1 / 45

Interpretation of more complex spectra

Interpretation of more complex spectra. 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.

gotzon
Download Presentation

Interpretation of more complex spectra

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. Interpretation of more complex spectra

  2. 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.

  3. Predict the 1H spectrum of CH2F2 spin of F is = ½ Next Predict the 1H spectrum of • H = 2 ppm • F = 99 ppm JHH = 5 Hz JHF = 40, 10 Hz JFF = 50 Hz

  4. 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.

  5. J12 = J34 = 5.1 Hz J13 = J24 = 1.3 J14 = 1.95 J15 = J46 = -0.42 J16 = J45 = -0.18 J23 = 1.95 J25 = J36 = 1.4 J26 = J35 =1.4 J5,6 = 0.1 H1 = H4 = 6.22 ppm H2 = H3 = 6.53 H5 = H6 = 5.85

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

  7. Chiral Shift Reagents

  8. Are the hydrogens on the CH2 group equivalent? CH3CH2OCH(CH3)2 Always equivalent? 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 Operational definition: If replacing one hydrogen by deuterium leads to a new asymmetric carbon atom, the two hydrogens are enantiotopic

  9. Non-First Order 1H NMR Spectra Cases where  (chemical shift) < 10 J (coupling constant) AX case  AB case  A2 case ?

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

  11. JAB = 247.5-231.5 = 16; JAB = 211.5-195.7= 15.8; JAB (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

  12. A2X A2B Case A3 A3 X A2 A2B

  13. Top and bottom 9 transitions can be observed, 1 weak Numbering the lines from the B nucleus A2B 8 7 6 5 4 3 2 1

  14. Top and bottom 9 transitions can be observed, 1 weak Numbering the lines from the B nucleus B = 3 = 9 Hz A = 1/2( 5 + 8) ½(31.9+39.1) = 35.5 JAB= 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? 8 7 6 5 4 3 2 1 40.7 39.1 32.5 31.9 15.6 9.0 6.7 0

  15. Raccoon B= 9 Hz A = 35.5; 35.5 Hz JAB= 7.9 JAA = ? A- B = 35.5-9 = 26.9

  16. Calculated A2B spectra for different values of JAB

  17. 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: JAB, JAX, JBX What does an ABX spectrum look like?

  18. AB X

  19. ignore step 6

  20. JAB 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 JAB = 15.45 7-4 = 94.6-79.3 = 15.3

  21. JAB 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 JAB = 15.45 7-4 = 94.6-79.3 = 15.3

  22. JAB = 15.5 The chemical shift of X is the average of all X lines X =(9+10+11+12)/4 = 183.9

  23. JAB = 15.5 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

  24. JAB = 15.5; X = 183.9 (A + B) = 157.5 6 5 4 8 2 8 ½ (JAX+JBX) = the separation of the centers of the two AB quartets (JAX+JBX) = 2[(8+6+5+2)/4-(7+4+3+1)/4]= 12.7

  25. JAB = 15.5 X = 183.9 5 6 4 (A + B) = 157.5 (JAX+JBX) = 12.7 8 2 8 2D+ = separation of the 1st and 3rd lines of first AB quartet; 2D- = separation of the 1st and 3rd 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

  26. JAB = 15.45 X = 183.9 6 4 (A + B) = 157.5 (JAX+JBX) = 12.7 D+ = 11.4 D- = 9.35 8 2 8 2M = (4D+2 –JAB2) ½ ; 2N = (4D-2 –JAB2) ½ M = ½(4*11.42 – 15.52) ½ = 8.36 N = ½(4*9.352 – 15.52) ½ = 5.23 1. A - B = M+N = 13.6; ½(JAX –JBX) = M-N = 3.13 2. A - B = M-N = 3.13; ½(JAX –JBX) = M+N= 13.6

  27. (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 (JAX+JBX) a + sign if ab+ is centered at a higher frequency than ab- or a – sign if the reverse is true

  28. (A - B) = 13.6; ½(JAX-JBX) = 3.13; (JAX –JBX) = 6.26 • (A + B) = 157.5 (JAX+JBX) = -12.7 • A = 85.65 JAX = -3.2; • B = 71.85 JBX = -9.5; • 2. (A - B) = 3.13; ½(JAX-JBX) = 13.6; (JAX –JBX) = 27.2 • (A + B) = 157.5 (JAX+JBX) = -12.7 • A = 80.3 JAX = 7.25; • B = 77.2 JBX = -19.95;

  29. All values in Hz 72.2 JAB =J12= 15.45 X = 3 =183.9 Solution 1 A = 1 = 85.65; JAX = J13 = -3.2 B = 2 = 71.85; JBX = J23 = -9.5 Solution 2 A = 1 = 80.3; JAX = J13 = 7.25 B = 2 = 77.2 ; JBX = J23 = -19.95

  30. Some special cases of ABX like systems Deceptively simple spectra: A = 452 Hz JAB = 8 Hz B = 452.5 Hz JAX = 0 HZ X = 473 Hz JBX = 3 HZ

  31. 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

  32. 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

  33. Deceptively complex spectra: Virtual coupling SupposeA = 100 Hz; B = 104 Hz; X = 200 Hz and JAX = 0 Hz; JBX = 8; JAB = 6 Hz; Predict the first order spectrum for X This is an example of an ABX system

  34. 2,6-dimethylquinone and 2,4-dimethylquinone 1 = 4 = 6 2 = 3 = 5 = 6 = 2 J12 = J13= J45=J46=6 J14 = 1; J15 = J16 = J24=J34 =2 J23 = J56 = -10 J25=J26= J35=J36 =0

More Related