1. Chapter 13 NMR Spectroscopy

2. Recall that electrons have two “spin states”: spin up (1/2) and spin down (-1/2). Similarly nuclei have spin quantum states…. Nuclei of interest. By coincidence, each of these has two states, ½ and – ½. By the way, note that 14 N has three states: -1, 0, 1. They each differ by 1.

3. The nuclear spin quantum number determines how many spin states there are Spectroscopy involves using energy to excite a system from one state (ground state) to another of higher energy (excited state). Nuclear spin quantum number Quantum numbers of spin states Number of spin states 001 1/2- ½, ½2 1-1, 0, 13 3/2-3/2, -1/2, 1/2, 3/24

4. Normally, nuclei in different spin states have the same energy. Can not do spectroscopy. We need to have a ground state and excited state. In a magnetic field they have different energies. Now we can do spectroscopy…. We apply a magnetic field and create a ground state and a higher energy excited state (perhaps more than one).

5. 5249CHEM51.TR9AKUNJAPPU148NE 5250CHEM51.TR9JHOWELL133NE 5251CHEM51.TR9CFANG133NE 5252CHEM51.TR9EKUNJAPPU148NE 5253CHEM51.TR9GHANS1310N 5255CHEM51.TR9BMETLITSKY1310N 5278CHEM51.TR9KLUSHTAK148NE 5279CHEM51.TR9DMOLLICA1310N Final Exam Schedule, Thursday December 18, 8AM

6. Apply a strong external field….. Both orientations have same energy if no magnetic field

8. Figure 13.3, p.497

9. Figure 13.4, p.499

10. Two kinds of hydrogens in methyl acetate: two peaks. (Peak at zero is tetramethyl silane to standardize the instrument. ) Chemical shift: where on horizontal axis the signal from a nucleus occurs. Question: What causes nuclei to appear with different chemical shift?? Answer: the sigma bonding electrons in a molecule will be set in motion to establish a magnetic field that opposes the external magnetic field. The nuclei are shielded. The shielded nuclei experience less of a magnetic field, closer energy states. The shielded nuclei require less energy to excite and their signal occurs to the right in the spectrum. More shielded, nuclei experience lesser magnetic field. Less energy to excite. Example of nmr spectrum: methyl acetate.

11. p. 481

12. Doing nmr spectroscopy: the magnetic field creates the energy difference between the spin states of the nucleus and Radio waves provide the energy needed to excite the nucleus from the lower energy state to the excited state. More Shielding Simplifying The energy supplied by the radio waves has to match the energy gap created by the magnetic field. We can vary either the magnetic field or the frequency of the radio waves to match the exciting radiation energy with the energy needed to reach the excited state.

13. Hold the external magnetic field constant, vary radio frequency. Less energy needed to excite the nuclei when more shielded. Hold excitation energy (radio waves) constant, vary magnetic field. Stronger magnetic field needed to overcome shielding. More Shielding More Shielded, Less energy needed from radio waves More shielded, stronger magnetic field needed to create the right energy difference. Since we control energy of excited state (magnetic field) and the energy being supplied by radiowaves: two ways for an nmr spectrometer to function: or Terminology based on this approach: downfield (lower ext field) on left; upfield on right

14. Remember that methyl acetate only gave two peaks in its spectrum. There were two sets of equivalent hydrogens. Equivalent hydrogens Hydrogens are equivalent if They are truly equivalent by symmetry. -or- They are bonded to same atom and that carbon atom can rotate freely at room temperature to interchange the positions of the hydrogens making the equivalent to the spectrometer.

15. Figure 13.6, p.500 Equivalence by Symmetry

16. Equivalent by rotation equivalent Note that if it were not for rotation the methyl hydrogens would not be equivalent. Two are gauche to the Cl and one is anti.

17. p.501 Some molecules which have only one type of hydrogen - only one signal

18. p. 484

19. Figure 13.7, p.503 Signal area: proportional to the number of hydogens producing the signal Looking at the molecular structure # Methyl hydrogens : # tert butyl hydrogens = 3:9 = 1:3 In the spectrum we find two peaks 23 : 67 = 1 : 2.91 Conclude smaller peak due to methyl hydrogens; larger due to tert butyl hydrogens.

20. p.504 Now return to chemical shift and factors affecting it. Look at two isomeric esters to get some feeling for chemical shift. The electronegative oxygens play the key role here. Most electron density around the H atoms, most shielded, upfield. Less shielded, more deshielded, downfield Most deshielded, furthest downfield. Sigma electrons pulled away by oxygen.

21. Figure 13.8, p.505 Chemical shift table…

22. Further left, downfield, Less shielded Relationship of chemical shift to electronegativity Less electrons density around hydrogens as ascend table.

23. Expect vinylic hydrogens to be deshielded due to hybridization but acetylenic (recall acidity) should be even more deshielded and they aren’t. Some other factor is at work. Magnetic induction of pi bonds. sp 3 sp sp 2 For C-H bond as the hybridization of the carbon changes sp 3 to sp 2 to sp the electronegativity of the C increases and expect to deshield (move left) the H peak.

24. Figure 13.9, p.507 Diamagnetic shielding Hydrogen on axis and shielded effectively. Hydrogen experiences reduced magnetic field. Less energy needed to excite. Peak moves upfield to the right.

25. Figure 13.10, p.507 Diamagnetic shielding Hydrogen off axis and external field increased. Hydrogen experiences increased magnetic field. More energy needed to excite. Peak moves downfield to the left.

26. Figure 13.11, p.508 In benzene the H atoms are on the outside and the induced magnetic field augments the external field.

27. Spin Spin Splitting If a hydrogen has n equivalent neighboring hydrogens the signal of the hydrogen is split into (n + 1) peaks. The spin-spin splitting hydrogens must be separated by either two or three bonds to observe the splitting. More intervening bonds will usually prevent splitting.

28. Example Expect the signal for this hydrogen to be split into seven by the six equivalent neighbors. Expect the peak for the methyl hydrogens to be split into two peaks by the single neighbor. Overall: small peak split into seven (downfield due to the Cl). larger peak (six times larger) split into two (further upfield).

29. p. 491 Attempt to anticipate the splitting patterns in each molecule.

30. p. 491

31. Spin-spin splitting. Coupling constant, J. Split into a group of 4 Split into a group of 2 The actual distance, J, between the peaks is the same within the quartet and the doublet.

32. More Shielding due to electrons at nucleus being excited. In preparation for discussion of origin of Spin-Spin recall earlier slide Due to shielding, less of the magnetic field experienced by nucleus, Lower energy needed to excite. Peaks on right are “upfield”. Reduced shielding, more of the magnetic field experienced, higher energy of excitation. Peaks are “downfield”.

33. Origin of spin-spin splitting In the presence of a external magnetic field each nuclear spin must be aligned with or against the external field. Approximately 50% aligned each way. Non-equivalent hydrogen nuclei separated by two or three bonds can “spin – spin split” each other. What does that mean? Consider excitation of a hydrogen H1. Energy separation of ground and excited states depends on total magnetic field experienced by H1. Now consider a neighbor hydrogen H2 (passive, not being excited) which can increase or decrease the magnetic field experienced by H1. Energy H1, being excited Here H2 augments external field, peak moved downfield. Here H2 decreases external field, peak moved upfield. The original single peak of H1 has been split into two peaks by the effect of the neighbor H2. The energy difference is J About 50% of the neighboring hydrogens will augment the applied magnetic field and about 50% will decrement it. Get two peaks, a double

34. Figure 13.14, p.511 3-pentanone The left side of molecule unaffected by right side. Coupling constant, J, in Hz Peak identification… Same as gap here.

35. Magnitude of Coupling Constant, J The magnitude of the coupling constant, J, can vary from 0 to about 20 Hz. This represents an energy gap (E = h ) due to the interaction of the nuclei within the molecule. It does not depend on the strength of the external field. J is related to the dihedral angle between bonds. J largest for 0 (eclipsed) or 180 (anti), smallest for 90, intermediate for gauche.

36. Rapid interconversion of enol structures from 2,4-pentadione Have mixture of keto (16%) and hydrogen bonded enol forms (84%). Spectrum shows both forms: Keto with two types of H at 2.24 and 3.60 Enol with three types (not four) indicating rapid interconversion of the two enol structures making the methyl groups equivalent.

37. Table 13.4, p.511 anti gauche vinyl systems

38. Now look at some simple examples. Examine the size of the peaks in the splitting. H b is augmenting external field causing a larger energy gap. H b decrementing external field causing a smaller energy gap. H a is being excited. H b is causing spin- spin splitting by slightly increasing or decreasing the magnetic field experienced by H a. Spin-Spin Splitting

39. Figure 13.15b, p.512 Two neighboring atoms assist external field. More energy needed to excite. Peak is “downfield”. One neighbor assists, one hinders. No effect. Both neighbors oppose. Less energy needed to excite, “upfield”. Again H a is flipping, resonating. The two H b are causing spin- spin splitting by slightly changing the magnetic field experienced by H a. Recall that for the two H b atoms the two states (helping and hindering the external field) are almost equally likely. This give us the 1 : 2 : 1 ratio.

40. Figure 13.15c, p.512 All H b augment Two augment, one decrement. One augment, two decrement. All decrement. H a being excited. Three equivalent H b causing spin spin splitting. Three neighboring H b ’s causing splitting when H a is excited.

41. Intensities of peaks in a multiplet corresponds to Pascal’s triangle.

42. Figure 13.17, p.513 Naturally if there are two non-equivalent nuclei they split each other.

43. Figure 13.19, p.513 Three nonequivalent nuclei. H a and H b split each other. Also H b and H c split each other. Technique: use a tree diagram and consider splittings sequentially.

44. Figure 13.20, p.514 More complicated system

45. Figure 13.21, p.514 Not equivalent (R 1 is not same as R 2 ) because there is no rotation about the C=C bond. Return to Vinyl Systems

46. Example of alkenyl system We will perform analysis of the vinyl system and ignore the ethyl group. Three different kinds of H in the vinyl group. We can anticipate the magnitude of the coupling constants.

47. J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 5 – 10 Hz, MIDDLE Each of these patterns is different from the others. Analysis Now examine the left most signal….

48. J AB = 11-18 Hz J AC = 0 - 5 J BC = 5 - 10 H a being excited. Both H b and H c are coupled and causing splitting. H b causes splitting into two peaks (big splitting, J AB ) H c causes further splitting into a total of four peaks (smallest splitting, J AC )

49. J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 5 – 10 Hz, MIDDLE Each of these patterns is different from the others. Look at it this way...This signal appears to have big (caused by trans H-C=C-H) and small (caused by geminal HHC=) splittings. The H being excited must have both a trans and geminal H. The H must be H a. Analysis in greater depth based on knowing the relative magnitude of the splitting constants. Aim is to associate each signal with a particular vinyl hydrogen.

50. J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 5 – 10 Hz, MIDDLE Each of these patterns is different from the others. And the middle signal. This signal appears to have big (caused by trans H-C=C-H) and middle (cis H-C=C-H) splittings. The H being excited must have both a trans and cis H. The H must be H b. Analysis in greater depth - 2.

51. J AB = 11-18 Hz, BIG J AC = 0 – 5 Hz, SMALL J BC = 5 – 10 Hz, MIDDLE Each of these patterns is different from the others. And the right signal. This signal appears to have small (caused by geminal HHC=) and middle (cis H-C=C-H) splittings. The H being excited must have both a geminal and cis H. The H must be H c. Analysis in greater depth - 3.

52. As with pi bonds, cyclic structures also prevent rotation about bonds Approximately the same vinyl system as before. No spin spin splitting of these hydrogens. Nothing close enough Non equivalent geminal hydrogens. Analyze this.

53. Figure 13.25, p.516 Note the “roof effect”. For similar hydrogens the inner peaks can be larger.

54. A triplet of triplets H a will be a triplet (two H b ); Likewise for H c. We analyze H b. Here H a and H c have same coupling with H b (J ab = J bc ),,, coincidental overlap: splits to 5, four equivalent neighbors. Coincidental Overlap: Non-equivalent nuclei have same coupling constant.

55. Recall heights in a triplet are 1 : 2 : 1 Analyze what happens as J ab becomes equal to J bc. First get peak heights when J ab does not equal J bc. 1 2 1 1 1 2 First split the H b by H a in ratio of 1:2:1. Each component is split by H c in ratio of 1:2:1. Result for each final peak is product of probabilities 1 x 11 x 22 x 1 2 x 2

56. Peak heights shown when J ab does not equal J bc. 1 2 1 2 4 2 1 2 1 Examine middle peak. Let J bc become larger until it equals J ab and add overlapping peaks together. 1+4+1

57. 1 2 1 2 4 2 1 2 1 Now adjacent peak. 2+2 1 4 6 4 1

58. Fast Exchange ethanol Expect coupling between these hydrogens. Three bond separation. There is no coupling observed especially in acid or base. Reason: exchange of weakly acidic hydrogen with solvent. The spectrometer sees an “averaged hydrogen”. No coupling and broad peak.

59. Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not? Are these two hydrogens truly equivalent? Seemingly equivalent hydrogens may be homotopic, enantiotopic, diastereotopic. How to tell: replace one of the hydrogens with a D. If produce an achiral molecule then hydrogens are homotopic, if enantiomers then hydrogens are enantiotopic, if diastereomers then diastereotopic. We look at each of these cases. Seem to be equivalent until we look at most stable conformation, the most utilized conformation.

60. Homotopic The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions.

61. Enantiotopic The hydrogens are enantiotopic and equivalent in the NMR unless the molecule is placed in a chiral environment such as a chiral solvent.. The hydrogens are designated as Pro R or Pro S This structure would be S Pro S hydrogen. Pro R hydrogen

62. Diastereotopic If diastereormers are produced from the substitution then the hydrogens are not equivalent in the NMR. Diastereotopic hydrogens. The hydrogens are designated as Pro R or Pro S This structure would be S Pro S hydrogen. (Making this a D causes the structure to be S.) Pro R hydrogen

63. Diastereotopic methyl groups (not equivalent), each split into a doublet by H c a and a’ Example of diastereotopic methyl groups.

64. 13 C NMR 13 C has spin states similar to H. Natural occurrence is 1.1% making 13 C- 13 C spin spin splitting very rare. H atoms can spin-spin split a 13 C peak. ( 13 CH 4 would yield a quintet). This would yield complicated spectra. H splitting eliminated by irradiating with an additional frequency chosen to rapidly flip (decouple) the H’s averaging their magnetic field to zero. A decoupled spectrum consists of a single peak for each kind of carbon present. The magnitude of the peak is not important.

65. 13 C NMR spectrum 4 peaks  4 types of carbons.

66. 13 C chemical shift table

67. Hydrogen NMR: Analysis: Example 1 1.Molecular formula given. Conclude: One pi bond or ring. 2. Number of hydrogens given for each peak, integration curve not needed. Verify that they add to 14! 3. Three kinds of hydrogens. No spin-spin splitting. Conclude: Do not have non- equivalent H on adjacent carbons. 4. The 9 equivalent hydrogens likely to be tert butyl group (no spin-spin splitting). The 3 equivalent hydrogens likely to be methyl group. The two hydrogens a CH 2. Fragments: (CH 3 ) 3 C-, -CH 2 -, CH 3 - 5. Have accounted for all atoms but one C and one O. Conclude: Carbonyl group! -(C=O)- 6. Absence of splitting between CH 2 and CH 3. Conclude: they are not adjacent.

68. Example 2, C 3 H 6 O 1. Molecular formula  One pi bond or ring 2. Four different kinds of hydrogen: 1,1,1,3 (probably have a methyl group). 3. Components of the 1H signals are about equal height, not triplets or quartets 4. Consider possible structures.

69. Possible structures

70. Figure 13.8, p.505 Chemical shift table… Observed peaks were 2.5 – 3.1 vinylic ethers Observed peaks were 2.5 – 3.1. Ether!

71. Possible structures

72. NMR example Formula tells us two pi bonds/rings Three kinds of hydrogens with no spin/spin splitting. What can we tell by preliminary inspection….

73. Now look at chemical shifts 2. From chemical shift conclude geminal CH 2 =CR 2. Thus one pi/ring left. 3. Conclude there are no single C=CH- vinyl hydrogens. Have CH 2 =C-R 2. This rules out a second pi bond as it would have to be fully substituted, CH 2 =C(CH 3 )C(CH 3 )=C(CH 3 ) 2, to avoid additional vinyl hydrogens which is C 8 H 14. X In CH 2 =CR 2 are there allylic hydrogens: CH 2 =C(CH 2 -) 2 ? 1. Formula told us that there are two pi bonds/rings in the compound.

74. Do the R groups have allylic hydrogens, C=C-CH? 1.Four allylic hydrogens. Unsplit. Equivalent! 2.Conclude CH 2 =C(CH 2 -) 2 3.Subtract known structure from formula of unknown… C 7 H 12 - CH 2 =C(CH 2 -) 2 ------------------------------------------ C 3 H 6 left to identify Remaining hydrogens produced the 6H singlet. Likely structure of this fragment is –C(CH 3 ) 2 -. But note text book identified the compound as