核磁共振图谱解析解析NMR.pptx

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1、NMR:原子核间的相互作用分子中的原子并不是孤立存在,它不仅在相互间发生作用也同周围环境发生作用,从而导致相同的原子核却有不同的核磁共振频率.Larmor 频率化学位移自旋-自旋偶合e.g.B0=11.7 T,w(1H)=500 MHzw(13C)=125 MHz化学位移 B0 kHz自旋-自旋偶合 Hz-kHz江南大学超值江南大学超值江南大学超值江南大学超值-划算划算划算划算-购物推荐群：购物推荐群：购物推荐群：购物推荐群：302284607302284607C-13 spectraC-13 spectra can be determine the number of nonequivale

2、nt carbons and to identify the types of C(CH3,CH2,aromatic,C=O)that 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 dow

3、nfield shift since the electronegativity atom is directly attached to the 13C atom.Fig.4-2.2 A Fig.4-2.2 A 1313C correlation chart for carbonyl and nitrile functional groups C correlation chart for carbonyl and nitrile functional groups 01234567891011（环）烷烃（环）烷烃 取代取代烷烃烷烃炔烃炔烃 单取代单取代烷烃烷烃 双取代双取代烷烃烷烃烯烯烃烃

4、芳芳烃、杂芳环烃、杂芳环羧酸羧酸醛醛1H020406080100120140160180200220季碳季碳伯碳伯碳羰基羰基芳芳烃、杂芳环烃、杂芳环13C仲碳仲碳叔碳叔碳炔烃炔烃烯烯烃烃化学位移化学位移B.Calculation of 13C Chemical Shiftsm-xylene,the base value for the C in a benzene ring is 128.5 ppm.ipsoorthometaparaCH38.90.7-0.1-2.9C1=base+ipso+meta=128.5+8.9+(-0.1)=137.3 ppmC2=base+ortho+ortho=

5、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 137.6,130.0,126.2 and 128.2 ppm.Spin-Spin CouplingCoupling Constants:Coupling Constants:n nJ J is a constant.is a constant.Homonuc

6、lear coupling,3J(1H-1H)=8Hz;heteronuclear coupling,1J(13C-1H)=156Hz;The electrons in the intervening bonds between the two nuclei transfer spin information from one nucleus to another by means of interaction between the nuclear and electronic spins.CH异核J-couplingJCHCHHC同核 J-couplingJHH同核J-偶合(Homonuc

7、lear J-Coupling)多重峰出现的规则:1.某一原子核与N个相邻的核相互偶合将给出(n+1)重峰.2.等价组合具有相同的共振频率.其强度与等价组合数有关.3.磁等价的核之间偶合作用不出现在谱图中.4.偶合具有相加性.例如:HaHbCCwawbJABHBHBHAHAJABHaHbCCHcAB,CBACAB,C是磁等价的核JAB=JAC同核J-偶合(Homonuclear J-Coupling)自旋-自旋偶合引起共振线的分裂而形成多重峰.多重峰实际代表了相互作用的原子核彼此间能够出现的空间取向组合.CHJCHCHJCHoriginal frequencyww-J/2w+J/2JCH异核J

8、-偶合(Heteronuclear J-Coupling)One-bond couplingsJ,100-250 HzHaHbCCHcB,C是化学不等价的核JAC=16 HzJAC=8 HzJBC=3 HzACBwAJACJAB同核J-偶合(Homonuclear J-Coupling)异核J-偶合(Heteronuclear J-Coupling)*CH*CH2*CH3CH1H2H3CH1H2CH1*CC由于一些核的自然丰度并非如此100%。因此谱图中可能出现偶合分裂的峰和无偶合的峰。氯仿中的氢谱是一个典型的例子。x100H-13CH-13C105 HzH-12C异核J-偶合(Heteron

9、uclear J-Coupling)proton-coupled spectra(nondecoupled spectra)Fig.4-2.4 Ethyl phenylacetate.(a)The proton-coupled Fig.4-2.4 Ethyl phenylacetate.(a)The proton-coupled 1313C C spectrum.(b)The proton-decoupled spectrum.(b)The proton-decoupled 1313C spectrumC spectrumQuartet,J=127 HzProton-coupled spect

10、ra 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.原子核间的偶合导致谱图的复杂化(“精细裂分”)，灵敏度下降。如果峰数不多，偶合的方式仍可分析出。但当很多锋出现时，偶合方式的分析就不是那么容易。直接偶合:1J(1

11、3C,1H)125-150Hz 长程偶合:nJ(13C,1H)1-10Hz*CH3-CH2-未去偶氢去偶去偶(Decoupling)Proton-Decoupled 13C SpectraThe decoupling technique obliterates all interactions between 13C-H;therefore,only singlets are observed in a decoupled 13C NMR spectrum.The decoupler simultaneously irradiates a second,tunable radiofrequen

12、cy 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 interactions are averaged to zero by the rapid chang

13、es.The 13C“senses”only one average spin state for the attached Hs rather than two or more distinct spin states.氢对碳的偶合作用可以通过对氢施加一个脉冲消除。此一技术称为去偶。对氢核的饱和照射，促使氢核的自旋状态快速的变换，临近的碳核无法感觉到氢核的自旋状态的取向而只感受到氢核两种取想的平均效果。具体的说，对氢核的饱和照射使碳核原来的两条共振线w-J/2和w+J/2合并平均而得到(w-J/2)+(w+J/2)/2=w。CHJCHCHJCHp-pulse on H这相当于使用一系列180

14、0脉冲快速照射氢核。C-HpHC-HpHpHpHC-HC-HC-HC-HpHw w+J/2w w-J/2w+w+J/2w w-J/2w w+J/2w w-J/2去偶(Decoupling)Fig.4-2.5 The proton-decoupled Fig.4-2.5 The proton-decoupled 1313C spectrum of 1-propanolC spectrum of 1-propanol氢去偶除简化碳谱还因为有核的Overhauser效应而增加信噪比。decoupledcoupledC-HC-H2*CH3-CH2-去偶(Decoupling)Nuclear Overh

15、auser Enhancement(NOE)The intensities of many of the C resonances in a proton-decoupled 13C spectrum increase significantly above those observed in a proton-coupled experiment.核核Overhauser效应效应（NOE）Overhauser等人及以后等人及以后的研究人员发现，当核外电子自旋或相邻磁性核的核的研究人员发现，当核外电子自旋或相邻磁性核的核自旋发生共振并达到饱和时，偶极偶极相互作用自旋发生共振并达到饱和时，偶极偶

16、极相互作用(弛弛豫豫)引起核自旋态分布变化，使得待观察的核的信号强引起核自旋态分布变化，使得待观察的核的信号强度增加。其空间作用距离度增加。其空间作用距离 CH2 CH CThe interaction of the spin-spin dipoles operates through space,not through bonds,and its magnitude decreases as a function of the inverse of r3(radial distance).rNOEs are sometimes used to verify peak assignments.

17、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,allowing the peaks to be assigned.结构测定结构测定 (空间位置关系空间位置关系)Dimethylformamideanti,31.3 ppmsyn,36.2 ppm4-2.6 Problems

18、with Integration in 13C SpectraIntegral 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 relaxation of 13C is quite variable,depending on the molecular environment of the

19、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.Fig.4-2.6 A typical FT-NMR Fig.4-2.6 A typical FT-NMR pulse sequence.pulse sequence.4-2.7 Molecula

20、r Relaxation ProcessRelax,excited excess 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 increas

21、ed further.Relaxation processes,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 in

22、efficient.Spin-spin 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 carb

23、ons)have 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.碳谱12645ethylbenzene145236

24、31313CC1 1H NMR(H NMR(1 1H decoupling)H decoupling)1264534-2.10 Some Sample Spectra-Equivalent CarbonsFig.4-2.8 The proton-decoupled Fig.4-2.8 The proton-decoupled 1313C NMR spectrum of 2,2-dimethylbutaneC NMR spectrum of 2,2-dimethylbutaneFig.4-2.9 The proton-decoupled Fig.4-2.9 The proton-decouple

25、d 1313C NMR spectrum of cyclohexanol C NMR spectrum of cyclohexanol Fig.4-2.10 The Proton-decoupled Fig.4-2.10 The Proton-decoupled 1313C NMR spectrum of cyclohexeneC NMR spectrum of cyclohexeneFig.4-2.11 The Proton-decoupled Fig.4-2.11 The Proton-decoupled 1313C NMR spectrum of cyclohexanone.C NMR

26、spectrum of cyclohexanone.4-2.11 Compounds with Aromatic Rings100-150ppm,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 weak peak due to a long relaxation time and a weak NOE.Fig.

27、4-2.12 The Proton-decoupled Fig.4-2.12 The Proton-decoupled 1313C NMR spectrum of toluene.C NMR spectrum of toluene.A symmetrically disubstitutedFig.4-2.13 The Proton-decoupled Fig.4-2.13 The Proton-decoupled 1313C NMR spectra of the three C NMR spectra of the three isomers of 1,2-dichlorobenzene is

28、omers of 1,2-dichlorobenzene 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.Two peaks200 180 160 140 120 100 80 60 40412,63,5OCH3CH2OH4-2.11 C-13 NMR Solvents-H

29、eteronuclear Coupling of C to DeuteriumMost FT-NMR spectrometers require the use of deuterated solvents because the instruments use the deuterium resonance signal as a“lock signal”,or reference signal,to keep the magnet and the electronics adjusted correctly.Multiplicity=2nI+1Fig.4-2.14 The Fig.4-2.

30、14 The 1313C peaks of two C peaks of two common solvents.(a)Chloroform-mon solvents.(a)Chloroform-d.(b)Dimethylsulfoxide-d(b)Dimethylsulfoxide-d6 677ppm39.5ppmExample4-2.1 C3H6O2Example4-2.2 C4H10OC5H8O2,an esterC7H8OC8H8ODEPT 极化转移技术DEPT:Distortionless Enhancement by Polarization TransferThe sample

31、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 complex

32、 pulses.提高杂核实验的灵敏度 DEPT：区分13C的级数4590 135 CH+CH2+-CH3+CHCHCH2CH3Fig.4-2.8(a)Standare Fig.4-2.8(a)Standare 1313C decoupled spectrum of ipsenol in CDClC decoupled spectrum of ipsenol in CDCl3 3,at,at 75.5MHz.(b)DEPT subspectra:DEPT 13575.5MHz.(b)DEPT subspectra:DEPT 135 CH and CH CH and CH3 3 up.CHup.C

33、H2 2 down.down.(C)DEPT 90 CH only.(C)DEPT 90 CH only.DEPT区分碳的级数65 4321sethylbenzene1452364-3.2 The Mechanism of CouplingWhen two spin-active nuclei prefer an opposed alignment,When two spin-active nuclei prefer an opposed alignment,J J is is usually positive.If the nuclei are parallel or aligned,usu

34、ally positive.If the nuclei are parallel or aligned,J J is usually is usually negative.negative.Couplings involving an odd number of intervening bonds(Couplings involving an odd number of intervening bonds(1 1J J,3 3J J)are positive,while those involving an even number of)are positive,while those in

35、volving an even number of intervening bonds(intervening bonds(2 2J J,4 4J J)are negative.)are negative.CCH2H4H1H3H5Spins of nucleusSpins of electronDirac vector modelA.One-Bond Couplings(1J)According to the Pauli Principle,pairs of electrons in the According to the Pauli Principle,pairs of electrons

36、 in the same orbital have opposed spins;therefore,the Dirac same orbital have opposed spins;therefore,the Dirac model predicts that the most stable condition in a bond is model predicts that the most stable condition in a bond is where both nuclei have opposed spins.where both nuclei have opposed sp

37、ins.It is unusual for coupling constants to depend on the It is unusual for coupling constants to depend on the hybridization of the atoms involved.hybridization of the atoms involved.1313C CHH13131313C C C CH HH H110-270 Hz110-270 Hz110-270 Hz110-270 Hzspspspsp3 3 3 3 115-125Hz,sp 115-125Hz,sp 115-

38、125Hz,sp 115-125Hz,sp2 2 2 2 150-170Hz,sp 240-270Hz 150-170Hz,sp 240-270Hz 150-170Hz,sp 240-270Hz 150-170Hz,sp 240-270Hz2J,two bond couplings,geminal coupling,JgemThe nuclei prefer to have parallel spins,resulting in a negative coupling constant.The amount of geminal coupling depends on the HCH angl

39、e,.Fig4-3.2 The Mechanism of Geminal CouplingFig4-3.2 The Mechanism of Geminal CouplingCHHCHHFig4-3.3 The dependence of the magnitude of Fig4-3.3 The dependence of the magnitude of 2 2J JHCHHCH,the,the geminal coupling constant,on the HCH bond angle geminal coupling constant,on the HCH bond angle .A

40、s the decreases,the two orbitals move closer,and the electron spin correlations become greater.Table 4-3.2 Variations in 2JHH with Hybridization and Ring SizeAs ring size decreases the absolute value of the coupling As ring size decreases the absolute value of the coupling constant constant 2 2J J a

41、lso decreases.also decreases.As the angle CCC in the ring becomes smaller(as As the angle CCC in the ring becomes smaller(as p p character character increases),the complementary HCH angle grows larger(increases),the complementary HCH angle grows larger(s s character increases)and consequently the ge

42、minal coupling character increases)and consequently the geminal coupling constant decreases.constant decreases.Hybridization is important and that the sign of the coupling Hybridization is important and that the sign of the coupling constant for alkenes changes to positive except where they have con

43、stant for alkenes changes to positive except where they have an electronegative element attached.an electronegative element attached.In many cases,no geminal HCH coupling(no spin-spin splitting)In many cases,no geminal HCH coupling(no spin-spin splitting)is observed,either the geminal protons are eq

44、uivalent by is observed,either the geminal protons are equivalent by symmetry or free rotation renders them equivalent.symmetry or free rotation renders them equivalent.The geminal coupling is often observed in cyclic compounds that The geminal coupling is often observed in cyclic compounds that are

45、 conformationally rigid-for instance,bicyclic compounds.are conformationally rigid-for instance,bicyclic compounds.Deuterium substitution experiments:if one of the Hs in a Deuterium substitution experiments:if one of the Hs in a compound which shows no spin-spin splitting is replaced by a D,compound

46、 which shows no spin-spin splitting is replaced by a D,geminal splitting with D is observed.geminal splitting with D is observed.3J,three-bond couplings,vicinal couplings,the Hs are on neighboring carbon atoms,Jvic.(H-C-C-H)Nuclear and electronic spin interactions carry the spin information Nuclear

47、and electronic spin interactions carry the spin information from one H to its neighbor.from one H to its neighbor.Since the Since the C-C bond is nearly orthogonal(perpendicular)to the C-C bond is nearly orthogonal(perpendicular)to the C-H bonds,there is no overlap between the orbitals,and the C-H b

48、onds,there is no overlap between the orbitals,and the electrons cannot interact strongly through the electrons cannot interact strongly through the bond system.bond system.They transfer the nuclear spin information via the small amount of They transfer the nuclear spin information via the small amou

49、nt of parallel orbital parallel orbital overlapoverlap that exists between adjacent C-H bond that exists between adjacent C-H bond orbitals.orbitals.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-bon

50、d couplings,regardless of atom types,are found to be positive.Fig4-3.4 The Method of Transferring Spin Fig4-3.4 The Method of Transferring Spin Information Between Two Adjacent C-H BondsInformation Between Two Adjacent C-H BondsThe spins of the H are paired and the spins of the electrons which are i