For the structure determination of the organic molecules from their spectra, the followingsome general approaches are helpful: •Perform all routine operations: Determine the molecular weight from the Mass Spectrum. Determine relative numbers of protons in different environments from the 1H NMR spectrum. Determine the number of carbons in different environments and the number of quaternary carbons, methine carbons, methylene carbons and methyl carbons from the l3CNMR spectrum (C, CH, CH2 and CH3). Examine the problem for any additional data concerning composition and determine the molecular formula if possible. From the molecular formula, determine the degree of unsaturation. Determine the molar absorbance in the UV spectrum, if applicable. •Examine each spectrum (IR, mass spectrum, UV, l3CNMR, 1H NMR) in turn for obvious structural elements: Examine the IR spectrum for the presence or absence of groups with diagnostic absorption bands e.g. carbonyl groups, hydroxyl groups, NH groups, C=C or C=N, C≡N, NO2, etc. Examine the mass spectrum for typical fragments e.g. PhCH2- (91), CH3CO- (45), CH3- (15), Ph (77), etc. Examine the UV spectrum for evidence of conjugation, aromatic rings, conjugated carbonyl, etc. Examine the lH NMR spectrum for CH3- groups, CH3CH2groups, aromatic protons, CH2-X, exchangeable protons etc. Write down all structural elements you have determined. Note that some are monofunctional (i.e. must be endgroups, such as -CH3, -C≡N, -NO2) whereas some are bifunctional (e.g. -CO-, -CHz-, -COO-), or trifunctional (e.g. CH, N). Add up the atoms of each structural element and compare the total with the molecular formula of the unknown. The difference (if any) may give a clue to the nature of the undetermined structural elements (e.g. ether oxygen). At this stage, elements of symmetry may become apparent. Try to assemble the structural elements. Note that there may be more than one way of fitting them together. Spin-spin coupling data or information about conjugation may enable you to make a definite choice between possibilities. •Return to each spectrum (IR, UV, mass spectrum, l3CNMR, lH NMR) in turn and rationalise all major features (especially all major fragments in the mass spectrum and all features of the NMR spectra) in terms of your proposed structure. Ensure that no spectral features are inconsistent with your proposed structure. Molecular formula determination: Elemental Analysis data and GC-MS may be used directly to determine the number of C, H, N and S atoms (% composition) in the molecular formula. Each molecule passing into the Mass Spectrometer is first converted by electron ionization (EI) to the molecular ion (M.+) in which the molecule has simply lost one electron. The ratio of intensities of M.+, (M+1).+ and (M+2).+ The M+2 peak % will generally be 1% or less than that of the M peak % for small to medium sized organic molecules containing only C, H, N and O. A sizeable M+2 peak indicates the presence of (A+2) elements, most notably Cl and Br. Unsaturation Number: For a molecule having molecular formula CxHyNzOn, The number of double bonds + number of rings = [(1 x x) – (0.5 x y) + (0.5 x z) + (0 x n) + 1]. This is a direct result of the known valences of these elements (C = 4, H = 1, N = 3, O = 2). For other elements, all that is needed is to know the valence of the element. Halogens can safely be said to have valence = 1 for the majority of organic compounds (so replace by H). The other elements like S and P having more than one valences so can be detected before calculation. 1H NMR (integration): The integration of the 1H NMR spectrum is the most reliable method for determining the number of protons in the molecular formula. Only two problems can interfere with this: Stray peaks in the 1H NMR spectrum may be mistakenly counted. Exchangeable protons (e.g., NH, OH) are broadened and may be exchanged with D2O present in some deuterated solvents (e.g. d6-acetone) thus giving a lower integral. Occasionally, these peaks (e.g., -COOH) may be overlooked altogether. 13C NMR: The number carbons should be checked with the number of lines in the 13C NMR. First, make sure that which peaks are from the deuterated solvent. Since 13C NMR spectra are generally acquired with broadband 1H decoupling, all resonances will appear as sharp singlets unless they are coupled to other nuclei such as 2D (I = 1), 19F (I = 1/2) or 31P (I = 1/2). Example: Deduce the structure of the organic molecule based on following spectra. The elemental analysis of molecule is C 57.74%, H 4.24%, N 6.34% and S 0.0%. Ans.: Structure determination: The molecular formula of the compound can be determined either from elemental analysis of by using C-13 rule. The GC-mass spectrum shows a highest ion pattern at m/z 209 which will be assumed to be basic molecular ion peak M+. From the intensity (~ 36% with respect to basic molecular ion peak) of the (M+2)+ peak, one Cl is clearly present in the molecule. •Molecular formula determination from elemental analysis: From the elemental data analysis, the number of carbon, hydrogen and nitrogen atoms present can be calculated as follows: C 57.34%, H 4.24%, N 6.34% Number of carbons = = 10.05 ≈ 10 Number of hydrogens = = 9.60 ≈ 8 or 9 Number of nitrogens = = 1.005 ≈ 1 However, from the NMR integration, it is clear that there are 8 H. The 13C NMR spectrum shows 10 lines, which is consistent for 10 C if the molecule has no symmetry. Adding up the groups so far, 10 x 12 + 8 x 1 +1 x 14 + 1 x 35 = 177. But molecule having mass 209 so remaining mass is from oxygen which is not determine by elemental analysis; 209 - 177 = 32 i.e. molecule containing two oxygen atoms. The molecular formula is C10H8NO2Cl. 1H •Molecular formula determination by using C-13 rule: The molecular formula is also determined by using C-13 rule. The mss spectra of the molecule give basic ion molecular peak at 209 and 211 (36% to molecular ion peak) indicate it contain one chlorine atom. Therefore molecular mass is 209 (Cl-35) and 211 (Cl-37). The molecular formula for corresponding hydrocarbon is determined as follows: Total number of carbon atoms present = = 16.0769. Total number of hydrogen atoms present = 16 + 0.0769 x 13 = 17. Therefore molecule formula for corresponding hydrocarbon is C16H17. But molecule containing one chlorine (replace 3C and add one H by Cl) and nitrogen atom (replace CH2 by one N) because mass of the molecule is odd which indicate it should contain odd number of nitrogen atom. Therefore molecular formula of molecule is C12H16NCl. But CMR spectra of molecule gives 10 singles indicates presence of 10 carbon atoms (only if molecule does shows any symmetry) while the intergral area under the peak in PMR spectra indicate it should contain 8 hydrogen atoms. Also IR spectra of the molecule shows strong absorption band in carbonyl region i.e. at 1732 cm-1 indicates molecule should contain ester carbonyl group which is shielded by conjugation. Therefore other possible molecular formula is (replace –CH4 by one O) C11H12NOCl or C10H8NO2Cl. •Determination of site of unsaturation: Replace Cl by one H, therefore molecular formula of corresponding hydrocarbon is C10H9NO2. Site of unsaturations = [(1 x x) – (0.5 x y) + (0.5 x z) + (0 x n) + 1] = 10 - 4.5 + 0.5 + 1 = 7. It indicates that molecule containing C=C, C=O, C=N, C≡C, C≡N, rings, -Ph, etc. •Identification of Unsaturated Functional Groups A cyano group (2254 cm-1) and a carbonyl group (1732 cm-1) are clearly present in the IR spectrum. The aromatic CH bond shows absorption at 3050 cm-1. The IR spectra does not shows any absorption in the –OH and NH2 region indicates these groups are absent. The carbonyl group present in the molecule is of either shielded ester or highly deshielded amide. The fact that 10 lines are found in the CMR spectrum for 10 C means that each line represents exactly one carbon of the structure. CMR spectra: 165.0, 133.6, 133.5, 131.8, 131.4, 130.2, 127.6, 118.0, 60.6, 17.9 ppm PMR spectra: 7.90 (ddd, 1H, J = 0.5, J = 2.0, J = 8.0 Hz), 7.59 (ddd, 1H, J = 2.0, J = 7.0, J = 8.0 Hz), 7.56 (ddd, 1H, J = 0.5, J = 2.0, J = 8.0 Hz), 7.47 (ddd, 1H, J = 2.0, J = 7.0, J = 8.0 Hz), 4.57 (t, 2H, J = 6.0 Hz), 3.04 (t, 2H, J = 6.0 Hz). Structure: Mass fragmentation: Example: Deduce the structure of the organic molecule based on following spectra. Example: Deduce the structure of the organic molecule based on following spectra. Example: Deduce the structure of the organic molecule based on following spectra. Example: Deduce the structure of the organic molecule based on following spectra. Example: Deduce the structure of the organic molecule based on following spectra. Example: Deduce the structure of the organic molecule based on following spectra.