1 / 52

NMR N uclear M agnetic R esonance

NMR N uclear M agnetic R esonance. Introduction. Index. NMR Basic Concept. http://faculty.ccc.edu/cabrams/. Orientation of the magnetic moment in absence/presence of a magnetic field. Random orientation. In presence of a magnetic field Magnetic moments precess and

ping
Download Presentation

NMR N uclear M agnetic R esonance

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. NMRNuclear Magnetic Resonance Introduction Index

  2. NMR Basic Concept http://faculty.ccc.edu/cabrams/

  3. Orientation of the magnetic moment in absence/presence of a magnetic field Random orientation In presence of a magnetic field Magnetic moments precess and Orient with or against the field

  4. Precession

  5. B0 a m = + ½ b m = - ½ Directional Quantization If a nucleus with angular momentum P and Magnetic Moment m is place in a magnetic field B0 The angular momentum takes an orientation along B0 such that it is an integral or half integral of h (Plank constant) m is the magnetic or directional quantum number it can take values: m = I, I-1 , …. , -I Pz = + ½ h Pz = - ½ h There are then 2I + 1 orientations in the field

  6. B0 a m = + ½ Pz = + ½ h Pz = - ½ h b m = - ½ Directional Quantization The Magnetic Moment m = m g hprecess around the z-axis (magnetic field B0) nL =gB0/2p The Energy of a magnetic dipole in the magnetic field B0 is: E = - m B0 Where m = m g h For a nuclei with (2I+1) orientations the energy of the individual states (Zeeman levels) is: E = - mg hB0

  7. Energy level and resonant transitions For spin I = 1 m = - 1 E = - 1 g hB0 m = - ½ m = 0 E = 0 m = + 1 m = + ½ E = + 1 g hB0 For spin I = ½ the energy gap between the 2 levels is: DE = g ħB0 = h n n = g B0 /2p E E = - ½ g hB0 DE E = + ½ g hB0

  8. Energy vs Field strenght E b DE = g ħB0 a B0 nL 60 MHz 200 MHz 300 MHz e.g. with B0=1.4 T (nL=60 MHz) DEH ~2.4 J/mol At T=300K Nb = .9999904 Na with B0=7.05 T (nL=300 MHz) Nb = .99995 Na Boltzmann Equation Nb = e (DE/kT) ~ 1 - DE/kT Na Where k=1.3805 x 10-23 J K-1 T is the temperature

  9. Field Dependance of a spectra

  10. Basic Concept review Even mass: # protons & # Neutrons Both Even : I=0 (4He, 12C …) # Protons & # Neutrons Both Odd : I=1, 2, …. (Integer) Odd mass: # protons odd & # Neutrons Even : I=1/2, 3/2, ...(Half-integer) # protons even & # Neutrons odd : I=1/2, 3/2, ...(Half-integer)

  11. Multinuclear NMR MHz

  12. Elements in Organic Chemistry There are 4 important elements in Org. Chemistry: (frequency at 2.35T) • H : Best NMR element • 1H , I = ½(sharp lines),a = 99.98%, high frequency (n =100 MHz) • 2H , I =1 (broad lines), a = 0.02%, n = 15.4 MHz, Q=0.038 • C : • 12C , I = 0 (no signal in NMR) • 13C ,I = ½(sharp lines),a = 1.1%, n =25.3 MHz • N : • 14N , I =1 (broad lines), a = 99.6%, n = 7.2 MHz, Q=1.0 • 15N, I = ½(sharp lines),a = 0.4%, n =10.1 MHz • O : • 16O , I = 0 (no signal in NMR) • 17O , I =5/2 (broad lines), a = 0.04%, n = 13.5 MHz, Q=-0.037

  13. Elements in Organic Chemistry • N : • 14N , I =1 (broad lines), a = 99.6%, n= 7.2 MHz Q=1.0 • 15N, I = ½(sharp lines),a = 0.4%, n=10.1 MHz Q=0 Although N-14 has large natural abundance, the presence of quadrupole moment and very low frequency make it impractical for NMR studies. The presence of quadrupole moment in that abundant element alter the shape of nearby protons and carbons N-15 is a better candidate but the low abundance & low frequency render it’s observation extremely difficult.

  14. Additional Elements in Organic Chemistry • Halogens: 19F, Cl, Br, I • 31P, S • B, 29Si • Na • Co, Cd, W, Pt, Hg…

  15. 1D NMR : Pulse ->FID Acquisition -> Fourier Transform

  16. RF Flip Angle (Nutation Angle)

  17. The Pulse

  18. Pulse Calibration

  19. FID

  20. Free Induction Decay : FID

  21. Sampling rate and Foldover

  22. O1 SW r.f. DIGITALLIZATION Digitallization: Convert FID (Volt/Time) in Digital form Digitallization process is limited by: • Accuracy • Speed Carrier OffsetorTransmitter Offsetor “O1” is the frequency of the irradiating field. It is also the “Reference” or “Rotating Frame” frequency The “Window” or “Spectral Width” also called “SW” define the range of frequencies that can be measured _1__2*SW The Sampling Rate => 2 Points/Cycle Dwell Time = DW=

  23. If Maximum Frequency to be sampled is fmax = SW We must sample at a rate no less than 2 * SWsec. Digital Resolution The amount of memory limit the accuracy of the signal to be recorded For a given # of memory(# Points -> TD (time domain)), one obtain: NP(real) andNP (Imaginary) 22 Digital Resolution = D.R. = Df (Separation between 2 points) D.R. = 2 * SWNP

  24. Digitallization : resolution and AcQuisition time

  25. Example At 200 MHz If: SW=2000 Hz (10 ppm) TD = 16,000 points (16K) What is the Digital Resolution: D.R. = 2*SW/TD = 4000 / 16,000 = 1 / 4 = 0.25 Hz What is the Acquisition Time AQ: AQ = TD * DW = TD / (2 * SW) = 4 seconds D.R. = 1 / AQ = 2 * SW / TD

  26. FID and Fourier Transform

  27. FID duration vs Line Width

  28. Relaxation Time

  29. Relaxation Effects T1: Spin-Lattice Relaxation: reestablish population equilibrium along Z Convert Spin Energy to thermal energy => spin transition from upper to lower state T2 : Spin-Spin Relaxation: Dephase nuclear dipole in the XY plane There are 5 mechanism for T1 Relaxation • Dipole-Dipole : most important => intramolecular interaction between nearby nuclei. Depends on intensity of magnetic moment and on distance between nuclei • Spin-Rotation • Anisotropy • Scalar coupling • Quadrupolar Relaxation (very large)

  30. NMR – From Spectra to Structures An Experimental approachSecond edition (2007) Springler-Verlag • Terence N. MitchellmBurkhardCostisella Integration Relaxation time at 300 MHz 26`C OH 0.4 s Aromatic 5.4 s CH 3.3 s OCH2 2.7 s CH3 2.9 s

  31. 1 1 pT2 pT1 T2 : Spin-Spin Relaxation From Classical NMR theory (Bloch Equation) Single peak fits Lorentz Curve (Lorentzian shape) Half width Dn1/2= ~ In fact the experimental line width measure the T2 influenced by inhomogeneity of the field ½

  32. Relaxation with unpaired electron Neutral Radical Line shape become extremely broad with unpaired electron relaxation (fast relaxation)

  33. Effect of Relaxation with Quadrupolar nuclei For nuclei having I > ½ , there is ellipsoidal charge distribution with 2I +1 orientations in the magnetic field. (below is representation of the 3 orientations for I = 1) Fluctuating field causes transitions among nuclear spin state => fast relaxation To be effective, quadrupolar relaxation require unsymmetrical charge distribution at nucleus and at electron cloud e.g. (CH3)4-N+ => no quadrupolar relaxation (sharp line 14N)

  34. Effect of Relaxation with Quadrupolar nuclei H-N Relaxation rate between 14N spin states is slow compared to JNH Relaxation rate between 14N spin states is moderately fast compared to JNH Relaxation rate between 14N spin states is very fast compared to JNH

  35. Chemical Exchange Me-NH Amine Fast chemical exchange, Fast 14N relaxation NH Me-NH Amide Slow chemical exchange moderate 14N relaxation NH

  36. Heteronuclear Coupling and decoupling Coupling to 14N broaden the lines

  37. Coupling with Deuterium Most solvents used in NMR are deuterated (usually to 99%) Therefore, in proton NMR, we observe residual protons with Deuterium coupling (when appropriate). HCD2OD HCD2SOCD3 H D D2 2nI + 1 lines

  38. gH gH gH JHH JHH JCH = = = JHD JHD JCD gD gD gD Size of Coupling with Deuterium When 2 protons are identical in chemical shift but you want to measure JHH, use deuterium substitution 1H : 100 MHz 2H : 15.4 MHz In DMSO: 2JHD = 1.8 Hz 2JHH = 12 Hz = 6.514 In 13C NMR: In DMSO-d6: 1JCD = 21.0 Hz 1JCH = 136.8 Hz = 6.514

  39. Pulse NMR spectrometer

  40. Absolute Frequency:  The absolute frequency is reported in Hertz respect to 1H nuclei, TMS is exactly 100 MHz (or 100,000,000 Hz)  Nucleus Compound 1H 100,000,000 Hz TMS 2H 15,350,650 Hz DMSO-d6 13C 25,145,004 Hz TMS 14N 7,223,876 Hz N(Me)4+I_ 15N 10,136,783 Hz MeNO2 19F 94,078,500 Hz C6F6 31P 40,486,455 Hz P(MeO)3 29Si 19,867,184 Hz TMS

  41. Relative Chemical Shift ref + - 0  - ref  (Hz) (Hz)  106  (ppm) = =  106 ref ref (MHz) (MHz) MeNO2 NH4+Cl- 0 -337.8  Since it is not easy to report and measure Absolute Frequency, Relative scale referencing is usually adopted in NMR In 1H, 2H, 13C, 29Si : ref. is TMS In 77Se : ref. is (Me)4Se In 119Sn : ref. is (Me)4Sn In 15N : ref. is Nitromethane

  42. Parameters in NMR • Chemical shift (d)  ppmStructural information nPEAK – nREF(Hz) d (ppm) = ------------------------------ = ppm Freq of the nuclei (MHz) e.g. at 200 MHz: If R-CH(OR)2 appear at 1,000 Hz from TMS (0) It’s chemical shift is: d = 1,000 Hz/200 MHz = 5 ppm d is dimensionless (independent from the applied field) n changes with the applied field

  43. Parameters in NMR • Coupling constant (J)  HzStructural and Conformational information • Area of peaksRelative proportion of nuclei • Distance between nucleiInformation is contained inrelaxation and NOE • Molecular MotionInformation is contained inrelaxation, NOE and Variable Temperature

  44. NMR Parameters

  45. Proton Coupling Constants

  46. Proton Chemical Shifts

  47. 1H NMR: C6H10O2 CH3 (Et) CH2O

  48. 1H NMR:C8H14O2 triplet triplet Roof effect #lines = n +1 quintet sixtet

  49. H1-NMR : C4H8O3 OH CH3 CH

  50. singlet quartet Triplet doublet C13-NMR: coupled to proton

More Related