html5-img
1 / 52

Nuclear Magnetic Resonance (NMR) Spectroscopy

Nuclear Magnetic Resonance (NMR) Spectroscopy. basic theory 1. properties of nucleus spin of nucleus nuclear spin quantum number I = n/2 n : integer atomic mass number number Z A I example 1 even even 0 12 C, 16 O, 28 Si, 56 Fe

darena
Télécharger la présentation

Nuclear Magnetic Resonance (NMR) Spectroscopy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Nuclear Magnetic Resonance (NMR) Spectroscopy basic theory 1. properties of nucleus spin of nucleus nuclear spin quantum number I = n/2 n : integer atomic mass number number Z A I example 1 even even 0 12C, 16O, 28Si, 56Fe odd even n : even 2H, 10B, 14N, 50V odd odd n : odd 1H, 13C, 19F, 55Mn * NMR properties of some nuclei with I = 1/2 * NMR properties of some quadrupolar nuclei (I > 1/2) number of possible spin states = 2I + 1 magnetic quantum number m = +I, +(I-1), ……., -I without a magnetic field, the spin states are degenerate nucleus I No. of states m values 1 1H 1/2 2 +1/2, -1/2 11B 3/2 4 +3/2, +1/2, -1/2, -3/2 12C 0 1 0 14N 1 3 +1, 0, -1 1

  2. NMR properties of some nuclei with I = 1/2

  3. NMR properties of some quadrupolar nuclei (I > 1/2)

  4. 2. nuclear Zeeman effect a nucleus with I≠0 in a magnetic field, 2I+1 spin states are not degenerate; they separate in energy with the largest positive m value corresponding to the lowest-energy state ex. I = 1/2 m = -1/2 Bo E DE m = +1/2 Bo spin state energy gh Ei = -miBo ——g : magnetogyric ratio 2p transition Dm = -1 for a nucleus with I = 1/2, the energy difference g h Bo DE = ———— 2p precession – some sort of uniform periodic motion, the magnetic moment wobble around the axis of applied field Lamar frequency w = gBo linear Lamar frequency u = w/2p = gBo/2p

  5. Boltzmann distribution Pm=-1/2ghBo ——— = e –DE/kTDE = ——— Pm=+1/2 2p If Bo = 2.35 T DE = 6.63 x 10-26 J Pm= -1/2 Pm=-1/2 =0.4999959 ———— = 0.999984 Pm= +1/2 Pm=+1/2 =0.5000041 experimental considerations sample solution solid (magic-angle spin) magnet radio-frequency transmitter spectrometer receiver decoupler recording device

  6. magnet : permanent magnet (1 – 2 T) electromagnet (1.8 – 2.3 T) superconducting magnet (up to 13 T) two important characteristics of magnet • stability – sensitive to temperature • homogeneity continuous wave experiment 1. frequency-sweep 2. field-sweep

  7. Fourier transform technique relaxation processes spin-lattice relaxation T1 -t/T1 Peq – P = (Peq – Po)e spin-spin relaxation T2 much faster than spin-lattice relaxation T2 < T1

  8. information from NMR spectrum (1) chemical shift the nuclei are screen from the magnetic field Bo, the net field effective at a nucleus is Beff = Bo (1 – s)s : the shielding constant each chemically distinct nucleus is associated with a characteristic frequency ex. B10H14 4 distinct B nuclei chemical shift d relative to a standard for the isotope concerned uobs - uref d = 106× —————————— spectrometer frequency unit: ppm a shift to higher frequency than standard ==> positive d decrease in shielding ≡ increase in chemical shift

  9. relative NMR frequency (MHz) standard compound common nucleus (B0 = 4.7 T) reference range (ppm) 1 1H 200.0 (CH3)4Si -30 – 20 13C 50.2 (CH3)4Si -100 – 400 19F 188.2 CFCl3 -200 – 200 29Si 39.8 (CH3)4Si -350 – 40 31P 81.0 85% aq. H3PO4 -100 – 250 77Se 38.2 (CH3)2Se -300 – 200 119Sn 74.5 (CH3)4Sn -1000 – 8000 195Pt 43.0 [Pt(CN)6]2- -200 – 15000 (2) intensity integration of the area not for 13C (3) spin-spin coupling non-equivalent magnetically active nuclei couple each other chemically equivalent magnetically equivalent

  10. notationDd >> JA, X, M, Q Dd smallA, B, C splitting pattern 2nI + 1 coupling constant J ex. 1H, 13C NMR spectra of H13CO2-   (i) first-order (ii) satellites (iii) second-order (4)exchange

  11. number of lines splitting determined by Pascal’s triangle number of equivalent name of coupling nuclei pattern ratio of integration 0 singlet1 1 doublet1 1 2 triplet1 2 1 3 quartet1 3 3 1 4 quintet1 4 6 4 1 5 sextet 1 5 10 10 5 1

  12. AX2 ?? classification of the nuclei • I = 1/2, 100%abundance 1H, 31P, 19F, 103Rh • I = 1/2, low abundance 13C, 15N, 29Si, 77Se, 109Ag, 119Sn, 125Te, 183W, 195Pt, 199Hg • I > 1/2, 100% abundance 14N, 27Al, 51V, 59Co • I > 1/2, low abundance 11B, 121Sb, 193Ir (I)1H

  13. 52 Hz Sn(CH3)4 1H 119Sn expanded 1H13C 54 Hz 1J119Sn-13C= 329 Hz 1J117Sn-13C= 317 Hz

  14. GeH4 Si2H6(29Si I = ½, 4.7%)

  15. CH3–CH2–S–PF2 K[BH4]

  16. 1H NMR spectrum of PF215NHSiH3 3JPH = 8 Hz, 3JHH = 4 Hz, 2JNH = 2 Hz, 4JFH = 2 Hz 1H{15N} NMR spectrum of PF215NHSiH3

  17. [HV(CO)5]2- doublet of doublet of triplet J Pt-H = ? J Pc-H = ?

  18. PMe2Ph for CH3 (or C(CH3)3) group in tertiary phosphine complexes, doublet 1H spectra indicate mutually cis arrangements and triplet spectra mutually trans PtCl2(PMe2Ph)2

  19. LKH MCl3‧xH2O MCl3L3 MHCl2L3 (I) EtOH, 1h (M: Rh, Ir) D (L: PR3, AsR3) MHCl2L3 (II) (i) Ir, PEt2Ph 158 Hz 19 Hz 18 Hz 12 Hz

  20. (ii) Rh, AsMePh2 4 Hz 9 Hz

  21. 206 Hz (iii) Rh, PMePh2 2JHP = 9 Hz 1JRhH = 4 Hz 2JHP = 14.5, 9 Hz 1JRhH = 13.5 Hz

  22. M-CO M=C (II) 13C

  23. 1JPt-C = 35 Hz 2JPt-C = 0 Hz 1JPt-C = 1936 Hz 2JPt-C = 180 Hz

  24. DEPT

  25. 20 lines 5 lines [Ti(13CO)6]2- 13C 47,49 Ti

  26. (III) 31P

  27. 31P NMR spectrum of P(OMe)3 31P NMR spectrum of [Cu(PMe)3]+ 31P NMR spectrum of PHF2(15NH2)2

  28. 31P NMR spectra of the mixed products from the reaction of trans-[PtCl4(PEt3)2] + trans-[PtBr4(PEt3)2] [PxFy]- x = ? y = ?

  29. (IV) 19F 19F NMR spectrum of the products from the reaction: UF6 + Me3SiCl (halogen exchange)

  30. (V) 29Si 29Si NMR spectrum of SiMe4 (VI) 195Pt

  31. 2-D NMR 1. correlated spectroscopy (COSY) provide information about couplings between nuclei of a single isotopes the off-diagonal peak at a frequency (f1, f2) implies that there is a coupling between the nuclei resonating at f1 and f2 ex. COSY 11B spectrum for B10H14 B(2), B(4) (-35 ppm) B(6), B(9) (11 ppm) B(1), B(3) (13 ppm) B(5), B(7), B(8), B(10) (1 ppm)

  32. ex. COSY 11B spectrum for B9H11NH  2. heteronuclear correlation spectroscopy (HETCOR or HCOR) ex. HETCOR 11B/1H spectra for B10H14

  33. 3. nuclear Overhauser effect spectroscopy (NOSEY) identify a NOE which arises from the proximity of the two nuclei in space heteronuclear NOSEY (HOSEY) ex. 2D 1H/6Li HOSEY spectrum for tmeda adduct of 2-lithio-1-phenylpyrrole 

  34. ex. homonuclear 2D 13C scalar coupling (COSY) and chemical exchange (NOSEY) spectra for [Os3H2(CO)10] ex. CH2CH2Br O=P OCH2CH2OCH2CH2 expanded 1H NMR spectra

  35. COSY (1H—1H) COSY (1H—13C)

  36. exchange reactions ex. 1 31P{1H} NMR spectrum of the products derived from [Rh4(CO)9{P(OPh)3}3] under 400 atm of CO at 300 K Rh4 cluster broke down to 2 dinuclear complexes

  37. ex. 2 19F NMR at 180 K chemical shift pattern intensity d 68 ppm doublet of triplets 2 of doublets d -61 ppm triplet of doublets 1 of narrow triplets d 68 ppm triplets of quartets 1 230 K two higher-frequency resonances broaden and lose detail 300 K coalesced to a single broad line the lowest-frequency peak remained unchanged

  38. ex. 3 13C{1H} spectrum of Rh5(13CO)15- under pressure of 13CO (5 bar)

  39. ex. 4 13C{1H} spectrum of the CO region of (h5-C5H5)2Rh2(13CO)3

  40. ex. 5 31P{1H} NMR spectra of [Ru2Cl5(PEtPh2)4•Ag(PEtPh2)]

  41. ex. 6 trans-[IrCl(CO)(PMe3)2] + SF4 ―→ Cl P F Ir CO P SF3 d 68 -61 -383 ppm (2F) (1F) (1F)

  42. ex. 7 variable-temperature 1H NMR spectra of [Ru3W(C5H5)(CO)11H3]

  43. ex. 8 EXSY 2D 1H NMR of Pt complex

  44. solid state NMR spectroscopy difficulties – immobility of the nuclei in solids (i) dipolar coupling are not averaged to zero ==> very broad resonance (ii) chemical shift anisotropy in solids is not averaged out ==> line broadening (iii) relaxation time T1 is very long ==> good signal-to-noise ratio is difficult to get solution: (i) magic angle sample spinning (MASS, MAS) technique an angle q = 54.7o to the magnetic field, the effect of chemically anisotropy can be averaged out (ii) cross-polarization (CP) technique overcome the problem of long relaxation time

  45. ex. 1 13C NMR spectra of 2Ca(CH3CO2)2•H2O ex. 2 119Sn chemical shift of Ph3SnOH in solution d –80 ppm ==> 4-coordinated, tetrahedral in solid phase d –298 ppm ==> 5-coordinated similar to Me3SnF

More Related