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NMR N uclear M agnetic R esonance

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  1. NMRNuclear Magnetic Resonance ProtonNMR: Symmetry Index NMR-basics H-NMR

  2. Homotopic protons: A2 spin System Chemical shift equivalence : Isochronous nuclei These nuclei are interchangeable by symmetry or rapid exchange Protons are equivalent in chiral and achiral environment

  3. Homotopic protons examples CH2Cl2 CH2F2 A2X2 A2 A4

  4. Enantiotopic Protons Plane of symmetry A2X3 A2X d =6.15 J = 53.6 Hz Enantiotopic protons are Equivalent in Achiral environment (like CDCl3) Non-equivalent in Chiral environment (optically active solvent)

  5. A2X2 spin system?? AA’XX’ H1 & H2 => same shift : chemically equivalent (homotopic) 3JH1-F4=>cis 3JH2-F4 =>trans The two protons are coupled to the same nuclei with different coupling! Magnetic Non-Equivalence

  6. JH1-H2 JH1-H4 Magnetic Equivalence The 2 Hgeminal to Fluorine are enantiotopic The 2 Hgeminal to Chlorine are enantiotopic The 2 H(1,3) and2 H(2,4)are chemically equivalent Is this an A2M2X2 spin system? These protons are chemically equivalent but are magnetically non-equivalent because they have different couplings with neighbors AA’MM’XX spin system

  7. No symmetry: Asymmetric center * ABX A B Protons A and B have different shifts: they are Diastereotopic Accidental overlap can occur producing deceptively simple spin system

  8. H1-NMR OH CH3 CH

  9. Dissymmetric center Plane of symmetry Enantiotopic groups  Hb1 = Hb2 Ha1 = Ha2 Diastereotopic protons A2B2X

  10. X AB AB AB X

  11. 2 dissymmetric centers in symmetrical molecule Me Me H HOOC H COOH H H H H H COOH H COOH Me Me Me Symmetrical : C2 axis Enantiotopic protons Symmetrical : s plane Diastereotopic protons CH CH2 HA HB Mixture of 2 isomers

  12. * * d H1 = d H2 d Me1 = d Me2 Equivalence, non-equivalence and symmetry d H1 d H2 d Me1 d Me2 d H1 d H2 AB d H3 d H4 AB d H3=d H4A2

  13. Example of dissymmetric spin system d B = 3.55 d A = 3.40 2JAB = 9.4 Hz 3JA-Me= 3JB-Me= 7.0 Hz ABX3 dq

  14. Chemical Shift Non-Equivalence over a distance Diastereotopic protons * AB AB 2 doublets

  15. HA1 = HA2 HB1 = HB2 HA1HB1 HA2HB2 Magnetic Equivalence Magnetic equivalence JA1-B1 JA2-B1 Enantiotopic protons: A1 and A2 are Magnetically different Diastereototopic protons: AA’BB’

  16. AA’BB’ SpinWorks => load AA’BB’

  17. AA’BB’ 2 sets of homotopic protons : magnetically non-equivalent

  18. AA’BB’: para SpinWorks => load AA’BB’-para

  19. AA’BB’: Ortho SpinWorks => load AA’BB’

  20. Spin System: Pople Notation • Each Chemical Shift is designated by a letter • Dn-> Difference in Shift in Hz • J-> Coupling in Hz • If the ratio Dn/J is Small(<8),Lettersused to designate the shift are closeAB, ABC …This represent case of second order spectra: These spectra must be simulated with the help of quantum mechanic equations. Such programs are available on Nuts or MestrecSpinWorks or Spectrometer software. This case is also called strongly coupled • If the ratio Dn/Jislarge(>8),Lettersused to designate the shift are farAM, AX …This case give rise to first order type spectra is also refer to as weakly coupled case

  21. Pople Notation A2X(if the shift difference of CH2 and CH is large compare to coupling). A2B(if the shift difference of CH2 and CH is small compare to coupling). 3 Spins AMX -> if the 3 spins have large chemical shift difference ABX -> if 2 spins are close and 1 is far away ABC -> if 3 spins are close When nuclei have identical shift but different magnetic coupling, prime symbol is used. For example: AA’BB’ or AA’XX’

  22. M X A X Jtrans Jcis M AA’BB’C JAX = Jcis = 10 Hz JAM = Jtrans = 17 Hz JMX = Jgem = 2 Hz A : dd

  23. Virtual Coupling Virtual coupling First order Same shift CH2-OH CH3 broad CH2b

  24. Me broad doublet Virtual Coupling A2B2CX3 Because of the close shifts of ABC protons we observe “virtual coupling”

  25. Virtual Coupling : Symmetrical chains 1 2 3 4 1) CO2Me – CH2 – CH2 – CO2Me A2A2A4 Singlet 1 2 3 4 5 2) CO2Me – CH2 – CH2 – CH2 – CO2Me A2 X2 A2 A4X2 Triplet, Quintet 1 2 3 4 5 3) CO2Me – CH2 – CH2 – CH2 – CH2 – CO2Me A2 X2X2 A2 A2A2’ X2X2’ Complex spectra Same shift, different J with A/A’ Virtual coupling

  26. Virtual Coupling 3) CO2Me – CH2 – CH2 – CH2 – CH2 – CO2Me

  27. Ph Br H Ph H1 H2 H Br Br H H1 H2 Ph H Br Ph Symmetrical Molecules with 2 chiral centers 1r, 3r; erythro 1r, 3s; Meso H1 = H2 H1 = H2 diastereotopic protons ABX2 Enantiotopic protons Magnetically non-equivalent AA’XX’ Due to fast rotation, J is average A2X2

  28. COOH COOH H1 H1 H1 H1 OH OH H H H2’ H2’ H2 H2 OH OH H H H3’ H3’ H3 H3 COOH COOH Chiral Centers in Symmetrical Molecules AB AB AB AA AB AB Erythro: axis of symmetry Meso: plane of symmetry H1 H1’ diastereotopic H1 H1’ diastereotopic H3 H3’ diastereotopic H3 H3’ diastereotopic Group1 = Group3 Group1 = Group3 H2 = H2’ enantiotopic H2  H2’ diastereotopic

  29. X HA X HA X HA X HA R R R R HB HB HB HB X HA HA X HA HA R R R R HB HB HB HB X X Chiral Centers in polymers Isotactic polymer AB Syndiotactic polymer A = B A2

  30. Calculating Shifts for simple aliphatic compounds d = 0.23 + SSi(d) CH3Cld(calc)=2.76 d(exp.)=3.1 CH2Cl2d(calc)=5.29 d(exp.)=5.3 CHCl3d(calc)=7.82 d(exp.)=7.27

  31. Calculating Shifts for aliphatic compounds d = 0.933 + SSi(d) 1 2 3 e.g. CH3-CO-CO-CH3 Subst. Effect value - C2-C3 +0.244 =O (at C2) +1.021 =O (at C3) +0.004 -CR3 (at C3) -0.038 SSi(d) +1.231 d = 0.933 + 1.231 = 2.164 Experimental = 2.23

  32. Calculating Shifts for olefinic compounds

  33. Ph OEt C C H1 H2 Calculating Shifts for olefinic compounds H1 d = 5.23 + Phgem + OEttrans + 1.35 + (-1.28) d = 5.32 H2 d = 5.23 + Phtrans + OEtgem + (-0.10) + 1.18 d = 6.33

  34. Z isomer effect Base 5.23 Ph (gem) 1.43 CN (cis) 0.78 COORconj *(trans) 0.33 Total 7.77 E isomer effect Base 5.23 Ph (gem) 1.43 CN (trans) 0.58 COORconj * (cis) 1.02 Total 8.26 Calculating Shifts for olefinic compounds (deciding which isomer) Experimental: 8.22 ppm Which one ?? * Double bond is further conjugated

  35. Calculating Shifts for aromatic compounds

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

  37. Aromatic substitution pattern: ortho AA’ XX’ Typical spectra for ortho (symmetrical)

  38. Aromatic substituent pattern: para

  39. Aromatic substituent pattern Calculated shifts 1 = 8.86 2 = 8.52 3 = 7.53 4 = 8.24 4 1 3 t J=8.1 t J=1.8 2 dt J=7.7, 1.5 ddd J=8.1, 2.2, 1.1

  40. Aromatic substituent pattern Calculated shifts 1 = 8.31 2 = 7.74 1 = 7.69 1 = 8.08 1 td J=7.4, 1.1 dd J=8.1, 0.7 ~td J=8.1, 1.5 dd J=7.7, 1.5

  41. C5H9NO4 3H 3H 2H 1H NO2 t d CH3 O CH C CH3 CH2 O q q

  42. C5H8O I = 5 – 8/2 + 1 = 2 CH3 CH2 CH CH t, J=7.4 CHO Trans J CHO, d J=8.1 Hz ddt, J=15.8, 8.1, 1.5 CH2 dt, J=15.8, 6.9

  43. C4H6O2 I = C – H/2 + 1 = 2 H CH3 O s C C C H H O dd 6.6, 1.5 dd 14.0, 1.5 Jtrans = 14.0 Jcis = 6.6 2Jgem = 1.5 dd 14.0, 6.6 CH=

  44. 6’ 3 400 MHz 5’ 6 4’ 6, d (9.2) 3, d(2.4) 5 3’ 5, dd 9.2, 2.4 6’, dt 7.9, 1.0 3’, ddd 5.0, 1.5, 0.9 5’, ddd 7.9, 7.4, 1.6 4’, ddd 7.4, 5.0 , 1.0 80 MHz

  45. NEXT Proton and Heteronuclear NMR Index NMR-basics H-NMR NMR-Symmetry Heteronuclear-NMR