核磁共振图谱解析解析NMR.ppt

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.,江南大学超值-划算-购物推荐群： 302284607,C-13 spectra,C-13 spectra can be determine the number of nonequivalent 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 inf

2、ormation 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 attached to the 13C atom.,Fig. 4-2.2 A 13C correlation chart fo

3、r carbonyl and nitrile functional groups,1H,13C,化学位移,B. Calculation of 13C Chemical Shifts,m-xylene, the base value for the C in a benzene ring is 128.5 ppm. ipsoorthometapara CH38.90.7-0.1-2.9,C1 = base+ipso+meta =128.5+8.9+(-0.1) =137.3 ppm C2 = base+ortho+ortho =128.5+0.7+0.7 =129.9 ppm C3=C1 C4

4、= base+ortho+para =128.5+0.7+(-2.9) =126.3 ppm C5 = base+meta+meta =128.5+(-0.1)+(-0.1) =128.3 ppm C6=C4 The observed values for C1, C2, C4, and C5 are 137.6, 130.0, 126.2 and 128.2 ppm.,Spin-Spin Coupling,Coupling Constants: nJ is a constant. Homonuclear coupling, 3J(1H-1H) = 8Hz; heteronuclear cou

5、pling, 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.,同核J-偶合(Homonuclear 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

6、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:,Ha,Hb,C,C,Hc,A,B,C,B,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.,original frequency,w,w-J/2,w+J/2,JCH,异核J-偶合(Heteronuclear J-Coupling),One-bond couplings J, 100-250 Hz,Ha,Hb,C,C,Hc,B,C是化学不等价的核,

7、JAC=16 Hz,JAC=8 Hz,JBC=3 Hz,A,C,B,wA,JAC,JAB,同核J-偶合(Homonuclear J-Coupling),异核J-偶合(Heteronuclear J-Coupling),*CH,*CH2,*CH3,*C,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。,x100,H-13C,H-13C,105 Hz,H-12C,异核J-偶合(Heteronuclear J-Coupling),proton-coupled spectra (nondecoupled spectra),Fig. 4-2

8、.4 Ethyl phenylacetate. (a) The proton-coupled 13C spectrum. (b) The proton-decoupled 13C spectrum,Quartet, J=127 Hz,Proton-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 large

9、r 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(13C,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 Spectra,The decoupling technique obliterates all interaction

10、s between 13C-H; therefore, only singlets are observed in a decoupled 13C NMR spectrum. The decoupler simultaneously 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.

11、 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 changes. 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

12、。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。,这相当于使用一系列1800脉冲快速照射氢核。,去偶(Decoupling),Fig. 4-2.5 The proton-decoupled 13C 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。,decoupled,coupled,C-H,C-H2,*CH3-CH2

13、-,去偶(Decoupling),Nuclear Overhauser 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Overhauser效应（NOE） Overhauser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自旋发生共振并达到饱和时，偶极偶极相互作用(弛豫)引起核自旋态分布变

14、化，使得待观察的核的信号强度增加。其空间作用距离56。 C with H directly attached are enhanced the most, and the enhancement increases (but not always linearly) as more Hs are attached.,The maximum enhancement,The NOE effect is general, showing up when one of two different types of atoms is irradiated while the NMR spectrum o

15、f the other type is determined. The effect can be either positive or negative, depending on which atom types are involved. In the case of 13C-1H, the effect is positive; irradiating the H increases the intensities of the C signals.,Cross-polarization,In H-decoupled 13C spectra, irradiated H saturate

16、d a distribution of spins very different from their equilibrium (Boltzmann) state. (more spins than normal in the excited state) Due to the interaction of spin dipoles, the spins of 13C “sense” the spin imbalance of H, begin to adjust themselves to a new equilibrium state, more spins in the lower st

17、ate. excess population In a H-decoupled 13C spectrum, the total NOE for a given C increases as the number of nearby H increases. CH3 CH2 CH C The 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 dis

18、tance) .,r,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, allowing the peaks to be assigned. 结构测定 (空间位置关系),

19、Dimethylformamide,anti, 31.3 ppm syn, 36.2 ppm,4-2.6 Problems with Integration in 13C Spectra,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 relaxation of 13C

20、 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.,Fig. 4-2.6 A typical FT-NMR

21、 pulse sequence.,4-2.7 Molecular Relaxation Process,Relax, 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

22、the upper spin state cannot be increased 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, rela

23、xation is most effective, small molecules, very inefficient. 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 i

24、ntensities. Quaternary carbons (including most carbonyl carbons) 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 an

25、d see the carbons with high T1 values.,碳谱,1,2,6,4,5,3,4-2.10 Some Sample Spectra-Equivalent Carbons,Fig. 4-2.8 The proton-decoupled 13C NMR spectrum of 2,2-dimethylbutane,Fig. 4-2.9 The proton-decoupled 13C NMR spectrum of cyclohexanol,Fig. 4-2.10 The Proton-decoupled 13C NMR spectrum of cyclohexene

26、,Fig. 4-2.11 The Proton-decoupled 13C NMR spectrum of cyclohexanone.,4-2.11 Compounds with Aromatic Rings,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 weak peak

27、 due to a long relaxation time and a weak NOE.,Fig. 4-2.12 The Proton-decoupled 13C NMR spectrum of toluene.,A symmetrically disubstituted,Fig. 4-2.13 The Proton-decoupled 13C NMR spectra of the three isomers of 1,2-dichlorobenzene,Polysubstitution,Most other polysubstitution patterns on a benzene r

28、ing yield six different peaks. When identical substituents are present, planes of symmetry may reduce the number of peaks.,Two peaks,4-2.11 C-13 NMR Solvents-Heteronuclear Coupling of C to Deuterium,Most FT-NMR spectrometers require the use of deuterated solvents because the instruments use the deut

29、erium resonance signal as a “lock signal”, or reference signal, to keep the magnet and the electronics adjusted correctly. Multiplicity = 2nI+1,Fig. 4-2.14 The 13C peaks of two common solvents. (a) Chloroform-d. (b) Dimethylsulfoxide-d6,77ppm,39.5ppm,Example4-2.1 C3H6O2,Example4-2.2 C4H10O,C5H8O2, a

30、n ester,C7H8O,C8H8O,DEPT 极化转移技术,DEPT: Distortionless Enhancement by Polarization Transfer The sample 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

31、to each carbon. Each type of C will behave slightly differently, depending on the duration of the complex pulses. 提高杂核实验的灵敏度,DEPT：区分13C的级数,CH,CH,CH2,CH3,Fig. 4-2.8 (a) Standare 13C decoupled spectrum of ipsenol in CDCl3, at 75.5MHz. (b) DEPT subspectra: DEPT 135 CH and CH3 up. CH2 down. (C) DEPT 90

32、CH only.,DEPT区分碳的级数,4-3.2 The Mechanism of Coupling,When two spin-active nuclei prefer an opposed alignment, J is usually positive. If the nuclei are parallel or aligned, J is usually negative. Couplings involving an odd number of intervening bonds (1J, 3J) are positive, while those involving an eve

33、n number of intervening bonds (2J, 4J) are negative.,Dirac vector model,A. One-Bond Couplings (1J),According to the Pauli Principle, pairs of electrons in the same orbital have opposed spins; therefore, the Dirac model predicts that the most stable condition in a bond is where both nuclei have oppos

34、ed spins. It is unusual for coupling constants to depend on the hybridization of the atoms involved.,13CH,13CH110-270 Hz sp3 115-125Hz, sp2 150-170Hz, sp 240-270Hz,2J, two bond couplings, geminal coupling, Jgem,The nuclei prefer to have parallel spins, resulting in a negative coupling constant. The

35、amount of geminal coupling depends on the HCH angle, .,Fig4-3.2 The Mechanism of Geminal Coupling,Fig4-3.3 The dependence of the magnitude of 2JHCH, the geminal coupling constant, on the HCH bond angle .,As the decreases, the two orbitals move closer, and the electron spin correlations become greate

36、r.,Table 4-3.2 Variations in 2JHH with Hybridization and Ring Size,As ring size decreases the absolute value of the coupling constant 2J also decreases. As the angle CCC in the ring becomes smaller (as p character increases), the complementary HCH angle grows larger (s character increases) and conse

37、quently the geminal coupling constant decreases. Hybridization is important and that the sign of the coupling constant for alkenes changes to positive except where they have an electronegative element attached.,In many cases, no geminal HCH coupling (no spin-spin splitting) is observed, either the g

38、eminal protons are equivalent by symmetry or free rotation renders them equivalent. The geminal coupling is often observed in cyclic compounds that are conformationally rigid-for instance, bicyclic compounds. Deuterium substitution experiments: if one of the Hs in a compound which shows no spin-spin

39、 splitting is replaced by a D, 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 from one H to its neighbor. Since the C-C bond is nearly ortho

40、gonal (perpendicular) to the C-H bonds, there is no overlap between the orbitals, and the electrons cannot interact strongly through the bond system. They transfer the nuclear spin information via the small amount of parallel orbital overlap that exists between adjacent C-H bond orbitals.,Since the

41、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 types, are found to be positive.,Fig4-3.4 The Method of Transferring Spin Information Between Two Adjacent C-H Bonds,The spi

42、ns 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-H bonds is the major factor determining

43、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.,Fig4-3.5 The definition of a dihedral angle ,The side-side overlap of the two C-H bond orbitals is at a maxim

44、um at 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.,Karplus equation,The bond length RCC, the valence angles 1 and 2, and the electronegativity of any substituents X attached to C.,Fig4-

45、3.6 The Karplus relationshipthe variation of the coupling 3J with the dihedral angle ,Fig4-3.7 Factors influencing the magnitude of 3JHH,Jcis=10-12Hz =0,Jtrans=7-8Hz =120,Jaa=10-14Hz =180,Jae=Jee=4-5Hz =60,Conformationally rigid compounds: cyclohexane derivatives,Table 4-3.1 Some Three-Bond Coupling

46、 Constants (3Jxy),6-15 8-11 5-7 2-4 0-2,4-3.3 Magnetic Equivalence,Chemically equivalent, two or more nuclei are equivalent by symmetry. In most cases, chemically equivalent nuclei have the same resonance frequency (chemical shift), do not split each other, and give a single NMR signal.Magnetically

47、equivalent.,化学等价质子与化学不等价质子的判断,可通过对称操作或快速机制（如构象转换）互换的质子是化学等价的。 不可通过对称操作或快速机制（构象转换）互换的质子是化学不等价的。 与手性碳原子相连的CH2上的两个质子是化学不等价的。,化学等价质子与化学不等价质子的判断,Magnetic equivalence has two strict requirements:,Magnetically equivalent nuclei must be isochronous; that is, they must have identical chemical shifts. Magneti

48、cally equivalent nuclei must have equal coupling (same J values) to all other nuclei in the molecule.,The Hs in each group experience identical average magnetic environments, mainly because of free rotation, and are magnetically equivalent. Because of rotation, the Hs in each group are equally coupl

49、ed to the Hs in the other groups. Without free rotation there would be no magnetic equivalence. Because of the fixed dissimilar dihedral angles, Jab and Jab would not be the same.,4-3.4 Nonequivalence within a groupthe use of tree diagrams when the n+1 Rule fails,(a) Free ratation The n+1 Rule appli

50、es,(a) Locked conformation A tree diagram is required,A=B JAC=JBC JAB=0,AB JACJBC JAB0,Ha,Hb,C,C,Hc,B,C是化学不等价的核,JAC=16 Hz,JAC=8 Hz,JBC=3 Hz,A,C,B,wA,JAC,JAB,同核J-偶合(Homonuclear J-Coupling),The ring blocks rotation, causing HA and HB to have different chemical shift values, they are chemically and mag

51、netically nonequivalent.,Fig4-3.8 The NMR Spectrum of Styrene Oxide,The n+1 Rule no longer applies to the nonequivalent protons.,Fig4-3.9 An analysis of the splitting pattern in Styrene Oxide,Small ring compounds, 3JBC(cis) 3JAC(trans),4-3.5 Is the n+1 Rule ever really obeyed?,Fig4-3.19 Construction

52、 of a quintet for a methylene group with four neighbors all coupled to the same extent.,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 carbons.,Fig4-3.20 Loss of the simple quintet when JAB JBC .,In many molecules JAB i

53、s only slightly different from JBC, this leads to peak broadening in the multiplet, since the lines do not quite overlap.,4-3.6 Alkenes,HA HB HC JAB JACJBC,cis 3J 6-15 Hz,trans 3J 11-18 Hz,Terminal methylene 2J 0-5 Hz,Fig4-3.13 The types of coupling present in alkenes.,13.21 ,Fig4-3.10 The NMR Spect

54、rum of trans-cinnamic acid (肉桂酸）,The protons on double bonds differ in that they are rarely 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

55、double bond does not show any cis or trans splitting, since the protons HA and HB are chemically and magnetically equivalent.,Fig4-3.11 The NMR Spectrum of vinyl acetate,Fig4-3.12 A graphical analysis of the splittings in vinyl acetate,4-3.7 Mechanisms of coupling in alkenes; allylic coupling,4J 0-3

56、 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 transmit the spin information from the nucleu

57、s 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 Hz,When the allylic C-H bond is perpendicular to the C=C plane (as is the bond) , the interaction assumes the maximum value, 4J=3 Hz.,Fig4-3.14 Geometric arrangements that

58、maximize and minimize allylic coupling,Allylic splitting is observed in compounds such as the following:,Fig4-3.15 The NMR Spectrum of compound 3.,c,b,a,Fig4-3.16 A graphical analysis of the NMR spectrum of compound 3.,Fig4-3.17 The NMR spectrum of 3-Br-1-propylene.,Fig4-3.18 The NMR spectrum of 4-a

59、llyloxyanisole,4-3.8 Long-range coupling,Only under special circumstances does coupling occur between protons which lie farther apart than in 3J. Long-range coupling,Fig4-3.51 Allylic coupling in acetylenes.,Homo-allylic coupling,5J = 0-1.6 Hz,In some alkenes 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.,Fig4-3