1. Lecture 2 1 H Nuclear Magnetic Resonance

2. Gas Chromatograph of Molecular Hydrogen at –100 °C Thermoconductivity Detector 12

3. If the two components are collected separately and re-injected, the following is observed:

4. Gas Chromatograph of Peak 1

5. Gas Chromatograph of Peak 2

6. Peak 1 + O 2  H 2 O; Peak 2 + O 2  H 2 O Peak 1 + Ni  H 2 Peak 2 + Ni  H 2

7. Whats going on?

8. Ortho Hydrogen Para Hydrogen O/P = 1:1 at 80 K

9. S N The magnetic moment of a spinning nucleus is colinear with the axis of spin For a nucleus with a spin of  1/2, only two orientations in a magnetic field are allowed

11. Processional frequency,, is proportional to the magnetic field strength, H =  H For 1 H For 13 C At 14,000 Gauss (1.4Tesla) = 60 MHz = 14 MHz At 70,500 Gauss (7.05 Tesla) = 300 MHz = 70 MHz At 141000 Gauss (14.1 Tesla) = 600 MHz = 140 MHz 1 Hz = 1 cycle/second

12.  E Energy difference between the two states 0 H  Magnetic field strength Since the energy of the two states are different, so must be their populations

13. For resonance to occur, the processional frequency of the nucleus,  must equal the radiofrequency,. In a continuous wave spectrometer (now hardly ever used), the following occurs: Considering the excess nuclei in the ground state only …

15. In a pulsed experiment, the frequency generator provides all frequencies simultaneously; only those frequencies present in the precessing nuclei interact causing those nuclei to to tip in the same manner. The angle of tip depends on the duration and the intensity of the rf signal. During this condition of resonance, some nuclei absorb rf energy and are transported into their excited state. Their contribution to the resulting net magnetic moment is lost. If the rf frequency is sufficiently intense, and short in duration, the protons respond by pressessing about the field generated by the rf field. Once the rf is turned off, these nuclei continue to precess about the external field. Suppose an intense rf field is maintained long enough to cause the protons to precess one quarter turn (90 °)about the rf field.

16. N S Before the rf pulse Consider a set of identical nuclei

17. rf generator signal coil N S Just after the 90 ° rf pulse

18. Resultant moment When the rf is turned off detector coil rf coil N S

19. z x y

20. Receiver coil output with time

21. The resulting magnetic moment of the protons will precess about the xy plane at some frequency . Suppose now we simply examine how the magnitude of the resulting magnetic moment of the protons varies with time. To do this we need to precess with the resulting moment about the xy plane at the same frequency. This is called the rotating frame of reference. This is like being on a merry go round on a stationary horse watching the other horses go up and down. To us the motion is simple an up and down motion. To an observer who is not on the merry go round, the motion of the horses going up and down is complicated by the fact that they are also moving in a circular direction.

22. Remembering that =  H, local field inhomogeneities about H will cause  to vary with time t = 0t = t 1 t = t 2

23.  Net magnetic moment detected by the detector coil No magnetic inhomogeneity With magnetic homogeneity

24. The rate by which the magnetic moment decays in the xy plane is first order and is described by a rate constant, k 2 (s -1 ). The reciprocal of k 2 is a time constant, T 2. It is the time necessary for H of the nuclei to decay to 1/e (1/2.71= 0.36) of the original value, H o. Generally the magnetization can be thought to be completely gone after 5T 2 First order process ln(H/H o ) = - k 2 t; when t = 1/k 2 ln(H/H o ) = -1 and H/H o = e -1 T 2 is also referred to as transverse relaxation and it refers to the loss of magnetization in the xy plane

25. Once magnetization is lost in the xy plane, how is it re-established in the z direction?

26. Equilibrium condition

27. The buildup of magnitization in the z direction is also a first order process. It is characterized by the rate constant k 1 ; the reciprocal of k 1 is T 1, the spin lattice relaxation time. It is the time necessary to build magnitization in the z direction to it equilibrium value. The process of tipping the nuclei 90 ° results in absorption of energy and this energy must be dissipated into the environment. T 1 is referred to as spin lattice relaxation because the nucleus in relaxing must transfer its energy to a component in the lattice with the same frequency. Buildup of magnitization is considered complete after 5 T 1 ln(H ze -H zo ) = -k 1 t; when t = 1/k 1 (T 1 ) ln(H ze - H zo ) = -1 (H ze - H zo ) = e -1

28. Magnitude of T 1 and T 2 T 1 values vary depending on the local environment of the nucleus in question. Typical values for protons is seconds or less; much longer times are observed for 13 C. T 2 values depend on the magnetic inhomogeneity T 1 values are generally much longer than T 2

29. T 1 are dependent on the type of nucleus. It can range from 10 -2 to 100 sec. For protons it is of the order of a few seconds but can be longer. For solids it can be of the order of days. For spin lattice relaxation to occur, there needs to be certain magnetic moments cause by spinning dipoles or other magnetic fileds that have a magnetic component with the same frequency as the processional frequency of the nucleus in question. This provides a mechanism for relaxation or energy transfer to occur. The magnetic field fall off rapidly with distance so these field need to be in close proximity.

33. Suppose now that instead of only one kind of nucleus, we have two kinds, with different processional frequencies. Let say that they are related in the following fashion: 1 =  ; 2 = 3 , but equal in number. Following the 90 ° pulse, the following results:

35. ω 3 ω

36. +   3 

37. Fourier analysis of this free induction decay results (fid) in the two frequencies that lead up to this beat pattern. In NMR the beat pattern is caused by magnetic moments precessing at different frequencies. Fourier transform simply identifies the frequencies needed to reproduce the fid signal. It does so by an algorithm called fast Fourier transform.

39. For a 1 H nucleus precessing at 300 MHz, what is the relative populations of nuclei in the ground (g)and excited states (e)? N e /N g = e -  H/RT -  H/RT = h /kT since h is the energy of a photon of frequency and R = k*(Avogadros Number) h /kT = 6.62*10 -34 Js *300*10 6 s -1 / 1.38*10 -23 J/K* 300 K h /kT = 0.000048; e -0.000048 = 0.999952/1 or 999952/10 6 the difference is 48 nuclei in a total of /2*10 6 nuclei 1 mmol of 1 H nuclei = 6*10 23 *0.001*24/10 6 = 144*10 14 excess nuclei in the ground state