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NMR N uclear M agnetic R esonance

NMR N uclear M agnetic R esonance. Proton NMR. Index. NMR-basics. Anisotropy of Aromatic compounds: in plane and above. d ring  7.27-6.95 ppm. d Me  -0.51 ppm. d ring  8.14-8.64 ppm. d Me  -4.25 ppm. d OUTSIDE  9.28 ppm. d INSIDE  -2.99 ppm. Anisotropy: Aromatic. Deshielded.

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NMR N uclear M agnetic R esonance

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  1. NMRNuclear Magnetic Resonance ProtonNMR Index NMR-basics

  2. Anisotropy of Aromatic compounds: in plane and above dring 7.27-6.95 ppm dMe -0.51 ppm dring 8.14-8.64 ppm dMe -4.25 ppm dOUTSIDE 9.28 ppm dINSIDE -2.99 ppm

  3. Anisotropy: Aromatic

  4. Deshielded Electronic effects 7.10 ppm 6.28 ppm 6.83 ppm 5.93 ppm 6.10 ppm 7.07 ppm 6.38 ppm 7.71 ppm 6.28 ppm

  5. Electronic effects: conjugation with carbonyl

  6. Electronic effects: conjugation with carbonyl deshielded

  7. Electronic effects: conjugation with heteroatom shielded 6.06 ppm 5.48 ppm 5.81 ppm 6.22 ppm 5.78 ppm 4.82 ppm

  8. Electronic effects: no conjugation with heteroatom

  9. Electronic effects: conjugation with heteroatom shielded

  10. Electronic effects: conjugation with carbonyl m o p deshielded deshielded

  11. Electronic effects: conjugation with heteroatom Shielded shielded o m p

  12. Electronic effects: conjugation with heteroatom Shielded m o p shielded

  13. Aromatic: inductive effect and resonance effect

  14. Hydrogen bond

  15. Protons on Heteroatoms • OH, NH, SH • Exchangeable (with D2O) • Hydrogen bonding • On Nitrogen (14N), as the spin state of that nuclei is 1, there can be partial coupling that produce broaden lines. There can be also full coupling that would produce 3 lines of equal intensity (I=1has 3 orientationsin a magnetic field)

  16. Protons on Heteroatoms • OH • Aliphaticd 0.5-4.0 ppm (depend on Concentration) • Intramolecular hydrogen bonding deshield OH and render it less sensitive to concentration • Usually OHexchange rapidly (no coupling with neighbors • In DMSO or Acetone, the exchange rate is slower => there is coupling with neighbors • Phenols : d 7.5-4.0 ppmIntramolecular bond  12-10 ppm • Carboxylic Acids: Exist as Dimers  13.2-10 ppm

  17. H2O signal moves with temperature H2O

  18. OH in DMSO CH3-CH2-OH CH2 qd OH (CH3)2 -CH-OH OH CH

  19. Protons on Heteroatoms • NH : 14N: I=1 => 2I+1 lines • NHhas different rate of exchange • 14Ncan relax quickly. Depending on relaxation rate, heteronuclear coupling will be visible or produce broadened peaks. • R-NH : Aliphatic amines => rapid exchange • Sharp singlets : no coupling to N: d~3-0.5 ppm • R-NH: Amides, Pyrroles, Indoles, Carbamates • NH broad • CHa shows coupling the NH

  20. NH

  21. Amide

  22. Protonated Amines

  23. Formamide H{14N}-NMR H-CO-NH2 H-NMR

  24. NH: Amide, PyrroleIndole d: 8.5-5.0 ppm In Amides: Slow rotationcan show different isomers In Amine Salt: • Moderate Rate of exchange => broad peaks ~ d 8.5-6.0 ppm • CHa => show coupling to NH+ Sometimes broad [NHx+]consist of 3 broad humpdue to 14N coupling 1JNH ~ 50 Hz

  25. SH • Slow exchangeSH couple to CHa • When shaken with D2O, SH Disapeard ~ 1.6 – 1.2 ppm Aliphatic SHd ~ 3.6 – 2.8 ppm Aromatic SH

  26. Chemical Shift and Coupling

  27. I = 10 + 1 – 12/2 = 5 2H 3H 3H 2H 2H O O An example: C10H12O2 O-CH2-CH3 J=7 Hz Me-C= J=7 Hz J=8 Hz Me-C=C X 4 = 12

  28. Scalar coupling: Coupling through bond2nI + 1 lines n = 0 1 2 3 4 5 6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 a doublet a b b septet

  29. Scalar coupling: Coupling through bond2nI + 1 lines n = 0 1 2 3 4 5 6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 o m b p a m o p 2 x triplet a b

  30. Scalar coupling: Coupling through bond2nI + 1 lines n = 0 1 2 3 4 5 6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 c 2 x triplet 1 quintet a b a c b

  31. Scalar coupling: Coupling through bond C7 H14 O2 I = 7 -14/2 + 1 = 12nI + 1 lines • (ppm) Intmult J (Hz) COMMENT • 0.9 3H triplet 7 CH3->(CH2) • 1.1 3H triplet 7 CH3->(CH2) • 1.35 2H sixtet 7 CH2 (CH3, CH2) • 1.55 2H quintet 7 CH2 (CH2, CH2) • 2.3 2H quartet 7 =C- CH2 (CH3) • 4.1 2H triplet 7CH2-O (CH2) n = 0 1 2 3 4 5 6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 2 x triplet CH2 2.3 O CH3 CH2 CH3 1.1 CH2 Roof effect 0.9 ppm CH2 1.35 3H Triplet: 3H O 2H Quartet: 2H Quintet: Sixtet: 2H 2H

  32. Scalar coupling: Coupling through bond2nI + 1 lines n = 0 1 2 3 4 5 6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 2 x triplet 6 1 Triplet: 4 Quartet: 5 Quintet: 3 Sixtet: 2

  33. Ha Hb Hb Ha Hb C C Hb Hb C C C Hb Hb Ha Hb Hb C C Hb Ha Hb C C C Ha Hb Hb C C Hb Common first order spin system 2nI + 1 lines

  34. Hc Ha Hb Hb C C C Hb Jab = Jab’ Ha Hb Hb’ C C Hb’ Hb’ Hc’ Ha Ha Ha Hb Hb Hb C C C C C C C C C Hb’ Hb Common first order spin system 2nI + 1 lines Jab Jab qd Jab’ td

  35. Geminal Coupling

  36. Vicinal Coupling 3J => Perch 3J => tool 1 3J => Mestrec tool

  37. Ha Using Vicinal Coupling to establish isomer Jab Jad Jac

  38. Long Range Coupling

  39. Long Range coupling 4JH1-H3 = 1.07 Hz 5JH1-H4 = 1.21 Hz 5JH1-H5 = 0.95 Hz 5JH4-H7 = 0.67 Hz 4JH-H = 9 Hz 5JH-H = 3 Hz 4JH-H = 1-2 Hz 4JH-H = 3 Hz 4JH-H = 1.1 Hz 5JH-H = 3 Hz

  40. Spin System in Pople notation Structural Unit Spin system Partial spectrum -CH2-CH3 A3X2 -CH-CH3 A3X CH2-CH2-CH3 A3M2X2 Each chemical shift is represented by a letter (far way letter for very large shift difference – compare with the size of the coupling)

  41. Dn=  (1-4) * (2-3) Dn Second Order spectra:AB instead of AX Dn / J J J 5.0 1 2 3 4 As the difference in shift become smaller- compare with the size of the coupling the outer peaks become smaller in intensity 4.0 3.0 nAandnB : center of gravity of doublet Chemical shift 2.0 1.0 SpinWorks => load AB 0.5

  42. AB-Spectra

  43. AMX C6 H4 O5 N2 I = 6 - 4/2 + 2/2 +1 I = 6 Phenyl = 4 I NO2 = 1 I

  44. A2X and A2B SpinWorks => load A2B

  45. AMX

  46. OMe 7.58 ppm 7.58 ppm OMe CHO 6.83 ppm 6.83 ppm 7.2 ppm 7.2 ppm CHO OMe OMe AMX Substituants : 2 OMe (~ 3.9 ppm) CHO (~ 9.8 ppm) J Meta Para J Ortho Meta J Ortho Para

  47. meta bromo nitro benzene Calculated shifts dHA=8.44 dHB=7.82 dHC=7.31 dHD=8.19 HA HB HC HD

  48. AFMX C5 H4 N Br I = 5 – 4/2 – 1/2 +1/2 +1 I = 4 (aromatic ring) d J

  49. Assignment of 1H NMR of: cartilagineal Me CHO

  50. Jcis=10.5 dd H-5 dd 3J4,5 = 8.5 4J3,5 = 1.0 H-3 ddd 3J3,4=15.5 4J3,5 =1.0 4J4,9=2.0 CHO-9 J = 2.0 Hz H-1(s) Jtrans=17 Hz 3J3,4=15.5 3J4,5 =8.5 H-4 dd

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