An Introduction to NMR Spectroscopy. The types of information accessible via high resolution NMR include:

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1 1 of 40 An Introduction to NMR Spectroscopy 1 NMR 13C NMR The types of information accessible via high resolution NMR include: 1. Functional group analysis (chemical shifts) 2. Bonding connectivity and orientation (J coupling) 3. Through space connectivity (Overhauser effect) 4. Molecular conformations, DNA, peptide and enzyme sequence and structure 5. Chemical dynamics (lineshapes, relaxation phenomena) 1 2

2 Atom Structure Electron spins:!electron waves (or particles) are characterized by 4 quantum numbers: n, l, m, s The electron spin quantum numbers s can assume only two values: +1/2 or 1/2. 2 of 40 Bo external magnetic field s = +1/2 s = -1/2 higher lower energy energy particle spinning on its axis spin angular momentum charged particles tiny magnetic moment 3 Nuclear Spin:!The proton is a spinning charged particle and has also a magnetic moment. B 1 : - nuclear spin quantum number m = 1/2 - such a nucleus is described as having a nuclear spin I of 1/2 m = +1/2 lower energy m = -1/2 higher energy Because nuclear charge is the opposite of electron charge, a nucleus whose magnetic moment is parallel to the magnetic field has the lower energy. 4

3 The difference in energy is given by:!δε = hγb0/2π 3 of 40 γ = magnetogyric ratio (a constant, typical for a nucleus, which essentially reflects the strength of the nuclear magnet) B0 = strength of the applied magnetic field h = Planckʼs constant (3.99 x kj s mol -1 ) 5 Note that as the field strength increases, the difference in energy between any two spin states increases proportionally.! nuclear spin quantum # m = -1/2 E i = -mh#b 0 /2$ 0 degenerate "! = -mµ N B 0 m = +1/2 0 B 0 [values of ±1/2 were picked for m, so that the difference in energy between two neighboring states will always be an integer multiple of B0 (γh/2π)]. 6

4 The number of nuclei in the low energy state (N) and the number in the high energy state (N) will differ by an amount determined by the Boltzmann distribution: 4 of 40 Nβ/Nα = e (-ΔE/kT)!!!!! k = x JK -1 When a radio frequency (RF) signal is applied, this distribution is charged if the radio frequency matches.! ΔE = hν = hγ B0/2 ν = resonance frequency = γ B0/2 ν is therefore dependent upon both the applied field strength and the nature of the nucleus. 7 1 nuclei (protons) exhibit two possible magnetic spin orientations. What about other nuclei with additional protons and neutrons? 8

5 5 of 40 Rules: - A nucleus with an even number of protons (Z) and neutrons (N) will have a nuclear spin I = 0 (for ex. 12 C, 16 O, 18 O, 32 S). - cannot be detected by NMR - both Z and N odd leads to integer values of I ( 2 (I =1), 10 B (I=3), 14 N (I=1)) - detectable by NMR - Z even, N odd, or Z odd, N even will have I values of n/2: 1 (I=1/2), 11 B (I =3/2), 13 C (I =1/2), 15 N (I =1/2), 17 O (I=5/2), 19 F (I=1/2), 31 P (I=1/2) - detectable by NMR 9 The total number of possible spin states (values of m) is determined directly by the number of I. Number of possible spin states: 2I +1 (-I, -I +1,...I -1, I) m Ε -1 ΔΕ = hγ B 0 /2π 0 B 0 +1 m 10

6 1 : in a 2.35 T field (earth magnetic field = T) = of 40 ~ 1 in 10 6!! (ν = 100 Mz) The difference in population of the two states is exceedingly small, in the order of few parts per million (even smaller in 13 C, because γ is smaller). Relatively low sensitivity of NMR compared to IR or UV Large Bo needed to increase the population difference (Bo is usually given in Mz of 1 resonance frequency)

7 SUMMARY - Nuclear spin is a property characteristic of each isotope and is a function of Z and N. 7 of 40 - Each isotope with I 0 has a characteristic magnetogyric ratio (γ) that determines the frequency of its precession in a magnetic field of strength B0 ν = γb 0 2π It is this frequency that must be matched by the incident electromagnetic radiation for absorption to occur When a collection of nuclei with I 0 is immersed in a strong magnetic field, the nuclei distribute themselves among 2I + 1 spin states, the relative population of which is determined by the Boltzmann distribution, usually being near unity N β N α = e (-ΔE/kT) - If two (or more) spin state populations become equal, the system is said to be saturated. 14

8 8 of 40 Obtaining an NMR Spectrum Magnet Source of RF radiation Detector + amplifier Plotter, sample 15 The magnet: permanent electromagnet superconducting cheap, stable, fixed field 1.4T more expensive, stronger, variable field expensive, stronger, variable field 18T (24T) Strength of magnetic field shifts: lock necessary (= substance with strong, defined NMR signal) Older: reference internal, external CDCl 3 TMS: 0.0 ppm singlett 16

9 Once a stable field is established, the question remains as to whether that field is completely homogeneous throughout the region between the pole faces of the magnet. 9 of 40 N S lines of magnetic flux not uniform Sample sample has to be placed near the center of the pole gap 17!For 2.35 T, to achieve a precision of +/- 1 z (10 ppb at 100 Mz) the field must be homogeneous to the extent of +/ x 10-8 T! Such a phenomenal uniformity, even at the center of the field, can be achieved only by means of two additional techniques: Spinning of the sample ("averages" out small inhomogeneities) Variation of the contour of the field by passing extremely small currents through shim coils wound around the magnet itself: Shimming (manually, automatically) Paradox: Large sample in order to have as many nuclei as possible, small sample to increase uniformity of the field. narrow bore tubes 18

10 The Pulsed Fourier Transform Technique Further advances in S/N ratio improvement had to await the development of faster computer microprocessors: ~1970ʼs. 10 of 40 - RF radiation is supplied by a brief but powerful pulse of RF current through the transmitter coil. The spectral width of the pulse is chosen to cover absorption of all nuclei of interest. RF intensity The duration of the pulse (tp) determines the frequency range covered (eisenberg's uncertainty principle: ΔΕ Δt h) >_ 2SW SW ~ tp -1 ; tp _ > (4SW) -1 Frequency ν o Optimum tp are obtained by trial and error and are usually in the order of 10 µs for α = 90 for best S/N ratio. 19 The next step in the PFT process is to monitor the induced AC receiver signal. Digital data collection gives us the modulated free induction decay (function of Mxy). FID because the current intensity decreases with time. This decay is the result of T 2 (spin-spin) relaxation. M 20

11 11 of 40 Voltage the microprocessor samples the voltage in the receiver coil at a regular interval, called dwell time, td. td > (2SW) -1 t d t 0 t 0 Time 21 The frequency of the cosine wave is unaffected by the exponential decay: ν = 1 t 0 = ν precession - ν o (in the date acquisition process νo is subtracted electronically from the observed signal prior to digitization). time to frequency transformation of the data. 22

12 In a set of nuclei with different ν precession and T1/T2, the digital FID curve becomes very complex: 12 of 40 C 3 f ( x) = a o + At this point it becomes necessary for the computer to recognize the patterns mathematically and extract the signal frequencies and relative intensities for each set of nuclei. This analysis is performed by a Fourier transformation of the FID date. n= 1 where: a 0 = constant; a n = amplitude; x = period; s o = fundamental frequency; x o = 1/so; and n = order of harmonic { a n cos2πns o x + b n sin 2πns o x} The parent function is constructed by summing together a series of sine waves. line width: uncertainty principle: ΔνΔt > 1 ν 1/2 > 1 T 2 * Nuclei that are slow to relax give sharp signals, nuclei that relax rapidly give broad signals (solids). Δν 1 Δν 2 Δν 3 0 paramagnetic residues line broadening frequency 23 Summary: A typical example of the generation of a PFT spectrum: 24

13 !t p :! pulse time (µsec)!t acq: the length of the time a given FID signal is actually monitored!!(resolution, the ability to distinguish two nearby signals, is inversely!! prop. to t acq. R = (t acq ) -1 3 sec 0.3 z 13 of 40 tw: delay time, to allow for equilibrium distribution tw = 3T 1 - t acq. (for 1 no waiting time) + dead time ( phasing necessary) phase correction adding up, FT, spectrum 25 26

14 Taking an NMR Practical Consideration 14 of 40 - Use 5 mm tube filled with ~ 0.5 ml of solution containing 1-5 mg of sample (1 NMR). - common deuterated solvents: CCl4, CDCl3, C6D6, DMSO-d6, D2O, CD3CN, CD2Cl2, d6-acetone, CD3OD (because of Cl formation, do not leave sample in CDCl3!) - peak listings in ppm and/or z. - paramagnetic metal ion broad peaks 27 Chemical shift ν depends on Bo therefore relative frequencies are reported: ν = ν act. ν TMS ν o = 83.4z 60x10 6 = 1.39x10 6 =1.39 ppm 1.39 ppm = δ = downfield from TMS. 2 ppm (δ) 1 0 downfield upfield deshielded shielded 28

15 15 of 40 Integration! Area under absorption peak ~ # of nuclei resonating at that ν But:!nuclei must relax to equilibrium between pulses, not generally true of 13C NMR! 29 Correlating Proton chemical Shifts with Molecular Structure Shielding and Deshielding!That actual magnetic field (Bo) experienced by a given nucleus is diminished slightly by an opposing local magnetic field (BN) which results from circulation of nearby electron, induced by Bo: B N Thus: shielding B o e deshielding The circulation of e - follows the right hand rule and magnetic field lines follow the left hand rule. 30

16 16 of 40 Consequence:!The greater the magnitude of BN, the less the magnitude of B required to effect excitation. In other words, the frequency of excitation correlates with the e- density surrounding the nucleus in question. Thus structural features which deplete e--density around a nucleus will cause the resonance to occur at relatively low field (downfield), and the nucleus is said to be deshielded. increase of e density upshield shift: nucleus is said to be shielded

17 Specific Effects 17 of 40 Inductive Effects: e - withdrawing groups move C, resonances downfield; e donating groups move C, resonances upfield

18 18 of 40 Ex. OC 3 C CO 2 -O-R -COO -Phenyl (Exp.: 4.8) ʼs attached directly to Si + Metals very upfield

19 Anisotropic Effects Non-spherical electron-distribution around bond. 19 of 40 Alkenes: C sp 2 -hybridized and more electronegative 37 C sp 2 -hybridized and more electronegative B N B o shielded upfield C C downfield C C δ c = ppm δ = 5.25 ppm for ethylene Lies in de-shielding region of local opposing magnetic field induced by B O /π e - interaction. Thus, both inductive withdrawal of e - density and anisotropic effects move downfield 38

20 20 of Resonance effects can be superimposed on this to explain relative chemical shifts. O a b Obs. a b c CO 2 C 3 c: 6.43 a: = 5.8 b: = 6.05 c: =

21 21 of 40 Benzene: Allylic & benzylic protons moved downfield B 0 > > > > > > deshielded region δ = 7.27 ppm δ C = ppm 41 42

22 First-Order Spin-Spin-Coupling 22 of From previous discussions one could have gotten the impression that a typical 1 NMR spectrum exhibits just one signal for each set of equivalent 1 -nuclei and that the same thing is true for 13 C spectra, as well as for spectra of any other isotope. owever, there are many more lines in a spectrum, and while these extra lines do make a spectrum more complex, they also offer valuable structural information that complements the chemical shift data

23 C3C2O:!!The spin states of two hydrogens (methylene group): 23 of 40 B 0 Total Magnetization M= Three spin states with population ratio of 1:2:1 For methyl hydrogens the net experienced field will depend on the magnetization of the neighboring methylene group! The methyl signal will be split into three lines with intensity ratio 1 : 2 : 1 (= spin-spin-coupling, homonuclear coupling because the coupling is between nuclei of the same isotope). Triplet. 45 Accordingly, for the methylene signal, the possible spin states of the methyl group determine its multiplicity (number of lines in the signal). The spin states of three hydrogens: M=3/2 M=-3/2 4 spin states with population ratio of 1 : 3 : 3 : 1 M=1/2 M=-1/2 Quartet 46

24 24 of 40 Accordingly, a doublet is observed for hydrogens that are coupled to a methine (C) proton. 47 The multiplicity of a given resonance = n+1 (n=# of neighboring equivalent nuclei). The relative intensities of the multiplet follow Pascalʼs triangle. 48

25 25 of J J J = spacing between lines. The slight difference in energy between the resonances is the coupling constant J [z]. Jʼs are independent of instrumental parameters! 50

26 26 of 40 The magnitude of J depends on the through-bond distance and the angle between a + b. a C b a b C C two-bond coupling 2 J "geminal" three-bond coupling 3 J "vicinal" Φ C (C) n C long range coupling n+3 J 51 Consider a 3-spin system with/without equivalent nuclei: a b c C C C a will be a doublet with 3 J ab b will be a doublet of doublets with 3 J ab and 3 J bc c will be a doublet with 3 J bc 52

27 27 of 40 J bc J ab J ab J bc a b c 53 54

28 28 of

29 eteronuclear Spin-Spin-Coupling 29 of 40 Any magnetic (I 0)(nuclear spin) nuclei can lead to spin-coupling interactions. 2NI + 1 = multiplicity 57 58

30 30 of Symmetry/Chirality Before we are able to understand and predict the appearance of homonuclear NMR spectra, we must be able to recognize when nuclei (and atoms) in a given structure will be distinguishable and when they will not. The test of distinguishability is based on symmetry relations among the nuclei. 60

31 31 of 40 Test of distinguishability: O C 3 C 3 heterotopic nuclei different constitutional environment 61 B C A Nuclei that are equivalent by virtue of a rotational axis are said to be homotopic. 62

32 32 of 40 Br C Cl D Nuclei that are equivalent by virtue of a reflection in the mirror plane are said to be enantiotopic. 63 If we have difficulty deciding whether two nuclei are related by a mirror plane, we can use the isotope substitution test. Br C Cl D Br D C Cl enantiomers 64

33 33 of 40 Cl C C 3 C Cl Cl D : C C 3 C Cl Cl D C C 3 C Cl diastereomers diastereotopic 65 homotopic A O A' but: different J's B NO 2 B' homotopic J AB J ABʼ 2 nd order effect AAʼBBʼ A/Aʼ are chemically equivalent, but not magnetically equivalent 66

34 34 of 40 Si OR 67 A set of nuclei that is magnetically equivalent must have identical coupling constants to all other nuclei! (Coupling between magnetically equivalent nuclei occurs, but does not show up in the spectrum). omotopic/enantiotopic :!chemically equivalent (same chemical shift); magnetically equivalent or not (identical or non-identical Jʼs) eterotopic/diastereotopic :!chemically and magnetically non-equivalent. Chemical equivalency (same chemical shift) is necessary but not sufficient for magnetic equivalency. 68

35 35 of 40 Accidental equivalence: C C C 3 δ = 1.80 It occasionally happens that two nuclei that are not symmetryequivalent in any way accidentally precess at exactly the same frequency, and, hence, give rise to a single NMR signal C NMR 12 C (98%) has I = 0 no NMR 13 C (1.1.%) has I = 1/2 NMR (Ε = hν = γhb o /2π gyromagnetic ratio such that ν obs 1/4 that of 1 (300 Mz 75!Mz) Observe typically ppm (rel. to TMS) 70

36 36 of 40 Coupling in 13 C NMR 1. Low natural abundance means 13 C- 13 C couplings are rare J C1 = 500 p; p = 0.25 sp 3 C [125] p = 0.33 sp 2 C [165] p = 0.50 sp C [250] 2 J C1 = z 3 J C1 = +/-1 z 71 Typically decouple the protons by saturating them with a second broadband RF puls (double resonance technique, second transmitter coil; white noise if the irradiating field is strong enough, not only will the 1 nuclei approach saturation, but virtually all the 1 magnetization will be tipped into the x 1 y plane. Since the 1 nuclei are no longer aligned with (or against) the applied field (which is along the z axis) they can no longer augment or diminish the magnetic field experienced by the carbons. As a result, the coupling interaction disappears, and each 13 C multiplet collapses to a singlet! (D coupling not effected!) causes all 13 C resonances to be singlets affords Nuclear Overhauser Effect makes integration of 13 C spectra unreliable 72

37 37 of C-NMR Chemical Shifts α C 4 C 3 -C 3 C 2 (C 3 ) 2 C(C 3 ) 3 C(C 3 ) β C 3 -C 3 C 3 -C 2 -C 3 C 3 -C(C 3 ) 2 C 3 -C(C 3 ) γ C 3 C 2 C 3 C 3 C 2 C 2 -C 3 C 3 C 2 C(C 3 ) 2 C 3 C 2 C(C 3 ) α effect: +9 ppm for each added C β - effect: +9 ppm for each added C γ - effect: -2.5 ppm for each added C 74

38 38 of Alkenes: C C C C C C C C * thus: δ C3 = = (exp.: 130.3) 76

39 39 of 40 Alkynes: 75 to 95 ppm C C Aromatics: not effected by ring current ppm substituted Cʼs are typically of lower intensity

40 Carbonyls: 40 of 40 O O O O O OEt O RO OR Cl ~ O O C N 3 I eavy atom effect. 79

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