J耦合、谱分析以及偶极耦合.ppt

J couplingSpin-spin Coupling•dipole-dipole interaction： magnetic dipole interaction between nuclei directly•Spin-spin Coupling：： It will occur between two magnetically active nuclei that are connected through chemical bonds. We can see it for two atoms directly connected, or for atoms that ‘see’ one another across several bonds. Couplings are perhaps the most important parameter in NMR, as they allow us to elucidate chemical structure.I SJ   0I SJ = 0ExampleEthanol: CH3CH2OH Three singlets The spectrum is a result of spin-spin coupling (scale coupling)J Coupling•Each nucleus act as a small magnet and will affect the magnetic moment of the other nuclei nearby (The energy levels of a nucleus will be affected by the spin state of nuclei nearby). The two nuclei that show this are said to be coupled to each other. •The two nuclei couple leading to splitting of the two lines associated with each nucleus. Note that the interaction is mutual, and the same splitting is observed for both resonances. •A splitting of a signal means that we have more energies involved in the transition of a certain nuclei.•The strength of the coupling is called the coupling constant, J – separation between the multiplet lines.•IA and IX are the nuclear spin vectors which are proportional to mA and mB, the magnetic moments of the two nuclei. JAX is the scalar coupling constant. It does not matter if we have a 60, a 400, or an 800 MHz magnet. The coupling constants are always the same!J Coupling• The resonance frequency of Ha depends on whether the adjacent Hb has spin up or down. • The frequency separation betweenthe two peaks is called J(scalar coupling constant) (typically 0 ～15 Hz). • The coupling constant is independent of Bo and is reported in Hz.J Coupling• The resonance frequency of Ha depends on whether the adjacent Hb has spin up or down. • The frequency separation betweenthe two peaks is called J(scalar coupling constant) (typically 0 ～15 Hz). • The coupling constant is independent of Bo and is reported in Hz.ExampleEthanol: CH3CH2OHSpin-spin coupling tells us how many nuclei there are near to a given nucleus.J CouplingJ CouplingPascal Triangle11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 1J CouplingIf a proton interacts with two different sets of equivalent protons, the multiplicity will be: — spin system, AnMmXp  (m+1)(p+1)Consider CH3CH2CH2I — H = 1.02; H = 1.86; H = 3.17-CH3; 1:2:1 triplet (from methylene CH2)-CH2; quartet of triplets – 12 lines (3+1)(2+1)-CH2; 1:2:1 triplet (from methylene CH2)Coupling between Sets of Spins:J CouplingFactors affect the size of J : • The numbers of bonds involved• The dihedral angleφ,for vicinal protons• The configurations • The conformationsKarplus EquationJ couplings provide an estimation of molecular conformation!J(θ)=Acos2(θ) + Bcos(θ) + CA, B, and C are empirically derived parameters. A1 = E4 - E2 = nA - 1/2 JAX A2 = E3 - E1 = nA + 1/2 JAX X1 = E2 - E1 = nX - 1/2 JAX X2 = E4 - E3 = nX + 1/2 JAX J CouplingWhen B0 paralells Z, the H of the system is:Analysis of spin-spin coupling:H= - νA IAz - νX IXz+ JAX IA ▪ ▪IX△m=±1H= - νA IAz - νB IXz + JAX IAz IXzE4 = 1/2 nA + 1/2 nX + 1/4 JAXE3 = 1/2 nA - 1/2 nX - 1/4 JAXE2 = - 1/2 nA + 1/2 nX - 1/4 JAXE1 = - 1/2 nA - 1/2 nX + 1/4 JAXB B核核跃迁迁频率率A A核核跃迁迁频率率E A X A XA1A2A1A2BoJ = 0J > 0E4E3E2E1J CouplingE A X A XA1A2A1A2BoJ = 0J < 0E4E3E2E1A1A2n n (A1) = n n (A2)A2A1n n (A2) n n (A1)A1A2n n (A1) n n (A2)J = 0J > 0J < 0Analysis of spin-spin coupling: Analysis of 1st Order SystemsCH3CH2CH3CH2J (Hz)4.5 ppm1.5 ppmEthylacetate (CH3COOCH2CH3) :Nuclei A is coupled to n identical nuclei X (of spin 1/2), A will show up as n + 1 lines in the spectrum.Since we have the same probability of finding the system in any of the states, and states in the same rows have equal energy, the intensity will have a ratio 1:2:1 for the CH3, and a ratio of 1:3:3:1 for the CH2.Pascal Triangle11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 1•In a spin system in which we have a certain nuclei coupled to more than one nuclei, all first order, the splitting will be basically an extension of what we saw before.•Say that we have a CH (A) coupled to a CH3 (M) with a JAM of 7 Hz, and to a CH2 (X) with a JAX of 3 Hz. We basically go in steps. First the big coupling, which will give a quartet:• Then the small coupling, which will split each line in the quartet into a triplet:• This is called a triplet of quartets (big effect is the last…). Analysis of 1st Order Systems7 Hz)3 Hz•Let’s finish our analysis of 1st order system with some pretty simple rules that we can use when we are actually looking at 1D 1H spectra.•To do that, say that again we have a system in with a CH (A) coupled to two CH’s (M and R) with a JAM of 8 Hz and JAR of 5 Hz, and to a CH2 (X) with a JAX of 6 Hz:• The first rule is that if we have a clear-cut first order system, the chemical shift of nuclei A is always the center of the multiples. • The second one is that no matter how complicated the pattern may end up, the outermost splitting is always the smallest coupling for nuclei A. This is the one we can measure without worrying of picking the wrong peaks from the pattern. Analysis of 1st Order Systems6 Hz8 Hz5 Hz5 Hzd dA Analysis of 2nd Order Systems1 2 3 4ABn nZn nAn nBWhat we have been describing so far is a spin system in which Dn >> J.B0 DnDn DnDn ~ J A very simple way to determine if we have an AB system is by looking at the roofing effect: coupled pairs will lean towards each other, making a little roof:ROOF As with an AX system, JAB is the separation between lines 1 and 2 or 3 and 4:| JAB | = | f1 - f2 | = | f3 - f4 |Now, the chemical shifts of nuclei A and B are not at the center of the doublets. They'll be at the center of mass of both lines. This can be calculated as follows. Being DnDn the n nA A - n nB chemical shift difference:DnDn2 = | ( f1 - f4 ) ( f2 - f3 ) |n nA = n nZ - DnDn / 2 n nB = n nZ + DnDn / 2I1I2I3I4I2 I3 | f1 - f4 | = = I1 I4 | f2 - f3 |Magnetic & Chemical Equivalence•As we saw last time, we can do a very simple first order analysis of this spin system, because we assumed that all CH2 protons were ‘equal’, and all CH3 protons were ‘equal’. •We can easily see that they are chemically equivalent. Additionally, we have free rotation around the bond, which makes their chemical shifts and couplings equal.•So in effect, the coupling of any proton in CH2 to any proton in the CH3 will be the same. Magnetic equivalence :An ethoxy group (-O-CH2-CH3) Chemical Equivalence :The nuclei groups whose chemical shifts are equal are chemical equivalence. d d(CH2) >> d d(CH3) A2X3 system• We have 2 magnetically equivalent 1Hs on the CH2, and 3 on the CH3.The 2JHH coupling is zero in this case, because the energies for any of the three protons is the same.• We use A to refer to the CH2 protons, and X to refer to the CH3 protons because they have very different s. We start with the letter A for the most deshielded spin.Magnetic & Chemical Equivalence• CH2F2is another example of an ‘AX’ type system. In this case, the 1Hs and the 19Fs are equal not due to rotation, but to symmetry around the carbon. It’s an A2X2 system.• For CH2F2, we can also compare the couplings to check the 1Hs and 19Fs are equivalent: JH1F1 = JH1F2 = JH2F1 = JH2F2. All due to their symmetry.Symmetry no Rotationd dHa = d dHb d dFa = d dFbdue to the geometry of this compoundJHaFa   JHaFb JHbFa   JHbFbThe energy levels for Ha and Hb are different (not degenerate anymore as in CH3), and we have JHaHb ≠ 0.For Ha, we have JHaFa, JHaFb, and JHaHb. For Hb, we have JHbFa, JHbFb, and JHbHa.CH2F2H2C=CF2A2X2AA'XX'Spectral Analysis1H-NMR Data: 3 components1.Chemical shift - each nonequivalent hydrogen gives a unique signal.2.J - Scale couplings produce splitting pattern correlated with molecule structure.3.Integration - peak areas are proportional to numbers of equivalent nuclei giving a signal.Two signals split intomultiple peaks havinga ratio of areas of 2:3.“quartet”“triplet”Interpretation of NMR SpectraIntegrationArea under the curveis relative to the numbers of protonsC3H6O2Numbers represent relative peak areas of multipletsC5H10O C4H8O2C10H14ODipolar coupling If two different nuclear spins, I1 and I2 are separated by a ditance rthen the energy of interaction between the two magnetic dipoles, µ1 and µ2 is given by:If the two dipoles have the same orientation with respect to r then:Where since they are parallelDipolar couplingDipolar couplingThe maximum dipolar coupling between two nuclear spins, I1 and I2, is given by:The measured value depends on the orientation of the vector between I1 and I2 and the magnetic field:Scalar and Dipolar CouplingØCoupling of nuclei gives information on structureThroughBondsThroughSpaceThank You !。