1. Interpretation of more complex spectra

5. Diasteriotopic nuclei: chemically different nuclei with different chemical shifts Two nuclei or groups attached to the same atom that are present in environments that are not related to each other by an axis of symmetry. Operational definition: If sequentially replacing the nuclei in question with an isotopic nucleus leads to diasteriomers, the nuclei in question are diasteriotopic.

6. Predict the 1 H spectrum of CH 2 F 2 spin of F is = ½ Next Predict the 1 H spectrum of  H = 2 ppm  F = 99 ppm J HH = 5 Hz J HF = 40, 10 Hz J FF = 50 Hz

8. Chemical equivalence Two nuclei related by an element of symmetry including a plane of symmetry in an achiral environment are chemically identical. Magnetic equivalence Two chemically identical nuclei may not necessarily be magnetically equivalent. Two chemically equivalent nuclei will have the same chemical shift but may become magnetically non- equivalent if they are coupled to the same nucleus with different coupling constants.

9. H 1 = H 4 = 6.22 ppm H 2 = H 3 = 6.53 H 5 = H 6 = 5.85 J 12 = J 34 = 5.1 Hz J 13 = J 24 = 1.3 J 14 = 1.95 J 15 = J 46 = -0.42 J 16 = J 45 = -0.18 J 23 = 1.95 J 25 = J 36 = 1.4 J 26 = J 35 =1.4 J 5,6 = 0.1

10. Chemical shift reagents Europium dipivaloylmethane Eu(dpm) La(fod)

14. Chiral Shift Reagents

17. Enantiotopic nuclei: Enantiotopic nuclei: two nuclei that are related to each other by a plane of symmetry but not an axis of symmetry. These nuclei have the same chemical shift in an achiral environment but have different chemical shifts in a chiral environment Are the hydrogens on the CH 2 group equivalent? CH 3 CH 2 OCH(CH 3 ) 2 Always equivalent? Operational definition: If replacing one hydrogen by deuterium leads to a new asymmetric carbon atom, the two hydrogens are enantiotopic

18. Non-First Order 1 H NMR Spectra Cases where  (chemical shift) < 10 J (coupling constant) AX case  AB case  A 2 case ?

21. Analysis of AB Case JABJAB J AB  4  3  2  1  = [(  1 -  4 )(  2 -  3 )] 1/2 = A - B ½( A + B ) = C C = (  1 +  2 +  3 +  4 )/4 C three unknowns: A, B, J AB

22. J AB = 247.5-231.5 = 16; J AB = 211.5-195.7= 15.8; J AB (avg)= 15.9 Hz 2C = 2(  1 +  2 +  3 +  4 )/4; 2C = (247.5+231.5+211.5+195.7)/2 =443.1 = A + B A - B = [(  1 -  4 )(  2 -  3 )] 1/2 = [(247.5-195.7)(231.5-195.7)] 1/2 = 32.2 A + B = 443.1; A - B = 32.2; A = 237.6; B = 205.5; ∆ = 32.2

23. A 2 X A 2 B Case A 3 A2A2 X A2BA2B A 3

24. Top and bottom 9 transitions can be observed, 1 weak Numbering the lines from the B nucleus A2BA2B 8 7 6 5 4 3 2 1

25. Top and bottom 9 transitions can be observed, 1 weak Numbering the lines from the B nucleus  8  7  6  5  4  3  2  1 B =  3 = 9 Hz A = 1/2(  5 +  8 ) ½(31.9+39.1) = 35.5 J AB = 1/3  (  1 -  4 +  6 -  8 )  = 1/3(0-15.6+32.5-40.7) = 7.9 How do you know that you have analyzed a spectrum correctly? 40.7 39.1 32.5 31.9 15.6 9.0 6.7 0

26. Raccoon B = 9 Hz A = 35.5; 35.5 Hz J AB = 7.9 J AA = ? A - B = 35.5-9 = 26.9

27. Calculated A 2 B spectra for different values of J AB

28. ABX spectrum What do you need to know to analyze an ABX spectrum? You need to know: 3 chemical shifts: A, B, X 3 coupling constants: J AB, J AX, J BX What does an ABX spectrum look like?

29. X AB

30. ignore step 6

31. J AB appears four times in the AB portion of the spectrum 8-6 = 98.9-83.7 = 15.2 5-2 = 80.4-64.7 = 15.7 3-1 =72-56.4 = 15.6 J AB = 15.45 7-4 = 94.6-79.3 = 15.3

32. J AB appears four times in the AB portion of the spectrum 8-6 = 98.9-83.7 = 15.2 5-2 = 80.4-64.7 = 15.7 3-1 =72-56.4 = 15.6 J AB = 15.45 7-4 = 94.6-79.3 = 15.3

33. The chemical shift of X is the average of all X lines X = (9+10+11+12)/4 = 183.9 J AB = 15.5

34. X = 183.9 The average of all AB lines is equal to ½( A + B ) ( A + B ) = 2*(1+2+3+4+5+6+7+8)/8 = 157.5

35. J AB = 15.5; X = 183.9 ( A + B ) = 157.5 ½  (J AX +J BX )  = the separation of the centers of the two AB quartets  (J AX +J BX )  = 2[(8+6+5+2)/4-(7+4+3+1)/4]= 12.7 8 2 8 6 4 5

36. J AB = 15.5 X = 183.9 2D + = separation of the 1st and 3 rd lines of first AB quartet; 2D - = separation of the 1st and 3 rd lines of second AB quartet; Chose 2D + as the larger value D + = ½(4-1) =11.45; ½(7-3)= 11.3; D +av = 11.4 D - = ½(6-2) =9.2; ½(5-8) = 9.5 D -av = 9.35 8 28 4 6 ( A + B ) = 157.5  (J AX +J BX )  = 12.7 5

37. J AB = 15.45 X = 183.9 2M = (4D + 2 –J AB 2 ) ½ ; 2N = (4D - 2 –J AB 2 ) ½ M = ½(4*11.4 2 – 15.5 2 ) ½ = 8.36 N = ½(4*9.35 2 – 15.5 2 ) ½ = 5.23 1. A - B = M+N = 13.6; ½(J AX –J BX ) = M-N = 3.13 2. A - B = M-N = 3.13; ½(J AX –J BX ) = M+N= 13.6 8 28 4 6 ( A + B ) = 157.5  (J AX +J BX )  = 12.7 D + = 11.4 D - = 9.35

38. (7,4,3,1) = ab + quartet because D + = 11.3; ab + centered at 75.6 (8,6,5,2) = ab - quartet because D - = 9.5; ab - centered at 81.9 Assign  (J AX +J BX ) a + sign if ab + is centered at a higher frequency than ab - or a – sign if the reverse is true

39. 1.( A - B) = 13.6; ½(J AX -J BX ) = 3.13; (J AX –J BX ) = 6.26 ( A + B ) = 157.5 (J AX +J BX ) = -12.7 A = 85.65 J AX = -3.2; B = 71.85J BX = -9.5; 2.( A - B ) = 3.13; ½(J AX -J BX ) = 13.6; (J AX –J BX ) = 27.2 ( A + B ) = 157.5 (J AX +J BX ) = -12.7 A = 80.3 J AX = 7.25; B = 77.2 J BX = -19.95;

40. J AB =J 12 = 15.45 X = 3 =183.9 Solution 1 A = 1 = 85.65;J AX = J 13 = -3.2 B = 2 = 71.85;J BX = J 23 = -9.5 Solution 2 A = 1 = 80.3; J AX = J 13 = 7.25 B = 2 = 77.2; J BX = J 23 = -19.95 All values in Hz 72.2

41. Some special cases of ABX like systems Deceptively simple spectra: A = 452 HzJ AB = 8 Hz B = 452.5 HzJ AX = 0 HZ X =473 HzJ BX = 3 HZ

42. When two chemical shifts are fortuitously very similar, an ABX case can give a deceptively simple spectrum. Only by simulating the spectrum can you convince yourself the the structure is more complex than the nmr suggests

43. Whenever the chemical shifts of the A and B nuclei are very similar, a deceptively simple spectrum can be obtained 100 MHz; 100, 101, 700, 8, 6 10 Hz

44. Suppose A = 100 Hz; B = 104 Hz; X = 200 Hz and J AX = 0 Hz;J BX = 8;J AB = 6 Hz; Predict the first order spectrum for X Deceptively complex spectra: Virtual coupling This is an example of an ABX system

45. 2,6-dimethylquinone and 2,4-dimethylquinone 1 = 4 = 6 2 = 3 = 5 = 6 = 2 J 12 = J 13 = J 45 =J 46 =6 J 14 = 1; J 15 = J 16 = J 24 =J 34 =2 J 23 = J 56 = -10 J 25 =J 26 = J 35 =J 36 =0