Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam:...

20
Name _________________________ Final Exam: CHEM/BCMB 4190/6190/8189 1 Final Exam: CHEM/BCMB 4190/6190/8189 (290 pts) Tuesday, 16 December, 2003 1). (15 pts) For ‘a’ and ‘b’ below, indicate relative intensities of signals in spectra, and indicate coupling constants (in terms of J CH ) where appropriate. Mark the Larmor frequencies on your spectra with ν H and ν C . Identify individual transitions (i.e. M C Hα , etc.) a. Sketch the 1D, 1 H NMR spectrum of chloroform, CHCl 3 (no decoupling). b. Sketch the 1D, 13 C NMR spectrum of chloroform (no decoupling). c. If you were going to perform an SPI experiment to enhance the intensities in the 13 C NMR spectrum, indicate (circle) one transition in one of the signals above that you might excite to accomplish this. In the SPI experiment, either the M H Cα or M H Cβ transition in the 1 H NMR signal (but not both!) could be excited to enhance the intensities of the components of the 13 C signal. I’ve arbitrarily circled one of them. J CH ν H 1 H- 12 C- M H Cβ M H Cα M C Hα M C Hβ J CH ν C

Transcript of Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam:...

Page 1: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

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

Final Exam: CHEM/BCMB 4190/6190/8189 (290 pts) Tuesday, 16 December, 2003

1). (15 pts) For ‘a’ and ‘b’ below, indicate relative intensities of signals in spectra, and indicatecoupling constants (in terms of JCH) where appropriate. Mark the Larmor frequencies on yourspectra with νH and νC. Identify individual transitions (i.e. MC

Hα, etc.)

a. Sketch the 1D, 1H NMR spectrum of chloroform, CHCl3 (no decoupling).

b. Sketch the 1D, 13C NMR spectrum of chloroform (no decoupling).

c. If you were going to perform an SPI experiment to enhance the intensities in the 13C NMRspectrum, indicate (circle) one transition in one of the signals above that you might excite toaccomplish this.

In the SPI experiment, either the MHCα or MH

Cβ transition in the 1H NMR signal (but not both!)could be excited to enhance the intensities of the components of the 13C signal. I’ve arbitrarilycircled one of them.

JCH

νH

1H-12C-

MHCβMH

MCHα MC

HβJCH

νC

Page 2: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 2

2). (20 pts) In the DEPT experiment, the phase angle (pulse width) of the third 1H pulse(applied along the ‘y’ axis) can be set to any value, in order to achieve the desired result. Shownin the figure below are the intensities of signals from CH, CH2, and CH3 groups as a function ofthe phase angle Θy:

a. The DEPT(135) spectrum for isopentyl acetate is shown below. Draw the DEPT(45)-DEPT(135) spectrum for isopentyl acetate. Make sure to identify (label) the signals in the samemanner that they are labeled in the DEPT(135) spectrum. Write a short explanation of why theDEPT(45)-DEPT(135) spectrum looks as it does.

With the phase angle Θy set at 45°, the intensities of –CH and –CH3 signals (DEPT(45)) arealmost identical to their intensities when Θy set at 135° (DEPT(135)). Thus in the DEPT(45)-DEPT(135) spectrum, signals from these groups will be absent. The intensities of signals from–CH2 groups in the DEPT(45) spectrum are very large and positive, whereas signals from –CH2

groups in DEPT(135) spectra are very large and negative. Subtracting the latter from theformer results in very large positive signals (DEPT(45)-DEPT(135)).

E D

O ||C1H3-C

2-O-C3H2-C4H2-C

5H-C6H3

| C7H3

isopentyl acetate

AC

B

DE

80 60 40 20 0 13C, δ (ppm)

DEPT(135)

80 60 40 20 0 13C, δ (ppm)

DEPT(45)-DEPT(135)

Page 3: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 3

b. Indicate in the table below whether each signal (A-E) arises from a –CH3, -CH2, -CH, orquaternary carbon by placing an ‘X’ in each row in the appropriate column.

signal -CH3 -CH2 -CH quaternaryA (21 ppm) XB (23 ppm) XC (25 ppm) XD (35 ppm) XE (65 ppm) X

c. In the table below, assign each signal in the DEPT(135) spectrum of isopentyl acetate byplacing the appropriate integer (1-7) in the ‘assignment’ column, corresponding to the correctcarbon atom in the structure of isopentyl acetate shown above. Give a concise but thoroughrationale for the assignment.

signal assignment rationale

A (21 ppm) 1

Signals from E and D must be from –CH2 groups (downin DEPT(135)). C must be from –CH (very small and upin DEPT(135)). A and B must be from –CH3. B is twiceas large as A, so must be due to equivalent –CH3 groups6 & 7. Thus, A is from –CH3 group 1.

B (23 ppm) 6 & 7

Signals from A and B are due to –CH3 groups (seeabove). B is twice as large a A, so B must be due toequivalent –CH3 groups 6 & 7

C (25 ppm) 5

C must be from –CH (very small and up in DEPT(135)).5 is only –CH in molecule

D (35 ppm) 4

Signals from E and D must be from –CH2 groups (downin DEPT(135)). 4 will be upfield from 3 because 3 nextto very electronegative group. Thus, signal D is from 4.

E (65 ppm) 3

Signals from E and D must be from –CH2 groups (downin DEPT(135)). 3 will be downfield from 4 because 3 isnext to a very electronegative group. Thus, signal E isfrom 3.

Page 4: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 4

3). (20 pts) Consider the effect of the normal spin-echopulse sequence (right, top) and the effect of this pulsesequence on heteronuclear and homonuclear J coupling,ignoring the effects of chemical shift evolution andmagnetic field inhomogeneity. Consider a simpleheteronuclear spin system (1H-13C, i.e. CHCl3) with aLarmor frequency equal to our reference frequency (νc = νrf), and a simple 1H-1H homonuclearspin system (A-X) with a Larmor frequency equal to our reference frequency (νH = νrf).

a. Using vector diagrams, show the effect of the spin echo pulse sequence on the heteronuclearspin system with τ=1/(4JCH). Discuss the effect of the spin-echo pulse sequence on heteronuclearJ coupling refocusing.

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

b. Using vector diagrams, show the effect of the spin echo pulse sequence on the homonuclearspin system with τ=1/(4JAX). Discuss the effect of the spin-echo pulse sequence on homonuclearJ coupling refocusing.

The 180° pulse reflects the vectors through x. This is in effect a pulse on the ‘A’ spin. The 180°pulse also exchanges the populations of the α and β states of the ‘X’ spin, i.e. this is the effect ofthe pulse on the ‘X’ spin. Both operations cause the vectors to move away from one anotherduring the second τ period, becoming antiphase to one another at ‘e’. Thus, this pulse sequencedoes not promote refocusing of homonuclear J coupling.

τ τ

a b c d e

180°x90°x

transverseplane

M0

y

z

x

90°x

yx

τMC

MCHβx

Θ = 90°y

MCHβ

MCHα

Θ = 90°x

y

180°x τy

z

x

a b c d e

MAXα

MAXΒ

transverseplane

M0

y

z

x

90°x

yx

τMA

MAXβx

Θ = 90°y

MAXα

MAXβ

Θ = 90°x

y

180°x τy

z

x

a b c d e

Page 5: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 5

4). (15 pts) The following diagram represents the pulse sequence used to measure T1 for 13Cnuclei:

At point ‘a’, just before the first (180°x) pulse, the magnitude of thelongitudinal component of the bulk 13C magnetization (Mz) is equal to themagnitude of the equilibrium magnitization (M0), as shown in the vectordiagram at the right.

a. For some value of τ, the magnitude of the longitudinal component of thebulk magnitization is zero (Mz=0) at point ‘c’, and the Fourier transform of the FID reveals nosignal. If T1 is 10 seconds, at what value of τ does this occur?

We assume, as did Bloch, that this relaxation process is first order, and Mz = M0(1− 2e− t / T1 ) .When Mz=0;0 = M0(1− 2e−τ / T1 ) 0 = M0 − 2M0e

−τ / T1 M0 = 2M0e−τ / T1 1= 2e−τ / T1 ln(1/2) = −τ /T1 T1 ln(2) = τ

b. You find that when τ is 10s, there is no signal at point ‘d’. What is T1?

From your answer to question ‘a’, you know that no signal is observed when T1ln(2) = τ. If τ =10 s, then T1 = 10 s/0.693 = 14.43 s.

c. In order to measure T1 properly, we measure the amplitude of the signal (Fourier transform ofthe FID) as a function of τ and fit the data to the appropriate first order equation. We shouldwait for a time equal to at least 5T1 between successive experiments in order to allow for re-equilibration of the bulk magnetization. If T1 is 10 s, and if we wait for 5T1 betweenexperiments, how complete is the re-equilibration (assume τ=0, the acquisition time=0 and the90° pulse width is also=0)?

For τ=0, from question ‘a’ we know that at point ‘d’ the vector diagram to the right describesthe bulk magnetization. In this case, the equation that governs the return to equilibriumis Mz = M0(1− e−t / T1 ). For t=5T1=50 s,Mz

M0

=1− e−50 /10 =1− 0.006737947 = 0.9933.

So, after 5T1, the equilibration process is > 99% complete

ba cd

τ

90°x180°x

broadband decoupling1H

13CFID

Mz=M0

y

z

x

Page 6: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 6

5). (30 pts) Consider the normal spin-echo pulsesequence (right) and the effect of this pulse sequence onheteronuclear J coupling, ignoring the effects of chemicalshift evolution and magnetic field inhomogeneity. We willconsider a single 13C nucleus with three attached protons(i.e. 13CH3Cl) with a Larmor frequency equal to ourreference frequency, νc=νrf.

a. Sketch the normal (coupled) 13C signal (multiplet) for this nucleus.

b. What are the theoretical relative intensities of the individual components of the multiplet?

1:3:3:1

c. In terms of the Larmor frequency, νc, and the coupling constant, JCH, what will be thefrequencies of each of the components of the multiplet?

νc – 3JCH/2, νc – JCH/2, νc + JCH/2, νc + 3JCH/2

d. What is the frequency difference between the outermost components of the multiplet? Whatis the frequency difference between the innermost components of the multiplet?

From ‘c’, the frequency difference between the outermost components is 3JCH, and the frequencydifference between the innermost components is JCH.

e. How large is the phase angle (degrees) between the two vectors corresponding to the twooutermost components of the multiplet after a time τ=1/4(JCH)? How large is the phase anglebetween the two vectors corresponding to the two innermost components of the multiplet after atime τ=1/4(JCH)? Please show your work.

outermost: phase angle, Θ = 2π(3JCH)τ = 2π(3JCH)/4(JCH) = 3π/2 radians = 270°innermost: phase angle, Θ = 2π(JCH)τ = 2π(JCH)/4(JCH) = π/2 radians = 90°

τ τ

a b c d e

180°x90°x

Page 7: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 7

f. Using vector diagrams, show the effect of the spin echo pulse sequence on this spin systemwith τ=1/(4JCH). Show diagrams corresponding to points ‘a’-‘e’ in the pulse sequence. Label thevectors corresponding to the individual multiplet components with the frequencies from youranswer to ‘c’. Be sure to label your axes. Be sure to show the direction of precession of eachcomponent of the multiplet relative to the reference (axes).

νc-3 JCH/2

νc+3 JCH/2

νc+JCH/2

νc-JCH/2

ztransverseplane

M0

y

z

x

yx

τ=1/(4JCΗ)y

x

90°x≡

180°x τ=1/(4JCΗ)

x

yνc+3 JCH/2

νc-3 JCH/2

νc+JCH/2

νc-JCH/2

x

y

‘a’ ‘b’ ‘c’

‘d’ ‘e’

‘a’

270°

90°

νc+3 JCH/2

νc-3 JCH/2

νc+JCH/2

νc-JCH/2

x

y

Page 8: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 8

6). (20 pts) Repeat part ‘f’ of the previous questionfor τ = 1/(2JCH), τ = 3/(4JCH), and τ = 1/(JCH). Inaddition, show the Fourier transformation of thesignal that you would get (for τ = 1/(4JCH), τ =1/(2JCH), τ = 3/(4JCH), and τ = 1/(JCH)) after thesecond τ period if you applied broadband 1Hdecoupling during acquisition only (as shown in thepulse sequence to the upper right). Finally, alsoshow the Fourier transformation of the signal thatyou would get if instead you used the second pulsesequence (lower right) where the broadbanddecoupling is applied immediately after the 180°pulse.

Θ = angle between νC+JCH/2 and νC-JCH/2Ω = angle between νC+ 3JCH/2 and νC- 3JCH/2

νc+JCH/2

νc-3 JCH/2

νc-JCH/2

νc-JCH/2

νc-JCH/2νc-3 JCH/2

νc+3 JCH/2νc+JCH/2

νc+3 JCH/2νc+JCH/2

transverseplane

νc-3 JCH/2νc+JCH/2

νc-JCH/2

x

13C

1H broadbanddecoupling

τ τ

a b c d e

180°x90°x

13C

1H broadband decoupling

τ τ

a b c d e

180°x90°x

τ=1/(4JCH) 180°x τy

z

x

τ=1/(2JCH)y

180°x τy

z

x

τ=3/(4JCH)

x

y

xy

180°x τy

z

x

τ=1/(JCH)

x

y

x-y

180°x τy

z

x

FT

FT

FT

FT

second pulsesequence

‘c’ ‘d’ ‘e’ first pulsesequence

Ω=270°

Θ=90°

νc+3 JCH/2

νc-3 JCH/2

νc+JCH/2

νc-JCH/2

x

yνc+3 JCH/2

νc-3 JCH/2

νc+JCH/2

νc-JCH/2

x

y

νc+3 JCH/2

νc-3 JCH/2νc+JCH/2

νc-JCH/2

x

y

νc+3 JCH/2

νc-3 JCH/2

νc+3 JCH/2

νc+JCH/2

νc-JCH/2

Θ=90°

Ω=270°

Θ=180°

Ω=540°

Θ=270°

Ω=810°

Θ=360°

Ω=1080°

νc-3 JCH/2

νc+3 JCH/2

Page 9: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 9

7). (10 pts) In the Friebolin text (page 95), he discusses H, H couplings in benzene compounds:

“In benzene and its derivatives, the ortho, meta, and para couplings are different,and by analyzing the aromatic region of the proton spectrum one can thereforedetermine the arrangement of the substituents. It is often possible to analyze thespectrum by first-order methods, particularly if the spectra have beenrecorded at high resonance frequency”.

Knowing that the chemical shift range of the (coupled) ring protons of benzene and derivatives issmall, please explain in detail the highlighted (bold) sentence.

If the frequency difference, ∆ν, between coupled nuclei is large compared with the couplingconstants, (i.e. if ∆ν >> J), then the spectra/signals are considered to be first order. As a result,they can usually be analyzed by first order methods and usually display first ordercharacteristics. Given the dependence of frequency on magnetic field strength, B0 (ν = γB0/(2π)),the frequency difference, ∆ν, between the ring protons in benzene and derivatives can beincreased by increasing the resonance frequency, i.e. by increasing the magnetic field strength.This of course has no effect on J, thus the difference between ∆ν and J is increased as is the first-order character.

Page 10: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 10

8). (20 pts) The COSY spectrum for menthol is shown (next page), along with the 1D 1Hspectrum of menthol, the structure of menthol and the correct 1H assignments (next page).Circled in the COSY spectrum is a crosspeak (and the symmetry related crosspeak) that remainsunassigned. Scheme IV (next page) shows coupling constants for the indicated 1H pairs incompounds 15-18.

a. Please give a concise explanation of Scheme IV. What is the point that this figure/scheme istrying to make? What does this have to do with COSY?

1H-1H couplings through more than 3 bonds (H-C-C-H) are normally very small. This is thereason that crosspeaks for protons more than 3 bonds away from one another normally do notappear in COSY spectra. However, as shown in Scheme IV, in some cases, couplings through 4bonds are substantial (1Hz or more). For instance, when the four bonds linking the nuclei forma fixed “W” conformation, the couplings can be pretty large (up to 4 Hz). In these cases,crosspeaks for coupled protons can be observed in COSY spectra.

b. Assign the circled crosspeak(s) and justify your assignment(s).

The chemical shifts of the crosspeaks indicate that one of the ‘8’ and one of the ‘7’ protons areinvolved. Given Scheme IV, which suggests that a substantial coupling should be observedbetween the equatorial positions across the ring, the best assignment for the crosspeak would bethat it arises from the equatorial protons at “8” and “7”.

Page 11: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 11

7

5 810 6

4

solvent

9

2

3

1

748

Page 12: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 12

9). (30 pts) The diagram for the INEPT pulse sequence is shown below.

Consider the effect of this pulse sequence on heteronuclear (-13C-1H) J coupling, ignoring theeffects of chemical shift evolution and magnetic field inhomogeneity. We will consider a simple1H-13C- spin system (i.e. 13CHCl3) with a Larmor frequency equal to our reference frequency,νH=νrf. We will assume that the delay τ is equal to 1/(4JCH).

a. What does INEPT stand for?

Insensitive Nuclei Enhanced by Polarization Transfer

b. Complete the vector diagrams below for points ‘c’, ‘d’, ‘e’, and ‘f’ in the pulse sequence. Besure to label the vectors (MH

Cα, MHCβ), to include arrows indicating the direction of precession for

the vectors in the rotating frame, and to indicate the angle (Θ) between the vectors at each point.

e

τ τ

g′h

a b c d f g

90x 180x 90y

180x 90x

1H

13C

Θ = 0°

180°x 1H

MHCβ

MHCα

MHCβ

MHCα

transverseplane

y

z

x

90°x

y

x

τMH

MHCβx

Θ = 90°y

cba

d

MHCβ

MHCα

Θ = 90°x

y

e

180°x 13C

MHCα

MHCβ

Θ = 90°x

y

τ

f

MHCα

MHCβ

Θ = 180°

x

y

Page 13: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 13

c. Explain the consequences of changing the phase of the 13C 180°x pulse to y.

The 180° pulse on 13C serves to exchange the 13C α and β spin populations. In this case,the axis (x or y) about which it is applied is (for the most part) irrelevant.

d. Explain the consequences of changing τ from 1/(4JCH) to 1/(2JCH) (for the simple 1H-13C- spinsystem). Your answer should include vector diagrams for points ‘c’ through ‘f’ for τ = 1/(2JCH)and an explanation of the results.

For the simple 1H-13C-system, the phase angle (Θ) between the two vector components MHCα and

MHCβ is equal to 2πJCHτ since the frequency difference between the two components is JCH. When

τ = 1/(4JCH), Θ = 2πJCH/(4JCH) = π/2 radians or 90°. If τ = 1/(2JCH), Θ = 2πJCH/(2JCH) = πradians or 180°. Thus from ‘b’ to ‘c’ (τ), the vectors rotate 180° apart from one another ratherthan 90°. The 1H 180°x pulse has no effect on the vectors, and the 13C pulse exchanges 13Cα and13Cβ spin populations and the direction of rotation. During the last τ period, the vectors thusmove 180° towards each other and are refocused at ‘f’.

MHCα

Θ = 180°Θ = 180°

180°x 13C

MHCβ

MHCα

x

y

180°x 1H

d e

MHCβ

x

y

τ

f

MHCα

MHCβ

x

y

τ

MHCα

MHCβ

x Θ = 180°y

c

Page 14: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 14

10). (50 pts) Again consider the INEPT pulsesequence, and the simple 1H-13C spin system (i.e.CHCl3). At point ‘g’ in the pulse sequence, theMH

Cα and MHCβ components

appear as shown below. Theenergy diagram for this systemis depicted (far right), where A1

and A2 are the 1H transitions,and X1 and X2 are the 13Ctransitions. We define ∆H asthe difference in the number ofspins in α and β states for 1H,and ∆X as the difference in thenumber of spins in α and β states for 13C.

a. How is ∆H related to ∆X? Quantitatively, what is their ratio, and how is this ratio derived?

The ratio ∆H/∆X is proportional to the ratio of the gyromagnetic ratios for 1H and 13C.γ1H/γ13C = 26.7519/6.7283 = 3.98 ≈ 4, so ∆H = 4∆X.

b. Which transition corresponds to the MHCα vector: A1, A2, X1 or X2?

As indicated in the above diagram (above, right), the first spin in the pairs correspondsto the 1H spin and the second corresponds to the 13C spin. Thus, as shown in the diagram, the 1Htransition (MH) with the 13C spin in the α state (Cα) is the A2 transition (αα → βα).

c. Complete the table below where N1-N4 are the populations of the spin states in the abovediagram, and A1, A2, X1 and X2 are the population differences for the A1, A2, X1 and X2

transitions respectively. Assume that N4 = N

at equilibrium at ‘g’N4 = N N4 = N + ∆HN3 = N + ∆X N3 = N + ∆XN2 = N + ∆H N2 = NN1 = N + ∆H + ∆X N1 = N + ∆H + ∆XA1 = N2 – N4 = ∆H A1 = N2 – N4 = -∆HA2 = N1 – N3 = ∆H A2 = N1 – N3 = ∆HX1 = N3 – N4 = ∆X X1 = N3 – N4 = ∆X - ∆H = -3∆XX2 = N1 – N2 = ∆X X2 = N1 – N2 = ∆X + ∆H = 5∆X

MHCβ

MHCα

y

z

x

g

13C1H

N2

N1

N3

N4 X1

αβ

βα

αα

ββ

X2

A1

A2

Page 15: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 15

d. The vector diagram corresponding to point g’ in the INEPT pulse sequence (see question 4) isshown below (right). The magnitudes of the 13C vectors are not drawn to scale, and the correctmagnitudes should be available to you from your table above. Label each vector properly withthe following information: MC

Hα or MCHβ, X1 or X2 (transition) and magnitude (in terms of ∆X).

e. At point g’ in the INEPT pulse sequence, a 13C 90°x pulse is applied to create transverse 13Cmagnetization that subsequently is recorded (the FID, which we will call FID ‘A’) and Fouriertransformed. Two signals are observed, corresponding to MC

Hα and MCHβ, of opposite phase,

whose magnitudes are indicated in your table in ‘c’ and in your answer to ‘d’ above. Now, if thephase of the last 1H 90°y pulse (point ‘f’) is changed to –y, and the FID (which we will call FID‘B’) is added to the FID from the first experiment (FID ‘A’ + FID ‘B’), after Fouriertransformation of the sum of the two FIDs, what will the signal look like (what will be themagnitude and phase of each component, how does this compare to the normal 13Cspectrum/signal)? Please be sure to show your work. What happens if we instead subtract FID‘B’ and FID ‘A’?

Changing the phase of the last 1H pulse to –y will result in opposite phases forthe vectors at ‘g’ with respect to the case where the pulse is y (see figure atright), and will result in the populations and population differences shownbelow:

N4 = N A1 = N2 – N4 = ∆HN3 = N + ∆H + ∆X A2 = N1 – N3 = -∆HN2 = N + ∆H X1 = N3 – N4 = ∆X + ∆H = 5∆XN1 = N + ∆X X2 = N1 – N2 = ∆X - ∆H = -3∆X

Thus, if we add the components:MH

Cα (X2) ‘y’ + MHCα (X2) ‘-y’ = 5∆X + (-3∆X) = 2∆X

MHCβ (X1) ‘y’ + MH

Cβ (X1) ‘-y’ = (-3∆X) + 5∆X = 2∆XThe signal will look like the normal 13C signal: a normal signal would show MH

Cα and MHCβ to be

of the same phase and each of magnitude ∆X. In this case, each signal is 2∆X, but we added twosignals together. Thus, we have achieved absolutely nothing by adding the signals.

However, if we subtract the components:MH

Cα (X2) ‘y’ - MHCα (X2) ‘-y’ = 5∆X - (-3∆X) = 8∆X

MHCβ (X1) ‘y’ - MH

Cβ (X1) ‘-y’ = (-3∆X) - 5∆X = -8∆XThe signal will show MH

Cα and MHCβ to be of opposite phase but each is eight times as large as

the normal 13C signal. Since we have used 2 FIDs, we have essentially enhanced the signalseach equally by a factor of 4.

MCHβ, X1, -3∆X

MCHα, X2, 5∆X

y

z

x

g’

MHCα

MHCβ

y

z

x

g

Page 16: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 16

11). (30 pts) Consider the populations N1,N2, N3 and N4 of the αα, αβ, βα and ββ statesrespectively for a 1H-1H spin system (spins‘A’ and ‘X’) without J coupling (no couplingbetween A and X). The energy diagram forthis system is depicted (right), where the A’srepresent the transitions of one of the 1Hnuclei, and the X’s represent the transitions ofthe other 1H nucleus. We will define ∆H asthe difference in the number of spins in α andβ states for A, and ∆H will also represent thedifference in the number of spins in α and βstates 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.

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

b. On the energy diagram above, draw dashed lines showing the W0 (zero quantum) and W2

(double quantum) relaxation pathways. Make sure to clearly, correctly and unambiguously labeleach pathway.

The W2 pathway connects the ββ and αα states, whereas the W0 pathway connects the αβ andβα states.

c. (fill in the blanks) The W0 and W2 pathways do not operate with equal efficiencies for allmolecules. For small molecules, the _____ pathway is more efficient, whereas for largemolecules the _____ pathway is more efficient.

The W0 and W2 pathways do not operate with equal effeciencies with all sizes of molecules. Forsmall molecules, the W2 pathway is more efficient, whereas for large molecules the W0 pathwayis more efficient.

W2

W0

XA

N2

N1

N3

N4 X

αβ

βα

αα

ββ

X

A

A

Page 17: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 17

d. For a particular medium-sized molecule, the efficiencies of the W0 and W2 pathways areidentical. If we perform a simple NOE experiment by selective saturation of the A transitions,what are the new values for N1-N4 and the population differences for the A and X transitionsafter saturation but before any relaxation takes place? What are the resulting values for N1-N4

and the population differences for the X transitions after relaxation via W0 and W2 take place(assume that the number of spins (δ) relaxing via the W0 pathway is equal to the number relaxingthrough the W2 pathway)?

Before any relaxation takes place, the values for N1-N4 and the population differences for the Aand 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

If we assume that δ spins relax via W2 (add δ to N1, subtract δ from N4) and W0 (add δ to N2,subtract δ from N3) then the new values for N1-N4 and new population differences for X become:

N4 = N + ∆H/2 - δN3 = N + 3∆H/2 - δN2 = N + ∆H/2 + δ X = N3 – N4 = N1 – N2 = ∆HN1 = N + 3∆H/2 + δ

e. Explain your result to question ‘d’ in terms of medium-sized molecules in general.

The result in ‘d’ suggests that when both W0 and W2 are operating with equal effeciencies, thepopulation differences for the X transitions following saturation do not differ from those atequilibrium, i.e. there is no NOE enhancement. Thus, in general, for medium-sized molecules,neither the W0 nor the W2 pathway predominates, but both are operative, and the balance of W0

and W2 prevents observation of measurable NOE effects.

Page 18: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 18

12). (30 pts) The HSQC pulse sequence is shown below.

a. The magnetization vectors present at point ‘h’ in the pulse sequence are shown, as are thevectors at point ‘i’ (following the first t1/2 period), for a simple spin system. The vector MC

precesses slower than the reference frequency, and MCHβ precesses faster than the reference

frequency, such that after t1/2, the vectors have moved ψ and φ degrees, respectively, away fromthe y axis. Complete the vector diagrams for points ‘j’ and ‘k’ below. Be complete.

b. Based on your knowledge of the HSQC pulse sequence and method, and based on youranswer to ‘a’ above, what is the purpose of the 1H 180°x pulse in the center of the t1 evolutionperiod? How does your answer to ‘a’ support your contention?

The 180° pulse in the middle of the t1 evolution period prevents net 1H-13C coupling evolution.As shown in the answer to ‘a’, both at the beginning (‘h’) and at the end (‘k’) of t1, the 13Cmagnetization components are 180° out of phase with one another, with no net phase changebetween them occuring during t1. Thus, the 180° pulse decouples 1H from 13C during t1.

i j

kh

τ

180x 90y

180x 90x

1H

13Cdecouple

90x 180x 180

18090

τ τ τ

t1/2 t1/2

90

φ

ψ

ψφ 180x

MCHβ

t1/2MC

y

x

MCHα

‘h’

y

x

MCHα

‘i’MC

y

x

MCHα

‘j’

MCHα

y

x

MCHβ

‘k’

t1/2

Page 19: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 19

c. Let’s compare the sensitivity of INEPT and HSQC. If, at point ‘h’ in HSQC, we simply detectthe 13C magnetiztion, this is essentially the INEPT experiment, and, as we know, the signals from13C nuclei are enhanced substantially by INEPT. On the other hand, if the magnetization from‘h’ is subjected to a reverse INEPT (this is HSQC) and the 1H magnetization is detected, thesignal is much more intense compared to INEPT. Assuming no additional magnetization lossesassociated with the reverse INEPT period, calculate the ratio of the intensity of HSQC to INEPT.

Sensitivity is proportional to the electromotive force (ε) induced in the receiver coil by the bulkmagnetic moment. The magnitude of ε is proportional to the rate of change in the magneticmoment: ε ∝dM/dt = γM0B

Remember, M0 =Nγ 2h2B0I(I +1)

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

so ε ∝γM0B =Nγ 3h2B0

2I(I +1)

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

dependent on γ2, the sensitivity is dependent on γ3.

The main difference between HSQC and HETCOR in this respect is that the magnetization isdetected on 1H in HSQC and on 13C in HETCOR.

Thus, ε1H

ε13C

=

Nγ1H3h2B0

2I(I +1)

3kBT

Nγ13C3h2B0

2I(I +1)

3kBT

=γ1H

3

γ13C3 =

(26.7519 ×107 rad/Ts)3

(6.7283×107 rad/Ts)3 = 62.856 So, all else being

equal, HSQC is about 63 times more sensitive than HETCOR.

Page 20: Nametesla.ccrc.uga.edu/courses/4190_03/problem_sets/finalexamanswers1.pdfName _____ Final Exam: CHEM/BCMB 4190/6190/8189 6 5). (30 pts) Consider the normal spin-echo pulsesequence

Name _________________________

Final Exam: CHEM/BCMB 4190/6190/8189 20

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

γ1H = 26.7519 x 107 rad/Ts, I = 1/2 γ15N = -2.7126 x 107 rad/Ts, I = 1/2γ10B = 2.8747 x 107 rad/Ts, I = 3 γ17O = -3.6280 x 107 rad/Ts, I = 5/2γ13C = 6.7283 x 107 rad/Ts, I = 1/2

M0 =Nγ 2h2B0I(I +1)

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

ε ∝dM/dt = γM0B =Nγ 3h2B0

2I(I +1)

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

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

m = (-I, -I+1, …, I-1, I)

η = γa / (2γx)

I = (1 + η) I0

ν = γB0/(2π) f ∝ 1/r6

∆δ =∆ν

observe frequency×106 I ∝ 1/r6

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

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 = -mγhB0

∆E = µzB0 = γhB0