NMR Nuclear Magnetic Resonance Spectroscopy p. 83. a hydrogen nucleus (a proton) has a charge, spread over the surface
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1 NMR Nuclear Magnetic Resonance Spectroscopy p. 83 a hydrogen nucleus (a proton) has a charge, spread over the surface a spinning charge produces a magnetic moment (a vector = direction + magnitude) along rot axis which produces a magnetic field, just like a bar magnet
2 p. 83 this moment interacts with an external field, just like a magnet B 0 N S BUT only certain orientations are allowed, which depends upon the nuclear spin quantum number I
3 The magnetic moment, is proportional to the spin quantum number I = I (h/2 ) p. 83 where = Magnetogyric ratio = magnetic moment angular momentum is different for each nucleus (a constant) I is allowed only certain values and these values can range from m I = I to +I in integer steps ONLY Our proton is simple as it has I = ½, so only two values of m I are permitted, +1/2 and -1/2
4 In bar magnet terms, these values of m I correspond to the magnet aligned with and opposed to the external field B 0 or = m I = +1/2-1/2 owever, these two states ONLY have different energy in a non-zero external field, B 0 E B 0 = 0 B 0 E = h p. 83/84
5 p. 84 B 0 E = h so, if we put our proton in a strong magnetic field, B 0, there are TWO nuclear spin states: lower energy = aligned with field (m I = +1/2) higher energy = opposed to field (m I = -1/2) A transition will occur at a frequency this is in the radiofrequency range, ~10 8 z (100 Mz) and this is the basis of the NMR (or MRI) experiment we make a nucleus flip its spin
6 owever, we do NOT really have little bar magnets!!! p. 84 But this frequency, is the same as what we have called the flip frequency NOTE: is proportional to B 0 more about that later
7 The spin number, I depends on the isotope p. 85 x E Y x = number of protons + neutrons = mass Y = number of protons = charge = Atomic # e.g. 1 1 has mass of 1 and charge of = D has mass 2, charge 1 mass 2 = 1 proton + 1 neutron IF x and y are both even, I = 0, NOT NMR ACTIVE e.g. 12 6C 16 8O 4 2e Common isotopes
8 p. 85 If charge (y) is odd and mass (x) is even, I = integer e.g. 2 1D and 14 7N both have I = 1 If mass is odd, I = Integer/2 1 1, 13 6C 19 9F 31 15P all have I = 1/2 IMPORTANT Cl I = 3/2; 17 8O I = 5/2 too difficult for now
9 Number of spin states p. 85 Nuclei have spin states of magnetic quantum number M I = +I, (I-1), (I-2)...-I where total number of states is then 2I+1 so if I = ½, # states = 2, and M I = +1/2 and -1/2 if I = 1, # = 3, M I = +1, 0, -1 case for Deuterium if I = 3/2, # = 4, M I = +3/2, +1/2, -1/2, -3/2 case for 11 B
10 p. 86 I = ½ 1 I = 3/2 11 B
11 E p. 86 B 0 E = h B 0 = 0 This frequency is the Larmor frequency, where: = B 0 2 so irradiation of nucleus with frequency causes a transition from lower to upper state which we call flipping the spin This frequency depends on B 0 (field) and (nucleus type)
12 = B 0 2 p. 86/87 Earth s magnetic field is ca. 0.5 G at surface For 1, = (radian Mz per Tesla) Then when B 0 = 2.35 Tesla, = 100Mz [1 Tesla = 10,000 gauss = 10kGauss] so for each Tesla of B 0, = Mz so for each 100 Mz you need 2.35 Tesla of B 0 Our largest departmental NMR is referred to as a 500 Mz instrument because this is the resonance frequency of 1 at its magnetic field of Tesla (cost ~ 500K$)
13 Magnetogyric ratios p. 86 nucleus rel F P B C Si (relative freq to 1 ) E 1 19 F 31 P 13 C 29 Si 2 D Magnetic field 2.35 Tesla , Mz
14 p. 87 For different nuclei: 1 = k 1 B 1 2 = k 2 B 2 k = 1/2 so 1 1 B 1 = 2 2 B 2 So if know five of these we can calculate other but note that B 1 = B 2 for the same instrument
15 NMR Instruments p. 87 Known by what PROTON frequency they operate at e.g. we have 300, 360 and 500 Mz instruments these all can run carbon spectra, and since rel for 13 C = 0.25, these same instruments operate at 75, 90 and 125Mz respectively when running carbon spectra [magnetic field strength is essentially fixed for each instrument] = B 0 2 If all protons absorbed at exactly the same frequency the NMR experiment would NOT be very useful!
16 Fortunately, this is not the case: electrons have an associated magnetic field too so small differences in the electronic environment can cause one 1 nucleus to resonate at a slightly different than another p. 87 = B 0 2 C C O alcohol Essentially, the applied field is slightly modified by the local electronic environment: is a very sensitive probe of the local field B and we can measure radiofrequencies very accurately B actual = B 0 + B local
17 p. 88 SIELDING: local electronic effects on B net B o B induced e- Electrons in bonds circulate in a magnetic field creating an opposing magnetic field B induced that cancels part of B o More e - means a bigger B induced and a smaller B net so to reach resonance, you would need to apply a higher B o (or, if B o is fixed, a lower ν is required)
18 CEMICAL SIFT, p. 88/89 Variations in caused by local effects are on the order of 1 part in 10 5 to 10 7 of the normal 1 resonant We COULD just quote the absolute resonant frequencies for a given 1 type BUT: a) The differences would be tiny and unwieldy to quote: e.g vs Mz b) Absolute frequencies vary linearly with B 0 : e.g vs on a 250 vs. 500 Mz machine
19 CEMICAL SIFT, p. 88 A better way: measure all resonances RELATIVE to a standard and divide by B 0 (or more simply the standard resonance frequency for a given nucleus at a particular B 0 ) to remove the field dependence The Standard: TMS = tetramethylsilane = Si(C 3 ) 4 = 0 = Sample peak - TMS peak Spectrometer in Mz = z Mz = ppm shift in z sample TMS
20 p. 89 CCl 3 = 7.2 ppm The shift from TMS is thus 7.2 x spectrometer operating frequency in Mz e.g. at 100Mz = 7.2 x 100 = 720 z at 300Mz = 7.2 x 300 = 2160 z at 500Mz = 7.2 x 500 = 3600 z Thus the absolute shift depends upon the instrument, but the chemical shift does not (7.2 ppm) igher frequency (larger field) instruments are more sensitive, and more expensive!
21 SUMMARY Nuclei have spin I (depends on # protons/neutrons) Spinning charges = magnetic field Nuclear spin I quantized: only m I states from +I,..,-I allowed, each with a different magnetic moment m I states same energy unless placed in an external magnetic field: + m I (spin aligned with ext. field) lower energy than - m I NMR experiment probes required to flip spin: cause nuclear spin state to change (eg. m I = 1/2 to -1/2) depends on ext. field strength but also on tiny variations in electron density (because e - are spinning charges too) in the locale of the nucleus: we can tell something about the nuclear environment from
22 p. 90 Deshielded Downfield Low field Br O TMS Cl O Cl
23 p. 90 Cl O Cl = integration = area under peak is proportional to NUMBER of s = 1:3 it is a RATIO, i.e 1:3 or 2:6 or 3:9 etc
24 p. 90 Br O TMS = integration = area under peak is proportional to NUMBER of s = 2:3 it is a RATIO, i.e 2:3 or 4:6 or 6:9 etc
25 p. 91 O C C 3 C C 3 C size ratio 3:1
26 p. 92 SPIN-SPIN COUPLING F---PCl 2 Both P and F have spin I = ½ (we can ignore Cl here) high low field high field low
27 TE COUPLING CONSTANT J p. 93 n J XY where n = number of bonds between coupled the nuclei X = nucleus that is being observed Y = nucleus that is coupling to X (often the order of XY or YX is arbitrary) 1 C 2 C 3 1 C F 2 C C 3 4 C 1 2 F P 3 J 4 J F 2 J PF
28 p. 93 = chemical shift = centre of all lines (ppm) J = coupling constant = separation of lines (z) s, d, t, q = quartet, pentet, sextet, septet, octet nonet, decet
29 Cl Br A C C B A = B Cl Br ** p. 93 **cannot see coupling between identical nuclei
30 More nuclei AX 2 e.g. PF 2 Cl The P can see both fluorines up (twice as likely) one up, one down both down so there are TREE lines, the middle one has twice the intensity
31 TREE DIAGRAMS p. 94 AX 2 A A split by X J AX split by X again J/2 J/2 J/2 J/2 Can always construct pattern as a tree, by splitting one nucleus at a time when nuclei are same, e.g. X above, lines fall on top of each other and simplify pattern
32 p AX AX 2 AX n Spectrum of A: (2nI + 1) lines AX The number of lines at any branch in the tree diagram represent the number of degenerate possibilities for the alignment of the nuclear spins. This translates to the area under the line in the multiplet. The quintet shown above would have relative areas of 1 : 4 : 6 : 4 : 1 I = spin of X; n = number of X atoms For I = ½, = n + 1 lines AX = 5 lines 1:4:6:4:1 intensities
33 Numerically, we can get intensities from Pascal s triangle: =6 =15 =20 =15 =6 p. 94 so, intensities of a septet are 1:6:15:20:15:6:1
34 Example A p neighbors, so 3+1=4 lines 1:3:3:1 Br----C C 3 2 neighbors, so 2+1 = 3 lines 1:2:1
35 Example B p. 95 CCl 2 CCl---CCl 2 1 neighbor, so doublet 2 identical neighbors, so triplet
36 Example C p. 96 If J s are different, best to draw a tree, e.g. PFCl You need to know J s, assume 1 J PF > 1 J P for now In 31 P Spectrum: 1 J PF P P split by F 1 J P 1 J P split again by so spectrum is a doublet of doublets
37 Example D p. 96 P 2 F 1 J PF > 1 J P P F P 2 F 31 P nmr P P P split by F 1 J PF 1 J P 1 J P split again by 2 1 J P 1 J P 1 J P 1 J P get a doublet of triplets
38 p. 96 For AXYZ draw tree by starting with largest J then next largest J then smallest J then lines will cross over each other the least! Consider CF 2 C 3 is triplet of quartets 13 CF 3 C 2 C 3 2 is doublet of quartets F 3 C C C 3 of quartets 2
39 SIZE OF COUPLING CONSTANTS Table, manual p. 97 Generally 1 J >> 2 J >> 3 J > 4 J > 5 J > 6 J Ball park figures!! 200z 50z 10z 3z 1z but there are plenty of exceptions * only P, F,... * For - = 0, except in pi-bonds >C A B ~15 =C A B ~2 0 z 1 z z z
40 Example E p. 98 C 5 8 F 4 O C 5 8 F 4 O - C 5 12 O = 0 DBE Integration = 1:1: total Peak at 6 is 1 t of t (triplet of triplets)
41 C 5 8 F 4 O Other signals are singlets Integration = 1:1: total so t of t must be caused by F s arranged There is a large coupling and a small coupling Large Coupling ~ 1ppm =60z = 2 J = -C-F 2 p. 98 small coupling ~0.1ppm =6z = 3 J = -C-C-F 2 so we have CF 2 -CF 2 - group not split by anything else
42 p. 98 C 5 8 F 4 O CF 2 CF 2 -- we have 3C s, O, 7 s to find Integration = 1:1: total 6 s are identical and are NOT split by neighbors X C(C 3 ) 2 Y X, Y not or F otherwise would see coupling 3 J One of these must be CF 2 CF 2
43 p. 98 C 5 8 F 4 O CF 2 CF 2 -- C C(C 3 ) 2 O Integration = 1:1: total now groups left to attach are and CF 2 CF 2 -- so CF 2 CF 2 C(C 3 ) 2 O we will learn later that O usually do not couple
44 You can now do ASSIGNMENT 4
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