Introduction to NMR spectroscopy

Transcription

1 Introduction to NMR spectroscopy Nuclei of isotopes which possess an odd number of protons, an odd number of neutrons, or both, have a nuclear spin quantum number, I, such that, I = 1/2n, where n is an integer 0,1,2,3...etc. Since atoms have charge, a spinning nucleus generates a small electric current which in turn creates a finite but small magnetic field. The magnetic dipole, of the nucleus varies with each element. Thus, any nucleus with a spin can be observed by NMR spectroscopy. When a nucleus with a spin is placed in a magnetic field the nuclear magnet experiences a torque which tends to align it with the external field just like a compass needle in the earth s magnetic field. For a nucleus with a spin of 1/2, there are two allowed orientations of the nucleus: parallel to the field (low energy) and against the field (high energy). Since the parallel orientation is lower in energy, this state is slightly more populated than the antiparallel, high energy state. spin = 1/2 Turn on mangetic field Field splits system into two allow spin states spin = +1/2 Direction of Applied Field Page 1

2 NMR spectroscopy The energy difference between the state is proportional to the magnetogyric ratio of the nucleus and the magnitude of the applied field as shown in the figure and equation below. Energy Applied Magnetic Field Planck's constant x10 14 Kcal/mol Magnetogyric ratio (26,753 radian/gausssec for protons) ΔE = h X ν 2π app Applied magnetic field (variable) gauss for a routine machine An applied magnetic field of roughly 7000 gauss corresponds to energy difference of 2.85 x 10 5 Kcal/mol for protons, an extremely small difference If you have 2,000,000 protons in such a field roughly 999,995 would be spin down aligned against the field and 1,000,005 would be spin up with the field. Since E = hν, the frequency of electromagnetic radiation needed to promote the nuclei from the lower energy to higher energy state, for the condition described in the equation above corresponds to a frequency of = 300 x 10 6 sec 1 or 300 Mz which is FM radio frequency Page 2

3 In the NMR experiment a radio frequency applied to molecules is applied to the nucleus in the magnetic field. When he RF radiation energy exactly matches the energy gap between the energy level for the spin aligned with the magnetic field and the spin antiparallel to the magnetic field the nuclei can absorb the energy and be promoted from the lower energy state to the higher energy state. Energy of Spin States spin = 1/2 spin = +1/2 ΔE apply RF corresponding to exactly ΔE and a spin is promoted to the higher energy state Intensity Peak corrsponding to absorption of energy Chemical Shift When this condition is met the nuclei are said to be in resonance with the applied field and this absorption of radiation leads to a peak in a spectrum. Page 3

4 NMR and chemical structure. Four types of information from an NMR spectra make it an extremely valuable technique for determining chemical structure. 1. The number of sets of peaks: This corresponds to the number of magnetically (chemically) different types of protons. 2. Chemical Shift: The key to NMR spectroscopy is that the exact energy at which this peak comes (which can be measured extremely accurately) is dependent upon the local environment of the proton. 3. Splitting: The number of peaks observed for a given proton is dependent upon the number and type of protons in the vicinity of the one that is being observed. 4. Integration: The area of the peak is proportional to the number of protons that are magnetically equivalent. The NMR spectrum of methyl isopropyl ether (C 3 OC(C 3 ) 2 ) is shown below: Note that there are three sets of resonances. One has seven peaks, one has one peak and one has two peaks. They show up in different places in the spectrum and they have different integrated areas under each of the resonances. It is the combination of all of this information that enables us to assign the structure to the molecule. Page 4

5 Chemical Shift Naively you might expect that all nuclei of a given type would undergo the spinflip transition at exactly the same applied frequency in a given magnetic field. owever, electrons in the molecule have small magnetic fields associated with them which tend to oppose the applied field. Electron Cloud app Induced Current Induced Magnetic field at nucleus These magnetic fields are said to screen the nuclei from the full strength of the applied field. The greater the electron density, the greater this 'shielding' will be, hence nuclei which are in electron rich environments will undergo transition at a higher applied field than nuclei in electron poor environments. Thus, the relevant field at the nucleus which determine the energy of the transition now becomes eff = app + shielding Remember that shielding is opposite in sign to app. The resulting shift in the NMR signal for a given nuclei is referred to as the chemical shift. For 1 the chemical shift is given relative to the absorption of tetramethyl silane ((C 3 ) 4 Si)(TMS). The chemical shift in z is proportional to the operating frequency and is given by Chemical shift (z) = ν ν TMS. Page 5

6 The delta scale for chemical shifts Chemical shifts are reported in a manner that is independent of frequency by normalizing both the frequency of the observed hydrogens and those for TMS hydrogens by the operating frequency of the instrument as defined by the equation shown below: Chemical Shift (δ) = (ν ν TMS (z))/operating frequency (z) x 10 6 ppm The factor of 10 6 is introduced into the equation to give a simple whole number scale for convenience. In this scale TMS is arbitrarily assigned a value of zero. TMS is used because its protons are more highly shielded than those observed in most common organic molecules and because it is chemically inert. For 1 NMR, the δ scale generally extends from 012 ppm and then each chemically inequivalent nucleus has it own chemical shift that is independent of the operating frequency of the NMR. The ppm scale is plotted from δ = 0 on the right and with increasingly deshielded nuclei (increasing δ) to the left. Peaks to the left are referred to a downfield of peak to the right since it would require a smaller applied field for them to be in resonance with a fixed excitation frequency. Electronegative groups are "deshielding" and thus protons on carbons in the vicinity of electronegative atoms are deshielded and appear at higher chemical shifts. The more electronegative the group the further down field the proton will be. If the electronegative group is directly attached to the carbon the effect is greatest and if it is attached to the α carbon it is somewhat damped and more damped when attached to the β carbon and so on. If multiple electronegative groups are attached then there is an additive (although not strictly additive) deshielding effect on the proton. Protons on oxygen or nitrogen have highly variable chemical shifts which are sensitive to concentration, solvent, temperature, etc. Page 6

7 Anisotropic Effects The πsystem of alkenes, aromatic compounds and carbonyls strongly deshield attached protons and move them "downfield" to higher δ values. This is due to an effect known as anistropic shielding and deshielding. app O app app app So now the total magnetic field is given by: eff = app + shielding + anisotropic shielding Thus, for aromatic rings, olefins and carbonyl compounds this additional magnetic field adds to app and for terminal acetylenes the field substracts from app. Page 7

8 Proton NMR Chemical Shifts for Common Functional Groups Page 8

9 Functional Group and NMR shifts δ Cyclopropane C Ethane C 3 C Ethylene C 2 =C Acetylene C C 2.88 Benzene C Propene C 2 =CC Propyne C CC Acetone C 3 COC Cyclohexane C Methyl chloride C 3 Cl 3.10 Methylene chloride C 2 Cl Chloroform CCl Ethanol C 3 C 2 O C 3 C 2 O C 3 C 2 O Acetic acid C 3 COO C 3 COO Acetaldehyde C 3 CO C 3 CO Diethyl ether (C 3 C 2 ) 2 O (C 3 C 2 ) 2 O Ethyl acetate C 3 COOC 2 C 3 C 3 COOC 2 C 3 C 3 COOC 2 C Trimethylamine N(C 3 ) Triethylamine N(C 2 C 3 ) Toluene C 6 5 C Benzaldehyde C 6 5 CO 9.96 Page 9

10 Chemical Shift Table II. Starting Value of Shift Positions: Methyl, δ = 0.90 methylene, δ =1.20 methine, δ =1.55 Alpha refers to the substituent being on the same carbon as the hydrogen and for β it is one removed. Bromine Iodine Aryl C=O and RC=O CO 2 and CO 2 R O and OAlkyl OAryl OCOAlkyl Amine (R=nitrogen) C N Functional Group R Type of ydrogen Alpha Shift Beta Shift C Chlorine C C C 3 C 2 C C 3 C 2 C C 3 C 2 C C 3 C 2 C C 3 C 2 C C=C C 3 C 2 C C 3 C 2 C C 3 C 2 C C 3 C 2 C C 3 C 2 C C 3 C 2 C C 3 NO 2 C 2 C C C C 3 C 2 C Page 10

11 Integration One further feature of the proton NMR is the fact that the intensity of the absorbance of a given class of nuclei (with a certain chemical shift) is proportional to the number of protons giving rise to the signal; that is, the area under a given peak (the integration) is directly proportional to the number of that type of proton in the molecule. Integrations are typically given as simplest wholenumber ratios, hence, acetic acid, C 3 COO, will have two peaks in the proton NMR, one at δ ~ 2, area = 3, and a second at δ ~ 12, area = 1. tbutyl, methylether, (C 3 ) 3 COC 3, will also have two peaks in the proton NMR, one at δ ~ 1, area = 3, and a second at δ ~ 3.5, area = 1 (the relative areas or both peaks are the same, but each one represents three hydrogens). Equivalence omotopic hydrogens: If you have a methylene group, of the form X C 2 X and you replace one of hydrogens of the C 2 with a dummy atom and then you independently replace the other hydrogen of the C 2 group with a dummy atom, then the two molecules thus created will be identical to each other, the hydrogens are said to be homotopic. The protons on a methyl group are homotopic. Enantiotopic: If you have a methylene group, of the form X C 2 Y and you replace one of the hydrogens of the C 2 with a dummy atom and then you independently replace the other hydrogen of the C 2 group with a dummy atom, the two molecules thus created will be enantiomers of each other, the protons are said to be enantiotopic. In a nonchiral environment enantiotopic protons are equivalent. owever, in a chiral environment such as a chiral solvent, they can, in principle, have different chemical shifts. Diastereotopic: If you have a methylene group, for example, in a chiral molecule X C 2 Y* (where the * indicates that Y is a chiral group) and you replace one of the hydrogens of the C 2 with a dummy atom and then you independently replace the other hydrogen of the C 2 group with a dummy atom the two molecules thus created will be diastereomers to each other thus, in principle, the two hydrogens should have different chemical shifts. Page 11

12 SpinCoupling/Splitting Proton NMR Diethyl ether, exhibits two sets of protons in the NMR spectrum; a C 3 group around δ = 1, and a C 2 group around δ 4 (shifted downfield by its proximity to the electronegative oxygen). The NMR spectrum of diethyl ether does not have simply two peaks, rather, has two sets of peaks, one with four lines of relative intensity 1:3:3:1 which in total integrate to two protons and one set of three lines of relative intensity 1:2:1 which in total integrate to three protons. This multiplicity is due to the phenomena known as spin coupling and arises because of interaction of among proton magnetic field mediated by bonding electrons. To understand this better let's first examine a simple case in which one has X 2 C a C b Y 2. Each proton has spin 1/2 and to first approximation there is an equal probability of its spin being aligned with (or aligned against) the applied magnetic field. Page 12

13 Coupling in X 2 C a C b Y 2 So if we observe a, for 50% of the molecules in the sample b is aligned with the field: thus the total field is slightly larger and thus for these molecules the resonance will occur at slightly higher frequency For the other 50% of the molecules in the sample b is aligned against the field: thus the total field is slightly smaller thus for these molecules the resonance will occur at slightly higher frequency As a result, we observed two peaks for a whose total integrated area adds to 1. Similarly we observed two peaks for b whose total integrated area adds to 1. The energy difference between the two peaks for a in z is the coupling constant (noted as J): because there are two peaks a would be called a doublet when two nuclei are coupled to each other the coupling constants are identical thus b would also be a doublet furthermore, the energy difference in z between the two peaks for b would equal that of a in z since J ab = J ba Thus, spin spin coupling is very useful to provide information about the connectivity in organic molecules since: protons on adjacent atoms can couple to each other and if there are multiple occurances of this happening in the spectra one can match up pairs of coupled protons by checking to see which set have identical coupling constants As we will see shortly, depending upon the distance and orientation of the protons and the hybridization of the carbons to which they are attached the coupling constants between protons have distinctive values. Page 13

14 The ethyl group Now lets examine a slightly more complicated case of an ethyl group as shown below: A A C B C O αspin βspin A B 2 Contribution to magnetic field felt by A from B 's J AB J AB Contribution to magnetic field felt by A from spectrometer Resultant net magnetic field felt by A A B A C C O A B 2 J BA Contribution to magnetic field felt by B from A 's J BA Contribution to magnetic field felt by B from spectrometer Resultant net magnetic field felt by B J BA In the spectrum for diethyl ether, the C 3 group is split by the two protons on the adjacent C 2 group into three peaks (a triplet). This results from the adjacent spin either being +1/2, +1/2 or +1/2, 1/2 or 1/2, +1/2 1/2, 1/2. or Notice that for the middle two spin possibilities, the magetic fields associated with each of the two spins exactly cancel, and that there are two possible ways for this two occur: +1/2, 1/2 or 1/2, +1/2. Likewise, the absorbance for the C 2 is split by the three protons on the methyl group into (n + 1) = 4 peaks (a quartet). The two ethyl groups in diethyl ether show as a single absorbance since the Page 14

15 molecule has a plane of symmetry, that is, both ethyl groups are chemically and magnetically equivalent: typical values for J are typically less than 20 z for simple splitting the number of peaks observed for a proton next to n equivalent protons is n + 1 the intensity of the lines for the observed resonance can be shown to be that as the prefactors from the binomial expansion, (1 + x) n, which follows Pascals triangle, shown below. n = 0 1 singlet s doublet d triplet t quartet q pentet p hextet h septet sp many mess multiplet m Page 15

16 Splitting Trees The effects of splitting are often described using a splitting tree which depicts the original absorbance being split by a coupling constant J into (n + 1) peaks. This is illustrated for the example below: note peak heights not to absolute scale C C 2 J ab J ab J ab C 2 peaks need to be 2X C 2 C A proton can be coupled to two inequivalent protons in which case the magnetic field coupling between each pair would not necessarily be equal. This is illustrated below: a C C b C c J ab J bc J ab J bc J bc J ab Jab C a C c C b In this case, you can see that for b coupling to a would yield a doublet with coupling constant J ab. When we consider the effect of coupling to c each of these lines are further split into two lines separated by an amount corresponding to J bc. Such a pattern is called a double of doublets. Page 16

17 More complex situations What happens to a doublet of doublet as J ab J bc decreases. This is illustrated below. J ab = J bc decreasing J ab J bc The simple coupling patterns we have seen so far assume the difference in chemical shift between the peaks (in z) is much larger than the coupling constants between the hydrogens. When this condition is met the spectrum is said to have firstorder coupling. owever, when Δδ ab is not much larger than J ab then the coupling is said to be nonfirstorder and the appearance of the spectrum changes. In some cases quite dramatically. J ab J ab J ab J ab J ab J ab Δδ ab >> J ab Δδ ab > J ab Δδ ab ~ J ab Δδ ab = 0 In particular, the inner lines of the pattern gain intensity at the expense of the outer line as shown below. The origin of this evolution is in quantum mechanics, which is beyond the scope of the course. These patterns are quite common and being able to recognize them coupled with a knowledge of which hydrogen atoms have similar chemical shift is a useful tool to make assignments, since you now have two pieces of information (i.e. the two hydrogen atoms are likely adjacent and have similar environments). Note that unlike chemical shift coupling constants are due to the internal magnetic field of the protons and thus coupling constants are independent of spectrometer frequency. Page 17

18 Things are not always ideal in the real world A splitting tree such as this is useful in understanding more complex splitting patterns, such as those that occur in BrC 2 C 2 C 2 OD, as shown below: As shown by the coupling tree, the spectrum is predicted to consist of 9 peaks centered around δ The actual NMR, as shown above, shows only five peaks in this region. This is because the central seven peaks are only separated by 2 z, and the peaks are not infinitely narrow. As a result some overlap and appear as single peaks. Page 18

19 Common 1 NMR Splitting Patterns The examples given above represent only the simplest and most common coupling patterns seen in the 1 NMR. A simple ethyl group displays a quartet and a triplet in the ratio 2:3; the chemical shift of the C 2 group is sensitive to the attached substituent and typically varies between δ 4 (for oxygen) to δ 2 (for a carbonyl). An isopropyl group displays a septet (7 peaks) and a doublet in the ratio 1:6; again, the chemical shift of the C group is sensitive to the attached substituent and typically varies between δ 4 (for oxygen) to δ 2 (for a carbonyl). A 1,4disubstituted aromatic compound displays two doublets in the ratio of 1:1, typically in the region around δ 7. Although the chemical shift of a monosubstituted aromatic compound should give rise to three sets of peaks (integration ratio 2:2:1), often only a single peak of integrated intensity 5 is observed around δ = in the 1 NMR. This is because the differences in chemical shift are typically small, as are the coupling constants. Page 19

20 Table of coupling constants between ydrogen (given in z) C 1220 C C 03.5 C C C C C C C C C C C C C C O C C C C C C 23 a e e J aa = 1013 J ae = 25 J ee = 25 a Page 20

21 Use of deuterium in NMR Although deuterium has a nuclear spin, deuterium NMR and proton NMR require different operating frequencies. Deuterium NMR absorptions are not detected under the conditions used for proton NMR. The coupling constants for protondeuterium splitting are very small. Deuterium substitution can be used to simplify NMR spectra and assign resonances. Solvents used in NMR spectroscopy must either be devoid of protons, or their protons must not have NMR absorptions that obscure the sample absorptions. NMR spectra of dynamic systems The spectrum of a compound involved in a rapid equilibrium is a single spectrum that is the timeaverage of all species involved in the equilibrium. The timeaveraging effect of NMR is not limited to simple conformational equilibria; the spectra of molecules undergoing any rapid process, such as a chemical reaction, are also averaged by NMR spectroscopy. The assignment of the O proton can be confirmed by adding a drop of D 2 O to the NMR sample tube shaking and taking the spectra. The O protons rapidly exchange with the protons of D 2 O to form OD groups on the alcohol and thus become invisible to NMR. NMR of alcohols and amines The chemical shift of the O proton in an alcohol depends on the degree to which the alcohol is involved in hydrogen bonding under the conditions that the spectrum is determined. The presence of water, acid or base causes collapse of the O resonance to a single line and obliterates all coupling associated with this proton. This effect is caused by a phenomenon called chemical exchange. This type of behavior is quite general for alcohols, amines, and other compounds with a proton bonded to an electronegative atom. Rapidly exchanging protons do not show spinspin splitting with neighboring protons. Acid and base catalyze this exchange reaction, accelerating it enough that splitting is no longer observed. In the absence of acid or base, this exchange is much slower, and splitting of the O protons and neighboring protons is observed. The assignment of the O proton can be confirmed by adding a drop of D 2 O to the NMR sample tube shaking and taking the spectra. The O protons rapidly exchange with the protons of D 2 O to form OD groups on the alcohol and thus become invisible to NMR. Page 21

22 Introduction to 13 C NMR Any nucleus with a nuclear spin can be studied by NMR spectroscopy; for a given magnetic field strength, different nuclei absorb energy in different frequency ranges. The NMR spectroscopy of carbon is called carbon NMR or CMR; the only isotope of carbon that has a nuclear spin is 13 C. The relative abundance of 13 C occurs in 1.1% natural abundance. The magnetogyric ratio for the 13 C nucleus is about 0.25 of the 1 nucleus and the sensitivity goes as this ratio to the third power which by itself would lead to a 13 C nucleus having the sensitivity of the 1 nucleus. Because of these two factors, 13 C spectra are about times as intense as 1 spectra. For weak signals, signal to noise is increased by averaging out the signal from multiple scans. The noise averages to zero and the signal increases. Signal to noise increases as (# of scans) 0.5. Given then relative sensitivity to 1 NMR you can see that many scans could be needed. Special instrumental techniques have been devised for obtaining such weak spectra. Coupling (splitting) between carbons generally is not observed. The reason is the low natural abundance of carbon (1.1%), making it unlikely that two 13 C nuclei will reside next to each other in a given molecule. Attached protons, however, with a spin of 1/2, will couple with the 13 C nucleus to generate spincoupling. The signal from the carbon will be split into (n + 1) peaks, where n is the number of attached protons. Page 22

23 13 C Chemical shfts The range of chemical shifts is large compared to that in proton NMR; trends in carbon chemical shifts parallel those for proton chemical shifts. Chemical shifts in CMR are more sensitive to small changes in chemical environment. Characteristics of chemical shifts for carbons: simple methyl carbons in alkanes, or attached to an alkene, of aryl group δ 1530 simple methylene, methine and quarternary carbons, carbons in alkanes attached to an alkene of aryl group δ 2060 characteristic chemical shifts for carbons attached to electronegative atoms oxygen (in ethers or the single bonded oxygen in an ester) δ 4080, nitrogen (in amines or the single bonded nitrogen in an amide) δ 2527 chlorine δ 2060 bromine δ 1050 iodine δ alkyne δ 6585, nitrile δ alkene carbons to δ aromatic carbons δ amide (C=O) δ ester (C=O) δ ketone and aldehyde (C=O) Page 23

24 13 C coupling and decoupling Splitting of 13 C resonances by protons ( 13 C 1 splitting) is so large relative to differences in chemical shifts that is can be difficult to distinguish between different peaks and peaks that are split by coupling. Spectra can be obtained in proton noisedecoupling mode wherein which the sample is irradiated at a second frequency which promotes all of the protons in the molecule to high spin states, disallowing the spincoupling process. All of the split 13 C peaks are thereby reduced to sharp singlets. While this operating mode provides less information than the nondecoupled mode, it is commonly used because it results in a significant signal enhancement due to a phenomena known as the "Nuclear Overhauser Effect" (NOE). NOE enhancement is not uniform, hence the integration of a 13 C NMR spectrum is typically meaningless. Spectra in which proton coupling has been eliminated are called protondecoupled carbon NMR spectra. The DEPT (distortionless enhancement by polarization transfer) technique yields separate spectra for methyl, methylene and methine carbons, and each line in these spectra corresponds to a line in the complete CMR spectrum; lines in the complete CMR spectrum that do not appear in the DEPT spectra are assumed to arise from quaternary carbons. CMR spectra are generally not integrated because the instrumental technique used for taking spectra, gives relative peak integrals that are governed by factors other than the number of carbons. Page 24