Understanding NMR Spectroscopy phần 6 pps

... derivative of Eq. [9] and then substituting for the derivatives of the populations gives 7– 16 As is shown in section 6. 9, such a double-quantum term evolves under the offset according to B BtBt y tI tI ... illustrated opposite. For NMR it is usual to display the spectrum with the absorption mode lineshape and in this case this corresponds to displaying the real part of S( ω ). 7 .6. 1.1...

Ngày tải lên: 14/08/2014, 09:21

... uses coherence transfer pathways to specify the desired outcome of the experiment. 9.2 Phase in NMR In NMR we have control over both the phase of the pulses and the receiver phase. The effect of ... Coherence Selection: Phase Cycling and Gradient Pulses † 9.1 Introduction The pulse sequence used in an NMR experiment is carefully designed to produce a particular outcome. For example, we...

Ngày tải lên: 14/08/2014, 09:21

... 90 180 180 180 6 90 –90 90 180 90 90 7 180 –180 90 180 0 0 8 270 –270 90 180 –90 270 9 0 0 180 360 360 0 10 90 –90 180 360 270 270 11 180 –180 180 360 180 180 12 270 –270 180 360 90 90 13 0 0 ... the coherence, column three, reduced to be in the range 0 to 360 ° by subtracting multiples of 360 ° (e.g. for step 3 we subtracted 360 ° and for step 4 we subtracted 720°). If we wished to select...

Ngày tải lên: 14/08/2014, 09:21

... ambient temperatures. c Diffusion constant taken as 0. 46 × 10 –9 m 2 s –1 . d Diffusion constant taken as 0.12 × 10 –9 m 2 s –1 . 9 .6. 6.1 Minimisation of Diffusion Losses The foregoing discussion ... we shall see, this alternative method is not without its own difficulties. 9 .6 Selection with field gradient pulses 9 .6. 1 Introduction Like phase cycling, field gradient pulses ca...

Ngày tải lên: 14/08/2014, 09:21

... 2002 Understanding NMR Spectroscopy James Keeler University of Cambridge, Department of Chemistry c  James Keeler, 2002 2–10 NMR and energy levels ααα αβα ββα βαα 1 23 4 ααβ αββ βββ βαβ 5 67 8 Fig. ... αβ−ν 0,1 − 1 2 J 12 + 1 2 J 13 6 8 ββ−ν 0,1 + 1 2 J 12 + 1 2 J 13 ααα αβα ββα βαα 1 2 3 4 ααβ αββ βββ βαβ 5 67 8 −ν 0,1 J 13 J 12 J 12 J 13 969 2 108104100 13 57 24 68 spin 2...

Ngày tải lên: 14/08/2014, 09:21

... most basic pulsed NMR ex- periment works, let alone the large number of subtle two-dimensional exper- iments which have been developed. To make any progress in understanding NMR experiments we ... = π/2 and the duration, t p is 12 × 10 6 s; therefore ω 1 = π/2 12 × 10 6 = 1.3 ×10 5 rad s −1 . The maximum offset is 2500 Hz, which is 2π × 2500 = 1 .6 × 10 4 rad s −1 . We see that the...

Ngày tải lên: 14/08/2014, 09:21

... situation? E 4 6 Explain why use of a sine bell weighting function shifted by 45 ◦ may enhance the resolution but use of a sine bell shifted by 90 ◦ does not. E 4–7 In a proton NMR spectrum the ... and then go on to explore some useful manipulations of NMR data such as sensitivity and resolution enhancement. 4.1 The FID In section 3 .6 we saw that the x and y components of the free indu...

Ngày tải lên: 14/08/2014, 09:21

... 5 6 Suppose that the spectral width is set to 100 Hz, but that a peak is present whose offset from the receiver is +60 Hz. Where will the peak appear in the spectrum? Can you explain why? 6 6 a ... t J y zz zz 1 212 12 2 21 2 12 1 2 12 12 1 2 12 2 ππ == →  →− 6. 4 Spin echoes It was shown in section 6. 2 .6 that the offset is refocused in a spin echo. In this section it will...

Ngày tải lên: 14/08/2014, 09:21

... / Hz a 30 Transferred to 30 b 30 and 60 transferred to 30 c 60 transferred to 30 and 60 d 30 transferred to 20, 30 and 60 e 30 and 60 transferred to 30 and 60 E7-3. What would the diagonal-peak ... section 6. 1.5. E6-2. Repeat the calculation in section 6. 1 .6 for a spin echo with the 180° pulse about the y-axis. You should find that the magnetization refocuses onto the –y axis....

Ngày tải lên: 14/08/2014, 09:21

... negative. 2– 16 NMR and energy levels enough to work out the Larmor frequencies using the following method; the idea is illustrated in Fig. 2. 16. J 12 J 12 D 1312 2434 D Fig. 2. 16 The quantities ... αβ−ν 0,1 − 1 2 J 12 + 1 2 J 13 6 8 ββ−ν 0,1 + 1 2 J 12 + 1 2 J 13 ααα αβα ββα βαα 1 2 3 4 ααβ αββ βββ βαβ 5 67 8 −ν 0,1 J 13 J 12 J 12 J 13 969 2 108104100 13 57 24 68 spin 2 spin 3...

Ngày tải lên: 27/06/2014, 08:20