UV Spectroscopy PDF

UNIT-II LECTURE-12 Electromagnetic Radiation Radiation is absorbed and emitted in photons. The defining characteristic of a photon is that its energy cannot be split into smaller pieces. Each photon’s energy is defined by its frequency (u) or wave length (l) or wave number...

UNIT-II LECTURE-12 Electromagnetic Radiation Radiation is absorbed and emitted in photons. The defining characteristic of a photon is that its energy cannot be split into smaller pieces. Each photon’s energy is defined by its frequency (u) or wave length (l) or wave number (Figure 15). Ephoton = hu = hc/l Two constants appear in these formulas i.e. h = plank’s constant, 6.63 x 10 -34 J s and c = speed of light, 3.00 x 108 m s-1 Wave number = 1/l Figure 15: Electromagnetic spectrum Beer-Lambert Law The Beer-Lambert law (or Beer's law) is the linear relationship between absorbance and concentration of absorbing specie. The general Beer-Lambert law is usually written as: A= *b*c where A is the measured absorbance, is a wavelength-dependent absorptivity coefficient, b is the path length, and c is the analyte concentration. Data are frequently reported in percent transmission (I/I0 * 100) or in absorbance [A = log (I/I0)]. Instrumentation: Experimental measurements are usually made in terms of transmittance (T), which is defined as: T = I / Io where I is the light intensity after it passes through the sample and I o is the initial light intensity as shown in Figure 16. The relation between A and T is: A = -log T = - log (I / Io) Figure 16: Absorption of light by sample solution and change in intensity. LECTURE-13 Electronic transitions The absorption of UV or visible radiation corresponds to the excitation of outer electrons. There are three types of electronic transition which can be considered; 1. Transitions involving σ, Π and n electrons 2. Transitions involving charge-transfer electrons 3. Transitions involving d and f electrons (not covered in this Unit) When an atom or molecule absorbs energy, electrons are promoted from their ground state to an excited state (Figure 17). In a molecule, the atoms can rotate and vibrate with respect to each other. These vibrations and rotations also have discrete energy levels, which can be considered as being packed on top of each electronic level. Figure 17: Absorbing species containing σ, Π and n electrons Absorption of ultraviolet and visible radiation in organic molecules is restricted to certain functional groups (chromophores) that contain valence electrons of low excitation energy. The spectrum of a molecule containing these chromophores is complex. This is because the superposition of rotational and vibrational transitions on the electronic transitions gives a combination of overlapping lines. This appears as a continuous absorption band. Possible electronic transitions of σ, Π and n electrons are; σ - σ* Transitions An electron in a bonding σ orbital is excited to the corresponding antibonding orbital. The energy required is large. For example, methane (which has only C-H bonds, and can only undergo σ - σ* transitions) shows an absorbance maximum at 125 nm. Absorption maxima due to σ - σ* transitions are not seen in typical UV-Vis. spectra (200 - 700 nm) as shown in Figure 18. Figure 18: Possible electronic transitions of σ, Π and n electrons Π - Π* Transitions Saturated compounds containing atoms with lone pairs (non-bonding electrons) are capable of n - Π* transitions. These transitions usually need less energy than n - Π* transitions. They can be initiated by light whose wavelength is in the range 150 - 250 nm. The number of organic functional groups with n - Π* peaks in the UV region is small. n - Π* and Π - Π* Transitions Most absorption spectroscopy of organic compounds is based on transitions of n or Π electrons to the Π* excited state. This is because the absorption peaks for these transitions fall in an experimentally convenient region of the spectrum (200 - 700 nm). These transitions need an unsaturated group in the molecule to provide the Π electrons. Molar absorptivity from n - Π* transitions are relatively low, range from 10 to100Lmol -1cm-1. Π - Π* transitions normally give molar absorptivity between 1000 and 10,000 L mol-1 cm-1. Solvent Effect: The solvent in which the absorbing species is dissolved also has an effect on the spectrum of the species. Peaks resulting from n - Π* transitions are shifted to shorter wavelengths (blue shift) with increasing solvent polarity. This arises from increased solvation of the lone pair, which lowers the energy of n orbital. Often the reverse (i.e. red shift) is seen for Π - Π* transitions. This is caused by attractive polarization forces between the solvent and the absorber, which lower the energy levels of both the excited and unexcited states. This effect is greater for the excited state, and so the energy difference between the excited and unexcited states is slightly reduced - resulting in a small red shift. This effect also influences n - Π* transitions but is overshadowed by the blue shift resulting from solvation of lone pairs. LECTURE-14 Application of UV-Spectroscopy  Detection of Impurities: UV absorption spectroscopy is one of the best methods for determination of impurities in organic molecules. Additional peaks can be observed due to impurities in the sample and it can be compared with that of standard raw material. By also measuring the absorbance at specific wavelength, the impurities can be detected. Benzene appears as a common impurity in cyclohexane. Its presence can be easily detected by its absorption at 255 nm.  Structure elucidation of organic compounds: UV spectroscopy is useful in the structure elucidation of organic molecules, the presence or absence of unsaturation the presence of hetero atoms. From the location of peaks and combination of peaks, it can be concluded that whether the compound is saturated or unsaturated, hetero atoms are present or not etc.  Quantitative analysis: UV absorption spectroscopy can be used for the quantitative determination of compounds that absorb UV radiation. This determination is based on Beer’s law which is as follows: A = log I0 / It = log 1/ T = – log T = abc = εbcbc Where εbc is extinction co-efficient, c is concentration, and b is the length of the cell that is used in UV spectrophotometer. Qualitative analysis: UV absorption spectroscopy can characterize those types of compounds which absorbs UV radiation. Identification is done by comparing the absorption spectrum with the spectra of known compounds. UV absorption spectroscopy is generally used for characterizing aromatic compounds and aromatic olefins.  Dissociation constants of acids and bases. pH = pKa + log [A-] / [HA] From the above equation the pKa value can be calculated if the ratio of [A -] / [HA] is known at a particular PH. and the ratio of [A-] / [HA] can be determined spectrophotometrically from the graph plotted between absorbance and wavelength at different PH values.  Chemical kinetics: Kinetics of reaction can also be studied using UV spectroscopy. The UV radiation is passed through the reaction cell and the absorbance changes can be observed.  Quantitative analysis of pharmaceutical substances: Many drugs are either in the form of raw material or in the form of formulation. They can be assayed by making a suitable solution of the drug in a solvent and measuring the absorbance at specific wavelength. Diazepam tablet can be analyzed by 0.5% H2SO4 in methanol at the wavelength 284 nm.  Molecular weight determination: Molecular weights of compounds can be measured spectrophotometrically by preparing the suitable derivatives of these compounds. For example, if we want to determine the molecular weight of amine then it is converted in to amine picrate. Then known concentration of amine picrate is dissolved in a litre of solution and its optical density is measured at λmax 380 nm. After this the concentration of themax 380 nm. After this the concentration of the solution in gm moles per litre can be calculated by using the following formula LECTURE-15 Infrared spectroscopy exploits the fact that molecules absorb frequencies that are characteristic of their structure. These absorptions occur at resonant frequencies, i.e. the frequency of the absorbed radiation matches the vibrational frequency. The energies are affected by the shape of the molecular potential energy surfaces, the masses of the atoms. In order for a vibrational mode in a sample to be "IR active", it must be associated with changes in the dipole moment. A permanent dipole is not necessary, as the rule requires only a change in dipole moment. A molecule can vibrate in many ways, and each way is called a vibrational mode. For molecules with N number of atoms, linear molecules have 3N – 5 degrees of vibrational modes, whereas nonlinear molecules have 3N – 6 degrees of vibrational modes (also called vibrational degrees of freedom). As an example H2O, a non-linear molecule, will have 3 × 3 – 6 = 3 degrees of vibrational freedom, or modes. Simple diatomic molecules have only one bond and only one vibrational band. If the molecule is symmetrical, e.g. N2, the band is not observed in the IR spectrum, but only in the Raman spectrum. Asymmetrical diatomic molecules, e.g. CO, absorb in the IR spectrum. More complex molecules have many bonds, and their vibrational spectra are correspondingly more complex, i.e. big molecules have many peaks in their IR spectra. The atoms in a CH 2X2 group, commonly found in organic compounds and where X can represent any other atom, can vibrate in nine different ways. Six of these vibrations involve only the CH2 portion: symmetric and asymmetric stretching, scissoring, rocking, wagging and twisting, as shown in Figure 19. Structures that do not have the two additional X groups attached have fewer modes because some modes are defined by specific relationships to those other attached groups. For example, in water, the rocking, wagging, and twisting modes do not exist because these types of motions of the H represent simple rotation of the whole molecule rather than vibrations within it. Figure 19: Bending vibrations in a molecule Summary of absorptions of bonds in organic molecules Finger Print Region: The region to the right-hand side of the diagram (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. It is much more difficult to pick out individual bonds in this region than it is in the "cleaner" region at higher wavenumbers. The importance of the fingerprint region is that each different compound produces a different pattern of troughs in this part of the spectrum. Using the fingerprint region: Compare the infra-red spectra of propan-1-ol and propan-2-ol. Both compounds contain exactly the same bonds. Both compounds have very similar troughs in the area around 3000 cm-1 - but compare them in the fingerprint region between 1500 and 500 cm-1. LECTURE-16 Applications of IR-Spectroscopy 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. Some of the major applications of IR spectroscopy are as follows: 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. 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-1and 685 cm-1 No other compound then benzaldehyde produces same IR spectra as shown above. 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. 5. Quantitative analysis The quantity of the substance can be determined either in pure form or as a mixure of two or more compounds. In this, characteristic peak corresponding to the drug substance is chosen and log I0/It of peaks for standard and test sample is compared. This is called base line technique to determine the quantity of the substance. Acetone Acetone. Notice the carbonyl (C=O) group which absorbs at 1711 cm-1. Acetone IR LECTURE-17 Raman Spectroscopy Raman spectroscopy (Named after Indian physicist Sir C. V. Raman) is spectroscopic technique used to observe vibrational, rotational, and other low-frequency modes in a system. Raman spectroscopy is commonly used in chemistry to provide a structural fingerprint by which molecules can be identified. Raman spectra relate to vibrational and /or roatational transitions in molecules but in different manner.In this case we measure the scattering and not the absorption of radiation.An intense beam of monochromatic radiation in the visible region is allowed to fall on a sample and the intensity of scattered light is observed at right angles to the incident beam. Raman spectroscopy reveals the chemical and structural composition of samples. Generally, all materials produce Raman spectra, with the exception of pure metals. Raman scattering Raman scattering occurs when light interacts with molecular vibrations. This is similar to the more widely known infrared absorption spectroscopy, but different rules apply. A change in molecular polarisability is required during the vibration for the Raman effect to occur. You will see some vibrations in the Raman spectrum that are not visible in the infrared spectrum, and vice-versa, because of the different selection rules. For example, Raman spectroscopy is superb for studying the carbon atoms that make up the structure of diamond, unlike infrared absorption spectroscopy. Scattered light The first step in producing a Raman spectrum is to illuminate your sample with a monochromatic light source, such as a laser. Most of the light that scatters off is unchanged in energy ('Rayleigh scattered'). A minute fraction— perhaps 1 part in 10 million—has lost or gained energy ('Raman scattered'). This Raman shift occurs because photons (particles of light) exchange part of their energy with molecular vibrations in the material. Where energy is lost the Raman scattering is designated as 'Stokes'; where energy is gained the Raman scattering is designated as 'anti-Stokes'. We rarely use anti-Stokes Raman light as it is less  intense than the Stokes, however it does represent equivalent vibrational information of the molecule. Vibrating atoms The change in energy depends on the frequency of vibration of the molecule. If it is very fast (high frequency)—light atoms held together with strong bonds—the energy change is significant. If it is very slow (low frequency)—heavy atoms held together with weak bonds—the energy change is small. Raman spectrometers It consist of:  single or multiple lasers, from UV (244 nm) to IR (1064 nm) – switch with a single click  high quality objective lenses, from high confocal 100× to long working distance and immersion options  custom designed motorised spectrometer lenses - automatically align for each configuration  laser-line-specific Rayleigh filters with a dual filter arrangement to optimise sensitivity  highest quality master diffraction gratings provide exceptional dispersion and longevity  thermoelectrically cooled (- 70 ºC) CCD detector – stable and sensitive  high specification multi-core PC for data collection and analysis Vibration frequencies The frequencies of vibration depend on the masses of the atoms involved and the strength of the bonds between them. Heavy atoms and weak bonds have low Raman shifts. Light atoms and strong bonds have high Raman shifts. We see the high frequency carbon-hydrogen (C-H) vibrations in the polystyrene spectrum at about 3000 cm-1. The low frequency carbon-carbon (C-C) vibrations are at around 800 cm-1. The C-H vibrations have a higher frequency than the C-C vibrations because hydrogen is lighter than carbon. We see the vibrations of two carbon atoms linked by strong double bonds (C=C) at around 1600 cm- 1. This is at a higher frequency than two carbon atoms lined by a weaker single bond (C-C, 800 cm-1). When a sample is illuminated by a laser, both Raman scattering and photoluminescence (PL) can occur. The latter can be many times stronger than the former and can prevent successful Raman analysis. PL comprises both fluorescence and phosphorescence processes and originates from an absorption/emission process between different electronic energy levels in the material. The amount and type of PL depends on which material you are studying and which laser wavelength you are using. Unwanted fluorescence interference can normally be avoided by choosing an appropriate laser wavelength.  Energy diagram showing absorption of light and the processes involved in the emission of light as fluorescence and phosphorescence LECTURE-18 Applications of Raman Spectroscopy 1 Carbon Materials  Purity of Carbon nanotubes(CNTs)  Sp2 and sp3 structure in carbon materials and Diamond like carbon coating properties  Defects and disorder analysis in carbon materials 2 Pharmaceuticals and Cosmetics  Compound distribution in tablets.  Polymorphic forms  Contaminant identification and Powder content and purity 3 Life Sciences  Bio-compatibility and DNA/RNA analysis  Drug/cell interactions and Disease Diagnosis  Bone Structure 4 Geology and Minerology  Gemstone and Mineral identification  Mineral behaviour under extreme conditions.  Phase Transitions and Fluid inclusion 5 Semiconductors  Alloy composition  Defect analysis  Contamination identification and Doping effects LECTURE-19 Microwave Spectroscopy (Rotational Spectroscopy) Free atoms do not rotate or vibrate. For an oscillatory or a rotational motion of a pendulum, one end has to be tied or fixed to some point. In molecules such a fixed point is the center of mass. The atoms in a molecule are held together by chemical bonds. The rotational and vibrational energies are usually much smaller than the energies required to break chemical bonds. The rotational energies correspond to the microwave region of electromagnetic radiation (3x1010 to 3x1012 Hz; energy range around 10 to100J/mol) and the vibrational energies are in the infrared region (3x10 12 to 3x1014 Hz; energy range around 10kJ/mol) of the electromagnetic radiation. For rigid rotors (no vibration during rotation) and harmonic oscillators (wherein there are equal displacements of atoms on either side of the center of mass) there are simple formulae characterizing the molecular energy levels. In real life, molecules rotate and vibrate simultaneously and high speed rotations affect vibrations and vice versa Rotational Spectra of diatomic molecules Fig. A rigid diatomic with masses m1 and m2 joined by a thin rod of length r = r1 + r2.The center of mass is at C. The two independent rotations of this molecule are with respect to the two axes which pass though C and are perpendicular to the “bond length” r. The rotation with respect to the bond axis is possible only for “classical” objects with large masses. For quantum objects, a “rotation” with respect to the molecular axis does not correspond to any change in the molecule as the new configuration is indistinguishable from the old one. The center of mass is defined by equating the moments on both segments of the molecular axis. (1) The moment of inertia is defined by (2) I= = = (3) = Since , r1 = m2r Therefore, Substituting the above equation in (3), we get (4) Where μ, the reduced mass is given by The rotation of a diatomic is equivalent to a “rotation” of a mass μ at a distance of r from the origin C. 2 (5) The kinetic energy of this rotational motion is K.E. = L /2I where L is the angular momentum, Iω where ω is the angular (rotational) velocity in radians/sec. The quantized rotational energy levels for this diatomic are (6) -1 The energy differences between two rotational levels is usually expressed in cm. The wave number corresponding to a given ∆E is given by -1 ν = ∆E /hc, cm The energy levels in cm-1 are therefore, The rotational energy levels of a diatomic molecule are shown in Fig. LECTURE-20 The selection rule for a rotational transition is, ∆ J = ± 1 ∆ J = + 1 absorption ∆ J = - 1 emission In addition to this requirement, the molecule has to possess a dipole moment,  μ  0 molecule gives a rotational spectrum only if it has a permanent dipole moment As a dipolar molecule rotates, the rotating dipole constitutes the transition dipole operator μ. Molecules such as HCl and CO will show rotational spectra while H2, Cl2 and CO2 will not. E For rigid rotor, J J + 1, J = BJ(J + 1) For the transition; J=0J=1 E j = E j =1 - E j =0 = 2B - 0 = 2B cm-1 = position of the first line in the spectrum = u j =0 j =1 For the transition; J = 1 J = 2 E j = E j =2 - E j =1 = 6B - 2B = 4B cm-1 = position of the second line in the spectrum = ujj = position of the third line in the spectrum = ujj Since, The allowed rotational energies are given by; u = BJ(J + 1) The wave numbers of the different rotational levels will be; 0, 2B, 6B, 12B, 20B, 30B (cm-1),… and so on. And the various lines in the rotational spectra will be equally spaced (separation between lines = 2B). And for two adjacent rotational states, the energy difference is given by; E j = E j +1 - E j = 2B(J +1) cm-1  Allowed Transition (i) The allowed rotational energy levels of a rigid (ii) Allowed transitions between the energy levels of diatomic molecule a rigid diatomic molecule and the spectrum