1. Ridwan Islam

2. Spectroscopy An instrumental technique for determining the concentration and structure of a substance by measuring the intensity of electromagnetic radiation it absorbs at various wavelengths. Absorption spectroscopy: A spectroscopic techniques that measure the absorption of radiation, as a function of frequency or wavelength, due to its interaction with a sample. The intensity of the absorption varies as a function of frequency, and this variation is the absorption spectrum. UV and visible spectroscopy: A type of absorption spectroscopy that uses the UV and visible part of the electromagnetic radiation.

3. Electromagnetic radiation: Electromagnetic radiation takes the form of self- propagating waves in a vacuum or in matter. EM radiation has an electric and magnetic field component which oscillate in phase perpendicular to each other and to the direction of energy propagation.

5. UV region: 200-400 nm and visible region: 400-700nm

6. Wavelength: is a linear distance from any point on the wave to the corresponding point on the adjacent wave. It is expressed by λ. Frequency: number of waves passing a given point in unit time. It is expressed by ν.

7. What happens when a substance absorbs electromagnetic radiation? Electromagnetic radiation is energy and when a substance absorbs electromagnetic radiation, it gains energy. The energy gained by the molecule in this way- may break bonds within the molecule (γ ray) may raise electrons to higher energy level (UV and visible spectroscopy) may bring about increased vibration and rotation of atom (IR spectroscopy) may change nucleus or electronic spin (microwave used by NMR)

8. Vacuum UV region The UV light below 200nm is relatively uninformative and difficult to measure. Since oxygen in the air also absorbs UV light in this region. Therefore in order to measure absorption of UV light below 200nm air must be removed from the instrument. The region is known as vacuum. The vacuum UV region ranges from 120nm to 200nm. This region excites the σ bond orbital. The UV light above 200nm causes excitation of electrons from π orbitals, particularly π conjugated system. This region can be readily measured and give informative spectra.

9. Basic principle of UV spectroscopy All atoms and molecules are capable of absorbing energy in accordance with their own structure variation and so the kind and amount of radiation absorbed by a molecule depend upon: The structure of the molecule. The number of molecules interacting with the radiation. When electromagnetic radiation is absorbed by a molecule, it undergoes transition from a state of lower to state of higher energy. If the molecule is monatomic, the energy absorbed can only be used to raise the energy levels of electrons. If the molecule consists of more than one atom, the radiation absorbed may bring about changes in electronic, rotational, vibrational or translational energy.

10. Electronic energy is associated with the motion of electrons around the nucli. Rotational energy is associated with the overall rotation of the molecule. Vibrational energy is associated with the movement of atoms within the molecules. Translational energy is associated with the motion of the molecule as a whole. Electronic transitions give absorption in the visible and ultraviolet regions of the spectrum where as translational, rotational and vibrational changes give absorption in the far and near infrared spectrum.

11. When a molecule absorbs visible or ultraviolet energy, an electron or electrons will be raised to a higher energy level if the energy requirement for that transition is equal to the energy of the incoming photon. The electrons in the inner shells of atom and those that are shared by two adjacent atoms not affected to the same degree by incoming radiation as those that can’t be localized within the molecule. Electrons of the latter type give rise to spectra in the UV and visible regions of the electromagnetic spectrum. In this way electromagnetic waves are absorbed by a compound which can be determined by the spectrophotometer and thus qualitative and quantitative analysis of compound can be done.

12. Types of transition states According to the molecular orbital theory when a molecule is excited by the absorption of energy (UV or Visible light), its electrons are promoted from a bonding to an anti- bonding orbital. The energy required for various transitions obey the following order:    *  n   *     *  n   * Fig: Electromagnetic energy level and various electronic transitions

13.    * : A transition in which a bonding  electron is excited to an anti-bonding  orbital is referred to as    * transition. It is a high energy processes because of  bonds are generally very strong. In organic compounds, all the valence shell electrons are involved in the formation of sigma bonds resulting do not show absorption in the normal ultra-violet region (180 – 400nm). For saturated hydrocarbons, like methane (CH 4 ) absorption occurs near 150nm (high energy). Consider    * transition in a saturated hydrocarbon: -   *  *

14. n   * : This type of transition takes place in saturated compounds containing one hetero atom with unshared pair of electrons (n electrons). Such transitions require comparatively less energy than that required for    * transitions. In saturated alkyl halides, the energy required for such a transition decreases with the increase in the size of the halogen atom (or decrease in the electronegativity of the atom.) n   *

15. For n   * transition in methyl chloride and methyl iodide. The absorption maximum for methyl chloride is 172 – 175nm whereas methyl iodide is 258nm because the electronegativity of chlorine is greater than iodide. Thus the excitation of chlorine atom is comparatively difficult than iodide. On the other hand methyl iodide has higher molar extinction coefficient than methyl chloride. n   * transitions are very sensitive to hydrogen bonding. Alcohols as well as amines form hydrogen bonding with the solvent molecules due to the presence of non bonding electrons on the hetero atom and thus transition requires greater energy.

16.    * : This type of transition occurs in the unsaturated centers of the molecule; i.e. in compounds containing double or triple bonds and also in aromatics. The excitation of  electron requires smaller energy and hence transition of this type occurs at longer wavelength. An  electron of a double bond is excited to  * orbital. Consider    * transition in an alkene: -   *  * π  π *

17. This transition requires still lesser energy as compared to n   * transition and therefore, absorption occurs at longer wavelengths. Absorption usually occurs within the region of ordinary ultra-violet spectrophotometer. In unconjugated alkenes, absorption bands appear around 170 – 190nm. In carbonyl compounds, the band due to    * transition appears around 180nm and is most intense, i.e. the value of extinction coefficient is high.

18. n   * : In this type of transition an electron of unshared electron pair on hetero atom gets excited to  * antibonding orbital. This type of transition requires least amount of energy out of all the transition and hence occurs at longer wavelength. Saturated aldehydes show both the types of transitions, i.e. low energy n   * and high energy    * occurring around 290nm and 180nm respectively.

19. UV terminology Chromophore: a group responsible for characteristic absorption of radiation by a molecule. Most chromophore contains one or more double bonds. For example: Here, -CO- is the chromophore Here, benzene is the chromophore

20. Auxochrome: a group which does not have the absorption properties of its own, but when attached to a chromophore alters both the position and intensity of the peak. All auxochromes contain an atom with lone pair of electrons. For example: -OH, -OR, -NHR, -NR 2, SH etc. λ max : is the wavelength at which the maximum fraction of light is absorbed by a solution.

21. Bathochromic shift: is the displacement of absorption maximum towards longer wavelength, due to- Effect of solvent or Presence of an auxochrome It is also known as red shift, since the absorption maximum is shifted towards longer wavelength. Hypsochromic shift: is the displacement of absorption maximum towards shorter wavelength, due to- Effect of solvent Removal of an auxochrome Removal of conjugation It is also known as blue shift, since the absorption maximum is shifted towards shorter wavelength. Hyperchromic shift: is an effect leading to increased absorption intensity. Hypochromic shift: Is an effect leading to decreased absorption intensity.

22. Transmittance: The ratio of the radiant energy transmitted to the total radiant energy incident on a given sample is known as transmittance. T = I/I 0 Absorbance: A measure of the capacity of a substance to absorb light of a specified wavelength. It is equal to the logarithm of the reciprocal of the transmittance. A = log (1/T) A = log (I 0 /I)

25. Table: Examples of the effect of auxochrome

26. Instrumentation UV visible spectrophotometer consists of the following parts- Radiation source Monochromator Sample compartment Detector Recorder

27. 1. Radiation source: The electrical excitation of deuterium or hydrogen at low pressure produces a continuous UV spectrum. Both Deuterium and Hydrogen lamps emit radiation in the range 160 - 375 nm. Quartz windows must be used in these lamps and quartz cuvettes must be used, because glass absorbs radiation of wavelengths less than 350 nm. Various UV radiation sources are as follows Deuterium lamp Hydrogen lamp Tungsten lamp Various Visible radiation sources are as follows Tungsten lamp Mercury vapor lamp

28. 2. Monochromator: All monochromators contain the following component parts; An entrance slit A collimating lens A dispersing device (a prism or a grating) A focusing lens An exit slit

29. Polychromatic radiation (radiation of more than one wavelength) enters the monochromator through the entrance slit. The beam is collimated, and then strikes the dispersing element at an angle. The beam is split into its component wavelengths by the grating or prism. By moving the dispersing element or the exit slit, radiation of only a particular wavelength leaves the monochromator through the exit slit.

30. 3. Sample compartment: The cell holding the sample should be transparent to the wavelength region to be recorded. Quartz or fused silica cuvettes are required for spectroscopy in the UV region. Silicate glasses can be used for the manufacture of cuvettes for use between 350 and 2000 nm. The thickness of the cell is generally 1 cm. cells may be rectangular in shape or cylindrical with flat ends.

31. 4. Detector: The photomultiplier tube is a commonly used detector in UV-Vis spectroscopy. It consists of a photoemissive cathode (a cathode which emits electrons when struck by photons of radiation), several dynodes (which emit several electrons for each electron striking them) and an anode.

32. A photon of radiation entering the tube strikes the cathode, causing the emission of several electrons. These electrons are accelerated towards the first dynode (which is 90V more positive than the cathode). The electrons strike the first dynode, causing the emission of several electrons for each incident electron. These electrons are then accelerated towards the second dynode, to produce more electrons which are accelerated towards dynode three and so on. Eventually, the electrons are collected at the anode. By this time, each original photon has produced 10 6 - 10 7 electrons. The resulting current is amplified and measured. Photomultipliers are very sensitive to UV and visible radiation. They have fast response times. Intense light damages photomultipliers; they are limited to measuring low power radiation.

33. 5. Recorder: The signal for the intensity of absorbance versus corresponding wavelength is automatically recorded on the graph. The more the absorbance the less the transmittance. The signal from the detector is normally proportional to the intensity of light incident on the detector and after amplification may be displayed as transmittance or absorbance.

34. Double beam UV visible spectrophotometer

35. Double beam spectrophotometer is a special type of spectrophotometer in which a single beam of radiation from a monochromator is split by a rotating disk into two beams. Among the two beams, one beam is passed through the sample solution and the other is passed through the blank solution. Then two half beams are recombined and passed through the detector.

36. Rotating disk: Light from the slit then falls onto a rotating disc. Each disc consists of different segments – A transparent section: If the light hits the transparent section, it will go straight through the sample cell, get reflected by a mirror, hits the mirrored section of a second rotating disc, and then collected by the detector. A mirrored section: If the light hits the mirrored section, gets reflected by a mirror, passes through the reference cell, hits the transparent section of a second rotating disc and then collected by the detector. An opaque black section: if the light hits the black opaque section, it is blocked and no light passes through the instrument, thus enabling the system to make corrections for any current generated by the detector in the absence of light.

37. Laws of spectrometry There are two laws which govern the absorption of light by the molecules. These are Beer’s law and Lambert’s law. Lambert’s law: When a beam of monochromatic radiation is passed through a solution of an absorbing substance, the rate of decrease in intensity of radiation with thickness of absorbing medium is directly proportional to the intensity of the incident radiation. -dI/db α I

38. -dI/db = K 1 I -dI/I = K 1 db Integrating the equation between the limits b=0 to b and I=I 0 to I, =  I = intensity of incident radiation.  dI = Infinitesimally small decrease in the intensity of radiation on passing through infinitesimally small thickness db, of the medium.  -dI/db = rate of decrease in intensity with the thickness b  K 1 = Proportionality constant or Absorption coefficient

39. -ln I/I 0 = K 1 b ln I 0 /I = K 1 b 2.303 log I 0 /I = K 1 b log I 0 /I = K 1 b/2.303…………………………………… (1) Beer’s law When a beam of monochromatic radiation is passed through a solution of an absorbing substance, the rate of decrease in the intensity of radiation with the concentration of the solute in the solution is directly proportional to the intensity of the incident radiation. -dI/dc α I

40. -dI/dc = K 2 I -dI/I = K 2 dc [c = concentration of absorbing medium -dI/dc = rate of decrease in intensity with the concentration c K 2 = absorption co-efficient]

41. Integrating the equation between the limits c=0 to c and I=I 0 to I, = -ln I/I 0 = K 2 c ln I 0 /I = K 2 c 2.303 log I 0 /I = K 2 c log I 0 /I = K 2 c/2.303…………………………….(2) Combining equation 1 and 2, we get log I 0 /I = (K 1 K 2 /2.303) × bc Or, A = abc

42. Where, A = absorbance (no units, since A = log I 0 / I) a = absorptivity. For a specific wavelength, absorptivity value is constant for a particular solute. The value of the constant depends on: the substance, the solvent, the wavelength, the units used for concentration and path length.

43. Beer’s – Lambert’s Law applies to a solution containing more than one kind of absorbing substances, provided there is no interaction among the various species. Thus for a multiple component system: [ Where, the subscripts refer to absorbing components 1, 2, ------------, n.] When c is expressed in Moles/liter, then absorptivity is called molar extinction co-efficient (ε).  = A/bc = a  M [b = the path length of the sample (Commonly b = 1 cm), M = molecular weight. c = the concentration of the compound in solution, expressed in mol L -1 = A 1 + A 2 + ------------ +A n = a 1 bc 1 + a 2 bc 2 + ----------------- + a n bc n A total

44. Limitation of the Beer’s – Lambert’s Law Actual limitation: If interaction occurs at higher concentration (  0.01M) Beer’s – Lambert’s Law will not be obeyed, due to electrostatic interactions between molecules in close proximity. Readings at high absorbance values are unreliable. Absorbance values tend to infinity as the transmittance tends to zero. At higher concentration, refractive index of solution changes, causing ε to change. Experimental conditions should be such as to keep absorbance readings below 1.5. The Beer’s – Lambert’s Law is rigorously obeyed when a single species gives rise to the observed absorption. If there are suspended particles in the sample, these will cause light scattering, thereby reducing the transmitted intensity.

45. Chemical limitations: The Beer’s – Lambert’s Law may not be obeyed: When different forms of the absorbing molecule are in equilibrium. When the solute in the solution associate, dissociate or react with the solvent. When there one fluorescent compound or compounds which are changed by irradiation. Example: Since the molar absorptivity values for the dichromate ion and the two chromate species are different at the wavelength of maximum absorption. Chromate solutions, when diluted with H 2 O deviate from Beer’s – Lambert’s Law. Cr 2 O 7 -2 + H 2 O = 2HCrO 4 - = 2H + + 2CrO 4 -

46. Instrumental deviation: Beer’s law is observed only when the radiation employed is monochromatic. Polychromatic beam may cause deviation from Beer’s law. Deviation caused by the solvent effects: The absorption spectrum of a drug depends on the solvent used to solubilize the substance. A drug may absorb a maximum of radiant energy at one wavelength in one solvent but will absorb little at the same wavelength in another solvent. These changes in spectrum are due to: The nature of the solvent The nature of the absorption The nature of the solute

47. Generally, the concentration of solution should be 10 mg/litre or, 10 microgram/ml. Solvent effect of the absorption occurs due to the following factors: Electromagnetic radiation absorptivity of the solvent. Impurities present in the solvent. pH change of the solvent. Polarity of the solvent. Reaction with drug.

48. So, we should use: Pure solvent Non-polar solvent Solvent where pH change does not occur. We should keep the following factors in mind: Sample should be 100% soluble in solvent. Solvent should be optically transparent at the experimental wavelength. Solvent must not react with the sample.

49. Concentration determination of unknown solution: If a compound follows the Beer’s – Lambert’s Law then its concentration in a supplied sample could be known from the calibration curve drawn for the various strength reference standard solution of that compound. Calibration curve of the Beer’s – Lambert’s Law means the curve in which absorbance or percent transmittance is plotted against concentration. If a compound follows the Beer’s – Lambert’s Law the calibration curve for this compound shows a straight line going through the origin.

50. Absor bance (A) Concentration (c) Absorption for the supplied sample (diluted) Concentration of the supplied sample (diluted)

51. Procedure: In this process the following steps should be operated for getting the concentration of the supplied sample: ----- Prepare a solution of concentration  0.01M by authentic sample of which concentration is to be determined by using a suitable optically transparent solvent. Then max for the compound must be determined and ensure 100% transmittance of the solvent at max. Various strength of the reference standard solution of that compound is made by dilution. E.g. 10  g/ml, 20  g/ml, 30  g/ml ---------- 100  g/ml. Now the absorption or percent transmittance for these various strength in the max is measured. These measured absorption are plotted against concentration of the reference standard solution and draw the “A vs c” curve.

52. If the curve gives a straight line going through the origin we could understand that the compound follows the Beer’s – Lambert’s Law. Then the unknown sample with proper dilution is supplied in cuvette to measure the absorption in the max. This value will indicate a point on the straight line drawn for the reference standard solution. Now a Vertical line drawn from this point on the Y-axis or the concentration axis of the curve will show the concentration of the compound of the supplied sample. Finally, the determined concentration is multiplied by the dilution factor and thus the actual concentration is measured.

53. Absor bance (A) Concentration (c) Absorption for the supplied sample (diluted) Concentration of the supplied sample (diluted)

54. Math: A tablet containing 100mg API. It was powdered and dissolved in 100 ml Ethanol. 1ml of this solution is again diluted to 100ml. The absorbance at 265nm is 0.64 and absorptivity is 66 in a 1 cm cell. Calculate the potency of the tablet.

55. We know, A = abc So, c = A/bc = 0.64/1×66 = 0.0097 mg/ml Total concentration = c × dilution factor = 0.0097 × 100 = 0.97 mg/ml Amount = concentration × volume = 0.97 × 100 = 97mg % potency = (97/100) × 100% = 97%

56. Choice of solvent: The choice of solvent to be used in ultraviolet spectroscopy is quite important: The first criterion for a good solvent is that it should not absorb ultraviolet radiation in the same region as the substance whose spectrum is being determined. Usually solvents which do not contain conjugated system are most suitable for this purpose. A second criterion is the effect of polar and non polar solvent on the fine structure of an absorption band. A non polar solvent does not form hydrogen bond with the solute and the spectrum of the solute closely approximates what it would be in a gaseous state. In a polar solvent the hydrogen bonding, form a solute solvent complex and the fine structure may disappear. Solvent must not contain trace impurities. Many impurities (e.g. Benzene in absolute alcohol) absorb radiant energy and complicate the analysis.

57. SolventUV absorbance cutoff (nm) Acetonitrile190 Water190 Hexane201 Methanol205 Ethanol205 Ether215 Chloroform240 Carbon tetrachloride257 Benzene280

58. Application of UV-Visible 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. Structure elucidation of organic compounds: spectroscopy is useful in the structure elucidation of organic molecules, such as, the presence or absence of unsaturation in the structure. 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 – Lambert’s Law.

59. Qualitative analysis: 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. 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. Detection of functional groups: This technique is used to detect the presence or absence of functional group in the compound. (But not as widely as IR Spectroscopy)

60. Quantitative analysis of pharmaceutical substances: 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. e.g. Diazepam tablet can be analyzed by 0.5% H 2 SO 4 in methanol at the wavelength 284 nm.

61. Question: Theoretically there should be sharp band in UV spectroscopy, but practically, broad bands are absorbed. Why? If a molecule absorbs UV radiation it usually causes electronic transition. In this case, the molecule absorbs radiation of a particular wavelength. So the UV spectrum should contain a sharp band. Inten sity (Wavelength, nm)  A B

62. at normal conditions, UV radiation causes vibrational and/or rotational transitions too. So molecules usually absorbed radiation at a relatively wide range. That is why, a broad band is observed. Question: What is the effect of increasing the number of double bonds of compounds on its UV-Visible spectrum?

63. Answer: Compounds having double or triple bonds contain electrons which are excited relatively easily. In molecules containing a series of alternating double bonds, the  electrons are delocalized and require less energy for excitation. So that the absorption occurs into higher wavelengths. When there are more conjugation, the compound will absorb less energy radiation.