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Applications of uv visible spectroscopy

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various applications of UV-Visible Spectroscopy in pharmaceutical chemistry.

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Applications of uv visible spectroscopy

  1. 1. SINDHOORA D First M.pharm Department of Pharmacology 1
  2. 2. CONTENT:  Introduction to Uv-visible spectroscopy.  Applications of Uv-visible spectroscopy.  Brief introduction to woodward –fieser rules. 2
  3. 3. Introduction to UV-Visible spectroscopy  UV-Visible spectroscopy is also known as electronic spectroscopy.In which the amount of light absorbed at each wavelength of UV and Visible region of electromagnetic spectrum is measured.  This absorption of electromagnetic radiation by the molecules leads to molecular excitation. 3
  4. 4.  This is called Electronic spectroscopy since it involves the promotion of electron from the ground state to the higher energy state.  It is very useful to measure the number of conjugated double bond and also aromatic conjugation within the various molecule.  It is also distinguishes between conjugated and non- conjugated system;ᾳ,β-Unsaturated carbonyl compound from β,ϒ-analogues; homoannular and heteroannular conjugated diens etc. 4
  5. 5.  Excitation occurs in the range 200-800 mᶙ. 5
  6. 6. APPLICATION OF UV- VISIBLE SPECTROCOPY  UV –Visible spectroscopy has been mainly applied for the , Detection of unknown compound. Detect the extent of conjugation Detection of polynuclear compounds by comparison etc. 6
  7. 7. Extent of conjugation :  The extent of conjugation in polyene can be estimated, R-(CH=CH)n-R .  Addition in unsaturation with the increase in the no.of double bond ( increase in the value of n)shifts the absorption to longer wavelength .  It is found that the absorption occurs in the visible region ,i.e at about 420mᶙ.If n=8 in the above polyene, such an alkene appears colouerd to human eye. 7
  8. 8. Distinguish in conjugated and non conjugated compounds  It is also distinguish between a conjugated and an non conjugated compounds. 8
  9. 9.  The forbidden n→π* band for the carbonyl group in the compound (i)will appear at longer wavelength compared to that for the compound (ii)  The alkyl substitution in an alkene causes a bathochromic shift. 9
  10. 10. Identification of an unknown compound:  A unknown compound can be identified by comparing its spectrum with the known spectra  If the two spectra coincide , the two compound must be identical.  If the two spectra do not coincide, then the expected structure is different from the known compound. 10
  11. 11. Examination of polynuclear hydrocarbon :  Benzene and polynuclear hydrocarbon have charactersitics spectra in the UV –Visible region .  Thus the identification of the polynuclear hydrocarbon can be made by comparision with the spectra of known polynucler compound .  The presence of substituent of on the ring , generally ,shifts the absorption maximum to longer wavelength. 11
  12. 12. Identification of a compound in different solvent:  Structure of the compound changes in the solvent .Chloral hydrate shows an absorption maxima at 290mᶙ in hexane while the absorption decreases in the aqueous solution.  The compound in the hexane contains a carbonyl group (a) where in aqueous solvent as (b)  (a)CCl3.CHO.H2O, (b)CCl3.CH(OH)2. 12
  13. 13. Elucidation of the structure of vitamin A and K:  Elucidation of the structure of vitamin K1& K2 and vitamin A1&A2.  The UV spectra of vitamin K1&K2 are due to the presence of the same chromophore ,i.e,2,3 dimethyl naphtha –quinone. The absorption maxima of this compound are 243,249,260,269 and 330mᶙ .  The elucidation of the structure of vitamin A1&A2 are possible by this method . 13
  14. 14.  Vitamin A1 absorbs at 325 mᶙ and for Vitamin A2 at 287 and 351 mᶙ.  The absorption maxima occurs at longer wavelength for Vitamin A2 due to presence of additional ethylenic bond. 14
  15. 15. 15
  16. 16. Preference over to tautomeric form:  Consider 2-hydroxy pyridine which exists in equilibrium with its tautomeric form,pyridone -2.  The spectra of these two compound were found to favour pyridone -2 which is an ᾳ,β-unsaturated ketone and clearly ,the equilibrium is shifted towards the right ,i.e., pyridone-2. 16
  17. 17. 17
  18. 18. Determination of configuration of geometrical isomers:  The results of absorption shows that cis-alkene absorbs at different wavelength as compared to their corresponding trans-isomer  The distinction is possible when one of the isomers is forced to be non coplanar by steric hindrence.Thus ,cis forms suffer distortion and absorption occurs at lower wavelength. 18
  19. 19. 19
  20. 20. Determination of strength of hydrogen bonding:  Solvents like water ,acetone forms a hydrogen bond with the n-electron of the carbonyl oxygen .Due to this the energy of n-electron in the ground state is lowered depending on the strength of hydrogen bond .Thus n→π* transition of carbonyl compound is shifted towards shorter wavelength . Hence, by measuring the λmax of the carbonyl compound in a non-polar and polar protic solvent, the strength of hydrogen bond can be determined. 20
  21. 21.  Example:  n→π* transition of acetone in hexane (at 279 nm)and that in water at 264.5 nm. Blue shift by 14.5nm corresponds to an energy of kcal/mol. The strength of hydrogen bond between water and acetone is 5kcal/mol. It is in fair agreement with the known strength of hydrogen bond. 21
  22. 22. Hindered rotation and conformational analysis  Any deviation from the co-planarity of double bonds will results in the reduction of π orbital overlap and thus λ max shifts to lower values of λ max and ϵ max.  Eg 2,3-di-tertbutyl-1,3-butadiene. In this case, there is little conjugation between 2 double bonds due to bulky tert-butyl groups. It shows absorption at lower wavelength (λ max= 180nm) compared to 2,3- dimethyl-1,3-butadiene (λ max= 225nm)in which there is no deviation from co-planarity. 22
  23. 23. 23
  24. 24. Detection of impurities  UV absorption spectroscopy is one of the best method 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 materials. By also measuring the absorbance at specific wavelength, the impurities can be detected. 24
  25. 25. Quantitative analysis  UV absorption spectroscopy can be used for the quantitative determination of compound that absorbs UV radiation. This determination is based on Beer’s law which follows, A= log I˳/ It = log 1/T = -log T = abc = ϵbc where ϵ = extinction coefficient c = concentration b = length of the cell that is used 25
  26. 26. Other methods are:  Calibration curve method  Simultaneous multi component method  Difference spectrophotometric method  Derivative spectrophotometric method 26
  27. 27. Molecular weight determination  Molecular weight of a compound can be measured spectrophotometrically by preparing the suitable derivatives of these compounds.  Eg: To determine the molecular weight of amine then its is converted to amine picrate. Then known concentration of amine picrate is dissolved in a litre of solution and its optical density is measured at λ max 380nm. After this the concentration of this solution in grams mol/litre can be calculated using C = (log I o / It )/ ϵ max x 1 27
  28. 28.  The various application follows  Woodward-Fieser rule for calculating λmax in Diens and Triens  Woodward-Fieser rule for calculating λmax in ᾳ,β- unsaturated compound  Woodward-Fieser rule for calculating λmax in Derivatives of Acyl benzene 28
  29. 29.  Woodward-Fieser rule for calculating λmax in Diens and Triens 29
  30. 30. Woodward-Fieser rule for calculating λmax in ᾳ,β-unsaturated compound a) The basic value of ᾳ,β-unsaturated ketone as 215 mᶙ CH-COX,If X=alkyl group is 215mᶙ,if X=H then 207mᶙ,if X=OH or OR then 193mᶙ. b)If the double bond and the carbonyl group are contained in a five membered ring,then the basic value is 202mᶙ 30
  31. 31. 31
  32. 32. 32
  33. 33. Woodward-Fieser rule for calculating λmax in Derivatives of Acyl benzene The basic value is 246mᶙ,if X=alkyl or alicyclic residue If X=H,then basic value becomes 250mᶙ If X=OH or OR,then 230 mᶙ 33
  34. 34. 34
  35. 35. References: 1) Elementary Organic spectroscopy –Y.R Sharma, Revised edition , S.Chand publisher, New Delhi 2) Instrumental and biomedical analysis, by Ravishankar ,Fifth edition 3) Quantitative analysis of Drugs in pharmaceutical formulations –P.D. Sethi, Third edition,CBS publisher ,New delhi 35
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