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1H-NMR of an Unknown Natural Product 13C-NMR of an Unknown Natural Product Too Little Information

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1H-NMR of an Unknown Natural Product 13C-NMR of an Unknown Natural Product Too Little Information 1H-NMR of an unknown natural product 13C-NMR of an unknown natural product Too little information 7.339 ppm 7.255 ppm Molecular Formulae Determining the structure of a molecule from NMR, FT-IR etc. usually requires also knowing the molecular formula. Molecular weight + Molecular formula Empirical formula Molecular weight alone… Rough estimates of the molecular weight are of limited use: Consider a molecule that is found to have a molecular weight of 84. There are many possibilities: C6H12 (isomers) O F N O S N O N F OH N F C3H4N2O C4H4O2 C4H4S C4H8N2 C5H8O C2H3F3 Combustion Analysis Leibig pioneered the determination of empirical formulae by carefully measuring the products obtained by burning organic compounds. This method is referred to as “combustion analysis”, Justus von Liebig “CHN analysis”, “microanalysis” or “elemental analysis” 1803-1873 excess O2 CxHyNz x(CO2) + y/2(H2O) + z(NO2) 1-5 mg sample CHN analysis Sample rapidly Inert carrier gas (e.g. He) carries volatile products from heated in O2 to induce combustion combusition chamber, adjusting the pressure to a standard pressure Sample with N2, H2O and CO2 passes through thermal nitrogen oxides (NO, Cl, P and S biproducts conductivity probe (TCP)-- NO2, etc) converted removed by solid measures total to N2; O2 removed substrate catalyst conduvitity CO2 removed and sample H2O removed and sample passed through 2nd TCP: passed through 3rd TCP: Difference in 1st and 2nd Difference in 2nd and 3rd measurement gives CO2 measurement gives H2O 3rd TCP measures N2 Schematic of CHN analysis Combustion -S, -P, -F, -Cl,-Br, -I CO2, O2, NOx, H2O S, P, halide impurities CO2, O2, NOx, H2O CO2, N2, H O NOx N2 -H2O TCP1 2 -O2 CO2, N2, N2, TCP2 -CO2 TCP3 Using the results… An unknown sample was found to have C, 64.30; H, 8.95; N, 12.51 remainder: (assume to be oxygen), 14.25 Convert from %weight to ratio of atoms; divide each number by atomic mass: C: 64.30/12.01= 5.35 H: 8.95/1.01 = 8.86 N: 12.51/14.01 = 0.89 O: 14.25/16.0 = 0.89 C5.35H8.86N0.89O0.89 C6.01H9.96N1.00O1.00 Using the results… • JOC standard: observed C, H and N should be within ±0.4% theoretical N C, 75.92; H, 6.37; N, 17.71 C, 75.60; H, 6.56; N, 17.50 pass C, 75.92; H, 6.37; N, 17.21 fail C, 75.60; H, 6.80; N, 17.51 fail C, 75.42; H, 6.67; N, 17.91 fail A refined picture… molecular weight of 84 S C4H4S C6H12 C, 85.63; H, 14.37 C, 57.10; H, 4.79; S, 38.11 N O N N N C3H4N2O C H N C, 42.86; H, 4.80; N, 33.32 4 8 2 C, 57.11; H, 9.59; N, 33.30 O F O F OH F C4H4O2 C2H3F3 C H O C, 57.14; H, 4.80 5 8 28.58; H, 3.60 C, 71.39; H, 9.59; Limitations of CHN C, 85.63; H, 14.37 (CH2)n Other problems: -incomplete burning -doesn’t “see” most elements -beware the solvent “fudge factor” Really a test of purity? A C, 85.63; H, 14.37 B C, 85.02; H, 14.98 2:1 mixture A:B C, 85.22; H, 14.78 Advantages of CHN Elemental analysis can “see” impurities that are invisible by other techniques. e.g. 5% LiCl in DMF has no effect on 1H-NMR (no protons in LiCl), and not evident in mass spectrometry. C, N and H all low by EA O CH H N 3 DMF CH3 Still too little information? 7.339 ppm C, 92.26%; H, 7.74% (CH)n C 49.02%; H, 2.74% 7.255 ppm (C3H2)n?? no N present (C3H2O2.7)n ?.
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