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NON-FIRST ORDER SPECTRA

p. 173. NON-FIRST ORDER SPECTRA. >C H ---C H 2 - t d. is only true if D  (in Hz) > 10J. This is easier to achieve with large fields:. d 1.5 1.3 D  = 0.2 ppm J = 7 Hz (constant with field). At 60MHz, 0.2ppm = 12Hz , so D  /J = 12 / 7 = ~2.

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NON-FIRST ORDER SPECTRA

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  1. p. 173 NON-FIRST ORDER SPECTRA >CH---CH2- td is only true if D (in Hz) > 10J This is easier to achieve with large fields: d1.51.3D = 0.2 ppm J = 7 Hz (constant with field) At 60MHz, 0.2ppm = 12Hz, so D/J = 12/7 = ~2 At 250MHz, 0.2ppm = 50Hz, so D/J = 50/7 = ~7 At 600MHz, 0.2ppm = 120Hz, so D/J = 120/7 = ~17

  2. p. 174 If D<10J, spectra are NOT as predictable 1st order not 1st order singlet when D=0

  3. p. 175 Brucine hydrate 60MHz

  4. p. 175 300MHz CH3O are now readily resolved

  5. p. 175 Higher field, more 1st order 600MHz 300MHz

  6. p. 176 Δ ~ 6 still triplets, but inside lines > outside lines

  7. p. 176 triplets hardly recognizable, outside lines v small

  8. p. 176 no longer recognizable

  9. p. 177 Spin system labels 1st order not 1st order Assumes ‘infinite’ resolution… actual lines will merge

  10. p. 178 NOMENCLATURE:Spin system notation If 1st order conditions, i.e. D> 10J, use letters well apart in alphabet (these are arbitrary but accepted) HF = AX; ClPF2=AX2 CH3CH2Br = A2X3 Different nuclei by DEFINITION satisfy the 1st order requirement! F2P(CH3) = AM2X3 If not 1st order, use letters close in alphabet ClCH2CH2OH =A2B2X NOTE: the order is arbitrary, e.g. A2M3X above and does not imply ANY additional information

  11. p. 179 CHEMICAL EQUIVALENCE: use same letter the two A atoms are the same and the three X atoms are same as each other, BUT A = X CH3CH2Br = A2X3 In this example, the two F are CHEMICALLY EQUIVALENT and the two H are CHEMICALLY EQUIVALENT BUT…

  12. p. 179 The two H’s (or the two F’s) are NOT MAGNETICALLY EQUIVALENT 3J = 2J Because: the coupling constant of H is not the same TO EACH F The F’s are MAGNETICALLY INEQUIVALENT

  13. p. 179 same argument for the H’s 2J Fx does not have same J to HA as to HA’ 3J so we call this an AA’XX’ system

  14. p. 179 TO TEST MAGNETIC EQUIVALENCE: A) hydrogens: Pick a different COUPLED SPINACTIVE NUCLEUS compare J of this atom to each of the atoms you wish to test JFH is not same as JFH’ so H and H’ are MAGNETICALLY INEQUIVALENT if you get a “YES” (equivalent), test ALL nuclei !!!

  15. p. 180 Test: HA and HA’ X HA HA’ Pick different spin nucleus, HC is JCA’ = JCA HB HB’ YES, so test again!!! HC NOW PICK HB: JBA is not same as JBA’ so HA and HA’ are NOT equivalent once you have one NO, no need to test again Test HB and HB’ NOT equivalent using HA or HA’ = AA’BB’C system

  16. p. 180 Pick one H d6.8 are J’s same to both H d7.8 NO, so H are not equiv again NO Likewise pick H and test H Dd ~ 1ppm (300Hz), Chem shifts: J=8, so (D/J) >10 so AA’XX’

  17. p. 181 Symmetry and its effects on NMR spectra What is symmetry? “A property of a physical system that allows the system to remain unchanged by a specific physical or mathematical transformation, such as translation or rotation” Or more plainly for a molecule: “A transformation of a molecule that duplicates the type and spatial arrangement of ALL atoms in space”

  18. p. 181 Transformations that meet this criterion are called ‘symmetry operations’ and the features of the molecule that they operate on are known as ‘symmetry elements’ Symmetry ElementSymmetry OperationSymbol 1. Identity ‘do nothing’ E 2. Inversion centre invert the molecule i through a point 3. Mirror plane reflect through a plane σ 4. Proper axis of rotation rotate about an axis by (360/n)° Cn 5. Improper axis of rotation rotate about an axis by (360/n)° Sn and reflect in a plane perpendicular to the axis All molecules possess the symmetry element E and many we commonly encounter also possess rotation axes (Cn) or mirror planes (σ); fewer contain inversion centres (i) or improper rotation axes (Sn).

  19. p. 182 Example 1 BF3 is a trigonal planar molecule with several symmetry elements including the BF3 plane itself (not shown): The C3 axis makes all 3 of the 19F nuclei equivalent and only one signal is observed in the 19F NMR.

  20. Example 2 Naphthalene is a flat aromatic molecule with rotation axes (shown), mirror planes that include the C2 axes (not shown for clarity) and an inversion centre marked with a star (*) : These symmetry elements result in only 3 unique 13C nuclei and 2 unique types of 1H nuclei. p. 182

  21. p. 182 Example 3 Water has the familiar bent structure and possesses a C2 axis as shown and two mirror planes: the plane of the page and one perpendicular to the page that bisects the H-O-H angle (not shown):

  22. p. 182 Example 4 Methane is tetrahedral and has mirror planes (σ), proper (Cn) and improper (Sn) axes of rotation, but NO inversion centre. Symmetry Elements: E, 6 σ, 4 C3, E, 3 σ, C3E, σE 3 C2, 3 S4

  23. p. 182 The molecule on the far right is recognizable as a CHIRAL moleculebecause it cannot be superimposed on its mirror image. If the molecule ITSELF had a mirror plane as a symmetry element then this would not be true. We therefore have a simple test for chirality based on symmetry elements: Chiral molecules cannot possess mirror planes, inversion centres or improper axes of rotation as symmetry elements!

  24. p. 183 However, chiral molecules CAN possess proper axes of rotation (Cn) as illustrated below:

  25. p. 183 How does symmetry affect NMR spectra? Basically, the presence of a symmetry element in a molecule reduces the number of resonances because carrying out the symmetry operations renders some atoms equivalent to others. Pay close attention to ANY reduction in the number of observed resonances. Symmetry can be an enormous help in solving structures if you understand the relationship between the symmetry elements present and the number of resonances. Make symmetry your friend!

  26. p. 183 Beware of CHIRAL compounds!!! A-HC-D B-H’ C-H B-DA-H’ so H and H’ are never the same, they have different chemical shifts, so are an AB or AX and couple to each other, JAX ~ 15Hz, they are called pro-chiral or diastereotopichydrogens

  27. A X Dn/J > 10 AB ‘quartet’ Dn/J < 10

  28. p. 184 If molecule frozen 3J values depend on dihedral angles and are different so would be AA’A’’XX’ But at room temp, there is free rotation, so all J values average to 7Hz, so A3X2 2

  29. p. 184-5 Inorganic systems, need to know geometry! axial F’s different from equatorial F’s If frozen, AX2Y3 but in fact fluxional, so AX5 Lower T 31P is triplet of quartets sextet

  30. ASIDE: Interconversion of 5-coordinate structures Berry Pseudo-rotation

  31. p. 184 In general, you need to know geometries to predict the spectra BF3 CH4 PtCl4 PF5 SF6 (square planar)

  32. p. 185 PF3(CH3)2 however has three isomers, which all have different spectra P = dtqq P = dt7 P = q7 F = dtqq F = ddqq F = dd7 F = dt7 F = d7 H = dq H = ddtq H = ddtq H = ddt

  33. p. 186 SPIN DILUTE SYSTEMS Pt has one isotope that has I = 1/2 , 195Pt = 34% abundant so ~2/3 Pt has no spin, 1/3 has spin and so couples

  34. p. 187-8 141 141 3

  35. ASSIGNMENT 7

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