chem 806 identification of organic and inorganic compounds by advance nmr techniques n.
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Chem-806 Identification of organic and inorganic compounds by advance NMR techniques
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  1. Chem-806Identification of organic and inorganic compounds by advance NMR techniques Tool box 2D-NMR: Homonuclear 2D-NMR: Heteronuclear 3D-NMR

  2. 2D NMR

  3. Homonuclear 2D correlation techniques • Through Bond: nJHH (scalar coupling) COSY : COrrelated SpectroscopY Directly coupled neighbors Relay-COSY : RELAY-COrrelated SpectroscopY Directly coupled neighbors and protons coupled to the coupled neighbors (relay transfer) TOCSY : TOtal Correlation SpectroscopY Directly coupled neighbors and protons coupled to the coupled neighbors (More efficient than Relay) • Through Space: Distance NOESY : NOE SpectroscopY ROESY : ROE SpectroscopYNOE in Rotating frame

  4. COSY

  5. Indirect chemical shift detection during t1

  6. COSY: evolution during t1

  7. Processing for Absolute value COSY: Sine-belland Pseudo-echo shaping

  8. COSY : explanation of cross peaks using Vector Model

  9. 1H NMR of glucose derivative

  10. 2D-COSY of glucose derivative 5 4 6 a/b 2 1 3 5 4

  11. C4H8O CH3 – CH2 – CH2 – OH

  12. C4H8O: COSY CH3 CH2 OH CH3 – CH2 – CH2 – OH CH2

  13. C5H8O2 I= 5 – 8/2 +1 = 2 CH2 - O CH2 - O O-C=O

  14. CH2 - O c d b a C5H8O2 4 b CH2 3 c CH2 a CH2 d CH2 C=O 2 O 5

  15. 3H C8H16O : I = 8 – 16/2 + 1 = 1 2H 3H 4H 2H CH2 – CO – CH2 2H Ketone C=O

  16. Me-1 5 4 3 2 C8H16O Me1 – CH22 – CH23 – CH24 CH25 C=O 5 Me-8 CH27 3 CH27/5 Me8

  17. C11H20O4 X 2 3H 3H CH2 – CH3 2H 2H O – CH2 – CH3 O-C O-C=O

  18. C11H20O4 CH2 – CH3 O – CH2 – CH3 O = C –O – CH2 – CH3 CH2 – CH3 CH3– CH2 C O = C –O – CH2 – CH3 Only missing quaternary carbon

  19. COSY-90: disaccharide H2 H1 H3 H5a H5b H4/2 H3? H1 H2

  20. COSY-45 J1,2 => negative J1,3 J2,3 => positive

  21. COSY-45 If we consider cross peak 2/1Passive couplings – involving passive nuclei 3 - (J1,3 and J2,3 ) have same Sign (positive slant) If we consider cross peak 3/1Passive couplings – involving passive nuclei2 - (J1,2 and J2,3 ) have different Sign (negative slant)

  22. COSY45: H2 and H4 overlap at ~ 4.7 ppm. Cross-peak in COSY-45 allow to assign H3 and H5a/b unambiguously as H4 is coupled to a geminal pair => different sign in J 2/4 5a 5b 3/5a/5b? 2 1

  23. H4 H2 H3 H6a/b H5

  24. H5’ H3’ H4 H2’ H2 H3 H4’ H6a H6b H5 Deshielded multiplet H1’ H6a/b H5 H6a’ H6b’ H4’

  25. Phase sensitive COSY

  26. GE-DQCOSY

  27. DQCOSY: Active couplingisantiphaseandthere isno intensity on central peak of a triplet (aspositiveandnegativepeaks cancel out)

  28. Soft-COSY

  29. Multiple Relay-COSY:As we have longer relay sequence => relaxation during transfer step attenuate the signal and Magnetization get weaker

  30. Relay-1 and Relay-2 COSY on disaccharide

  31. TOCSY or HOHAHA

  32. 1D-HOHAHA

  33. NOE and distance 1 F (tc) NOE r6 NOE is a consequence of cross-relaxation between 2 spins close to each other in space. NOE is a consequence of modulation of the Dipole-Dipole couplingby motion of the molecule in solution. The NOE intensity is related to the internuclear distance r and is a function of the correlation timetc

  34. Relaxation and tumbling rate Relaxation is caused by fluctuating magnetic field generated by neighboring dipole. If the rate at which the fluctuation occur in the transverse plane matches the frequency of double quantum transition, positive NOE will be observed. If the fluctuation is slower, zero quantum transition will produce negative NOE.

  35. NOE is related to dipole relaxation • For small molecules, translation and rotational motion occur at high frequency. • The vector linking two nuclei rijchange orientation more frequently in small molecule than in larger molecule (Small molecules tumble at rates around 1011 Hz, Larger molecules such as proteins tumble at rates around 107 Hz), dissipating the energy between different spin states • In small molecules the frequency of motion can occur frequently at Larmor frequency o and twice 2xo(W2) dissipating energy between single quantum and double quantum state. (this produce Positive NOE) • In Large molecule, only low frequency transition like Zero quantum (W0) can dissipate energy (aibj  biaj ) (Negative NOE)

  36. NOE: applying gB2 to the A of an AX spin system bb bb X2 A2 ba ba ab ab A1 X1 aa aa X1 Dp = 2 X1 Dp = 2 X2 Dp = 2 X2 Dp = 2 X2 A2 {A} A1 X1 Immediately after irradiation, there is NO change in the intensity of X Turning on the Decoupler do not change population of the X transition

  37. NOE: relaxation with double quantum pathway W2 probability (positive NOE) bb bb X2 A2 X2 A2 {A} ba ba ab ab A1 A1 X1 X1 aa aa X1 Dp = 2 X2 Dp = 2 bb bb X2 A2 X2 A2 ba ba ab ab A1 A1 X1 X1 aa aa X1 Dp = 3 X2 Dp = 3 delay … T1 Dec. continue W2 After W2 relaxation, there is a net increase in the intensity of X (50%) Relaxation takes time to establish a new equilibrium: T1 process

  38. NOE: Relaxation with zero quantum pathway W0 probability (negative NOE) bb bb X2 A2 X2 A2 {A} ba ba ab ab A1 A1 X1 X1 aa aa X1 Dp = 2 X2 Dp = 2 bb bb X2 A2 X2 A2 ba ba ab ab A1 A1 X1 X1 aa aa X1 Dp = 1 X2 Dp = 1 delay … T1 Dec. continue W0 W0 After W0 relaxation, there is a net decrease in the intensity of X (50%)  negative NOE Relaxation takes time to establish a new equilibrium: T1 process

  39. NOE: summary of relaxation pathways   bb  X2 A2  ba ab A1 X1 aa W1: probability of single quantum relaxation do not create nOe W2 W0 A new population ditribution is generated by relaxation through dipole-dipole relaxation : double quantum and zero quantum pathway W2 and W0 If W2 is efficient (small molecule – fast motion  large frequency ) Level  increase  level  increase also with decoupler continuing W2 pathway yield positive nOe If W0 is efficient (large molecule – slow motion small freq. Diff.) Level  increase  level  increase also with decoupler continuing W0 pathway yield negative nOe

  40. Maximum NOE vs correlation time (nuclei interacting with 1H) hC = 4/2 = 2 0.5 Fast tumbling maximum NOE : Line intensity : Ii = 1 + hi {H} hN = 10/-2 = -5 wtc ~1 Large molecule Small molecule

  41. NOE vs Distance and motion If B is irradiated, nucleus A should show the largest NOE (closest to nucleus B) (the relative distances are shown as A to B = 1, B to C and C to D = 2). Nucleus C is relaxed by nucleus Das well as B, so it shows a smaller NOE. Nucleus D has an indirect NOE from nucleus B. Indirect effects usually give rise to negative NOEs. Note that as the tumbling rate decreases all other parameters become irrelevant and the NOEs tend towards -100%. The notation fA{B} means the NOE enhancement of spin A when spin B is saturated.

  42. NOE difference: nOe-d Me Cl C C Me H d1 AQ irr Dec on frq irr d1 AQ Dec off frq NOE is a kinetic effect: need delay ~ T1  It take time to develop  It takes time to decay control nOe difference

  43. Choosing a structure by nOe {OH} {OMe} H3 H6 H5

  44. Assigning the relative stereochemistry