Understanding NMR Spectroscopy phần 4 doc

... ppm range of proton shifts runs from 40 0.000 MHz to 40 0.0 04 MHz. If we set the receiver reference frequency at 40 0.002 MHz and then subtract this from the NMR frequencies we end up with a range ... required attenuation would be: power ratio in dB = 20 log 10  15 25  = 4. 4dB. We would therefore need to increase the attenuator setting by 4. 4dB. Phase shifted pulses In section 3.9...

Ngày tải lên: 14/08/2014, 09:21

... ααα 3 2 + 1 2 ν 0,1 + 1 2 ν 0,2 + 1 2 ν 0,3 + 1 4 J 12 + 1 4 J 13 + 1 4 J 23 2 αβ α 1 2 + 1 2 ν 0,1 − 1 2 ν 0,2 + 1 2 ν 0,3 − 1 4 J 12 + 1 4 J 13 − 1 4 J 23 3 βαα 1 2 − 1 2 ν 0,1 + 1 2 ν 0,2 + 1 2 ν 0,3 − 1 4 J 12 − 1 4 J 13 + 1 4 J 23 4 ... − 1 2 − 1 2 ν 0,1 − 1 2 ν 0,2 + 1 2 ν 0,3 + 1 4 J 12 − 1 4 J 13 − 1 4 J 23 5 ααβ 1 2 + 1 2 ν 0,1 + 1 2 ν 0,2 − 1 2 ν 0,3 + 1 4...

Ngày tải lên: 14/08/2014, 09:21

... 90 ◦ ”(wetakean 2–18 NMR and energy levels 2.7 Exercises E 2–1 In a proton spectrum the peak from TMS is found to be at 40 0.135705 MHz. What is the shift, in ppm, of a peak which has a frequency of 40 0.136305 ... of the shift? E 2–2 Two peaks in a proton spectrum are found at 1. 54 and 5. 34 ppm. The spec- trometer frequency is quoted as 40 0.13 MHz. What is the separation of these t...

Ngày tải lên: 14/08/2014, 09:21

... 4. 9 Exercises 4 17 4. 9 Exercises E 4 1 In a spectrum with just one line, the dispersion mode lineshape might be ac- ceptable ... 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 ... Fig. 4. 8. If there were only one line in the spectrum...

Ngày tải lên: 14/08/2014, 09:21

... Wn n t Wn Wn SI SI IS IS SI 2 1 2 2 202 1 1 2 40 3 3 1 3 2 303 1 1 2 40 2 4 2 4 2 =− − − + + + =− − − + + + =− − () ( ) () ( ) () ( ) () ( ) () () 44 24 2 3 2 221 −+ + + () () Wn Wn Wn Wn SI All of ... IS S =−+− =− − − + + + ++ +−− − − () () () () () () () ( ) 132 4 1 1 1 121 1 2 1 3 24 1 3 2 303 1 1 2 40 2 11 2 2 202 1 1 2 40 3 2 4 2 42 4 2 3 2 221 () ( ) () ( ) () () () () −...

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 ... of pairs of spins. The effect is of enormous value, therefore, in structure determination by NMR. 8 .4. 1 Transient experiments A simple experiment which reveals the NOE is to invert just th...

Ngày tải lên: 14/08/2014, 09:21

... –180 180 360 180 180 12 270 –270 180 360 90 90 13 0 0 270 540 540 180 14 90 –90 270 540 45 0 90 15 180 –180 270 540 360 0 16 270 –270 270 540 270 270 In the first four steps the phase of the second ... advanced in steps of 45 °, can be used to select the refocusing of double- quantum coherence in which the transfer is from p = +2 to –2 (i.e. ∆p = 4) or vice versa. 9.5 .4 Combining phase...

Ngày tải lên: 14/08/2014, 09:21

... For example, the sequences (a) g RF t 1 t 2 g RF t 1 t 2 (a) (b) 9 46 and then combined in the manner described in section 9 .4. 4.2 to give a frequency discriminated absorption mode spectrum. In ... for steps 2 and 4 only gives step 1st pulse 2nd pulse 3rd pulse receiver 10000 2 90 90 0 180 3 180 180 0 0 4 270 270 0 180 This is exactly the cycle proposed in section 9.5.5 .4. 9.5.9 Furt...

Ngày tải lên: 14/08/2014, 09:21

... that the pathways with ∆p = –5, 4, –2, 3 and 4 are rejected]. Confirm that, as expected for this four-step cycle, the selected values of ∆p are separated by 4. (c) Complete the following table ... radio-frequency field produced by typical NMR probes, but in such a case the optimum dephasing rate is obtained by spin locking off-resonance so that tan – , 1 1 54 ωΩ kl () ≈°. (e) Dephasi...

Ngày tải lên: 14/08/2014, 09:21

... is usual to plot these with zero in the middle. 2–8 NMR and energy levels αα αβ ββ βα 1 23 4 −ν 0,1 −ν 0,2 frequency 13 24 12 34 24 13 34 12 spin 1flips flips spin 2 αβ αβ Fig. 2.7 On the left, ... coupling 2–15 342 4 1213 12 34 0 20 10 040 60 80 frequency (Hz) –ν 0,1 ν 0,2 = –90 ν 0,2 = –50 ν 0,2 = –20 ν 0,2 = –10 –ν 0,2 –ν 0,2 –ν 0,2 –ν 0,2 J 12 J 12 ~ Fig. 2....

Ngày tải lên: 27/06/2014, 08:20