Final Exam: CHEM/BCMB 4190/6190/8189 (265 pts)...

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 1

Final Exam: CHEM/BCMB 4190/6190/8189 (265 pts) Thursday, 16 December, 2004

1). In the DEPT experiment, the phase angleof the third 1H pulse ( y), applied along the ‘y’axis, is set to any desired angle in order toachieve the desired result. Shown in the figure(right) are the intensities of signals from CH,CH2, and CH3 groups as a function of the phaseangle y.

a). Four NMR spectra of the terpene andrographolide are shown. One of them is a decoupled13C spectrum, and three of them are DEPT spectra with differing values of the phase angle y orcombinations of DEPT spectra (for instance, DEPT (45) – DEPT (135)). Label each of thespectra to indicate what each is, including the phase angle(s) for DEPT spectra (i.e. “decoupled13C spectrum”, “DEPT (xx)”, “DEPT (xx) –DEPT (yy)”, etc.). (8 points).

b). In the spectrum at the bottom, there are three signals, at approximately 130 ppm, 150 ppm,and 172 ppm, that do not appear in the other spectra. Circle the carbon atoms in theandrographolide molecule that give rise to these peaks (it is not necessary to indicate whichcarbon atom corresponds to which peak). (4 points)

CH CH3

CH2

decoupled 13C spectrumandrographolide

Name _________________________


2). In the 1D, 1H spectrum of menthol, the multipletcorresponding to proton 9 appears to be composed of two tripletssurrounding a quartet (below). For the questions below, assumethat the coupling constants between proton 9 and the 4axial and4equatorial protons are equivalent. Also assume first order behavior.

Explain the observed multiplet structure for the signal from proton 9.

a). Your explanation should include a description of the protons that proton 9 is coupled to andhow each of these affects (splits) the signal from proton 9 (4 points).

Proton 9 is coupled to the 4 protons, which split the signal from 9 into a triplet. Proton 9 iscoupled to proton 5, which splits the signal into a doublet. Proton 9 is coupled to proton 1,which splits the signal into a doublet.

b). Your explanation should include a diagram and discussion describing the expected multipletpattern based on the expected splittings (4 points).

Thus, the triplet resulting from coupling to the 4 protons issplit into a doublet of triplets by coupling to proton 5, andthis doublet of triplets is split into a doublet of doublets oftriplets by coupling to proton 1.

c). Finally, your explanation should include a description of how the observed multipletstructure results from the expected splitting pattern, including a diagram of how this occurs (4points).

In menthol, the coupling constants between 9 and 5 and between9 and 1 are such that the two center triplets overlap: one outercomponent of each of these triplets overlaps with the centercomponent of the other resulting in what looks like a quartet.

HO

12

3

44

5

6 77

8

8

910

Name _________________________


3). Shown (right) is the pulse sequence forthe heteronuclear two-dimensional J-resolved experiment.

a). What parameter is measured in thedirectly detected dimension? (2 points)

13C chemical shift

b). What parameter is measured in the indirectly detected dimension? (2 points)

JCH

c). What is the role of the 13C 180°y pulse? Why is it necessary (or is it?) to include this pulse?For full credit you will have to use vector diagrams to illustrate you point(s). (10 points)

The 13C 180° pulse is necessary for chemical shift refocusing. As shown above (top), whenincluded, this pulse reflects the individual vectors through the ‘y’ axis so that during the secondt1/2 period the chemical shifts refocus, in this case along the ‘y’ axis. If not included, as shownabove (bottom), the chemical shifts do not refocus at ‘d’, and so the individual peaks in thespectrum are out of phase.

The 13C 180° pulse also removes the effects of magnetic field inhomogeneity. During the first t1/2period, because of magnetic field inhomogeneity, the spins that comprise both MC

H and MCH

start to fan out from the bulk because they experience slightly different magnetic fields. At ‘b’,the 13C 180°y pulse is applied. This pulse reflects all vectors through the ‘y’ axis. The fanningout of the spins due to magnetic field inhomogeneity is, then, refocused during the second t1/2period, as this is unaffected by the decoupler. Thus, at d, the effects of magnetic fieldinhomogeneity have also been refocused

a b c d

t1/2

180°y90°x

t1/2t2

bb decouplingbb decoupling1H

13C

x x

c2

c2

= 90°

t1/2

transverseplane

y

x

MCH

MCH

x

= 90°

y

180°y bby

a b c

y

d

MC2H

MC2H

t1/2c2

MCH

MCH

x

yMC2

H

MC2H

= 90°

x xc2

= 90°

t1/2

transverseplane

y

x

MCH

MCH

x

= 90°

y

no180 bb

y

a b

y

d

MC2H

MC2H

t1/2

c2

MCH

MCH

x

y

MC2H

MC2H

c

c2

Name _________________________


4). The 13C and 1H chemical shifts and 1JCH coupling constants for isopentyl acetate are shownbelow:

carbon 13C (ppm) 1H (ppm) 1JCH (Hz)-C1H3 21 2.1 120

-C2 171 - --C3H2 63 4.1 130-C4H2 37 1.5 130-C5H 25 1.7 140-C6H3 23 0.9 120-C7H3 23 0.9 120

Draw the heteronuclear two-dimensional J-resolved spectrum of isopentyl acetate. Label theaxes properly, and indicate which is F1 (indirectly detected dimension) and which is F2 (directlydetected dimension). (10 points)

The two methyl groups labeled ‘6’ and ‘7’ are equivalent, and so give rise to a single signal inF2 at 23 ppm. The quaternary carbon, whose 13C chemical shift is 171 ppm (F2), shows up as asinglet in F1 (no large coupling with any protons).

O ||C1H3-C

2-O-C3H2-C4H2-C

5H-C6H3

| C7H3

isopentyl acetate

JCH/2 = 60 HzJCH/2 = 65 Hz

JCH/2 = 70 Hz

100

100

75

50

25

0

25

50

75

13C (ppm) F2

70 60 50 40 30 20 10

F11JCH (Hz)

170

Name _________________________


5). Draw the heteronuclear two-dimensional 1H, 13C-HSQC spectrum of isopentyl acetate (seequestion 4). Label the axes properly, and indicate which is F1 (indirectly detected dimension)and which is F2 (directly detected dimension). (10 points)

6). Draw the two-dimensional 1H, 1H-COSY spectrum of isopentyl acetate (see question 4).Label the axes properly, and indicate which is F1 (indirectly detected dimension) and which is F2

(directly detected dimension). (10 points)

70

60

50

40

30

20

5.0 4.0 3.0 2.0 1.0 0.01H (ppm) F2

13C (ppm) F1

5.0 4.0 3.0 2.0 1.0 0.01H (ppm) F2

0.0

1.0

2.0

3.0

4.0

5.0

1H (ppm) F1

Name _________________________


7). The pulse sequence for the two-dimensional HETCORexperiment is shown (right). Consider the effect of thispulse sequence on a simple heteronuclear spin system (1H-13C, i.e. CHCl3).

a). Draw the two-dimensional HETCOR spectrum of CHCl3 that you would observe using thepulse sequence shown. Label the axes properly. Distinguish between the directly detecteddimension (F2) and indirectly detected dimension (F1). Indicate the Larmor frequencies of the13C and 1H nuclei. Indicate the coupling constant (1JCH) (6 points)

b). After the 90º 13C pulse, the components of the 13C magnetization that we areinterested in are oriented along the ‘y’ and ‘-y’ axes (right). At this point, webegin to acquire data (t2). We of course would like to turn on 1H-decouplingduring acquisition to remove the 1JCH coupling in F2, but the signal will disappear ifwe do so. Why? Please provide a complete explanation. (6 points)

At the beginning of acquisition, as shown in the vector diagram, the magnetization vectors MCH

and MCH are 180 degrees out of phase with respect to one another. Turning on the decoupler

maintains this phase difference, because the coupling is removed and both vectors rotate at theLarmor frequency. Because these vectors are 180 degrees out of phase, they sum to zero, andbecause the signal that we would observe is the sum of these two vectors, no signal is observed.

13C

1H

90x

90x90x

t1

t2

JCH

C

H

13C (ppm) F2

1H (ppm) F1

JCH

filled and opencircles representpeaks of oppositephase

MCHMC

H

y

z

x

Name _________________________


c). The pulse sequence for a modified HETCOR pulsesequence is shown (right). A delay ( 2) has beenadded that will allow decoupling during t2. Completethe vector diagram below to explain the effect ofadding 2 (4 points) and then explain what the valueof 2 should be and why (8 points).

For a –CH group, we know that the phase angle between the vectors MCH and MC

H will beequal to 2 JCH 2. When 2 is equal to 1/(2JCH), then the angle is radians or 180º. Because theangle between the vectors is 180º before 2, then after 2 the angle between them will be zerodegrees: in other words the vectors will be refocused. Thus, when 2 is equal to 1/(2JCH), thenthe vectors will be refocused just before the acquisition period and broadband decoupling can beapplied during acquisition without loss of signal.

MCHMC

H

y

z

x

2

y

z

x

MCH

MCH

13C

1H

90x

90x90x

2

t2

t1 decoupling

Name _________________________


8). In menthol, the distance between proton 10 and proton 4a (axial) is2.48 Å, whereas the distance between proton 10 and proton 4e(equatorial) is 3.65 Å. In a NOESY experiment, if the intensity of thecrosspeak between 10 and 4a is 100 (arbitrary units), what would be theexpected intensity for the crosspeak between 10 and 4e? (6 points)

We know that the intensity of a crosspeak in a NOESY spectrum (assuming that our mixing timewas reasonable) is proportional to 1/r6, where r is the internuclear (interproton) distance. Inthis case, the longer distance is 3.65/2.48 = 1.47 times longer than the shorter distance, and thusthe ratio of NOE intensities would be expected to be (1.47)6 = 10. Because the intensity of theNOE crosspeak for the shorter distance is 100, the expected intensity for the longer distancewould be 100/10 = 10.

9). You synthesized the phosphonate compound (right) in yourlaboratory using routine methods. You are collecting 13C NMR spectraon this compound and would like to remove the 13C-31P coupling(splitting) by performing broadband 31P decoupling. You also realizethat there will be an enhancement in S/N (signal-to-noise) due to the13C-31P NOE.

a). Calculate the percentage increase in S/N due to the 13C-31P NOE (for the 13C nucleus bound tothe phosphorous atom) when applying broadband 31P decoupling versus no decoupling. (6points)

For small molecules, for an AX spin system, the maximum fractional NOE enhancement, , isequal to a/(2 x) where a is the gyromagnetic ratio for the nucleus that is saturated (decoupled)and x is the gyromagnetic ratio for the nucleus that is observed. Thus,

= a

2 x

=10.8394 107

2 6.7283 107= 0.8055, or a 80.55% maximal enhancement.

4a

10

4e

Name _________________________


b). Without any decoupling, what will the signal from the carbon bonded to the phosphorousatom look like in a normal 1D, 13C spectrum? Draw the signal and explain the multipletstructure. (6 points)

The carbon nucleus in question will be coupled to itsdirectly bonded proton, resulting in a doublet. Thedoublet will be split into a doublet of doublets bycoupling to the 31P (spin 1/2) nucleus. Thus, the signalwill be a doublet of doublets, assuming a naturalabundance sample (phosphorous is 100% 31P, whereascarbon is only ~1% 13C – so there is no splitting of the13C signal by the neighboring carbon because it is 12C).

c). In a normal, 1D, 31P spectrum (no decoupling), what will the signal from the 31P nucleus looklike? Draw the signal and explain the multiplet structure. (6 points)

The 31P signal is a singlet at natural abundance. Onemight see 13C “satellite” signals flanking the singlet,corresponding to the doublet resulting from the naturalabundance (1%) 13C and the 13C-31P coupling.

d). Is 31P a ‘good’ NMR nucleus? Explain in detail. (6 points)

31P is a spin 1/2 nucleus. Its natural abundance is 100%! Its gyromagnetic ratio is 10.8394 x107 rad T-1 s-1, which is a bit less than 1/2 that of a proton but still substantially larger than 13C.Its relative sensitivity (for equal numbers of nuclei) is about 5 times better than 13C. Thus, it is aquite good nucleus: pretty high sensitivity, spin 1/2, and very high natural abundance.

JCP

P

31P-12C-

MPCMP

C

JCH

MCP MC

P

JCP

C

Name _________________________


10). Consider the populations N1, N2, N3 andN4 of the , , and states respectivelyfor a 1H-1H spin system (spins ‘A’ and ‘X’)without J coupling (no coupling between A andX). The energy diagram for this system isdepicted (right), where the A’s represent thetransitions of one of the 1H nuclei, and the X’srepresent the transitions of the other 1Hnucleus. We will define H as the differencein the number of spins in and states for A,and H will also represent the difference in thenumber of spins in and states for X.

a). Write down the equilibrium values for N1-N4 and the equilibrium population differences forthe A and X transitions (please show how you calculate the A and X transition populationdifferences). Assume that N4 = N. (5 points)

N4 = NN3 = N + H A = N2 – N4 = N1 – N3 = HN2 = N + H X = N3 – N4 = N1 – N2 = HN1 = N + 2 H

b). Explain what “W2” and “W0” in the figure represent, and describe any dependencies of W2

and W0 on molecular properties. (4 points)

W2 and W0 represent the double and zero quantum relaxation pathways, respectively. The W0

and W2 pathways do not operate with equal efficiencies with all sizes of molecules. For smallermolecules, the double quantum pathway is generally more efficient, whereas for largermolecules the zero quantum pathway is more efficient.

W2

W0

X A

N2

N1

N3

N4 X

X

A

A

Name _________________________


c). For a very large molecule, if you perform a simple NOE experiment by selective saturationof the A transitions, what are the new values for N1-N4 and the population differences for the Aand X transitions after saturation but before any relaxation takes place? After relaxation takesplace, what are the resulting values for N1-N4 and the population differences for the Xtransitions? If you introduce any new terms or symbols, please define them. (11 points)

If we saturate the A transitions, we equilibrate the N2 and N4 states, and we equilibrate the N1

and N3 states. Thus, before any relaxation takes place, the values for N1-N4 and the populationdifferences for the A and X transitions are:

N4 = N + H/2N3 = N + 3 H/2 A = N2 – N4 = N1 – N3 = 0N2 = N + H/2 X = N3 – N4 = N1 – N2 = HN1 = N + 3 H/2

Because the molecule is large, only the W0 pathway is operative. If we assume that spins relaxvia W0 (add to N2, subtract from N3) then the new values for N1-N4 and new populationdifferences for X become:

N4 = N + H/2N3 = N + 3 H/2 - N2 = N + H/2 + X = N3 – N4 = N1 – N2 = H - N1 = N + 3 H/2

Name _________________________


11). Consider the populations N1, N2, N3 andN4 of the , , and states respectivelyfor a 1H-13C spin system (spins ‘A’ and ‘X’)without J coupling (no coupling between A andX). The energy diagram for this system isdepicted (right), where the A’s represent thetransitions of the 1H nucleus, and the X’srepresent the transitions of the 13C nucleus. Wewill define H as the difference in the numberof spins in and states for A, and C willrepresent the difference in the number of spinsin and states for X.

a). Write down the equilibrium values for N1-N4 and the equilibrium population differences forthe A and X transitions (please show how you calculate the A and X transition populationdifferences). Assume that N4 = N. (6 points)

N4 = NN3 = N + C A = N2 – N4 = N1 – N3 = HN2 = N + H X = N3 – N4 = N1 – N2 = CN1 = N + H + C

b). For a very small molecule, for a directly bonded 1H-13C pair, J coupling is removed bybroadband 1H decoupling, and a change in the intensity of the 13C signal is observed due to theheteronuclear NOE (saturation of the A transitions). What are the new values for N1-N4 and thepopulation differences for the A and X transitions after saturation of the A transitions bydecoupling but before any relaxation takes place? After relaxation takes place, what are theresulting values for N1-N4 and the population differences for the X transitions? If you introduceany new terms or symbols, please define them. (11 points)

If we saturate the A transitions, we equilibrate the N2 and N4 states, and we equilibrate the N1

and N3 states. Thus, before any relaxation takes place, the values for N1-N4 and the populationdifferences for the A and X transitions are:

N4 = N + H/2N3 = N + H/2 + C A = N2 – N4 = N1 – N3 = 0N2 = N + H/2 X = N3 – N4 = N1 – N2 = CN1 = N + H/2 + C

Because the molecule is small, only the W2 pathway is operative. If we assume that spins relaxvia W2 (add to N1, subtract from N4) then the new values for N1-N4 and new populationdifferences for X become:

N4 = N + H/2 - N3 = N + H/2 + CN2 = N + H/2 X = N3 – N4 = N1 – N2 = C + N1 = N + H/2 + C +

W2

W0

X A

N2

N1

N3

N4 X

X

A

A

Name _________________________


12). If we compare 11B and 39K nuclei at natural abundance;

a). What is the ratio of the bulk/macroscopic magnetization (M0 11B/M0 39K) produced for equalnumbers of B and K nuclei at natural abundance? (6 points)

Remember, M0 =N 2h2B0I(I +1)

3kBT (“h” here is “h-bar”, Planck’s constant divided by 2 ):

M0 11B

M0 39K

=

N 11B2h2B0I11B(I11B +1)

3kBT

N 39K2h2B0I39K (I39K +1)

3kBT

=

80.42 11B2 3232

+1

93.1 39K2 3232

+1

=80.42 (8.5847 107 rad/Ts)2

93.1 (1.2499 107 rad/Ts)2= 40.75

b). What is the ratio of the 11B and 39K signal intensities generated in the receiver coil ( 11B/ 39K)by equal numbers of B and K nuclei at natural abundance? (6 points)

Sensitivity is proportional to the electromotive force ( ) induced in the receiver coil by the bulkmagnetic moment. The magnitude of is proportional to the rate of change in the magneticmoment ( dM/dt = M0B).

Remember, M0 =N 2h2B0I(I +1)

3kBT (“h” here is “h-bar”, Planck’s constant divided by 2 ),

so M0B =N 3h2B0

2I(I +1)

3kBT. So, whereas the magnitude of the bulk magnetization was

dependent on 2, the sensitivity is dependent on 3.

11B

39K

=

N 11B3h2B0I11B(I11B +1)

3kBT

N 39K3h2B0I39K (I39K +1)

3kBT

=

80.42 11B2 3232

+1

93.1 39K2 3232

+1

=80.42 (8.5847 107 rad/Ts)3

93.1 (1.2499 107 rad/Ts)3= 279.87

c). In order to achieve the same signal-to-noise, for each scan in a 11B experiment, how manyscans will have to be recorded in a 39K experiment (equal sample concentrations, naturalabundance)? (4 points)

In NMR, signal-to-noise (S/N) improves with the square root of the number of scans (N1/2). So,since the 11B sensitivity – S/N – from is 279.87 times better than 39K, it will take (279.87)2 =78,327 39K scans for every 11B scan to get equivalent S/N for equivalent natural abundancesamples.

Name _________________________


13). Consider the spin-echo pulse sequence(right). Ignoring relaxation and magnetic fieldinhomogeneity, and considering a single -13C-Hgroup with c= rf and =1/(4JCH), we see that thesignal at points ‘e’ and ‘f’ is the same (below). Ifwe were to perform a Fourier transform of thesignal either at ‘e’ or ‘f’, we would see the samesignal. Why is the second period necessary? You should consider a ‘real’ molecule with morethan one -13C-H group, use vector diagrams to prove your points, and explain (do not just state)clearly why the second period is necessary (do not consider –CH2 or –CH3 groups). (10points)

In a ‘real’ molecule, with more than one 1H-13C- group, at least one of the 13C frequencies willnot be equal to rf ( c2 rf). If we consider a nucleus with a Larmor frequency ( c2) that is notequal to our reference frequency ( c2 rf), we see that at ‘e’ the chemical shifts of the two nucleiare not refocused, but that after the second tau period, at point ‘f’, they are. Thus, the secondtau period is necessary for proper chemical shift refocusing.

d e

13C

1H broadband decoupling

a b c f

180°x90°x

x x

x x

c2c2c2

= 90°

transverseplane

y

x

MCH

MCHx

= 90°

y

MCH

MCH

= 90°

x

y

180°x bby

b c d e

y

f

transverseplane

y

x

MCH

MCH

x

= 90°

y

MCH

MCH

= 90°

x

y

180°x bby

z

b c d e

y

z

f

MC2H

MC2H MC2

H

MC2H

Name _________________________


14). Using complete sentences only, and without using any symbols, equations, drawings, etc.,explain the concept of directional quantization as it relates to NMR. (10 points)

Directional quantization refers to the behavior of a nucleus in a magnetic field. For a nucleuswith a nuclear magnetic moment (spin quantum number not equal to zero), in a magnetic fieldthe angular momentum of the nucleus will orient so that its z component (aligned along thedirection of the field) is an integral or half-integral multiple of a constant, with the number ororientations determined by the spin quantum number. Thus, rather than a continuum oforientations, the number of orientations and direction of the orientation are quantized.

15). Fill in the blanks with the correct word or words (only real words – no symbols oracronyms): (14 points)

a). Nuclei with spin quantum number, I, > 1/2, are said to be _quadrupolar_ nuclei.b). The contribution to the shielding of a nucleus by the shell or shells of electrons is called the_diamagnetic_ shielding term.c). Following a 90° pulse, the redistribution or randomization of the magnetic dipoles about thez axis (loss of phase coherence) is called _transverse_ relaxation.d). Indirect spin-spin coupling, or __scalar____ coupling, results from interactions ofneighboring magnetic dipoles via chemical bonds.e). Chemical shift __anisotropy__ refers to the dependence of the chemical shift of a nucleus onthe orientation of the nucleus relative to the magnetic field, B0.f). The dependence of the vicinal coupling between two protons on adjacent sp3 orbitals on thetorsion angle between the protons is called the ___Karplus___ relationship.g). In a magnetic field, the nuclear dipoles (magnetic moments) of a particular nucleus precessabout the applied field at the ___Larmor___ frequency of the nucleus.

Name _________________________


16). For this question, consider only carbon atom 1 andcarbon atom 2 in isopentyl acetate (right). Sketch the signalcorresponding to carbon atom 1 and the signal correspondingto carbon atom 2 when the pulse sequences shown below (a-d)are used. The signals should be drawn so that the relativeintensities of the two signals and the multiplet components arecorrect. You will have to justify your sketches, explaining the multiplet structures andintensities. (16 points)

a).

Both signals are singlets of equal intensity. The decouplingscheme applies decoupling only during acquisition, so, theonly significant effect is removal of the multiplet structure.So, the signals are of equal intensity since there are nocouplings to either, no resultant splittings, and no NOEenhancements.

b).

The broadband decoupling scheme shown removes allcouplings as decoupling is applied during acquisition.Therefore, both signals are singlets. The relative intensityof C1 will be much larger that C2 because it has several attached protons, and the heteronuclearNOE generated by decoupling these protons during the relaxation delay will be substantial,causing the large increase in intensity.

O ||C1H3-C

2-O-C3H2-C4H2-C

5H-C6H3

| C7H3

isopentyl acetate

13C

1H

90°x

bb decoupling

FID

13C

1H

90°x

bb decoupling

FID

C2C1

C2C1

Name _________________________


c).

This sequence does not apply decoupling during acquisition,so the splitting of the signal from C1 (into a quartet) is apparent, but the decoupling duringacquisition allows for NOE enhancement of the intensity of the signal from C1 because of theattached protons. So, the signal intensities of C1 and C2 will be comparable: the C2 signal isnot enhanced in any way, but the signal at C1 is enhanced substantially via the NOE.

d).

This is just a simple 13C pulse sequence without decoupling,so all couplings will be active and all splittings observed. The quartet from C1 will be muchlower in intensity than the singlet at C2 (no attached protons at C2, no splitting). Compared toC2, the largest component of the quartet is about 40% as intense, the smallest 10%.

13C

1H

90°x

bb decoupling

FID

13C

90°x

FIDC2

C1

C1

C2

Name _________________________


17). Your friend just acquired a 1H, 13C-HSQC spectrum on a sample and showed you thespectrum. You noticed that the digital resolution in the indirectly acquired dimension was verypoor. You decided to give your friend some advice on improving the digital resolution in theindirectly acquired dimension. What advice did you give to your friend. Please explain. (6points)

To improve the digital resolution in the indirectly acquired dimension, normally one simplyeither increases the number of points acquired in that dimension (the number of different t1

values) or decreases the spectral width in that dimension (increase the dwell time).

18). The following diagram (below, left) represents the normal inversion-recovery pulsesequence (experiment) used to measure T1 for 13C nuclei. Initially (point “a”), the bulkmagnetization is aligned along +z, and is inverted by the 180°x pulse (-z, point “b”). The vectordiagrams for these points are shown below.

a). For a particular 13C nucleus in your molecule, you find that when is equal to 7 seconds, youobserve no signal. What is T1 for the 13C nucleus? (6 points)

Mz = M0(1 2e / T1 ) 0 = M0(1 2e / T1 ) 1= 2e / T1

1/2 = e / T1 ln(1/2) = /T1 T1 = /ln(2) = 7 /0.693 =10.1 s

ba cd

90°x180°x

broadband decoupling1H

13CFID

point “a”

Mz=-M0

180°xMz=M0

y

z

x

y

z

x

point “b”

Name _________________________


b). For 2,2,4-trimethylpentane (right), the measured 13C T1 values for thecarbon nuclei are 9.8, 23, 68, 9.3, 9.8, 9.3, 13, and 9.3 seconds, in noparticular order. Fill in the table below with the correct T1 for eachcarbon, and justify the relative magnitude of each T1 for each carbon. (6points)

carbon T1 justification1 9.3 Carbon atoms 1, 6, and 7 have the shortest T1 values. Of course the number of

directly bonded protons decreases T1, and for these methyl groups, the numberof neighboring methyl groups increases the total –H density, thus the T1 valuesare somewhat lower that the methyl groups of carbons 8 and 5.

2 68 This is a quaternary carbon, so its T1 is quite long

3 13 This methylene has two directly bonded protons, and is in an area of very highproton density, therefore the T1 is rather low

4 23 This carbon has a single bound proton, so its T1 is higher than all other T1

values fot the carbon nuclei in this molecule except for the quaternary carbon.

5 9.8 See carbon 1

6 9.3 See carbon 1

7 9.3 See carbon 1

8 9.8 See carbon 1

c). For toluene (right), the measured 13C T1 values for carbon nuclei 1-4 are 24, 17,89, and 24 seconds, in no particular order. Fill in the table below with the correct T1

for each carbon, and justify the relative magnitude of each T1 for each carbon. (4points)carbon T1 justification

1 89 This is a quaternary carbon, so its T1 is quite long

2 24 Compared to carbon 4, carbon 2 and carbon 3 experience increased mobilitybecause the most efficient rotational motion of the molecule is most likely alongthe long (4-1) axis. Thus, the mobility is somewhat increased as is the T1.

3 24 Same as carbon 2

4 17 This T1 is shorter than 2 or 3 because the rotation of the molecule is probablymost efficient along the long (4-1) axis, so that this carbon experiences adecreased mobility and thus a shorter T1

1

CH3

2

3

4

C6H3 C8H3

| |

C1H3-C2-C3H2-C

4H-C5H3 |

C7H3

Name _________________________


19). Consider the effect of the normal spin-echo pulsesequence (right, top) and the effect of this pulsesequence on heteronuclear J coupling and chemicalshift evolution, ignoring magnetic fieldinhomogeneity. Consider simple heteronuclear spinsystems (1H-13C, i.e. CHCl3).

On a single vector diagram, show the effect of the spin echo pulse sequence on two 1H-13C spin systems, each with a different Larmor frequency, when =1/(4JCH). Label each vectorcomponent appropriately, as well as the angles between the components when necessary. Forone spin system, assume that the Larmor frequency of the 1H spin is equal to our referencefrequency ( H = rf). For the other, assume that the Larmor frequency of the 1H spin is fasterthan the reference frequency by an amount 4 J ( H = 4 J + rf). Discuss the effect of the spin-echo pulse sequence on heteronuclear J coupling refocusing and chemical shift refocusing. (8points)

The 180° pulse reflects the vectors through x, and the second period permits them to refocusalong –y. Thus, the pulse sequence promotes refocusing of heteronuclear J coupling andchemical shift refocusing.

= 90°

transverseplane

M0

y

z

x

90°x

yx

MHC

MHCx

= 90°

y

MHC

MHC

= 90°

x

y

180°x

y

z

x

a b c d e

a b c d e

180°x90°x1

MHC

MHC

= 90°

MHC

MHC

M0

Name _________________________


20). The HSQC pulse sequence is shown below.

The magnetization vectors present at point ‘h’ in the pulse sequence are shown, as are the vectorsat points ‘i’, ‘j’, and ‘k’ for a simple spin system. The vector MC

H precesses slower than thereference frequency, and MC

H precesses faster than the reference frequency, such that after t1/2,the vectors have moved and degrees, respectively, away from the y axis.

The 180° pulse in the middle of the t1 evolution period prevents net 1H-13C coupling evolution.Both at the beginning (‘h’) and at the end (‘k’) of t1, the 13C magnetization components are 180°out of phase with one another, with no net phase change between them occuring during t1. Thus,the 180° pulse decouples 1H from 13C during t1.

If we replace the 1H 180°x pulse with a13C 180°x pulse, we get the same result:the 13C magnetization components are180° out of phase with one another, withno net phase change between themoccuring during t1.

So, the question is, why can’t we replacethe 1H 180°x pulse with a 13C 180°x pulsein the pulse sequence (you can use thenext page to answer)? (10 points)

i j

kh

180x 90y

180x 90x

1H

13Cdecouple

90x 180x 180

18090

t1/2 t1/2

90

180x

MCH

t1/2MC

H

y

x

MCH

‘h’

y

x

MCH

‘i’

MCH

y

x

MCH

‘j’

MCH

y

x

MCH

‘k’

t1/2

13C 180x

MCH

y

x

MCH

‘j’

MCH y

x

MCH

‘k’

t1/2

Name _________________________


The 13C 180°x pulse refocuses the chemical shifts so that there is no chemical shift evolution. Asshown in the second vector diagrams, regardless of the chemical shift, the magnetization willrefocus. Thus there will be no chemical shift evolution, which is the point of the t1 delay period

Name _________________________


HO

12

3

44

5

6 77

8

8

910

solve

1 5 8 7 4

6

2

31

9

84 7

Name _________________________


You may find some of the information in this table useful:

Name _________________________


You may find some of the following information or equations useful:

1H = 26.7519 x 107 rad/T/s, I = 1/2 13C = 6.7283 x 107 rad/T/s, I = 1/2

10B = 2.8747 x 107 rad/T/s, I = 3 15N = -2.7126 x 107 rad/T/s, I = 1/2

11B = 8.5847 x 107 rad/T/s, I = 3/2 17O = -3.6280 x 107 rad/T/s, I = 5/2

M0 =N 2h2B0I(I +1)

3kBT (“h” here is “h-bar”, Planck’s constant divided by 2 )

dM/dt = M0B =N 3h2B0

2I(I +1)

3kBT (“h” here is “h-bar”, Planck’s constant divided by 2 )

S/N N/N1/2=N1/2 (signal-to-noise improves with (number of scans)1/2)

m = (-I, -I+1, …, I-1, I)

L = | /(2 )| B0

=observe frequency

106

Mz = M0(1 e t / T1 )Mz = M0(1 2e t / T1 )tzero=T1ln(2)

= a / (2 x)I = (1 + ) I0

I 1/r6

B0

(Tesla, T)Resonance frequencies

(MHz)1H 13C

9.4 400 100.611.74 500 125.714.09 600 150.918.79 800 201.2

E = µzB0 = -m hB0

E = µzB0 = hB0

Final Exam: CHEM/BCMB 4190/6190/8189 (265 pts)...

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Transcript of Final Exam: CHEM/BCMB 4190/6190/8189 (265 pts)...