核磁共振圖譜解析解析ppt課件

1、分子中的原子并不是孤立存在,它不僅在相互間發(fā)生作用也同周圍環(huán)境發(fā)生作用,從而導(dǎo)致一樣的原子核卻有不同的核磁共振頻率.0BELarmor 頻率化學(xué)位移自旋-自旋巧合e.g. B0=11.7 T, w(1H)=500 MHzw(13C)=125 MHz化學(xué)位移 B0 kHz自旋-自旋巧合 Hz-kHz;C-13 spectraC-13 spectraC-13 spectra can be determine the number of nonequivalent carbons and to identify the types of C (CH3, CH2, aromatic, C=O) tha

2、t may be present in a compound. C-13 NMR provides direct information about the carbon skeleton of a molecule. ;Electronegativity, hybridization, and anisotropy all affect 13C chemical shifts. The electronegative element produces a large downfield shift since the electronegativity atom is directly at

3、tached to the 13C atom. ;01234567891011環(huán)烷烴環(huán)烷烴 取代烷烴取代烷烴炔烴炔烴 單取代烷烴單取代烷烴 雙取代烷烴雙取代烷烴烯烴烯烴芳烴、雜芳環(huán)芳烴、雜芳環(huán)羧酸羧酸醛醛1H020406080100120140160180200220季碳季碳伯碳伯碳羰基羰基芳烴、雜芳環(huán)芳烴、雜芳環(huán)13C仲碳仲碳叔碳叔碳炔烴炔烴烯烴烯烴化學(xué)位移化學(xué)位移;B. Calculation of 13C Chemical Shiftsm-xylene, the base value for the C in a benzene ring is 128.5 ppm. ipsoorthom

4、etaparaCH38.90.7-0.1-2.9C1 = base+ipso+meta =128.5+8.9+(-0.1) =.3 ppmC2 = base+ortho+ortho =128.5+0.7+0.7 =129.9 ppmC3=C1C4 = base+ortho+para =128.5+0.7+(-2.9) =126.3 ppmC5 = base+meta+meta =128.5+(-0.1)+(-0.1) =128.3 ppmC6=C4The observed values for C1, C2, C4, and C5 are .6, 130.0, 126.2 and 128.2

5、ppm.CH3CH3123456;Spin-Spin CouplingSpin-Spin CouplingCH異核J-couplingJCHCHHC同核 J-couplingJHH;多重峰出現(xiàn)的規(guī)那么:1. 某一原子核與N個(gè)相鄰的核相互巧合將給出(n+1)重峰.2. 等價(jià)組合具有一樣的共振頻率.其強(qiáng)度與等價(jià)組合數(shù)有關(guān).3. 磁等價(jià)的核之間巧協(xié)作用不出如今譜圖中.4. 巧合具有相加性. 例如:HaHbCCJABHBHBHAHAJABobserved spincoupled spinintensityAB1B1BA1A1;HaHbCCHcAB,CBACAB,C是磁等價(jià)的核JAB=JAC;自旋-自旋

6、巧合引起共振線的分裂而構(gòu)成多重峰.多重峰實(shí)踐代表了相互作用的原子核彼此間可以出現(xiàn)的空間取向組合.CHJCHCHJCHoriginal frequency-J/2+J/2JCHOne-bond couplingsJ, 100-250 Hz;HaHbCCHcB,C是化學(xué)不等價(jià)的核JAC=16 HzJAC=8 HzJBC=3 HzACBAJACJAB;*CH*CH2*CH3CH1H2H3CH1H2CH1*CC;由于一些核的自然豐度并非如此100%。因此譜圖中能夠出現(xiàn)巧合分裂的峰和無巧合的峰。氯仿中的氫譜是一個(gè)典型的例子。x100H-13CH-13C105 HzH-12C;proton-coupled

7、 spectra (nondecoupled spectra) Quartet, J=127 HzProton-coupled spectra for large molecules are often difficult to interpret. The multiplets from different C commonly overlap because the 13C-H coupling constants are frequently larger than the chemical differences of the C in the spectrum.原子核間的巧合導(dǎo)致譜圖

8、的復(fù)雜化(“精細(xì)裂分)，靈敏度下降。; 假設(shè)峰數(shù)不多，巧合的方式仍可分析出。但當(dāng)很多鋒出現(xiàn)時(shí)，巧合方式的分析就不是那么容易。 直接巧合: 1J(13C,1H) 125 - 150Hz 長(zhǎng)程巧合:nJ(13C,1H) 1 - 10Hz*CH3-CH2-未去偶?xì)淙ヅ?The decoupling technique obliterates all interactions between 13C-H; therefore, only singlets are observed in a decoupled 13C NMR spectrum. The decoupler simultaneously

9、 irradiates a second, tunable radiofrequency which causes the protons to become saturated, and they undergo rapid upward and downward transitions, among all their possible spin states. These rapid transitions decouple any spin-spin interactions between 13C-H being observed. In effect, all spin inter

10、actions are averaged to zero by the rapid changes. The 13C “senses only one average spin state for the attached Hs rather than two or more distinct spin states. ;氫對(duì)碳的巧協(xié)作用可以經(jīng)過對(duì)氫施加一個(gè)脈沖消除。此一技術(shù)稱為去偶。對(duì)氫核的飽和照射，促使氫核的自旋形狀快速的變換，臨近的碳核無法覺得到氫核的自旋形狀的取向而只感遭到氫核兩種取想的平均效果。詳細(xì)的說，對(duì)氫核的飽和照射使碳核原來的兩條共振線w-J/2和w+J/2合并平均而得到(w-

11、J/2)+(w+J/2)/2=w。CHJCHCHJCHp-pulse on H這相當(dāng)于運(yùn)用一系列1800脈沖快速照射氫核。C-HpHC-HpHpHpHC-HC-HC-HC-HpH+J/2+J/2-J/2-J/2+J/2+J/2-J/2-J/2+J/2+J/2-J/2-J/2;氫去偶除簡(jiǎn)化碳譜還由于有核的Overhauser效應(yīng)而添加信噪比。decoupledcoupledC-HC-H2*CH3-CH2-;The intensities of many of the C resonances in a proton-decoupled 13C spectrum increase signific

12、antly above those observed in a proton-coupled experiment. 核Overhauser效應(yīng)NOE Overhauser等人及以后的研討人員發(fā)現(xiàn)，當(dāng)核外電子自旋或相鄰磁性核的核自旋發(fā)生共振并到達(dá)飽和時(shí)，偶極偶極相互作用(弛豫)引起核自旋態(tài)分布變化，使得待察看的核的信號(hào)強(qiáng)度添加。其空間作用間隔 CH2 CH CThe interaction of the spin-spin dipoles operates through space, not through bonds, and its magnitude decreases as a fu

13、nction of the inverse of r3 (radial distance) . HCr)1(=NOE3rf;NOEs are sometimes used to verify peak assignments.The syn methyl group is closer to the aldehyde hydrogen. Irradiation of the aldehyde H leads to a larger NOE for the carbon of the syn methyl group than for that of the anti methyl group,

14、 allowing the peaks to be assigned. 構(gòu)造測(cè)定 (空間位置關(guān)系)CONHCH3CH3Dimethylformamideanti, 31.3 ppmsyn, 36.2 ppm;Integral information derived from 13C spectra is usually not reliable unless special techniques are used to ensure its validity. NOE is not the same for every carbon. The time required for relaxat

15、ion of 13C is quite variable, depending on the molecular environment of the particular atom.Collection of the FID signal may have already cease before all of the 13C have relaxed. Some atoms have strong signals, since they have relaxed completely; while others have weaker signals. ;Relax, excited ex

16、cess nuclei at the upper spin state return to the lower spin state and equilibrium, they generate the FID signal. Saturation, all of the excess nuclei absorb energy, the populations of both spin states are equal. The population of the upper spin state cannot be increased further. Relaxation processe

17、s, excited nuclei return to their ground state and by which the Boltzmann equilibrium is reestablished. Spin-lattice relaxation, longitudinal, (T1). For 13C, if H directly bonded, T1 is fastest; larger molecules tumble slowly, relaxation is most effective, small molecules, very inefficient. Spin-spi

18、n relaxation, transverse, (T2).;T2T1, spin=1/2 and a solvent of low viscosity, T2 and T1 are usually very similar.T1 values are quite important to 13C NMR spectra, they are much longer for C and can dramatically influence signal intensities. Quaternary carbons (including most carbonyl carbons) have

19、long relaxation times because they have no attached Hs. ;To obtain a decent spectrum of this compound, it would be necessary to extend the data acquisition and delay periods so as to determine the entire spectrum of the molecule and see the carbons with high T1 values. ;1451401351301250 102030405060

20、7080901001101203025201505101520253035404550556012645ethylbenzene14523632030405060708090100110120130140ppm126453;100-150ppm, relatively few other peaks appear in this range, a great deal of useful information is available when peaks appear here. Monosubstituted, four peaks; the ipso C, has a very wea

21、k peak due to a long relaxation time and a weak NOE. ;A symmetrically disubstituted;PolysubstitutionMost other polysubstitution patterns on a benzene ring yield six different peaks.When identical substituents are present, planes of symmetry may reduce the number of peaks. XXXTwo peaksCH2OHOCH3123456

22、200 180 160 140 120 100 80 60 40412, 63, 5OCH3CH2OH;Most FT-NMR spectrometers require the use of deuterated solvents because the instruments use the deuterium resonance signal as a “l(fā)ock signal, or reference signal, to keep the magnet and the electronics adjusted correctly.Multiplicity = 2nI+177ppm3

23、9.5ppm;Example4-2.1 C3H6O2COCH3OCH3;Example4-2.2 C4H10OCO HC H3C H3H3CC HC H3O HC H2C H3C HC H3C H3C H2H O;C5H8O2, an esterCOOCH3CH2CCH3;C7H8OO HC H3O C H3;C8H8OCOC H3C HOC H2;CHH3CH3CH2CNH2959220;129128132OOCH2CH2CH2CH313016766302013;DEPT: Distortionless Enhancement by Polarization TransferThe samp

24、le is irradiated with a complex sequence of pulses in both the 13C and 1H channels. the 13C signals for the C atoms in the molecule will exhibit different phases, depending on the number of Hs attached to each carbon. Each type of C will behave slightly differently, depending on the duration of the

25、complex pulses. 提高雜核實(shí)驗(yàn)的靈敏度; DEPT：區(qū)分：區(qū)分13C的級(jí)數(shù)的級(jí)數(shù)4590 135 CH+CH2+ -CH3+ +CHCHCH2CH3;2030405060708090100110120130140ppm65 4321sethylbenzene145236;CCH2H4H1H3H5Spins of nucleusSpins of electronDirac vector model;nCH1 sptype ionhybridizat for )1n1(500Hz)(J+;2J, two bond couplings, geminal coupling, JgemTh

26、e nuclei prefer to have parallel spins, resulting in a negative coupling constant. The amount of geminal coupling depends on the HCH angle, .CHHCHH; As the decreases, the two orbitals move closer, and the electron spin correlations become greater. ;Table 4-3.2 Variations in 2JHH with Hybridization a

27、nd Ring Size;3J, three-bond couplings, vicinal couplings, the Hs are on neighboring carbon atoms, Jvic. (H-C-C-H);Since the interacting nuclei are spin-paired in the favored arrangement, three-bond H-C-C-H couplings are expected to be positive. In fact, most three-bond couplings, regardless of atom

28、types, are found to be positive. The spins of the H are paired and the spins of the electrons which are interacting through orbital overlap are also paired, is expected to represent the lowest energy and have the favored interactions. The spin interaction between the electrons in the two adjacent C-

29、H bonds is the major factor determining the size of the coupling constant. ;The actual magnitude of the coupling constant between two adjacent C-H bonds can be shown to depend directly on the dihedral angle between these two bonds. The side-side overlap of the two C-H bond orbitals is at a maximum a

30、t 0, where the C-H bond orbitals are parallel, and at a minimum at 90, where they are perpendicular. At =180, overlap with the back lobes of the orbitals occurs.5, 1, 72coscos3-+CBACBAJHH;The bond length RCC, the valence angles 1 and 2, and the electronegativity of any substituents X attached to C.

31、;HaHbHaHbJcis=10-12Hz =0Jtrans=7-8Hz =120HAHBHAHBJaa=10-14Hz =180Jae=Jee=4-5Hz =60Conformationally rigid compounds: cyclohexane derivatives;Table 4-3.1 Some Three-Bond Coupling Constants (3Jxy)6-15 8-11 5-7 2-4 0-2 HHHHHHHH;Chemically equivalent, two or more nuclei are equivalent by symmetry. In mos

32、t cases, chemically equivalent nuclei have the same resonance frequency (chemical shift), do not split each other, and give a single NMR signal.Magnetically equivalent. CH3CH3OPlaneCOOHHCOOHHAxisCl CH2CH2ClPlaneOHCH3CH3HAxisOHCH3HCH3Plane;化學(xué)等價(jià)質(zhì)子與化學(xué)不等價(jià)質(zhì)子的判別化學(xué)等價(jià)質(zhì)子與化學(xué)不等價(jià)質(zhì)子的判別可經(jīng)過對(duì)稱操作或快速機(jī)制如構(gòu)象轉(zhuǎn)換互換的質(zhì)子是可經(jīng)過對(duì)

33、稱操作或快速機(jī)制如構(gòu)象轉(zhuǎn)換互換的質(zhì)子是化學(xué)等價(jià)的?；瘜W(xué)等價(jià)的。不可經(jīng)過對(duì)稱操作或快速機(jī)制構(gòu)象轉(zhuǎn)換互換的質(zhì)子是不可經(jīng)過對(duì)稱操作或快速機(jī)制構(gòu)象轉(zhuǎn)換互換的質(zhì)子是化學(xué)不等價(jià)的。化學(xué)不等價(jià)的。與手性碳原子相連的與手性碳原子相連的CH2上的兩個(gè)質(zhì)子是化學(xué)不等價(jià)的。上的兩個(gè)質(zhì)子是化學(xué)不等價(jià)的。對(duì)稱操作對(duì)稱操作對(duì)稱軸旋轉(zhuǎn)對(duì)稱軸旋轉(zhuǎn)其他對(duì)稱操作其他對(duì)稱操作 如對(duì)稱面如對(duì)稱面等位質(zhì)子等位質(zhì)子化學(xué)等價(jià)質(zhì)子化學(xué)等價(jià)質(zhì)子對(duì)映異位質(zhì)子對(duì)映異位質(zhì)子非手性環(huán)境為化學(xué)等價(jià)非手性環(huán)境為化學(xué)等價(jià)手性環(huán)境為化學(xué)不等價(jià)手性環(huán)境為化學(xué)不等價(jià);化學(xué)等價(jià)質(zhì)子與化學(xué)不等價(jià)質(zhì)子的判別化學(xué)等價(jià)質(zhì)子與化學(xué)不等價(jià)質(zhì)子的判別;Magnetic equi

34、valence has two strict requirements:1.Magnetically equivalent nuclei must be isochronous; that is, they must have identical chemical shifts. 2.Magnetically equivalent nuclei must have equal coupling (same J values) to all other nuclei in the molecule.CClHHHCOCH3CH2CH2CH3;The Hs in each group experie

35、nce identical average magnetic environments, mainly because of free rotation, and are magnetically equivalent. Because of rotation, the Hs in each group are equally coupled to the Hs in the other groups. Without free rotation there would be no magnetic equivalence. Because of the fixed dissimilar di

36、hedral angles, Jab and Jab would not be the same. ;(a) Free ratation The n+1 Rule applies(a) Locked conformation A tree diagram is requiredCCHAHBHCCCHAHBHC;HaHbCCHcB,C是化學(xué)不等價(jià)的核JAC=16 HzJAC=8 HzJBC=3 HzACBAJACJAB;The ring blocks rotation, causing HA and HB to have different chemical shift values, they

37、 are chemically and magnetically nonequivalent. The n+1 Rule no longer applies to the nonequivalent protons. ;OHHH(a)(b)(c)(d)Small ring compounds, 3JBC(cis) 3JAC(trans);In a linear chain, the n+1 Rule is strictly obeyed only if the vicinal interproton 3J are the same for every successive pair of ca

38、rbons. ;In many molecules JAB is only slightly different from JBC, this leads to peak broadening in the multiplet, since the lines do not quite overlap. ;HBRHAHCHA HB HCJAB JACJBCcis3J 6-15 Hztrans3J 11-18 HzTerminal methylene2J 0-5 Hz;13.21 ;The protons on double bonds differ in that they are rarel

39、y magnetically equivalent, and they often give rise to splitting patterns that cannot be explained by the n+1 Rule. A molecule which has a symmetry element (a plane or axis of symmetry) passing through the C=C double bond does not show any cis or trans splitting, since the protons HA and HB are chem

40、ically and magnetically equivalent. ;4J 0-3 Hz 4J 0-3 Hz In addition to these three types of coupling, alkenes often show small couplings between protons substituted on carbons to the double bond and those on the opposite end of the double bond. The electrons of the double bond apparently help to tr

41、ansmit the spin information from the nucleus to the other. ;When all nuclei are coplanar, there is no interaction of the allylic C-H bond orbital with the system, and 4J=0 HzWhen the allylic C-H bond is perpendicular to the C=C plane (as is the bond) , the interaction assumes the maximum value, 4J=3

42、 Hz. ;Allylic splitting is observed in compounds such as the following:;cbaSO2C H2(a)(c)(b);HHHBrH2CabcdOCH3OCH2HHH(a)(c)(b)(d);Only under special circumstances does coupling occur between protons which lie farther apart than in 3J. Long-range coupling;Homo-allylic coupling5J = 0-1.6 HzIn some alken

43、es coupling can occur between the C-H bonds on either side of the double bond.This type of coupling is generally very small or even nonexistent in most molecules, but it sometimes appears in NMR spectra. ;The long-range coupling that occurs in aromatic rings must also transmit the spin information t

44、hrough the system of the ring. Similar types of coupling exist in all of the aromatic heterocycles. The spin information is also transmitted through the system of the ring. ;Long-range coupling in compounds without systems are less common but do occur in special cases. ;Angular methyl groups in ster

45、oids, or those at the ring junctions in a decalin system, often exhibit peak broadening due to W coupling with several H of the ring. ;A.First-order and Second-order spectra10/Jv, typical first-order spectraJv/approaches unity, second-order spectra;B. Spin system notationLARGEJABJAB A B v = ( A B)AB

46、, AMX, AX, A2, AACCHAHB; J A B A B , A X ; J A B A B , A B ; JAB/AB, A2. Most obvious is the decrease in intensity of the outer peaks of the doublets, with a corresponding increase in the intensity of the inner peaks. The chemical shifts of HA and HB dont correspond to the center point of each doubl

47、et.;3241-BA;C HC H2 v/J10.05.03.01.51.00.3AX2AB2;CH2CH2A2X2A2B2 v/J10.06.03.01.51.00.5;The ring protons of a benzenoid system: 7.3;Electron-withdraw ring substituents (-NO2, -CN, -COOH, -C=O), move the resonance of these protons downfield; Electron-donating ring substituents (-OCH3, -NH2) move the r

48、esonance of these protons upfield. Table 4-3.3 1H chemical shifts in p-disubstituted benzene compounds;A. Monosubstituted RingsAlkylbenzenesCH2CH3;Electron-Donating GroupsThe ortho/para hydrogens, upfieldThe meta hydrogens, downfieldOCH3OCH3OCH3OCH3-OHORNHNHRNR R,2OCH3AABBC;Anisotropy-Electron-Withd

49、rawing GroupsNR,N,NOCONHR,COOH,COOR,COR,CHO,2-;N H2HaHaHbC lHba four-line pattern, crudelySince Ha and Ha (Hb and Hb ) are chemically equivalent and not magnetically equivalent. B. para-disubstituted Rings;Jaa, JabJab;A single aromatic resonance integrating for 4 protons could easily represent a par

50、a-disubstituted ring.As the chemical shifts of Ha and Hb approach each other, the p-disubstituted pattern becomes similar to that of p-methoxyaniline.The inner peaks move colser together, and the outer ones become smaller or even disappear. Ultimately, when Ha and Hb approach each other colsely enou

51、th in chemical shift, the outer peaks disappear and the two inner peaks merge into a singlet. ;HbXHaHbHaXWhere the two ortho substituents are identical, leading to a plane of symmetry. ClCl;HHHHHHortho meta para 3J = 7-10 Hz 4J = 2-3 Hz 5J = 01 HzC. Other substitution;O M eAHBHCN O2HDN O23JBC 8 orth

52、o4JCD 2 meta 5JBD 0 para;NH2NO2NH2NO2NH2NO2;OHCHCH3CH2CH3;A. Diastereotopic methyl groups: (S)-(+)-3-methyl-2-butanol;The two methyl groups have slightly chemical shifts because the adjacent carbon atom is a chiral center.The two methyl groups are always nonequivalent in this molecule, even in the p

53、resence of free rotation. There are no planes of symmetry in any of these conformations; neither of the methyl groups is even enantiomeric. ;Diastereotopic fluorines: 1-Br-2-Cl-1,1,2-trifluoroethane;Under the usual conditions of determining the NMR spectrum of an alcohol, and for most alcohols, no c

54、oupling is observed between the hydroxyl hydrogen and H on the carbon atom to which the hydroxyl group is attached (-CH-OH). Whether or not spin-spin splitting involving the hydroxyl hydrogen is observed in a given alcohol depends on several factors, including temperature, purity of the sample, and

55、the solvent used. These factors are all related to the rate at which hydroxyl protons exchange with one another (or the solvent) in the solution. Under normal conditions, the rate of exchange of protons between alcohol molecules is faster than the rate at which the NMR spectrometer can respond. R-OH

56、a+R-OHbR-OHb+R-OHa;Rapid chemical exchange decouples spin interactions, and the NMR spectrometer records only the average environment it senses for the exchanging proton. J=5Hz;CH3-CH2-OH;Ultrapure ethanol : if purifies a sample of ethanol to eliminate all traces of impurity (especially of acid and

57、water, thereby slowing the proton exchange rate), the hydroxyl-methylene coupling can be observed in the form of increased complexity of the spin-spin splitting patterns. ;A. Acid/Water and Alcohol/Water Mixtures;CH3COOHa+H-OHbCH3COOHb+H-OHaWhen two compounds, each of which contains an O-H group, ar

58、e mixed, one often observes only a single NMR absorption (between the H positions of the pure substances) due to O-H. Exchange of the OH protons between two compounds occurs so rapidly that the NMR “sees the OH protons only in an averaged environment intermediate between the two extremes of the pure

59、 substances. The exact position of the O-H resonance depends on the relative amounts of acid and water. If there is more acid than water, the resonance appears closer to the pure acid OH resonance. With the addition of more water, the resonance moves closer to that of pure water. ;B. Deuterium Excha

60、ngeAcids, phenols, alcohols, and amines are the functional groups that exchange most readily. The result of each D exchange is that the peaks due to the exchanged H “disappear from the 1H NMR spectrum. The “l(fā)ost H generate a new peak, that of the H in DOH. RCOOH+D2ORCOOD+DOHArOH+D2OArOD+DOHROH+D2ORO