#2
Chem Exam 728
Spring 2002
Name:
** This examination is open book, but is to be worked on independently. You may not discuss or otherwise communicate any aspect of the exam with anyone other than C. Martin. This includes any discussions with anyone after you are done with the exam, but before the exams due date and time. This is very important. Due in class, 9:30am, Thursday, April 4
Show your work for full credit. Be concise, but complete. Avoid long rambling answers which indicate that you dont really understand the question. 1. (5 points) Show, on the laboratory frame spin vector diagrams below, what will happen for each pulse sequence (ie. what is the state of the single magnetization vector precessing at wo immediately after the sequence?). Note: tD = p 2 wo The applied field lies along the z-axis, as usual.
a)
- 90x - 90x -
y x y x y x y x y x
x x x
pointing down
y
x
y
b)
- 90x - tD - 90x -
pointing up
c)
- 90x - 2tD - 90x -
y
x
y
d)
- 90x - tD - 180y -
e)
- 90x - 2tD - 180y -
y
Page 1 of 6
Exam #2
Chem 728
Spring 2002
Name:
2. (5 points) In a 1H NMR experiment with an RF-generated pulse Hxy = 0.5 Gauss, for what length of time should one apply the RF field to achieve a 90 pulse? a 180 pulse? How long would one leave on the power to achieve 90 and 180 pulses for 13C?
From before we know that q = g Hxy t. Therefore t = For 1H: For 13C: t(90) = t(90) = q g Hxy t(180) = 0.23 msec
/2 = 0.12 msec 26753 rad G-1 s-1 0.5 Gauss /2 6728 rad G-1 s-1 0.5 Gauss
= 0.47 msec t(180) = 0.93 msec
3. (10 points) In protein NMR, even when the experiment is carried out in D2O, there is always a very large signal due to water protons. If this signal is too large, it overwhelms the rest of the spectrum. Fortunately, one can use pulse sequences to exploit relaxation effects to remove most of the water signal. Noting that water protons typically relax (T1) much more slowly than protein protons, explain how the sequence <180-tD-90-t-180-t-acquire> can remove the signal due to water protons, leaving only the protein signals. How would you adjust the delay tD to optimize this effect? Think about what is happening to the bulk magnetization vectors of the two classes of protons.
The 180 pulse orients all spins along the negative z axis. Relatively quickly, the spins from the protein reequilibrate to Boltzmann values (its magnetization vector returns to the positive z axis. In the meantime, the magnetization vector for water is slowly making its way back to the positive z axis. If we wait the right amount of time, the bulk magnetization from water will be half-way along, in other words, it will be 0! If at this point we start a regular <90-t-180-t-acquire> pulse sequence, we will get a regular spectrum from the protein, but nothing from the water. Cute.
Page 2 of 6
Exam #2
Chem 728
Spring 2002
Name:
4. (20 points) The energy levels for a singlet 1H split by a singlet 13C are shown at right. The difference in populations between the levels is illustrated qualitatively. For each of the following situations (a-d), draw: 1) Pictures illustrating the qualitative population distribution among the four energy levels immediately after the treatment indicated. 2) A view, in the rotating xy frame, of the magnetization vectors for each 1H, just after a 90-acquire sequence, which is preceded by the indicated treatment 3) The qualitative 1H NMR spectrum you would expect to see, including relative intensities. x) (example) No pre-treatment.
13 1 H b 13 C C
b a
a
b
a
1
H a
13 13
C C
b a
b Rotating frame at (a+b)/2
First, before any of the treatments below, we have the energy levels at their Boltzmann levels. To collect an NMR spectrum, we do our simple 90 pulse, followed by collection of the FID, to produce the spectrum shown at right.
a) Selective saturation (decoupling) of transition a (only).
In this case, the populations of the levels connected by b are made equal. Consequently, when we start our NMR experiment, the spins in those levels have no net bulk magnetization. We see nothing rotated into the xy-plane for those spins. We then see no NMR signal for that transition (in the old electronic way of looking at things - transitions are induced equally up and down, so there is no net absorption of energy in the classical scanning NMR experiment.
Page 3 of 6
Exam #2
Chem 728
Spring 2002
Name:
b) A very short 180 pulse along x (with no subsequent delay).
The prior execution of a broad frequency 180 pulse, inverts the z-magnetization. Thus, when we go to take our spectrum the spins get flipped to the negative y-axis. Their phasing is thereby 180 out from our normal spectra - this manifests itself as an inverted NMR spectrum. Again viewing things from the old way, the excited state is preferentially populated, such that application of an RF field induces emissive transitions.
c) Immediate pretreatment of the system with a pulse sequence of: 90(x) - t - 90(x) with a relatively long pulse centered at n = b, where 1/t = 2J13C-1H . Show the rotating frame at n = b.
a 13 C 1 H b 13 C J/2 b a b a b b a
Rotating frame at (b) z b a Set up for spectrum x b a
2nd 90
1
H a
13 C
13 C b
a
y
Since are we now in a different reference frame, a appears to move at twice its previous rate and b does not move at all. After the first 90 pulse, a precesses to exactly opposite b, such that the second 90 pulse brings it back to its original position along z, while b is inverted (its population is inverted). Now when we go to collect our spectrum , one spin system is inverted (as above), but the other is normal. The spectrum shows b with an inverted transition and a with a normal one.
Page 4 of 6
Exam #2
Chem 728
Spring 2002
Name:
d) Immediate pretreatment of the system with a pulse sequence of: 90(x) - t - 90(y) with a relatively long pulse centered at frequency (a+b)/2, where 1/t = 2J13C-1H . Show the rotating frame at (a+b)/2.
In this case, we return to the rotating reference frame of the chemical shift. This time, in the same time period, the two spins each execute a 90 rotation. Now a 90 pulse along the y-axis achieves a selective spin inversion.
5. Answer the following questions: a) (5 points) in the figure at right, match the FIDs with the resulting NMR spectral peaks.
A == n B == m
A
B
m
n
b) (5 points) in the application of post-collection processing of the FID, one can distort the FID (as above) to either increase or decrease the linewidth of the resulting NMR peak. What happens to the spectral noise associated with the resulting peak? Assuming that A and B are manipulations of the same original FID, which spectrum at right (m or n) will show higher signal-to-noise ratios (ie. less noise). The pictures, are ideal of course, and show no noise.
In A, later time (lower signal-to-noise) data is amplified. Thus there will be more noise in the spectrum that results from the FID In B, later time (lower signal-to-noise) data is dampened (decreased). Thus there will be less noise in both.
Page 5 of 6
Exam #2
Chem 728
Spring 2002
Name:
6. (50 points) You have been given an unknown pentapeptide. On the following pages are COSY and NOESY spectra taken in partial D2O. Amino acid analysis shows that the peptide contains one residue each of Asp, Ile, Leu, Ser, and Thr. a) (20 points) Assign the signals in each spectrum to individual amino acids On each spectrum, circle each cross peak group and write its assignment (eg. circle a multiplet and label it Asp-CaH in one direction and CbH in another). b) (20 points) Determine the sequence of the peptide, to the extent that you can. Comment on any amibuities.
NH2-
Ser-Asp-Thr-Ile-Leu-COOH
c) (5 points) Some protons do not show up in any spectra. In your peptide which protons fall into which of these classes and why?
In the presented spectra, the Ser NH2 and Leu COOH protons dont show up. They are in rapid exchange with D2O, as is the Asp carboxylate proton. Finally, the Thr OH proton does not show up. This latter result is perhaps unrealistic, under conditions where amide protons are observed.
d) (5 points) What new cross peaks might you expect if this short peptide formed a stable a-helix (if for example, it were run in deuterated trifluoroethanol which forces a peptide into an a-helix)? Draw them in on the appro...