Final Exam: CHEM/BCMB 4190/6190/8189 (276 pts) Thursday, 15 December, 2005

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1 Final Eam: CHEM/BCMB 4190/6190/8189 (276 pts) Thursda, 15 December, ). The two-proton spin sstem at right gives rise to the 1 H NMR spectra shown (spectra 1 and 2, below) as the ratio of /J changes (/J is the frequenc difference between the signals divided b the coupling constant between them): CH A CH B a). What are the relative magnitudes of /J for spectra 1 and 2 (which spectrum reflects a higher or lower value for /J)? Please eplain. (4 points) For a first order spectrum, the relative intensities of the peaks for both doublets would be approimatel equal, as is the case in spectrum 2. This occurs when the frequenc difference,, between the signals is large compared to the coupling constant, J, between them (large /J). When the frequenc difference becomes smaller compared to J, higher order spectra result such as spectrum 1 (small /J), and the intensities are no longer equal. b). If ou were going to acquire a spectrum of this compound, what could ou do to ensure that the spectrum looked like 2 instead of 1? Please eplain our answer. (4 points) As the magnetic field strength increases, the difference in frequenc,, between the signals will increase, whereas the coupling constant, J, will not change. Thus, a higher field will increase the ratio /J and make the spectrum appear first order (like spectrum 2). So, to ensure spectrum 2 rather than 1, use the highest magnetic field available. c). Draw what the spectrum will look like when /J = 0 (sketch it above, on spectrum 3). Eplain wh it looks as it does. (4 points) As gets smaller (as /J approaches ero), the signals move closer together, and the outer components of each of the doublets get smaller. When /J = 0, the frequenc difference between the peaks is ero and the outer components of the doublets are so small that the are not observable. Thus, one observes onl a singlet, in this case in the ver center of the spectral region shown. Final Eam: CHEM/BCMB 4190/6190/8189 1

2 2). For the compound shown (below, right), draw/sketch each signal that ou would observe in a 1D, 1 H-NMR spectrum: - ou should label the nuclei of the molecule and indicate accordingl which signal corresponds to which proton(s) -make sure to draw the appropriate splitting patterns -indicate what the multiplet structure of each signal is (singlet, doublet of doublets, or whatever) -for each signal, eplain wh ou observe the splitting pattern that ou do (ou do not have to specif what ou think the chemical shifts are) (12 points) a b c d b e CH 3 OH a c CH 3 C C CH 3 d H H f e f a doublet (the signal from three equivalent protons is split to a doublet b the single neighboring proton d) b doublet (the signal from three equivalent protons is split into a doublet b the single neighboring proton d) a and b are not equivalent and give rise to discrete signals because of the neighboring chiral center c doublet (the signal from three equivalent protons is split into a doublet b the single neighboring proton f) d - quartet of quartet of doublets (signal is split into a quartet b the three methl protons a, and this is split into a quartet of quartets b the three methl protons b, and, finall, all peaks are split into doublets b the neighboring proton f. The appearance will depend on the relative magnitudes of the coupling constants) e broad singlet (no coupled protons to the hdrol proton) f quartet of doublets (the signal is split into a doublet b the neighboring proton d, and these are split into quartets b the three equivalent neighboring hdrogens c) Final Eam: CHEM/BCMB 4190/6190/8189 2

3 3). Shown (right) is the structure of 4-hdro cinnamic acid and the 1 H chemical shifts for hdrogen nuclei in the molecule. a). Wh is the chemical shift of d so much lower than c? Please provide a complete eplanation. (6 points) ( 1 H) a 6.38 b 7.66 c 7.59 d 6.93 b H O OH OH H a c d The average 1 H chemical shift for the ring protons of benene is about The chemical shift for d is much lower, and is lower than the other ring hdrogen (7.59) in this compound. If one considers the electron donating abilit of the hdrol group on the ring, resonance structures can be drawn where the carbon net to that bearing the hdrol (the 3 position) bears a negative charge. Thus, this position is highl shielded, so the chemical shift of the hdrogen at this position, d, is ver low. O O H OH O O H OH b). Wh is the chemical shift of b so high? Please provide a complete eplanation. (6 points) There are two reasons. First, the ring current in the benene ring causes groups at the peripher to eperience deshielding. The b hdrogen is in the deshielding region and will eperience an overall deshielding, thus increasing its chemical shift. Second, one can draw a resonance structure where the electrons in the carbonl group of the carbolic acid are localied to the electronegative ogen, thus creating a (partial) positive charge at the carbon bearing b. This would further deshield this O OH OH hdrogen. Thus, conbined, these two effects lead to a significant deshielding of b. O + OH OH Final Eam: CHEM/BCMB 4190/6190/8189 3

4 4). For a newl discovered element Z, the 297 Z and 299 Z isotopes have spin quantum numbers of 1/2 and 5/2, respectivel. You have discovered that the ratio of the bulk magnetiation produced for equal numbers of 297 Z and 299 Z spins is 1.37, and ou know that the gromagnetic ratio for 297 Z is rad T -1 s -1. a). What is the gromagnetic ratio for 299 Z? Make certain to show our work. (6 points) M 0 297Z M 0 299Z = N 2 297Zh 2 B 0 I 297Z (I 297Z +1) T N 2 299Zh 2 B 0 I 299Z (I 299Z +1) T ( ) ( ) = 297Z = 2 297Z I I Z 297Z 2 299Z I 299Z I 299Z Z 2 ( ) ( ) =1.37 I 297Z I 297Z +1 I 299Z I 299Z +1 = ( ) Z ( +1 2 ) ( ) = ( ) Z = Z = ( ) Z = b). The natural abundance of 297 Z is 50% and the natural abundance of 299 Z is 25%. Because of this difference in abundance, and because of the differences in gromagnetic ratios, the relative sensitivities of signals induced in the NMR coil for these nuclei are different. If the temperature is 50 C (323 K) for measuring the sensitivit of 297 Z, what temperature is necessar for measuring the sensitivit of 299 Z so that the ratio of the sensitivities is 10 ( 297Z 299Z =10)? Please show our work. (6 points) Sensitivit is proportional to the electromotive force () induced in the receiver coil b the bulk magnetic moment. The magnitude of is proportional to the rate of change in the magnetic moment ( dm/dt = M 0 B). Remember, M 0 = N 2 h 2 B 0 I(I +1) T 297Z 299Z = N 297Z 3 297Zh 2 B 2 0 I 297Z (I 297Z +1) T 297Z = N 299Z 3 299Zh 2 B 2 0 I 299Z (I 299Z +1) T 299Z, so M 0 B= N 3 h 2 B 2 0 I(I +1). T N 297Z 3 297ZI 297Z (I 297Z +1) T 297Z = N 299Z 3 299ZI 299Z (I 299Z +1) T 299Z 50 ( ) ( ) 3 T 299Z 1 ( 1 +1) 2 2 =10 +1) 5 2 ( 5 2 T 299Z = 5 2 ( ) 1 2 ( ( ) 3 +1) 50 ( ) = o K = 21.5 o C Final Eam: CHEM/BCMB 4190/6190/8189 4

5 5). In phenanthrene (molecules A and B, right), the 13 C T 1 values for the carbon atoms 11 and 12 were measured to be 51 and 59 seconds, respectivel ( A ). When the hdrogens at positions 4 and 5 were replaced with deuterium ( 2 H, D ), the T 1 of one of the carbons (11 or 12) was found to be 59 seconds whereas the T 1 of the other (11 or 12) was found to be 80 seconds ( B ). Which carbon (11 or 12) has the shorter T 1 and which has the longer T 1 in B. Eplain wh. (6 points) A The dominant mechanism here for promoting 13 C B relaation involves dipole-dipole interactions between 13 C and 1 H nuclei. Directl bonded 1 H nuclei have ver large effects on relaation rates, but non-bonded 1 H nuclei also affect relaation. In phenanthrene, the 1 H nuclei at positions 4 and 5 contribute to the T 1 relaation of carbon atoms 11 and 12, as is evidenced b the fact that the T 1 values for 13 C nuclei at these positions increase when 1 H atoms at positions 4 and 5 are replaced with 2 H. Because the 1 H atoms at positions 4 and 5 are nearer in space to carbon atom 12, its T 1 value is increased more (to 80) than the T 1 at position 11 (T 1 increased to 59). 6). Below are two signals (each a singlet) for two separate 1 H nuclei in a particular molecule. One of the signals has a short T 2 and the other has a long T 2. Which is which? Eplain our answer. (4 points) Because the magnetic field homogeneit is the same for both signals (the signals arise from nuclei in the same molecule), its contribution to the observed line widths is the same for each signal. Therefore, the difference in line width is due to T 2 differences. As judged b the line width at height, the signal on the left results from the nucleus with the longer T 2 (narrower peak) and the signal on the right results from the nucleus with the shorter T 2 (broader peak). Final Eam: CHEM/BCMB 4190/6190/8189 5

6 7). Below are NMR spectra of the molecule shown at the right. The spectra are labeled A-E. Match each spectrum with the description below (put the correct letter net to the description of the spectrum). (10 points) C refocused INEPT spectrum with BB decoupling E 13 C spectrum with gated BB 1 H decoupling A refocused INEPT spectrum B 13 C spectrum with BB 1 H decoupling D INEPT spectrum A B C D E Final Eam: CHEM/BCMB 4190/6190/8189 6

7 8). Consider the simple spin-echo pulse sequence (right), ignoring the effects of magnetic field inhomogeneit. Consider a single 13 C nucleus with two attached protons (i.e. 13 CH 2 Cl) with a Larmor frequenc equal to our reference frequenc ( C = rf ). 13 C a b c d e a). Sketch the normal (coupled) 13 C signal that ou would observe from the 13 C nucleus in 13 CH 2 Cl in a normal 13 C NMR spectrum (no decoupling). Label each multiplet component with the relative intensit of that component compared to the others in the multiplet. (4 points) b). In terms of the Larmor frequenc, C, and the coupling constant, J CH, what are the frequencies of each of the components of the multiplet? (3 points) C J CH C C + J CH c). How large is the phase angle (, in degrees) between the two vectors corresponding to the two outermost components of the multiplet after a time =1/(4J CH )? Show how ou arrive at our answer. (3 points) The phase angle = 2(2J CH ) = 2(2J CH )/(4J CH ) = radians = 180 d). Using vector diagrams, show the effect of the spin echo pulse sequence on this spin sstem with =1/(4J CH ). Show diagrams corresponding to points a - e in the pulse sequence. Label the vectors corresponding to the individual multiplet components with the frequencies from our answer to b. Be sure to label our aes. Be sure to show the direction of precession of each component of the multiplet relative to the reference (aes). (6 points) M 0 a a b c d 90 transverse plane c -J CH c +J CH e =1/(4J CH ) = = 180 =1/(4J CH ) Final Eam: CHEM/BCMB 4190/6190/8189 7

8 e). Repeat d with a modified version of the spin-echo pulse sequence where the 180 pulse has been removed. In addition, sketch the signal that ou would observe using the normal eperiment (signal from part d above) and the signal that ou would observe with the modified pulse sequence. (6 points) M 0 a a b c transverse plane 90 d c -J CH c +J CH =1/(4J CH ) e = = 180 =1/(4J CH ) normal eperiment modified eperiment Final Eam: CHEM/BCMB 4190/6190/8189 8

9 9). If we consider a spin nucleus in a magnetic field, where N and N are the numbers of spins (nuclei) in the and states, respectivel: a). What is the relationship between N and N at equilibrium in the magnetic field? (3 points) At equilibrium in a magnetic field, N is slightl greater than N, but the difference is small. b). If we appl a ver long electromagnetic pulse at the Larmor frequenc of this nucleus, what will be the relationship between N and N at the end of the pulse? (3 points) N and N will be equal. c). If we appl a 180 pulse at the frequenc of this nucleus, what will be the relationship between N and N at the end of the pulse, and how do these populations relate to the equilibrium populations? (3 points) The equilibrium N and N populations will invert, and N will now be greater than N. d). Fill in the blanks to make the statement true: For a spin nucleus at equilibrium in a magnetic field, the beta () state has a magnetic quantum number of -1/2, its energ is greater than that of the alpha () state, and the component of its associated magnetic moment, μ, is aligned anti-parallel to the magnetic field. (3 points) 10). Eplain, for coupled spin nuclei, in general how the multiplicit of a signal can be predicted and how the relative intensities of the peaks in a multiplet can be predicted (assume first order spectra). Give a specific eample. (8 points) The multiplicit, M, for a signal from a nucleus or a group of equivalent nuclei, is equal to 2nI+1where I is the spin quantum number and n is Pascal s Triangle the number of coupled nuclei. For a spin nucleus, M=n+1. 1 singlet The intensities of the multiplet components can be predicted 1 1 doublet from Pascal s Triangle (below). So, for a methl group, triplet for instance, bonded to a methlene group, the signal from the quartet (equivalent) methl protons will be split into a triplet b the quintet two (n=2) neighboring methlene protons (M=2+1=3). The intensities of the triplet will be 1:2:1, according to Pascal s triangle. Final Eam: CHEM/BCMB 4190/6190/8189 9

10 11). In the DEPT eperiment, the phase angle (pulse width) of the third 1 H pulse (applied along the ais) can be set to an value, in order to achieve the desired result. Shown in the figure below are the intensities of signals from CH, CH 2, and CH 3 groups as a function of the phase angle : O C 1 H 3 -C 2 -O-C 3 H 2 -C 4 H 2 -C 5 H-C 6 H 3 C 7 H 3 isopentl acetate a). A DEPT spectrum, collected with set to a single value between 0 and 180, for isopentl acetate is shown below. What is the angle? Please eplain wh. Also, assign the signals A- E to the groups 1-7 in isopentl acetate and justif our choices. (8 points) DEPT(135) E D C A 13 C, (ppm) B The pulse angle must be ~135, and the spectrum is phased 180 differentl than normal. At set at 135, the two methlene groups, which will be shifted downfield due to the electron withdrawing effect of the ester, appear opposite in phase to the methine and methl groups (as epected for = 135 ). The largest peak, B, would then correspond to the equivalent methl groups 6 and 7, peak A would correspond to the remaining methl group 1, and the methine, which should give the smallest signal for = 135, is peak C. Peak E will be the methlene closest to the ester (3), and D would then be methlene 4. Final Eam: CHEM/BCMB 4190/6190/

11 12). The HSQC pulse sequence is shown below. 1 H C t 1 /2 t 1 /2 decouple a). The magnetiation vectors just after the second C pulse are shown below. Complete the vector diagrams for the remaining steps in the HSQC sequence (assume a simple 1 H- 13 CCl 3 sstem and a of 1/(4J CH ). Make sure to label properl the vectors, indicate the directions that the vectors are moving, and indicate important angles between the vectors or the vectors and the aes. (6 points) 1 H 90 1/(4J CH ) H C /(4J CH ) decouple b). For the previous eample, a simple 1 H- 13 CCl 3 sstem, a value of 1/(4J CH ) is appropriate for in the HSQC eperiment. If, instead, ou were considering a methlene group ( 13 CH 2 -Cl 2 ), what would be the appropriate value of? Please provide a detailed eplanation. (6 points) The appropriate value for is still 1/(4J CH ). The signal from each 1 H nucleus is alwas split into a doublet b 1 J CH, thus 1/(4J CH ) alwas assures that the angle between the vectors is appropriate (for instance, 180 after the third period, as demonstrated above). Also, the values of 1 J CH for methine, methlene, and methl groups are prett similar ( H), and so an average value for can be used with good results. Final Eam: CHEM/BCMB 4190/6190/

12 c). Which is more sensitive, HETCOR or HSQC? Wh? To answer this question ou should clearl state which ou think is most sensitive, give a general statement as to wh, and then perform a simple calculation that will serve to demonstrate the approimate relative sensitivities of the techniques. (8 points) Both techniques transfer magnetiation/polariation from 1 H to 13 C, and, thus, effect enhancement of the insensitive 13 C nucleus. In the HETCOR technique, however, the 13 C magnetiation created is detected directl. On the other hand, in the HSQC technique, this 13 C magnetiation is transferred back to 1 H, so that the 1 H magnetiation thus created can be detected. Because detection of 1 H magnetiation is much more sensitive than detection of 13 C magnetiation, overall the HSQC technique is much more sensitive (ignoring relaation and other affects). This can be shown b the following calculation of sensitivities: Sensitivit is proportional to the electromotive force () induced in the receiver coil b the bulk magnetic moment. The magnitude of is proportional to the rate of change in the magnetic moment: dm/dt = M 0 B Remember, M 0 = N 2 h 2 B 0 I(I +1), so M 0 B= N 3 h 2 B 2 0 I(I +1). So, whereas the T T magnitude of the bulk magnetiation is dependent on 2, the sensitivit is dependent on 3. The main difference between HSQC and HETCOR in this respect is that the magnetiation is detected on 1 H in HSQC and on 13 C in HETCOR. Thus, 1H 13C = N 3 1H h 2 B 2 0 I(I +1) T N 3 13C h 2 B 2 0 I(I +1) T = 3 1H = ( rad/ts) 3 = So, all else 13C ( rad/ts) 3 being equal, HSQC is about 63 times more sensitive than HETCOR. Final Eam: CHEM/BCMB 4190/6190/

13 13). Below is a region of the 1 H, 1 H-COSY spectrum for the ao-sugar whose structure is also shown below. a). Label correctl the diagonal peaks (for instance 3,3 to indicate the diagonal peak for the proton on carbon 3), and label correctl the crosspeaks on one side of the diagonal (for instance, 5,2 for a crosspeak resulting from protons on carbons 5 and 2). Make sure to discriminate between methlene protons if necessar. The 1D spectrum is shown above the COSY spectrum. Use the numbering as shown on the molecular structure below. (10 points). 2,6b 2,6a 2,5 3,6b 3,6a 3,5 5,5 6a,6a 6b,6b 6b,6a 2,4 4,4 5,4 6a,5 6b,5 3,3 4,3 2,2 3,2 b). On the other side of the diagonal, draw additional crosspeaks that ou might epect to be observed in a TOCSY spectrum. Label these additional peaks appropriatel. Also, what does TOCSY stand for? (8 points) Total Correlation Spectroscop Final Eam: CHEM/BCMB 4190/6190/

14 14). Consider the populations N 1, N 2, N 3 and N 4 of the,, and states respectivel for a 1 H- 1 H spin sstem (spins A and X ) without J coupling (no coupling between A and X). The energ diagram for this sstem is depicted (right), where the A s represent the transitions of one of the 1 H nuclei, and the X s represent the transitions of the other 1 H nucleus. We will define H as the difference in the number of spins in and states for A, and H will also represent the difference in the number of spins in and states for X. A X N 2 W 0 A W 2 N 4 X N 1 X A N 3 a. Write down the equilibrium values for N 1 -N 4 and the equilibrium population differences for the A and X transitions (please show how ou calculate the A and X transition population differences). Assume that N 4 = N. (6 points) N 4 = N N 3 = N + H N 2 = N + H N 1 = N + 2H A = N 2 N 4 = N 1 N 3 = H X = N 3 N 4 = N 1 N 2 = H b. On the energ diagram above, draw dashed lines showing the W 0 (ero quantum) and W 2 (double quantum) relaation pathwas. Make sure to clearl, correctl and unambiguousl label each pathwa. (4 points) The W 2 pathwa connects the and states, whereas the W 0 pathwa connects the and states. c. (fill in the blanks) The W 0 and W 2 pathwas do not operate with equal efficiencies for all molecules. For small molecules, the W 2 pathwa is more efficient, whereas for large molecules the W 0 pathwa is more efficient. (2 points) The W 0 and W 2 pathwas do not operate with equal effeciencies with all sies of molecules. For small molecules, the W 2 pathwa is more efficient, whereas for large molecules the W 0 pathwa is more efficient. Final Eam: CHEM/BCMB 4190/6190/

15 d. For a particular medium-sied molecule, the efficiencies of the W 0 and W 2 pathwas are identical. If we perform a simple NOE eperiment b selective saturation of the A transitions, what are the new values for N 1 -N 4 and the population differences for the A and X transitions after saturation but before an relaation takes place? What are the resulting values for N 1 -N 4 and the population differences for the X transitions after relaation via W 0 and W 2 take place (assume that the number of spins () relaing via the W 0 pathwa is equal to the number relaing through the W 2 pathwa)? (12 points ) Before an relaation takes place, the values for N 1 -N 4 and the population differences for the A and X transitions are: N 4 = N + H/2 N 3 = N + 3H/2 A = N 2 N 4 = N 1 N 3 = 0 N 2 = N + H/2 X = N 3 N 4 = N 1 N 2 = H N 1 = N + 3H/2 If we assume that spins rela via W 2 (add to N 1, subtract from N 4 ) and W 0 (add to N 2, subtract from N 3 ) then the new values for N 1 -N 4 and new population differences for X become: N 4 = N + H/2 - N 3 = N + 3H/2 - N 2 = N + H/2 + N 1 = N + 3H/2 + X = N 3 N 4 = N 1 N 2 = H e. Eplain our result to question d in terms of medium-sied molecules in general. (6 points) The result in d suggests that when both W 0 and W 2 are operating with equal efficiencies, the population differences for the X transitions following saturation do not differ from those at equilibrium, i.e. there is no NOE enhancement. Thus, in general, for medium-sied molecules, neither the W 0 nor the W 2 pathwa predominates, but both are operative, and the balance of W 0 and W 2 prevents observation of measurable NOE effects. Final Eam: CHEM/BCMB 4190/6190/

16 15). In a particular molecule, the distance between H a and H b is known to be 4.17 angstroms. In a NOESY spectrum, the intensit of the crosspeak between these hdrogens is 3163 (arbitrar units). In the same NOESY spectrum, the intensit of the crosspeak between H c and H d in the same molecule is What is the distance between H c and H d? (4 points) The intensities of crosspeaks in NOESY spectra are proportional to 1/r 6 where r is the internuclear distance. In this instance, the intensit increased b a factor of 12127/3.163=3.834, and thus r decreased b a factor X where X 6 =3.834, or X=1.25. So 4.17/1.25=3.336, so the H c -H d distance is angstroms. 16). For a small molecule at natural abundance, we observe that the 13 C signal from carbon atom A is 100 units (arbitrar units). We then saturate the 13 C frequenc corresponding to a directl bonded carbon atom B. What will be the new intensit of the 13 C signal of carbon A? (6 points) For NOE enhancements of small molecules, we know that the intensit of the enhanced signal, I will be equal to the original intensit (I 0 ) multiplied b 1+ where is the maimum enhancement (I=(1+)I 0 ). The maimum enhancement is defined as a /2 where a is the gromagnetic ratio of the saturated nucleus and is the gromagnetic ratio of the observed nucleus. In the case of an NOE enhancement of a 13 C nucleus in a small molecule at natural abundance, we have to remember, however, that, statisticall, onl about 1.1% (natural abundance) of directl bonded carbon atoms will be 13 C. Thus, the maimal enhancement in this case is 0.011(1/2)=0.0055, and the new signal intensit will be ( )100 = ). Fill in the blanks with the correct word or words (onl real words, no smbols or acronms): (12 points) a). In the NOESY eperiment, during m, polariation is transferred from one spin (nucleus) to another via cross-relaation (W 0 and W 2 ). b). In two-dimensional NMR eperiments, the evolution period introduces an indirectl detected frequenc dimension. c). For nuclei in a magnetic field, the projection of angular momentum of the nuclei onto the - ais of the laborator frame results in 2I+1 equall spaced energ levels that are referred to as the Zeeman levels or states. d). After a 90 pulse on a population of nuclei in a magnetic field, a fraction of the nuclear spins are bunched together in phase as the precess about the field direction. This is known as phase coherence. e). When the NMR signal is digitied, the dwell time is the time difference between two adjacent digitied points. f). The Ernst angle is the optimum pulse angle for a given repetition time or for a given relaation dela. Final Eam: CHEM/BCMB 4190/6190/

17 18). Consider the HETCOR pulse sequence, and the energ diagram for the simple 1 H- 13 C spin sstem (i.e. CHCl 3 ) A 1 N 4 X 1 N 3 1 H t 1 13 C 90 t 2 N 2 A 2 Just before the second 1 H 90 pulse, for a particular t 1 value, the and vectors happen to be positioned along the and aes, respectivel. X 2 1 H N 1 13 C a). Define the populations for the individual states N 1 -N 4 and the population differences for the A 1, A 2, X 1 and X 2 transitions both before the beginning of the pulse sequence and after the second 1 H 90 pulse (assuming the particular value of t 1 mentioned above, and no relaation losses during t 1 ). Assume that N 4 =N, and H is the difference in the number of spins in and states for 1 H, and X is the difference in the number of spins in and states for 13 C. Also, define the relationship between H an X. (8 points) After the second 1 H pulse, the vector will be inverted from its equilibrium position (- ), whereas the vector will be in the same position as it was at equilibrium (). Thus, the N 2 and N 4 populations are inverted after the second 1 H pulse. Also, the ratio of H and X is the ratio of their gromagnetic ratios, which is ~4: equilibrium (beginning of pulse sequence) N 4 = N N 3 = N + X N 2 = N + H N 1 = N + H + X A 1 = N 2 N 4 = H A 2 = N 1 N 3 = H X 1 = N 3 N 4 = X X 2 = N 1 N 2 = X after second 1 H pulse N 4 = N + H N 3 = N + X N 2 = N N 1 = N + H + X A 1 = N 2 N 4 = -H A 2 = N 1 N 3 = H X 1 = N 3 N 4 = X H = -3X X 2 = N 1 N 2 = X + H = 5X Final Eam: CHEM/BCMB 4190/6190/

18 b). Draw the vectors C and C that are present just before the 13 C pulse. Make sure to label each with the correct name ( C or C ) and with the proper magnitude (epressed in terms of X). (4 points) Ha 5X H -3X c). In the modified HETCOR eperiment (right) an additional pulse ( 13 C 180 ) and an additional dela () have been added in order to remove the 1 J CH splitting in the t 1 dimension (decouple). For the simple 1 H- 13 C spin sstem (i.e. CHCl 3 ), at point a in the sequence, the and vectors are positioned as shown below (in the transverse plane). For an arbitrar value of t 1 /2, show the results at points b - d. Then eplain wh decoupling is achieved. (8 points) 1 H 13 C t 1 /2 t1 /2 t 2 a b c d e a t 1 /2 b 13 C 180 c t 1 /2 d e MH Decoupling is achieved because the 1 J CH coupling is refocused during the t 1 /2-13 C 180t 1 /2 period. At d, the and vectors are refocused, and the net precession to this point for each vector is the same. d). The dela has to be set appropriatel in order for the pulse sequence to operate properl. What value should be set to? Please eplain our answer both b showing the results at e and b eplaining wh the value of that ou selected is appropriate. (6 points) The dela should be set to 1/(2 1 J CH ). This allows the refocused and vectors at d to move apart from one another b 180 so that the are opposite in phase at e. In this wa, the 1 H 90 pulse at e then converts the - and components of these vectors equall to or - magnetiation, respectivel, giving maimum polariation transfer and maimum sensitivit. Final Eam: CHEM/BCMB 4190/6190/

19 19). Consider the spin-echo pulse sequences (right) and the effect of these pulse sequences on heteronuclear J coupling (ignoring chemical shift evolution and magnetic field inhomogeneit.). Consider a 13 C nucleus with three attached protons (i.e. 13 CH 3 Cl) with a Larmor frequenc equal to our reference frequenc, c = rf. Complete the vector diagrams for points c - e to show the effect of the first (top) sequence on this spin sstem with =1/(4J CH ), =1/(2J CH ), =3/(4J CH ), and =1/(J CH ). Label the vectors corresponding to the individual multiplet components with the frequencies as shown below (right). Be sure to show the direction of precession of each component of the multiplet relative to the reference (aes). In addition, show the Fourier transformation of the signal that ou would get for both the top (first) and bottom (second) sequences. (20 points) = angle between C +J CH /2 and C -J CH /2 = angle between C + 3J CH /2 and C - 3J CH /2 a M 0 1 H broadband decoupling 13 C 13 C a b c d e 1 H broadband decoupling a b c d e a 90 b transverse plane c -3 J CH /2 c +3 J CH /2 c -J CH /2 c +J CH /2 transverse =90 plane c -3 J CH /2 =1/(4J CH ) = =270 c +3 J CH /2 =180 c -J CH /2 =1/(2J CH ) =540 c +J CH /2 =270 c -J CH /2 =3/(4J CH ) =810 c +J CH /2 c d c +J CH /2 =270 c +3 J CH /2 c -J CH /2 c -3 J CH /2 c +J CH /2 c -3 J CH /2 c -J CH /2 c -J CH /2 c +J CH /2 180 c -3 J CH /2 c +3 J CH /2 c +3 J CH /2 c +3 J CH /2 c -3 J CH /2 c -J CH /2 180 c -3 J CH /2 c +3 J CH /2 c +J CH /2 e decoupler on FT FT FT first pulse sequence second pulse sequence =360 =1/(J CH ) c -3 J CH /2 c -J CH /2 c +J CH /2 =1080 c +3 J CH / Final Eam: CHEM/BCMB 4190/6190/ c -3 J CH /2 c -J CH /2 c +J CH /2 c +3 J CH /2 FT

20 20). The upper pulse sequence (right) is used to measure J-couplings (heteronuclear J-resolved spectroscop) in a two-dimensional manner. The lower pulse sequence is not suitable for this purpose. For the questions below, consider a simple 13 C- 1 H sstem ( 13 CHCl 3 ) and ignore the effects of magnetic field inhomogeneit. 1 H 13 C 1 H 13 C bb decoupling bb decoupling bb decoupling t 1 /2 t 1 /2 a b c d t 1 /2 t 1 /2 a b c d t 2 bb decoupling t 2 a). Using vector diagrams, show how the first (upper) pulse sequence is useful for measuring 1 J CH coupling constants. Make sure to write an accompaning eplanation. (6 points) M 0 90 a t1/2 transverse plane b H = H c c/d 180 t 1 /2 = H H decouple The angle that separates the H and H vectors for a given value of t 1 /2 depends on the magnitude of 1 J CH (in the vector diagram it is the arbitrar angle, determined b 1 J CH for this particular value of t 1 /2). Thus, as t 1 /2 changes, the final signal intensit is modulated b 1 J CH (in the eample this particular value of t 1 /2 results in a small signal for the value of 1 J CH represented). Because both H and H are allowed to evolve (are not decoupled) during the first of t 1 (first t 1 /2 period), both of these frequencies will be represented in the Fourier transformation of t 1. Because the evolve onl during of t 1, the coupling (frequenc difference between the signals from H and H ) observed will be of its actual value. Final Eam: CHEM/BCMB 4190/6190/

21 b). Wh will the lower pulse sequence not be useful for measuring 1 J CH coupling constants. Please provide an eplanation along with the appropriate vector diagrams. (6 points) M 0 90 a t1/2 transverse plane b H = H c 180 = t 1 /2 H H d In this case, regardless of the value of 1 J CH, the vectors will be refocused at d. Thus, this pulse sequence will refocus the heteronuclear J coupling and the signal will not be modulated b 1 J CH. c). You can modif the lower pulse sequence so that it will be useful for measuring 1 J CH. Show how ou would modif it, and show with vector diagrams and an eplanation wh it now is useful (ou will not receive credit if the modifications simpl convert it to the upper sequence). (10 points) H bb decoupling bb decoupling 13 C 90 t 1 /2 180 t 1 /2 t 2 a t1/2 transverse plane b H = H 13 C 180 c H = 1 H 180 H a b c d One modification is to add a 1 H 180 pulse simultaneousl with the 13 C 180 pulse. As shown in the vector diagrams, this then causes the H and H vectors to continue to move apart from each other according to 1 J CH during the entire t 1 period, and the signal intensit is modulated b 1 J CH. Thus, the frequencies of both H and H will be observed in the Fourier transformed signal and the frequenc difference between them will be 1 J CH. c H H t1/2 H H d = 2 Final Eam: CHEM/BCMB 4190/6190/

22 You ma find some of the information in this table useful: Final Eam: CHEM/BCMB 4190/6190/

23 You ma find some of the following information or equations useful: B 0 (Tesla, T) Resonance frequencies (MH) 1 H 13 C H = rad/t/s, I = 1/2 13C = rad/t/s, I = 1/2 10B = rad/t/s, I = 3 15N = rad/t/s, I = 1/2 11B = rad/t/s, I = 3/2 17O = rad/t/s, I = 5/2 M 0 = N 2 h 2 B 0 I(I +1) T dm/dt = M 0 B= N 3 h 2 B 0 2 I(I +1) T S/N N/N 1/2 = N 1/2 (signal-to-noise improves with (number of scans) 1/2 ) m = (-I, -I+1,, I-1, I) E = μ Z B 0 = mhb 0 E = μ Z B 0 = hb 0 L = /(2) B 0 SW=1/(2DW)=Nquist frequenc ( NQ ) AQ=DW*TD DR=2SW/TD (TDNP) = observe frequenc 106 M = M 0 e t /T 1 M = M 0 (1 e t /T 1 ) M = M 0 (1 2e t /T 1 ) 1/2 = 1 T 2 * 1 T 2 * = B T 2 t ero =T 1 ln(2) = a / (2 ) I = (1 + ) I 0 I 1/r 6 =2J Final Eam: CHEM/BCMB 4190/6190/