Chapter 4 NMR Background: In this chapter we will discuss the interaction of molecules with a magnetic field. * Nuclear Spin Angular Momenta - recall electrons & spin -- our spin functions are and which are adhere to the following: S ˆ S ˆ ˆ ˆ z S Sz -- they are also orthonormal to each other * * * * d d d d 0 - since electrons are charged they act like magnetic dipole in the presence of a magnetic field - nuclei also have magnetic dipole and intrinsic angular momentum, I, but unlike electrons they are not restricted to a spin ½ designation - the analogous spin eigenequations for protons are I ˆ I ˆ ˆ ˆ z I Iz - the relationship between the magnetic dipole,, and the angular momentum is q gn I gnni I mn where gn is the nuclear g factor, N is the nuclear magneton (q/mn), mn is the mass of the nucleus and = gnn is the magnetogyric ratio -- g is unitless and both g and are characteristic of the identity of the nucleus -- the larger the the more easily the nucleus may be detected * Magnetic Moments and Fields - the potential energy of a magnetic dipole s interaction with a magnetic field, B is V B - if we take the field to be in the z-direction we obtain V zbz IzBz - this is equivalent to the spin amiltonian operator of a single isolated nucleus ˆ IzBz - so, the corresponding Schrödinger equation is ˆ IzBz E where E mibz - the difference between a proton that is aligned with the field, mi = +½, and one that is against the field, mi = -½, is completely dependent on the strength of the applied magnetic field E E m E m B I I z
-- this why the larger magnet is able to provide a much higher resolution for sample -- using E hv we get the relationship btwn the transition frequency and the Bz field: v Bz * Shielding and NMR - the equations above work fine for a bare nucleus however rarely is this the case - the applied magnetic field induces an opposing magnetic field in the electrons, Belec -- this induced field is proportional to the applied field Belec B0 -- (-) sign accounts for the fact that the field is opposing the applied field -- sigma is our proportionality constant aka shielding constant - the shielding is completely dependent upon the local field -- this allows us to differientiate protons in the same molecule since they experience different local fields -- the total field any nucleus feels is given by Bz B0 B0 B0 - therefore, the relationship btwn the total field and transition frequency is v B0 * Chemical Shift, - reference/standard TMS, Si(C3)4 has equiv -atoms and is farely nonreactive - our good buddy chemical shift, -- we use this idea in order to compare the results of the same system from different spectrometers recall field strength will give different results v vtms 6 0 which is in ppm vspec - back to shielding -- the larger the electron density around the nucleus the larger the shielding -- the larger the shielding the larger the field required to produce a transition, spec -- therefore, the larger the shielding the more upfield we would expect to detect the nucleus -- if we reduce the electron density around the nucleus we will shift the absorption downfield --- we can do this by adding electron-withdrawing groups C4 C3Cl CCl CCl3 (ppm) 0.3 3.05 5.33 7.6 -- furthermore, the more electronegative the added functional the more downshifted the nucleus will appear C3I C3Br C3Cl C3F (ppm).6.68 3.05 4.6 * Spin Spin Coupling
- neighboring nuclei also have an induced magnetic field - the effect of these neighboring nuclei is to split the signal of a given nucleus into multiplets this interaction is referred to as a spin-spin interaction - quantitative explanation: consider a system with two -atoms in different environments -- let j be the shielding of the jth nucleus -- we can then write the amiltonian as ˆ B I ˆ B I ˆ 0 z 0 z -- to account for the interaction btwn the two nuclei we need to include a spinspin coupling term, I ˆ ˆ I ˆ ˆ 0 0 ˆ B I ˆ ˆ z B Iz I I --- J is the spin-spin coupling constant --- h is included to give J the units of z -- we can treat this using first-order perturbation theory ˆ ˆ ˆ B0 I ˆ ˆ zb0 Iz I I ˆ 0 ˆ --- our two-nuclei system gives rise to 4 wavefunctions 3 4 --- the energy is given by E 0 E * d d ˆ ˆ 0 0 where E j j j j j j j --- since Iˆ j j zj ˆ 0 ˆ 0 B Iˆ B Iˆ z z 0 0 ˆ B0 Iˆ ˆ z B0 Iz ˆ 0 B0 B0 ˆ 0 0 B0 B0 E 0 E B0 --- the remaining wavefunctions give 0 B0 0 B0 0 E E3 E4 B0 --- now we need to take care of the perturbation term 0
* ˆ ˆ ii dd i IIi ---- we treat our dot product as: I ˆ I ˆ I ˆ ˆ ˆ ˆ ˆ ˆ xix IyIy Iz Iz our Iˆ ˆ z I z is pretty straightforward ˆ ˆ ˆ ˆ IzIz Iz Iz 4 * * ˆ ˆ * * z, dd II d d 4 * * z, d d 4 4 ---- similarly z, z,33 z,44 4 4 --- it can be shown: ˆ ˆ ˆ i ˆ i Ix Ix Iy Ix --- so, ˆ ˆ ˆ ˆ ˆ ˆ Ix Ix Ix Ix Ix Ix 4 ˆ * * * *, ˆ ˆ x d d II d d 0 4 --- therefore, the x and y terms do not contribute to our first-order case --- our energies with a first-order correction are 0 0 B0 E B0 E 4 4 0 B0 0 E3 E4 B0 4 4 B0 where v0 - the selection rule for transitions between nuclear spin states allow only one nucleus at a time to undergo a transition
J v v0 J 3 v 3 v0 J 4 v4 v0 J 3 4 v3 4 v0 - we can express these resonance frequencies as: J J v v0 v v0 -- so, we expect to see our 4 transitions to appear as a pair of two closely spaced lines or doublets * What happens for equivalent protons? - when we have equivalent protons we they absorb at the same frequency but do NOT cause a split in the signal let s see why this happens - for this system we have the following amiltonian ˆ ˆ ˆ B0 ˆ ˆ A Iz Iz I I ˆ 0 ˆ - our wavefunctions will be 3 4 - our energies are given by 0 Ei Ei Ei where i is a spinwave state 0 * * E dd B0 ˆ ˆ A Iz I z ˆ ˆ IzIz 0 * * E B d d B 0 A 0 A
* * E dd I xix IyIy IzIz ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Ix Ix IyIy IzIz 4 4 4 * * E d d 4 4 4 the first terms will disappear due to orthogonality * * E d d 4 4 this leads to E B0 A 4 - similarly we get energies for our other 3 states: 3 E E3 E4 B0 A 4 4 4 - now, why do we only see transitions? -- not only does only one spin undergo a transition, but also only transitions with the same symmetry are allowed -- therefore, the only transitions are from 3 and 3 4 E3 E B 4 0 A v 3 v3 4 v0 A h -- hence, we only see one transition * Intensities of transitions in a multiplet - obviously a singlet only has one intensity and so doesn t need to be addressed - for a doublet: the absorption peak for the b--atom in,-dichloroethane shows an intensity pattern of : -- the two peaks are equivalent -- the spin pattern for the adjacent nucleus is either spin up or spin down -- the probability of either case is the same so we should see equivalent peaks - triplet: the absorption peak for the b--atom in chloroethane shows an intensity
pattern of :: -- the spin pattern of the two adjacent -atoms can be the following: -- while once again the probability for any of the states is equivalent we have degenerate states and so the intensity should be twice that of the other states - quartet: the absorption peak for the a--atom in chloroethane shows an intensity pattern of :3:3: -- the spin pattern is * Second-Order Spectra - when our spin-spin coupling constant are not small relative to the separation of multiplets we need to use second-order spectra - for example:,,3-trichlorobenzene Figure 4.9 from text -- (a) is the multiplet which results from a 60 Mz NMR -- (b) is the first-order-like spectra we get from a 70 Mz NMR - Second-order spectra can be solved exactly using variational theory -- let s consider a molecule with nonequivalent -atoms -- our spin amiltonian: ˆ ˆ 0 0 ˆ B I ˆ ˆ z B Iz I I -- there are four possible spin wave functions (just like before) 3 4 -- we calculate the energy levels of this spin system exactly using a linear combo: cc c33 c44 where our c s are variational parameters we can use to minimize the system
E ˆ * d d d d * -- if we minimize this expression we obtain the following secular determinant E where 3 4 E 3 4 E 3 3 33 34 E 4 4 34 44 d dˆ * ij i j -- after we determine all of the ij values we obtain d d E 0 0 0 4 0 d d E 0 4 0 d d E 0 4 0 0 0 where d hv and d hv 0 0 d d E 4 -- this leads to the following energy expressions h E hv0 E v0 J 4 4 h E3 v0 J E4 hv0 4 4 -- I leave it up to you to look over the remaining 5 pages for your own edification 0