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CHEM 344

CHEM 344. Organic Chemistry Lab. Structural Determination of Organic Compounds Lecture 3 – More NMR. January 26 th & 27 th 2009. Review of Lecture 2. NMR – detailed info on molecular structure Different protons within the molecule give different signals in the 1 H-NMR spectrum

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CHEM 344

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  1. CHEM 344 Organic Chemistry Lab Structural Determination of Organic Compounds Lecture 3 – More NMR January 26th & 27th 2009

  2. Review of Lecture 2 • NMR – detailed info on molecular structure • Different protons within the molecule give different signals in the 1H-NMR spectrum • # different protons = # signals • Integration = # protons giving the signal • Chemical shift value influenced by local structure of molecule (e.g. electronegative groups) • Equivalent protons = same chemical shift

  3. Typical 1H-NMR spectrum from Lecture 2 2.14 CH3 5.80 CH2 3 2 CHCl3 SiMe4

  4. Consider spin-spin splitting…….. 1.58 4.42 CH3 CH2 Triplet 3 Quartet 2 n + 1 rule

  5. 1.12 CH3 12 CH 3.64 2

  6. Coupling constant J (Hz) – indicates strength of coupling J ~ 7 Hz for alkyl (sp3) systems 3.64 20 Septet 15 15 1:6:15:20:15:6:1 6 6 1 1

  7. Pascal’s Triangle 1 Singlet 1 1 Doublet n = 1 1 2 1 Triplet n = 2 1 3 3 1 Quartet n = 3 Quintet 1 4 6 4 1 n = 4 1 5 10 10 5 1 Sextet Septet 1 6 15 20 15 6 1

  8. 1.12 1:1 Doublet 12 3.64 2

  9. From Lecture 1……… Why isn’t Hb a doublet? 2.29 Equivalent H’s do not split each other 7.04 Ha Hb 6 4

  10. From Lecture 1……… Why is Hb a singlet and not a triplet? 0.87 H’s on heteroatoms (O,N,S) do not couple Ha 9 2.42 NH2 1.04 Hb 2 2

  11. 1.18 6.63 7.02 2.79 3.51

  12. 1.18 CH3 Septet & doublet NH2 Typical iso-propyl group pattern 6 CH 3.51 2 2.79 1

  13. Why is Ha more shielded than Hb? Ha Hb 6.63 7.02 Consider the substituents Consider resonance structures 2 2

  14. Ha more shielded than Hb in structures A and C NH2, OMe etc. are activating groups Activating groups direct e- density to the o and p positions

  15. 3.91 OMe 3

  16. Hb Ha 8.18 6.97 Why is Ha more shielded than Hb? Consider the resonance structures 2 2

  17. Hb deshielded relative to Ha in structures A and C -NO2, -NR3+, -CF3, -CO2R etc. are deactivating groups Deactivating groups reduce e- density at the o and p positions Remember that the OMe group will direct e- density to the Ha protons

  18. Coupling constants in aromatic systems Jab Jab = Jortho= 6 – 12 Hz

  19. Coupling constants in aromatic systems Jab ≈ Jad ≈ Jbd = Jmeta= 1-3 Hz Jac = Jpara= 0 - 1 Hz Jortho > Jmeta> Jpara Jbc ≈ Jcd = Jortho= 6 – 12 Hz

  20. 3.89 OMe 3

  21. 7.48 Ha 7.71 Hc Hb Hd 7.79 7.23 1 1 1 1

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