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See all Premium member Presentation Transcript Slide 1: U.V-Visible Spectroscopy QUANTITATIVE ESTIMATION PRESENTED BY : Rohit Kaswala Slide 2: CONTENT Selection of analytical wavelength Direct comparison method Simultaneous determination Wood wards fieser rule’s Slide 3: QUANTITATIVE ESTIMATION FOR STRUCTURE DETERMINATION Quantitative method is for determining an unknown concentration of a given species by absorption spectrophotometry, here the first step is to choose the absorption band at which absorbance measurements are to be made. Limitation: The absorption band should not overlap absorption bands of the solvents or possible contaminants, including excess reagents that might be in the sample. 1.Selection of analytical wavelength : 1.Selection of analytical wavelength Also called as CALIBRATION CURVE method. Calibration curve is plotted using Concentration Vs Absorbance value of 5 or more standard solution. A straight line is drawn either through maximum no. of points or in a such a way that there is equal magnitude of positive and negative errors. This method is called as Line of best fit. From the absorbance of the sample solution and using the calibration curve, the concentration of drug, amount and the % purity can be calculated 2.Direct comparison method : 2.Direct comparison method Absorbance of std. solution of known concentration and a sample solution is measured. The concentration of unknown can be calculated using the formula A1 = ε C1t A2 = ε C2t A1, A2 = absorbance of standard and sample C1, C2 = Concentration of standard and sample ε = Mol. Ext. coefficient t = Pathlength (1cm) On divided, we get C2 = C1 X A1/A2 3.Simultaneous determination : 3.Simultaneous determination When no region can be found free from overlapping spectra of two chromophores or group of compound, it is possible to devise a method based on measurements at two or more wavelength. Two dissimilar compound have different power of radiation absorption at some or several points in their absorptive spectra. Therefore, measurements are made on an unknown solution at two wavelength where absorptivities of the two components are different, by which it is possible to set up two independent equation & solve them simultaneously for the two unknown concentration. Slide 7: 4. Woodward-Fieser Rules Robert Burns Woodward : Robert Burns Woodward U.V absorption rules developed by Robert Burns Woodwards (1917-1979) First (empirical) rules to predict the UV absorption λmax for certain classes of compound (Polyenes and ketones) Rules developed by Woodward in 1941 then expanded by Feiser in 1948. Woodward’s Early Observations : Woodward’s Early Observations Extension of conjugation in polyenes increases λmax and εmax Dienes : Dienes heteroannular diene(both double bonds are not in the same ring) homoannular diene(both double bonds are in the same ring) CH2 Enones : Enones Acyclic Enones Acyclic Dienone O 6-member cyclic O 5-member cyclic Double bond extending conjugation : Double bond extending conjugation O OH OCH3 2 double bond extending conjugation Slide 13: These are the influences of a neighboring saturated carbon on a double bond or at a site of unsaturation Ring residue and alky substitute Slide 14: Most acyclic dienes can rotate about their central single bond to give either a CISOID or a TRANSOID conformation S-Cis S-Trans S-Cis : Absorbs at longer wavelength , Lower intensity than S-Trans Slide 15: CISOID (HOMOANNULAR) TRANSOID (HETEROANNULAR) Slide 16: Homoannular diene (Cisoid) : cyclic diene having conjugated double bonds in the same ring. Heteroannular diene (Transoid) : cyclic diene in which double bond in conjugation are present in different rings. Slide 17: Exocyclic double bond Endocyclic double bond Endocyclic double bond: Double bond present ring Exocyclic double bond: A double bond that is connected to and external to a ring structure Woodward-Fieser Rules for conjugated Dienes, Polyenes, etc. : Woodward-Fieser Rules for conjugated Dienes, Polyenes, etc. Slide 19: CH3 H3C Parent value for Homoannular diene = 253 Two alkyl substitue, 2 x 5 = 10 Two ring residues, 2 x 5 = 10 Calculated Value: 273 Slide 20: C-CH=CH-CH3 CH3 Parent Value Acyclic Conjugated Diene = 215 2 Alky substitute = 2 x 5 = 10 2 Ring residue = 2 x 5 = 10 Exocyclic double bond = 5 _______________ Calculated Value = 240 Slide 21: OCH2 CH3 CH3 Homoannular ? Cyclic ? Acyclic ? Heterocyclic ? No. of Ring residue? 1 ? 2 ? 3 ? Exocyclic ? Endocyclic? Other Group ? Heteroannular = 215 3 Ring residue= 3 x 5 = 15 Exocyclic = 5 OR = 6 Total value: 241 Slide 22: OH Homoannular ? Cyclic ? Acyclic ? Heterocyclic ? No. of ring residue? 1 ? 2 ? 3 ? Exocyclic ? Endocyclic? No. Of alkyl substitute? 1? 2? 3? Acyclic = 215 2 Ring Residue = 2 x 5 = 10 Double Extending conjuagation = 30 Alkyl group= 1 x 5 = 5 CH3 Total Value = 260 Slide 23: Homoannular ring : = 253 4 Ring residue: 4 x 5 = 20 2 Exocyclic double bond : 2 x 5 = 10 253 + 20 + 10 = 283 nm Slide 24: COOH Homoannular ? Cyclic ? Acyclic ? Heterocyclic ? No. of ring residue? 1 ? 2 ? 3 ? Exocyclic ? Endocyclic? No. Of alkyl substitute? 1? 2? 3? Homoannular diene = 253 3 ring residue = 3 x 5 = 15 One exocyclic = 5 1 alkyl substitute = 5 Total value = 278 Slide 25: Wood wards fieser rules for α, β- unsaturated carbonyl compounds Slide 26: Parent value α, β unsaturated 6 membered cyclic ketone= 215 1 ring residue at α position= 10 2 ring residue at β postion = 2 x 12 = 24 Double bond exocyclic = 2 x 5 = 10 Total value= 259 Slide 27: Parent value = 215 nm α Ring residue = 10 δ Ring residue = 18 1 exocyclic double bond = 5 1 double bond extending conjugation = 30 1 homoannular diene = 39 Total value = 317 Slide 28: Parent value = 215 2 ring residue at β position = 2 x 12 = 24 1 exocyclic double bond = 5 Total value = 244 Slide 29: COCH3 O Parent value = 215 2 ring residue at β position = 2 x 12 = 24 1 exocyclic double bond = 5 Total value = 244 nm Slide 30: Parent value = 215 1 ring residue at β position= 12 1 ring residue at δ position = 18 Extending conjugation = 30 Exocyclic double bond = 5 Total value = 280 Slide 31: COCH3 O Five membered? Six membered ring ketone? No. of ring residue? 1 ? 2 ? 3 ? Exocyclic ? Endocyclic? No. Of extending conjugation? Six membered = 215 β = 1 ring residue, higher position = 2 ring residue , 12 + 18 +18 2 exocyclic = 10 2 x 30 = 60 Total value = 333 Slide 32: THANK YOU You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.