR spectra which will corresponds to the characteristic Functional groups & bonds present in the chemical substance. 6
- 7. THEORY Absorption in the IR region is due to the changes in the Vibrational & Rotational levels In IR spectroscopy the absorbed energy brings about predominant changes in the Vibrational energy which depends on : 1. Masses of the atoms present in the molecule 2. Strength of the bond 3. The arrangement of atoms within the molecule. 7
- 11. VIBRATIONAL FREQUENCY : The value of the Stretching Vibrational frequency of a Bond can be calculated by using Hooke’s Law : V = 1 / 2 C [ K(m1+m2) / m1 . m2 ]1/2 V = 1 / 2 C [ K / m1 . m2 / (m1+m2) ]1/2 Where …. C = velocity of radiation (2.998×1010cm/sec) K = force constant = strength of the bond i.e.… for single bond = 5 × 105g/sec2 for double bond = 10 × 105g/sec2 for triple bond = 15 × 105g/sec2 m1 & m2 = Mol. Wt. of atoms m1 & m2 11
- 12. DEGREE OF FREEDOM Polyatomic molecules exhibits more than one fundamental Vibrational absorption bands these bands is related to degrees of freedom in a molecule. Each atom has 3-degrees of freedom corresponds to the 3 co- ordinates (X,Y,Z) which is necessary to describe the position relative to other atom in a molecule. When a atom combine to form a molecule, no degree of freedom is lost, i.e., the total no. of degree of freedom of a molecule will be Equal to 3n where…n = no. of atoms in a molecule. 12
- 13. A molecule of finite dimension is made up of Rotational, Vibrational & Translational degree of freedom. So… 3n degree of freedom = Translational + Rotational + Vibrational For LINEAR molecules : of n atoms ….. Total degree of freedom = 3n Translation degree of freedom = 3 Rotational degree of freedom = 2 Hence Vibration degree of freedom = 3n – 3 – 2 = 3n – 5 13
- 14. Each Vib degree of freedom corresponds to the fundamental mode of Vib & each fundamental mode corresponds to the band. Hence theoretically for linear molecules their will be 3n-5 fundamental bands Ex : for Linear molecule : CO2 No of atoms = 3 Total degree of freedom = 3 × 3 = 9 Translational degree of freedom = 3 Rotational degree of freedom = 2 Hence Vibrational degree of freedom = 9 – 3 – 2 = 4 Hence theoretically CO2 should have 4 fundamental bands. 14
- 15. For NON-LINEAR molecules : of n atoms ….. In non-linear molecules, there are three degree of rotation as the rotation about all three axes (X,Y,Z) will result in change in position. Hence it can be calculated as follows : Total degree of freedom = 3n Translation degree of freedom = 3 Rotational degree of freedom = 3 Hence Vibration degree of freedom = 3n – 3 – 3 = 3n – 6 15
- 16. Ex : for Non-Linear molecule : C6H6 No of atoms = 12 Total degree of freedom = 3 × 12 = 36 Translational degree of freedom = 3 Rotational degree of freedom = 3 Hence Vibrational degree of freedom = 36 – 3 – 3 = 30 Hence theoretically C6H6 should have 30 fundamental bands. 16
- 17. FACTORS INFLUENCING VIBRATIONAL FREQUENCIES Coupled vibrations & Fermi Resonance Electronic Effect Hydrogen Bonding Bond Angles 17
- 18. COUPLED VIBRATIONS & FERMI RESONANCE In case of isolated C-H bond there will be one stretching vibration is observed, But… in case of Methylene (-CH2-) two absorption occur, i.e., one is Symmetric & another is Asymmetric. In such cases asymmetric vibrations always occur at higher wave no compared with the symmetric vibrations. These are called Coupled Vibrations since these occur at different frequencies than that rq for an isolated C-H stretching. 18
- 19. Similarly in methyl(-CH3-) group coupled vibrations takes place at different frequencies compared to -CH2- groups some times there may be two different vibrations which have nearly same energy. If such vibrations belongs to same species then a mutual perturbation of energy may occur, This resulting shift of one towards lower frequency and another towards higher frequency. Hence there will be increasing intensity of respective band. Some times it may happen that the energy of overtone level may coincide with fundamental mode of different vibrations. Hence there may be a resonance occurs, this type of resonance is called Fermi resonance. 19
- 20. Here, a molecule transfer its energy from fundamental to overtone and back again. Hence resonance pushes the two levels apart and mixes their character so that each level becomes partly fundamental and overtone in character. This type of resonance gives rise to a pair of transitions of equal intensity For Eg : n-butyl vinyl ether…… the overtone of the fundamental vibrations is at 810/cm, their may be chance to coincide with the band at 1640/cm. Hence their will be mixing of the two bands in accordance with Fermi resonance & gives 2-bands of equal intensity at 1640/cm & 1630/cm 20
- 21. ELECTRONIC EFFECT The frequency shifts are due to the electronic effect which includes Under the influence of these effects , the force constant / the bond strength changes & it’s absorption frequency shifts from normal value. For Ex : Acetone(CH3COCH3) = 1715/cm Chloroacetone(CH3COCH2Cl) = 1725/cm Dichloroacetone(CH3COCHCl2) = 1740/cm The introduction of electronegative group causes –I effect which results in bond order to increase, thus the force const increases & hence the wave no of absorption increases. Inductive effect Mesomeric effect Field effect 21
- 22. Introduction of alkyl group causes +I lengthening / weakening of the bond absorption at lower wave no. In most of cases , Mesomeric effect works along with Inductive effect Conjugation lowers the absorption frequency of C=O stretching whether the conjugation is due to α, β - unsaturation / due to an aromatic ring. In some cases Inductive effect dominates over Mesomeric effect. Mesomeric effect causes lengthening / the weakening of a bond leading in the lowering of absorption of frequency 22
- 23. HYDROGEN BONDING Hydrogen bonding brings about remarkable downward frequency shift. Stronger the Hydrogen bonding, greater is the absorption shift towards lower wave no than the normal value. In IR technique Hydrogen bonding are of 2 types : 1. Intermolecular H2 bonding : give rise to broad bands. 2. Intramolecular H2 bonding : give rise to sharp & well defined bands. INTERMOLECULAR H2 BONDING : These are conc. dependent. On dilution , the intensity of such bands are independent of conc. The absorption frequency difference b/w free & associated molecule is higher 23
- 24. INTRAMOLECULAR H2 BONDING : These are observed in dil soln of di- & polyhydroxy compounds in CCl4. Where no intermolecular H2 bonds are formed. In such condition it was observed that a no of cyclic & acyclic diols have 2 bands & other have single band in the O- H stretching mode region. The spectrum of glycol in dil CCl4 shows 2-bands at wave no’s : 3644/cm & at 3612/cm. Here the band at 3644/cm is due to free O-H bond….. where as the band at 3612/cm is due to O-H…O bonding 24
- 25. BOND ANGLES In cyclobutanones the frequency is rises due to C=O bond, this can be explained by using Bond angles. If the bond angle is reduce below the normal angle 1200C it will leads to increase in s-character in C=O bond. Greater s-character causes shortening of C=O bond thus C=O str occurs at higher frequency If the bond angle is above 1200C then their will be opp effect. Due to this reason, di-tetra-butyl absorbs at (1697/cm) low frequency 25
- 26. References….. 1. Elementary Organic Spectroscopy by Y.R.Sharma. 2. Text book of Pharmaceutical analysis by Dr.S.Ravi shankar. 3. Instrumental Methods of Chemical Analysis. G. R. Chatwal. 4. Internet source 26
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