Chapter 14: NMR

Problem 14.1: From the mass spectral and elemental analysis data, you have determined the formula of

an unknown compound from a petroleum distillate to be C4H8O. You have limited the possibilities to the

four different structures shown below, which all have the formula C4H8O.

(a) Describe how an IR spectrum would aid in narrowing down your choices. Be specific.

From the IR, we could either confirm or exclude the first two (carbonyl) by the presence or absence of a

peak near 1700 cm-1. Likewise we might be able to infer the presence or absence of the third molecule

(alkene) by the presence or absence of a peak in the range of 1620 – 1680 cm-1.

(b) Use the hydrogen NMR spectrum of your unknown (shown) to determine which of these isomers you

have.

The simplicity of the NMR spectrum, having a well resolved triplet, a quartet and a singlet, leads us to

look for three non-identical carbons that are attached to hydrogens. Because the compound is known to

have four carbons, we are also looking for a carbon in the compound that has no hydrogens bonded to it.

Quickly we can survey the structures to see that the only compound that meets those criteria is the first

compound, the ketone. The second compound, the aldehyde, had four unique carbons with hydrogens

attached to all four; likewise does the third compound (the ether). The cyclical compound, because of

symmetry, as only two unique carbons. Thus the first compound, 2-butanone, is correct.

Determine if the nuclear spin (I) for each of the following nuclides is zero, an integer, or a multiple of ½. (See Example 14.1)

(a) 2H

(b) 3H

(c) 12C

(c) 13C

(e) 15N

(f) 31P

To determine I, the nuclear spin, we need to employ the rules as described section 14.2, which are based on the numbers of protons and neutrons.

a. Deuterium, 2H, has atomic number 1 and therefore has 1 proton. It has atomic mass of 2, and therefore has a total of 2 protons plus neutrons; and therefore has 1 neutron (2-1=1). Rule 3 states that “if the number of protons is odd (yes, in this case) and the number of neutrons is odd (yes, in this case), then I =1 or an integer value of 1 such as 2,

3, and so on.” Tritium, 3H, has atomic number 1 and therefore has 1 proton. It has atomic mass of 3,

and therefore has a total of 3 protons plus neutrons; and therefore has 2 neutrons (3-1=2). Rule 2 states that “if the number of protons plus neutrons is odd (yes, in this case) then I = 1/2 or an integer value of ½ such as 3/2, 5/2, and so on.”

Complete sections c through f using a similar approach.

Determine the nuclear magnetic quantum number (m) for each of the following nuclides.

(a) 2H (b) 3H

(c) 12C (d) 13C

(e) 15N (f) 31P

To determine the values for m, the nuclear magnetic quantum number, we need to employ equation 14.1. And in order to determine the values for m, we will need to first calculate I, the nuclear spin. Using the results for I, from problem 14.3, apply equation 14.1 to determine the values for m.

a. I = 1, and m = (2I +1) = 3; there are 3 values for m. They are 1, 0, and -1.

b. I = ½, and m = (2I +1) =2; there are 2 values for m. They are ½ and -1/2

c. I =0, and m = (2I +1) = 1; there is 1 value for m. It is 0.

Complete sections d through f using a similar approach.

Problem 14.4: Identify the nuclides from Problem 14.3 that are NMR active. Explain your choices.

NMR active nuclides have non zero values for I. All except 12C are NMR active since all the

others have non-zero spin quantum numbers.

For each nuclide below, use Equations 14.4 and 14.5 and the data in Table 14.1 to determine the Larmor frequency in a 4.7 T magnetic field.

(a) 1H (b) 13C (c) 31P (d) 19F

The Larmor frequency () can be calculated directly from Equation 14.3 (- Bo) or by algebraic

substitutions in Equations 14.4 and 14.5. To complete the substitutions, you will also need EI = h

(b) 13C, = 6.728 x 107 T-1s-1, therefore - 6.728x107 T-1s-1)4.7 T) = 3.16 x 108 s-1

(c) 31P, = 10.84 x 107 T-1s-1, therefore - 10.84x107 T-1s-1)4.7 T) = 5.09 x 108 s-1

Complete sections a and d using a similar approach.

Problem 14.6: For each nuclide below, use Equations 14.4 and 14.5 and the data in Table 14.1 to

determine the Larmor frequency in a 9.4 T magnetic field.

(a) 1H (b) 13C (c) 31P (d) 19F

This problem is completed analogously to Problem 14.5, using 9.4 T as the Bo value.

Use Equation 14.6 and the equation for the Boltzmann distribution to determine the relative population distribution between m = –½ and m = +½ for a proton at room temperature for the following NMR spectrometers:

(a) 60 MHz (b) 200 MHz (c) 300 MHz (d) 900 MHz

(b) A 200 MHz field means that ν = 200x106 s-1

Convert radio frequency to angular frequency:

ω = 2πν = 2π(200x106 s-1) = 1.256 x 109 s-1

Next, use Eqn 14.3 to calculate Bo

ω=γBo

And substitute, using the correct value for the gyromagnetic constant (from Table

14.1) for the 1H nucleus, to determine Bo:

1.256x109 s-1 = (26.75 x 107 T-1s-1) Bo

Bo = 4.698 T

Next use Eqn 14.6 to determine the energy gap (∆𝐸)

∆𝐸 = (ℎ

2𝜋) 𝛾𝐵𝑜 =

6.626𝑥10−34𝐽𝑠

2𝜋 (26.75 𝑥 107𝑇−1𝑠−1)(4.698𝑇) = 1.325 x 10-25 J

Finally, using the equation for the Boltzmann distribution, calculate the Ng/Ne ratio

for the energy gap

𝑁𝑒

𝑁𝑔= (1)𝑒−(

∆𝐸

𝑘𝑇) = 𝑒

−(1.325 𝑥 10−25 𝐽

(1.38 𝑥 10−23 𝐽/𝑠)(298.15 𝐾) = 0.9999678

Taking the inverse: Ng/Ne = 1.000032.

This tells us that there is a 32 ppm excess in ground state

(d) ν=900x106 s-1 ω=2πν = 2π(900x106 s-1) = 5.655 x 109 s-1

ω=γB 5.655x109 s-1 = (26.75 x 107 T-1s-1) B B = 21.14 T

∆𝐸 = (ℎ

2𝜋) 𝛾𝐵 =

6.626𝑥10−34𝐽𝑠

2𝜋 (26.75 𝑥 107𝑇−1𝑠−1)(21.14𝑇) = 5.962 x 10-25 J

𝑁𝑒

𝑁𝑔= (1)𝑒−(

∆𝐸

𝑘𝑇) = 𝑒

−(5.962 𝑥 10−25 𝐽

(1.38 𝑥 10−23𝐽𝑠

)(298.15 𝐾))

= 0.9998551

Ng/Ne = 1.000145, so 145 ppm excess in ground state

Complete sections a and c using a similar approach.

Problem 14.8: Determine the relative sensitivity of the spectrometers from Problem 14.7.

We can determine sensitivity in terms of their relative population excess. The

sensitivity of the 900 MHz vs. the 200 MHz is the ratio of 145/32. The 900 MHz

instrument is 4.53 times more sensitive than the 200 MHz instrument.

Using the data from Table 14.1, derive a megahertz-to-tesla conversion for:

(a) 13C (b) 31P (c) 19F

This shortcut can be arranged similarly to Problem 14.7. Convert radio frequency to

angular frequency:

ω = 2πν = 2π(N Hz)

Using Eqn 14.3: ω=γBo

Substitute for ω:

2π(N Hz) = γBo

And substitute, using the correct value for the gyromagnetic constant (from Table

14.1) for the nuclide of interest to determine the conversion factor for megahertz to

tesla for that nuclide:

(a) For 13C

2π(N Hz) = (6.728 x 107 T-1s-1) Bo

Bo Tesla = 2π(N)/(6.728 x 107)

Bo Tesla = 0.09339 N (where N is in MHz)

Sections b and c are determined using a similar approach.

The Boltzmann equation (see “Compare and Contrast—Population Distribution for Common Spectroscopic Methods”) indicates that one can also affect the relative population of states by adjusting the temperature. Relative to room temperature, at what temperature would you need to be able to collect your spectrum in order to obtain the same signal enhancement achieved by switching from a 60 MHz NMR to a 200 MHz NMR? Discuss the implications.

First, use the results of Problem 14.7 a and b to determine what the population

enhancement is from 60MHz to 200MHz. Here we will assume the calculation is for

the 1H nucleus.

For the 60 MHz field: ν = 60x106 s-1

Convert radio frequency to angular frequency:

ω = 2πν = 2π(60x106 s-1) = 3.7699 x 108 s-1

Next, use Eqn 14.3 to calculate Bo

ω=γBo

And substitute, using the correct value for the gyromagnetic constant (from Table

14.1) for the 1H nucleus, to determine Bo:

3.7699x108 s-1 = (26.75 x 107 T-1s-1) Bo

Bo = 1.409 T

Next use Eqn 14.6 to determine the energy gap (∆𝐸)

∆𝐸 = (ℎ

2𝜋) 𝛾𝐵𝑜 =

6.626𝑥10−34𝐽𝑠

2𝜋 (26.75 𝑥 107𝑇−1𝑠−1)(1.409𝑇) = 3.976 x 10-26 J

Finally, using the equation for the Boltzmann distribution, calculate the Ng/Ne ratio

for the energy gap

𝑁𝑒

𝑁𝑔= (1)𝑒−(

∆𝐸

𝑘𝑇) = 𝑒

−(3.976 𝑥 10−26 𝐽

(1.38 𝑥 10−23 𝐽/𝑠)(298.15 𝐾) = 0.999990337

Taking the inverse: Ng/Ne = 1.000009664.

Previously, in Problem 14.7, we showed that the Ng/Ne = 1.000032 at 200 MHz.

Thus the signal enhancement is: 32/9.664 = 3.311

To determine the temperature at which one could obtain the same population

enhancement (at 60MHz), consider that at 298.15 the population distribution is:

𝑁𝑒

𝑁𝑔= (1)𝑒−(

∆𝐸

𝑘𝑇) = 𝑒

−(3.976 𝑥 10−26 𝐽

(1.38 𝑥 10−23 𝐽/𝑠)(298.15 𝐾) = 0.999990337

And Ng/Ne = 1.000009664. A 3.311x enhancement would be 9.664 x 3.311, or Ng/Ne

= 1.000032. Take the inverse and use substitution and algebra to solve for T

𝑁𝑒

𝑁𝑔= (1)𝑒−(

∆𝐸

𝑘𝑇) = 𝑒

−(3.976 𝑥 10−26 𝐽

(1.38 𝑥 10−23 𝐽/𝑠)(𝑇 𝐾) = 0.9999678

-(3.976 x 10-26 J)/[(1.38 x 10-23 J/s)(T K)] = ln 0.9999678

-(3.976 x 10-26 J)/[(1.38 x 10-23 J/s)(T K)] = -0.000032201

1/(T K) = 0.000032201 ((1.38 x 10-23 J/s)/ (3.976 x 10-26 J)

1/T = 0.3331832 K

T = 89.476 K

Problem 14.11: As of the publication of this textbook, the highest commercial field NMR used in research

laboratories is a 900 MHz NMR (for 1H?). What is the magnetic field strength in units of tesla?

A 900 MHz field means that ν = 900x106 s-1

Convert radio frequency to angular frequency:

ω = 2πν = 2π(900x106 s-1) = 5.655 x 109 s-1

Next, use Eqn 14.3 to calculate Bo

ω=γBo

And substitute, using the correct value for the gyromagnetic constant (from Table

14.1) for the 1H nucleus, to determine Bo:

5.655x109 s-1 = (26.75 x 107 T-1s-1) Bo

Bo = 21.14 T

Determine the nuclear spin quantum number (I) and the nuclear magnetic quantum number (m) for each of the following nuclides. Identify the nuclides that are NMR active and explain your choices.

(a) 19F (b) 119Sn (c) 16O (d) 27Al (e) 23Na

As with Problem 14.2, to determine I, the nuclear spin, we need to employ the rules as described section 14.2, which are based on the numbers of protons and neutrons.

(a) 19F has an atomic number of 9 and an atomic mass of 19. There are 9 protons and 10 neutrons in this isotope. Rule 2 applies because the sum of protons and neutrons is odd. Therefore the spin quantum number is ½ or an integer of ½. The nuclear magnetic quantum number will have 2 values (2I +1) which will be +½ and – ½.

Sections b – e will be solved using a similar strategy.

Problem 14.13: What is the Bo field (in tesla) for a 1H nucleus with a Larmor frequency of 800 MHz?

A 800 MHz field means that ν=800x106 s-1

ω=2πν = 2π(800x106 s-1) = 5.0265 x 109 s-1

ω=γBo

5.0265x109 s-1 = (26.75 x 107 T-1s-1) Bo

Bo = 18.79 T

What is the Bo field (in tesla) for a 1H nucleus with a Larmor frequency of 500 MHz?

Problem 14.14 will be solved using a similar approach as is shown in Problems 14.11 and 14.13.

Using Figure 14.8 as a model, create a field-splitting diagram for 13C showing the field splitting (E) in units of MHz for a 2.34 T, 4.73 T, 7.07 T, and 9.46 T magnetic field.

The resulting diagram would look like Figure 14.8, same x and y axes; however the splitting (E values)

between -1/2 and +1/2 would have different values. At 2.34 T, where on Figure 14.8 for 1H nucleus is 200

MHz, the diagram for 13C would have a value calculated as shown in Example 14.2.

E = ℏ(Bo)

E = (6.626𝑥10−34𝐽𝑠 /2 (6.728 x 107 T-1s-1) 2.34 T

E =1.66 x 10-26 J

E = h

1.66 x 10-26 J = h

.66 x 10-26 J / 6.626𝑥10−34𝐽𝑠

2.51 x 107 s-1 = 25.1 MHz

To finish the problem, calculate the values for the other magnetic field strengths using a similar approach

and complete the labels on the diagram.

Problem 14.16: There are two fundamental reasons why the signal strength for 13C-NMR is much weaker

than the signal strength for 1H-NMR on any given instrument. Explain what those two reasons would be.

(Hint: See Table 14.1 and the results for Problem 14.15.)

The 13C magnetogyric ratio is much lower and the relative isotopic abundance is only

1%.

Using Example 14.4 as a guide, determine the resonance frequency difference in Hertz between 1H signals at 1.5 ppm and 3.5 ppm as measured on an instrument with a Bo field strength of:

(a) 200 MHz (b) 400 MHz (c) 900 MHz

(b) 𝑝𝑝𝑚 = 𝜈𝑠𝑖𝑔− 𝜈𝑟𝑒𝑓

𝜈𝑟𝑒𝑓 𝑥 106

1.5 𝑝𝑝𝑚 = 𝜈𝑠𝑖𝑔− 400𝑥106

400𝑥106 𝑥 106 νsig = 400000600 Hz

3.5 𝑝𝑝𝑚 = 𝜈𝑠𝑖𝑔− 400𝑥106

400𝑥106 𝑥 106

The resonance frequency difference would be (40001400 – 400000600) = 800 Hz You could also approach this by determining that the difference between 1.5 and 3.5 ppm is 2 ppm. Knowing that on a 400 MHz instrument, for 1H nucleus, 1 ppm = 400 Hz; therefore 2 ppm = 800 Hz. Parts a and c will be solved using a similar strategy.

Assume you have a detector with a baseline spectral resolution of 20 Hz. How far apart (in ppm) must two peaks be to have baseline resolution for a spectrometer with Bo equal to:

(a) 60 MHz? (b) 200 MHz? (c) 300 MHz? (d) 900 MHz?

Using the approach laid out in Example 14.5, the 1.0 ppm and 1.1 ppm peaks would have frequencies at 200 and 220 Hz, respectively. Thus, at 0.1 ppm difference, they would just have baseline resolution at 20 Hz. In comparison, the 1.0 ppm and 1.1 ppm peaks would have frequencies at 900 and 990 MHz, respectively, on the 900MHz spectrometer. Thus, at 0.1 ppm difference, they would have significant resolution, separated by 90 Hz. On this instrument, to get down to the 20 Hz resolution limit, the peaks could get as close as

0.1/90 = x/20 0.022 ppm at baseline resolution on the 900 MH instrument.

Determine the ratio of spin active 1H to 13C nuclei in each of the following molecules:

(a) Benzene (C6H6) (b) Hexane (C6H14) (c)Chloro(methyl)amine (CH3NHCl)

(a) In benzene the ratio of C:H is 6:6 or 1:1. The relative abundances of the 1H and 31C isotopes are then taken into account. The relative abundance of 1H isotope is 98.89%. The relative abundance of 13C is 1.0108%.

#𝐻 𝑥 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐴𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒

#𝐶 𝑥 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝐴𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒=

1 𝑥 98.89%

1 𝑥 1.0108%=

0.9889

0.010108= 97.83

In benzene, the 1H outnumber the 13C by almost 100 to 1, that is, 97.83 to 1. Solutions for parts b and c are obtained using the same strategy.

Problem 14.20: A typical 1H-NMR experiment will incorporate n = 16 data sets into a single signal-

averaged experiment. Assuming that the noise is the same and using Example 14.7 as a guide, how many

signal-averaged data sets would you need to collect in order to achieve approximately the same S/N

ratio in your 13C-NMR spectrum as you would in your 1H -NMR spectrum?

𝑆

𝑁= √𝑛 (

𝑆

𝑁)

𝑜

𝑆

𝑁= 4 (

𝑆

𝑁)

𝑜

400 = √𝑛 1

EXERCISE 14.1: Using the n + 1 rule reviewed in Section 14.1, explain the peak splitting pattern for the

compound ethyl crotonate. The structure of trans-ethyl crotonate (ethyl trans-2-butenoate) is shown

alongside the 1H-NMR spectrum.

There are three 1H nuclei whose resonance signal occurs at 1.30 ppm. These are from a methyl group. These nuclei have two 1H neighbors, therefore the peak at 1.30 ppm has a splitting pattern of (2+1), a triplet. There is another methyl group with a resonance signal at 1.71 ppm. This methyl has only one 1H neighbor and therefore the splitting is (1+1) a doublet. There is a methylene group with a resonance signal at 4.19 ppm. The methylene has three 1H neighbors; therefore the splitting is (3+1) a quartet. There are two single protons on the trans-double bond. One with a resonance signal at 5.83 ppm and the other is at 6.88 ppm. The single proton at 5.83 ppm has a strong coupling to the proton at 6.88 ppm and therefor will split (1+1) into a doublet, but also is coupled to the methyl group on the double bond and therefore has a more complicated splitting pattern (a multiplet) as the doublet is further split . Likewise the proton at 6.88 ppm is strongly coupled to the methyl group on the same carbon and also strongly coupled to the trans-proton; again generating a multiplet (a doublet of quadruplets).

EXERCISE 14.2: From the mass spectral and elemental analysis data, you have determined the formula of

an unknown compound from a petroleum distillate to be C3H6O2. You have limited the possibilities to the four different structures shown. Each has the formula C3H6O2.

(a) Describe how an IR spectrum would aid in narrowing down your choices. Be specific.

There would be a carbonyl stretch for B, C, and D. If the spectrum doesn’t have that, then A is would be the only choice. The OH groups would show up for A and C, but not for B and D.

(b) Use the 1H-NMR spectrum of your unknown shown to determine which isomer you have.

The spectrum shows only two NMR active protons between 1 and 3 ppm and a small peak around 12 ppm. The 1H signal at 1.2 ppm is split into a triplet, which means that it has two identical proton neighbors. A, B, and C have adjacent –CH2- groups, D does not. Thus D is eliminated as a possible structure. The 1H signal at 2.4 ppm is split into a quadruplet, demonstrating that it has three identical proton neighbors. Although B, C, and D have methyl

groups, D was previously eliminated and B’s methyl group is not adjacent to other protons. Therefore C is the only remaining candidate and also has the broad signal around 12 ppm for the isolated OH group of the carboxylic acid.

What is the Larmor frequency of 13C in a 7.07 T magnetic field?

Solved as in Problem 14.5. The Larmor frequency () can be calculated directly from Equation 14.3

(- Bo) or by algebraic substitutions in Equations 14.4 and 14.5. To complete the substitutions, you

will also need EI = h

13C, = 6.728 x 107 T-1s-1, therefore - 6.728x107 T-1s-1)7.07 T) =4.76 108 s-1

What is the Larmor frequency of 13C in a 19 T magnetic field?

Solved in the same way as Exercise 14.3.

For each nuclide below, determine the Larmor frequency in a 14.14 T magnetic field. (a) 1H (b) 13C (c) 31P (d) 19F

Solved in the same way as Exercise 14.3. Be sure to get the correct value for the gyromagnetic constant for each nuclide.

d) ω=γB = (25.18 x 107 T-1s-1)(14.14 T) = 3.56 GHz

EXERCISE 14.6: For each nuclide below, determine the Larmor frequency in a 21.21 T magnetic field.

(a) 1H (b) 13C (c) 31P (d) 19F

Solved in the same way as Exercise 14.3. Be sure to get the correct value for the gyromagnetic constant for each nuclide.

What is the magnetic field strength in units of tesla for a 600 MHz NMR instrument?

Since the problem doesn’t specify a nuclide, assume 1H nucleus.This Exercise is solved in an analogous manner to Problem 14.11. See solution to Problem 14.11

What is the magnetic field strength in units of tesla for a 700 MHz NMR instrument?

Since the problem doesn’t specify a nuclide, assume 1H nucleus.This Exercise is solved in an analogous manner to Problem 14.11. See solution to Problem 14.11

EXERCISE 14.9: Using Figure 14.9 as a model, create a field-splitting diagram for 31P showing the field

splitting (E) in units of MHz for 2.34 T, 4.73 T, 7.07 T, and 9.46 T magnetic fields.

Solved as in Problem 14.15. Be sure to use the correct value for the gyromagnetic constant for 31P.

Using Figure 14.9 as a model, create a field-splitting diagram for 19F showing the field splitting (E) in units of MHz for 2.34 T, 4.73 T, 7.07 T, and 9.46 T magnetic fields.

Solved as in Problem 14.15. Be sure to use the correct value for the gyromagnetic constant for 19F.

There are two fundamental reasons why the signal strength for 13C-NMR is much weaker than the signal strength for 1H-NMR on any given instrument. Explain what those two reasons would be. (Hint: See Table 14.1 and the results for Problem 14.15.)

Students should actually be able to come up with three reasons. First, intrinsic difference in gyromagnetic ratio (1H is more than four times larger) and second, isotopic abundance of 1H is much higher than 13C and third, population and energy differences between ground and excited states.

EXERCISE 14.12: Determine the resonance frequency difference between two resonances in an NMR

spectrum taken on a 300 MHz instrument. If the two 1H signals of interest are at 2.4 ppm and 3.0 ppm,

what is the frequency difference between the two peaks?

Solved similarly to Problem 14.17. The difference between 2.4 and 3.0 ppm is 0.6 ppm. Knowing that on a 300 MHz instrument, for 1H nucleus, 1 ppm = 300 Hz; therefore (proportionally) 0.6 ppm = 180 Hz difference in frequency.

Determine the resonance frequency difference between two resonances in an NMR spectrum taken on a 500 MHz instrument. If the two 1H signals of interest at 7.75 ppm and 7.90 ppm, what is the frequency difference between the two peaks?

Solved similarly to Problem 14.17 and Exercise 14.12.

Determine the resonance frequency difference in hertz between 1H signals at 2.20 ppm and at 2.30 ppm as measured on an instrument with a Bo field strength of: (a) 300 MHz (b) 600 MHz (c) 900 MHz

Solved similarly to Problem 14.17 and Exercise 14.12.

Determine the resonance frequency difference in hertz between 1H signals at 4.45 ppm and at 4.60 ppm as measured on an instrument with a Bo field strength of: (a) 100 MHz (b) 500 MHz (c) 900 MHz

Solved similarly to Problem 14.17 and Exercise 14.12.

Assume you have a detector with a baseline spectral resolution of 15 Hz, comment on the resolution of the peaks in Problem 14.17 at each field strength (a–c).

Solved similarly to Problem 14.18.

Assume you have a detector with a baseline spectral resolution of 10 Hz, comment on the resolution of the peaks in Problem 14.17 at each field strength (a–c).

Solved similarly to Problem 14.18.

Determine the relative population distribution between m = –½ and m = +½ for a proton at room temperature for the following NMR spectrometers: (a) 90 MHz (b) 400 MHz (c) 600 MHz (d) 800 MHz

Solved as shown in Problem 14.7.

EXERCISE 14.19: Determine the relative population distribution between m = –½ and m = +½ for a proton

at a temperature of 280K, for the following NMR spectrometers:

(a) 90 MHz (b) 400 MHz (c) 600 MHz (d) 800 MHz

Solved similarly to Problem 14.7, vary the temperature (in Kelvin).

EXERCISE 14.20: Determine the relative population distribution between m = –½ and m = +½ for a proton

at a temperature of 310K, for the following NMR spectrometers:

(a) 90 MHz (b) 400 MHz (c) 600 MHz (d) 800 MHz

Solved similarly to Problem 14.7, vary the temperature (in Kelvin).

Determine the ratio of spin-active 1H to 13Cnuclei in each of the following molecules: (a) Pentane (C5H12) (b) Ethylene diamine (C2H4(NH2)2) (c) Acetaldehyde (C2H4O)

Solved as shown in Problem 14.19.

Determine the ratio of spin-active 1H to 13C nuclei in each of the following molecules: (a) Glycerol (C3H8O3) (b) Heptane (C7H16) (c) Lysine (C6H14N2O2)

Solved as shown in Problem 14.19.

What is the S/N ratio enhancement when 1H-NMR experiment incorporates n = 64 data sets into a single signal-averaged experiment rather than n = 16 data sets?

Solved as shown in Problem 14.20.

What is the S/N ratio enhancement when 1H-NMR experiment incorporates n = 128 data sets into a single signal-averaged experiment rather than n = 64 data sets?

Solved as shown in Problem 14.20.

Repeat the calculations from Exercises 14.12 and 14.13 with 13C-NMR. Describe how the enhancement effect compares between 1H and 13C.

Solved similarly to Problem 14.17. Better not to use the shortcut, but start as shown in Problem 14.17 solution with the equation:

𝑝𝑝𝑚 = 𝜈𝑠𝑖𝑔− 𝜈𝑟𝑒𝑓

𝜈𝑟𝑒𝑓 𝑥 106

Calculate the frequencies (vsig and vref) using the Bo values for 13C.

If you were interested in observing nuclei other than 1H or 13C in an MRI, which of these biologically available elements would be possible candidates? (a) Oxygen (b) Calcium (c) Selenium

Which of the nuclides are NMR active? Using Problem 14.4 as a model, determine which element(s) are NMR active.

If you were interested in observing nuclei other than 1H or 13C in an MRI, which of these biologically available elements would be possible candidates? (a) Iron (b) Sulfur (c) Nitrogen

Which of the nuclides are NMR active? Using Problem 14.4 as a model, determine which element(s) are NMR active.

In an MRI, the patient is placed inside a central opening, the bore, of the magnetic field. The patient’s body is aligned with the magnetic field (Bo) along the z axis. The magnetic field runs from head to toe. Is this the same relative position as an NMR tube inside an NMR? Explain using diagrams. Be sure to consider the gradient coils and their function in your answer.

As shown in Figure 14.15, the magnetic field in the NMR lies across the NMR tube (not from head to toe). The z axis runs perpendicular to the long axis of the NMR tube. The gradient coils in the MRI set up field gradients across the sample (body or organ tissue); while spinning of the NMR tube helps to assure field homogeneity.

EXERCISE 14.29: In an MRI instrument, there are gradient coils that modify the external magnetic field,

creating gradients in the field. These gradients can be set up to image sections of the body along the

sagittal, transverse, or coronal planes. Use the internet to find MRI images of tissues that represent these

sections and describe the perspective of the image on the tissue. This article, from the National High

Magnetic Field Laboratory at Florida State University, may help you get started; you can find it at http://www.magnet.fsu.edu/education/tutorials/magnetacademy/mri/fullarticle.html.

Internet research, student answers will vary in terms of details.