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Chapter 13

Chapter 13. NMR Spectroscopy. 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 .

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Chapter 13

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  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 14N has three states: -1, 0, 1. They each differ by 1.

  3. Spectroscopy involves using energy to excite a system from one state (ground state) to another of higher energy (excited state). The nuclear spin quantum number determines how many spin states there are

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

  6. Figure 13.4, p.499

  7. Example of nmr spectrum: methyl acetate. More shielded, nuclei experience lesser magnetic field. Less energy to excite. • 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.

  8. p. 481

  9. More Shielding • 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. • 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.

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

  11. 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.

  12. Equivalence by Symmetry Figure 13.6, p.500

  13. Equivalent by rotation 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. equivalent

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

  15. p. 484

  16. 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. Figure 13.7, p.503

  17. 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. p.504

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

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

  20. For C-H bond as the hybridization of the carbon changes sp3 to sp2 to sp the electronegativity of the C increases and expect to deshield (move left) the H peak. sp3 sp sp2 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.

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

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

  23. 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.

  24. 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).

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

  26. p. 491

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

  28. In preparation for discussion of origin of Spin-Spin recall earlier slide More Shielding due to electrons at nucleus being excited. 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”.

  29. 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. About 50% of the neighboring hydrogens will augment the applied magnetic field and about 50% will decrement it. Get two peaks, a double The original single peak of H1 has been split into two peaks by the effect of the neighbor H2. The energy difference is J Energy H1, being excited Here H2 augments external field, peak moved downfield. Here H2 decreases external field, peak moved upfield.

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

  31. 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 = hn) 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.

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

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

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

  35. Three neighboring Hb’s causing splitting when Ha is excited. All Hb augment Two augment, one decrement. One augment, two decrement. All decrement. Ha being excited. Three equivalent Hb causing spin spin splitting. Figure 13.15c, p.512

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

  37. Three nonequivalent nuclei. Ha and Hb split each other. Also Hb andHc split each other. Technique: use a tree diagram and consider splittings sequentially. Figure 13.19, p.513

  38. More complicated system Figure 13.20, p.514

  39. Return to Vinyl Systems Not equivalent (R1 is not same as R2) because there is no rotation about the C=C bond. Figure 13.21, p.514

  40. 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.

  41. Analysis Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE Now examine the left most signal….

  42. Ha being excited. Both Hb and Hc are coupled and causing splitting. Hb causes splitting into two peaks (big splitting, JAB) Hc causes further splitting into a total of four peaks (smallest splitting, JAC) JAB = 11-18 Hz JAC = 0 - 5 JBC = 5 - 10

  43. 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. Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE 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 Ha.

  44. Analysis in greater depth - 2. Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE 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 Hb.

  45. Analysis in greater depth - 3. Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE 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 Hc.

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

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

  48. Coincidental Overlap: Non-equivalent nuclei have same coupling constant. Ha will be a triplet (two Hb); Likewise for Hc. We analyze Hb. A triplet of triplets Here Ha and Hc have same coupling with Hb (Jab = Jbc), ,, coincidental overlap: splits to 5, four equivalent neighbors.

  49. Analyze what happens as Jab becomes equal to Jbc. First get peak heights when Jab does not equal Jbc. Recall heights in a triplet are 1 : 2 : 1 2 1 1 2 1 1 First split the Hb by Ha in ratio of 1:2:1. 2 x 2 Each component is split by Hc in ratio of 1:2:1. Result for each final peak is product of probabilities 1 x 1 1 x 2 2 x 1

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