_________________________ Name Exam 2: CHEM/BCMB 4190/6190/8189 (166 pts) 1). In the example (right), the effect of a 90 ( /2) pulse applied along the "x" axis (90x) is shown for a bulk magnetization vector (M0) at equilibrium (on the `z' axis). For a-f below, the affects of the indicated pulses on the bulk magnetization vectors are shown. In each case, indicate where the bulk magnetization vector originated before the indicated pulse. (12 points) z M0 y x x 90x y Tuesday, 25 October, 2005 z
a.
z 90-x y x x z 90-y, 90x y x x z 450-y y x x
z y
d.
z 3 /2 -x y x x z 225-x y x x z -540 y y x x
z y
b.
z y
e.
z y
c.
z y
f.
z y
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________
2). The DEPT spectrum and the 13C spectrum with broadband 1H decoupling of the compound citronellol (right) are shown below. Also shown (far right) is a plot of the phase angle of the third 1H pulse ( y), applied along the `y' axis, versus the expected intensities of the signals for the various groups in DEPT spectra.
CH
CH3
CH2
a). What is y for the DEPT spectrum shown? Please explain how you came to this conclusion. (4 points) The value of y is 135. As shown in the plot of intensities versus y, when y is 135 we expect the signals from the CH2 groups to have the opposite phase (down in the spectrum) to the CH and CH3 groups (up in the spectra). This is apparently what we see in the DEPT spectrum. There are 5 signals phased up, corresponding to the -CH and CH3 groups, and 4 signals down, corresponding to the four CH2 groups. The small peak at ~ 0.0 ppm is most likely due to added TMS (see `d' below). b). The peak at ~132 ppm in the 13C spectrum is not observed in the DEPT spectrum. What carbon atom gives rise to this signal? Why does the signal not appear in the DEPT spectrum? Please explain your answers. (4 points) The peak at 132 ppm is due to carbon atom 7. This quaternary carbon will not be observed in the DEPT spectrum. This is due to the fact that in DEPT, polarization is transferred from 1H to 13C, and ultimately we observe the 13C magnetization created. For quaternary carbons, there are no attached 1H spins from which to transfer the polarization.
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________
c). Which carbon atom gives rise to the signal at ~125 ppm? Please explain in detail how you arrived at your answer. (4 points) There are two 13C nuclei that we would suspect would have chemical shifts that are far downfield. These would be carbons 6 and 7, the ethylenic carbons. Also, the signal is phased up (positive intensity) in the DEPT spectrum, so it is due to either a CH or CH3 group, and the signals for CH3 groups would be expected to be far upfield. Because we know that carbon 7 is quaternary and does not give a signal in the DEPT spectrum (see `b' above), we then know that the signal at 125 ppm is due to carbon 6. d). What gives rise to the signal at 0 ppm? Please explain. (4 points) Most likely this is due to added TMS because the 13C chemical shift of the carbon atoms in TMS is 0 ppm and because the signal intensity is much lower than any of the other signals in the spectra (indicating that there is only a small amount of this nucleus, i.e. only a small amount of added TMS). e). Compare the signals in the two spectra at 24 ppm. Explain what is occurring and why the signals appear as they do. (4 points) In the 13C spectrum, there is apparently only a single signal at 24 ppm, but in the DEPT spectrum it is clear that there are two nuclei that have the same chemical shift (24 ppm), one is a CH2 group (phased down, negative intensity) whereas the other is either a CH or CH3 group (phased up, positive intensity). Most likely the signal from the CH2 group is from carbon 1, which would be the most downfield shifted due to the attached OH group. f). The signal at ~77 ppm in the 13C spectrum is due to the natural abundance 13C in CDCl3. It appears as a triplet, where all of the peaks are of equal intensity. Explain in detail why the signal is a triplet and why all of the peaks in the triplet are of approximately equal intensity. (5 points) The 13C signal is split by coupling to the attached (quadrupolar) 2H (D) nucleus. We know that when a 13C nucleus is attached to a 1H nucleus it is split into a doublet, where the individual peaks in the doublet are of equal intensity. This is because the attached 1H (I= ) nucleus can be in either the (low energy) or (high energy) state. Because the and states are nearly equally populated, the intensities of the doublet components in the 13C spectrum are nearly identical. We know also that for a 2H (D) nucleus, where I=1, there will be three possible energy states, E = 1 hB0 , E = 0 hB0 , and E = 1 hB0 , and that they will be nearly equally populated. Thus, these will split the attached 13C nucleus into a triplet with nearly equal intensities for the three peaks. g). What does DEPT stand for (in relation to NMR spectroscopy)? (2 points) Distortionless Enhancement by Polarization Transfer 3
Exam 2: CHEM/BCMB 4190/6190/8189
Name _________________________ 3). For a pair of coupled hydrogens in a molecule, explain how application of a long pulse at the Larmor frequency of one of them leads to decoupling and loss of splitting in NMR spectra? Please use complete sentences and no diagrams or equations. (6 points) Application of an applied field at the Larmor frequency, i.e. the resonance condition, causes both absorption and emission of energy to occur with equal probability, i.e. nuclei in the spin state are converted to the state and vice versa. If the pulse is applied for a long time, the average lifetime for nuclei in any given spin state will be short, as it will undergo continuous interconversion between spin states. Coupling of nuclei requires that the lifetimes of the spin states of the coupled nuclei are long compared to the inverse of their coupling constant. Thus, when the lifetime is shortened by the long pulse, the nuclei become decoupled and the splitting is not observed.
4). The 13C NMR spectrum of natural abundance O 2-methylcyclohexanone (right) acquired using 1 1 broadband H decoupling (far right) consists of 6 6 2 5 3 signals. One signal is very intense. Of the 4 remaining signals, the intensities are similar with one being noticeably smaller. There is no chemical shift overlap between any of the signals, so therefore a signal from one of the carbon nuclei is so small that it is not observed.
1
H
bb decoupling 90x FID
7
13
C
a). Explain the intensities of the observed signals and why the intensity of one signal is so small that it is not observed. A complete answer will assign the signals to specific carbon nuclei. (6 points) Because the 13C spectrum is broadband decoupled during both the relaxation delay and during acquisition, the intensities of the signals will be enhanced due to the 1H-13C NOE. The signal from the methyl carbon (7) will be enhanced the most, because it has the most directly attached protons. Thus, the largest signal in the spectrum is from the methyl carbon. Likewise, of the remaining 5 observed signals, the least intense will be from the methine carbon (2), as it has only a single attached 1H. The signals from the methylene carbons (3, 4, 5, and 6) will be of intermediate intensity. There will be little enhancement of signals from quaternary 13C nuclei because they have no attached protons. Thus, the carbonyl carbon (1) signal will be the smallest signal in the spectrum. Because it is not observed, there must be some other contributor besides the lack of an NOE that is contributing to its small intensity. Because quaternary carbon atoms have very long relaxation times (T1), most likely the relaxation delay used for the experiment was too short to permit full relaxation of the carbonyl carbon during the delay, resulting in such poor signal-to-noise for the signal from this carbon that it is not observed.
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________
b). What are the multiplet structures of the 6 observed signals? (4 points) Broadband 1H decoupling collapses all of the 13C signals to singlets.
c). A small amount of the paramagnetic relaxation agent chromium(III)acetylacetonate [Cr(acac)3] is added to the sample. This compound very efficiently decreases the effective relaxation times of all of the 13C nuclei (to less than 1 second). If a second 13C spectrum, acquired precisely as before, is now recorded on the sample with the Cr(acac)3 added, how will this spectrum differ from the spectrum without the Cr(acac)3? (4 points) The Cr(acac)3 will decrease the relaxation times, which will have the largest effect on the signal from the quaternary carbon (1). In the spectrum with the Cr(acac)3 added, the expectation is that this signal will now be observed. It will probably be the smallest signal in the spectrum due to the lack of directly bonded protons and, therefore, the lack of an NOE enhancement (due to the broadband 1H decoupling) in the signal intensity. d). How would you collect a 13C spectrum, in a limited amount of time, on the natural abundance 2-methylcyclohexanone sample in such a manner such that you could accurately integrate all of the seven signals? Describe how you would attempt this. Sketch any pulse sequence diagrams that you would use if different than that shown above. Assume that digital resolution is not an issue. (6 points) Adding Cr(acac)3 as above would permit faster acquisition, as the relaxation delay could be considerably shortened. This would therefore eliminate T1-dependent signal intensity attuenuation, permitting more accurate integrals. It is 1 bb decoupling also necessary to eliminate NOE-dependent signal H enhancement, because CH, -CH2, and CH3 groups 90x are not equally enhanced. It is also necessary to FID eliminate multiplets by decoupling. Thus, inverse 13 gated decoupling would be used. In this experiment, C the broadband decoupling is only applied during acquisition, so that the multiplets are collapsed to singlets and there is very little overall NOE enhancement.
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________ 5). Consider the effect of the normal spin-echo pulse sequence (right) and the effect of this pulse sequence on heteronuclear JCH coupling and 13C chemical shift evolution, ignoring magnetic field inhomogeneity. Consider a simple heteronuclear spin system (13C-1H, i.e. CHCl3). 90x
13
180x
C
a b c d e
a. Using vector diagrams, show the effect of the spin echo pulse sequence on the 13C spin of this heteronuclear spin system with =1/(4JCH). Label each vector component appropriately (MCH , etc.), as well as the angle between the component vectors when necessary. Assume that the Larmor frequency of the 13C spin is equal to our reference frequency ( C = rf). Discuss the effect of the spin-echo pulse sequence on heteronuclear J coupling refocusing. (4 points) a z M0 y x x 90x y x MCH y b c
transverse plane
d MCH
= 90 180x = 90
e z y y x
MCH
MCH
x
The 180 pulse reflects the vectors through x, and the second period permits them both to refocus along y. Thus, the pulse sequence promotes refocusing of heteronuclear J coupling. b. Now, consider a 13C-13C (homonuclear) spin system. We'll call one of the 13C spins `A', and the other one `X'. We'll assume that =1/(4JAX). Using vector diagrams, show the effect of the spin echo pulse sequence on spin `A'. Label each vector component appropriately (MAX , etc.), as well as the angle between the component vectors when necessary. Assume that the Larmor frequency of the `A' spin is equal to our reference frequency ( A= rf). Discuss the effect of the spin-echo pulse sequence on homonuclear J coupling refocusing. (6 points) a b c d e z M0 y x x 90x y x
transverse plane
MAX
MAX = 90 180x y = 90 MAX x
z
MAX y
y x MA
X
MAX
The 180 pulse reflects the vectors through x. This is the effect of the pulse on the `A' spin. The 180 pulse also exchanges the populations of the and states of the `X' spin, i.e. this is the effect of the pulse on the `X' spin. Both operations cause the vectors to move away from one another during the second period, becoming antiphase to one another at `e'. Thus, this pulse sequence does not promote refocusing of homonuclear J coupling. 6
Exam 2: CHEM/BCMB 4190/6190/8189
Name _________________________ Below 6). is shown a 1H signal (singlet) for a particular 1H nucleus in a molecule that is acquired with poor magnetic field homogeneity. Sketch (on top of the signal shown) the signal that would be observed if the spectrum were acquired again in identical manner, except that the magnetic field homogeneity is improved. (4 points)
improved homogeneity
The signal acquired with better magnetic field homogeneity will be narrower and more intense than the other signal.
26
7). Would you expect the effective correlation times ( c) measured at C26 and C27 in cholesterol (right) to be larger or smaller than the effective correlation times measured at C19 and C18? Why (please explain)? (6 points)
18 19
27
The effective correlation time is roughly proportional to 1/T1, and is a measure of the mobility (rate of positional change or rotation) of a molecule or part of a molecule. We would expect the sidechain of cholesterol to experience less conformational and motional restriction and be more mobile than the ring structure. Thus, we would expect that the 13 C T1 values for C26 and C27 will be larger than those for C18 and C19, and the correlation times for C26 and C27 will be smaller than those for C18 and C19.
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________ 8). The following diagram (below, left) represents the inversion-recovery pulse sequence (experiment) used to measure T1 for 13C nuclei. Initially (point "a"), the bulk magnetization is aligned along +z, and is inverted by the 180x pulse (-z, point "b"). For a-e, for the indicated values of , indicate the orientation and approximate magnitude of the bulk magnetization vector on the coordinates axes at points "c" and "d". Also, sketch the signal obtained after Fourier transform of the FID. Assume the molecule we are studying has a single carbon atom whose 13C T1 is 20 s. (3 points each for a-e, 15 points total) 1 point "a" point "b" broadband decoupling H 180x
13
90x FID
z Mz=M0 y 180x
z y Mz=-M0
C
a b c
x
d
x
a. =0s
point "c" z 90x y
point "d" z y x z 90x y y x 90x y y x 90x y y x 90x y y z z z
signal (Fourier transform of FID)
b. = 300 s
x z
c. = 20 s
x
z
d. =6s
x
z
e. = T1ln(2) s
x
z
x x f. If T1 = 20 s, how long does have to be before 90% of the bulk magnetization at point "b" is returned to +z at point "c"? (4 points) Mz M z = M 0 (1 2e / T1 ) =0.9 = 1 2e / T1 0.1 = 2e / T1 0.05 = e / T1 2.9957 = /T1 = 59.91 s M0
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________
9). Consider the populations N1, N2, N3 and N4 of the , , and states respectively for a 1H-13C spin system. The energy diagram for this system is depicted (right), where A1 and A2 are the 1H transitions, and X1 and X2 are the 13C transitions. We will define H as the difference in the number of spins in and states for 1H, and X as the difference in the number of spins in and states for 13C.
N4
X1
N3
A1
N2
A2
N1
1
X2
H
13
C
a. Write down expressions for the equilibrium values of N1, N2, N3 and N4. Assume N4=N. Write down expressions for the population differences for the transitions A1, A2, X1 and X2. (4 points) A1 = N2 N4 = H N4 = N N3 = N + X A2 = N1 N3 = H X1 = N3 N4 = X N2 = N + H N1 = N + H + X X2 = N1 N2 = X b. Which transition, A1, A2, X1 or X2, corresponds to the MCH vector? Explain how you arrived at your answer. (4 points) As indicated in the diagram above (above, right), the first spin in the pairs corresponds to the 1H spin and the second to the 13C spin. Thus, the 13C transition (MC) with the 1H ). spin in the alpha state (MCH ) is the X2 transition ( c). What is the relationship between H and X. Quantitatively, what is their ratio, and how is this ratio derived? (2 points) The ratio H/ X is proportional to the ratio of the gyromagnetic ratios for 1H and 13C. 1 H/ 13C = 26.7519/6.7238 = 3.98 4.0, so H = 4 X d). In the SPI experiment, to enhance the signals for the 13C transitions, normally either the A1 or A2 1H transition is excited with a 180 pulse. You decide to find out what happens if you excite the A1 transition using a 90 pulse instead. Write down expressions for the populations N1, N2, N3 and N4 after exciting (90 pulse on) the A1 transition. (4 points) N4 = N + H/2 N3 = N + X N2 = N + H/2 N1 = N + H + X
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________
d. Now, write down expressions for the population differences for the transitions A1, A2, X1 and X2 (after exciting the A1 transition with the 90 pulse). For the X transitions, the final expressions should be written in terms of X, and for the A transitions the final expressions should be written in terms of H. (4 points) A1 = N2 N4 = 0 H A2 = N1 N3 = H X1 = N3 N4 = X - H/2 = - X X2 = N1 N2 = H/2 + X = 3 X e. Draw vector diagrams depicting the bulk magnetization vectors corresponding to the A and X transitions before and after selective excitation of the A1 transition by a 90 pulse. Make sure to label properly the individual vectors and indicate their magnitudes. (4 points) A transitions before A transitions after
z
A2, MHC H
A1, MHC H
A2, MHC H
z
A1, MHC 0 H
Because the axis on which the 90 pulse was applied was not specified, this vector could be anywhere in the transverse plane
y x x
y
X transitions before
X transitions after
z
X2, MCH X
X1, MCH X
z
X2, MCH 3 X
y
X1, MCH X
y x
x
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________ f. For the simple two spin system (1H-13C), draw the 1H and 13C signals that you would expect in 1D 1H and 13C NMR spectra for a compound at natural isotopic abundance (for instance, CHCl3) before and after selective excitation of the A1 transition by a 90 pulse. Make sure to label properly the individual peaks in the spectra/signals and indicate their magnitudes. Also indicate Larmor frequencies and coupling constants. (4 points) X2, MCH 3 X X
1 C
X2, MCH X
JCH
X1, MCH X before
13
C after
X1, MCH
JCH
C
JCH
1
before H-12C-
H after
JCH
1
H-12C-
MHC H
H
MHC H
MHC H
H
MHC 0 H
Exam 2: CHEM/BCMB 4190/6190/8189
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Name _________________________
10). The diagram (right) shows the INEPT pulse sequence. Consider the effect of this pulse sequence on heteronuclear (-13C-1H) J coupling, ignoring the effects of magnetic field inhomogeneity. We will consider a simple 1H-13C- spin system (i.e. 13 CHCl3) with a Larmor frequency equal to our reference frequency, H= rf. We will assume that the delay is equal to 1/(4JCH).
90x
1
180x
90y
H
a b c
180x
d
f
90x
g
13
C
e g h
a. Complete the vector diagrams below for points `c', `d', `e', and `f' in the pulse sequence. Be sure to label the vectors (MHC , MHC ), to include arrows indicating the direction of precession for the vectors in the rotating frame, and to indicate the angle ( ) between the vectors at each point. (4 points) MH2C z MHC
a MH2C
MH
C
transverse plane
b
90x y
MHC MH2C
= 0
c
MH2
= 90
H2 C
MHC
= 90
y x MH MH2C
C
y MH2
C
x
MHC x
MHC 180x 1H
= 90
d
MH2C MH2C 180x 13C y
= 90
MHC
= 90
e
MH2C y
= 90
C x MH2
f
MH2C MHC y
= 180
MHC
x
MHC
x MHC MH2C
b). On the same vector diagrams (a-f), include the vectors for an additional -13C-1H spin system where the 1H nucleus has a Larmor frequency that is faster than the reference frequency by an amount 3 JCH ( H2 = ...