Chapter 12, Problem 12A.3P

(a)

Interpretation Introduction

Interpretation:

The level of enrichment required for ${}^{15}\text{N}$ signal to have same intensity as that from ${}^{13}\text{C}$ with its natural abundance has to be stated.

Concept introduction:

Many nuclei and electrons have spin, due to this spin magnetic moment arises.  The energy of this magnetic moment depends on the orientation of the applied magnetic field.  In NMR spectroscopy, every nucleus has a spin.  There is an angular momentum related to the spin.  The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus.  Tetramethylsilane is taken as reference.

Answer to Problem 12A.3P

The level of enrichment required for ${}^{15}\text{N}$ signal to have same intensity as that from ${}^{13}\text{C}$ with its natural abundance is $6.8\%$.

Explanation of Solution

The natural abundance of ${}^{13}\text{C}$ is $1.108\%$.

The value of ${\gamma }_{\text{N}}$ for ${}^{13}\text{C}$ is $6.73\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}$.

The value of ${\gamma }_{\text{N}}$ for ${}^{15}\text{N}$ is $-2.7126\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}$.

The negative sign can be neglected in the calculation of a number of nuclei.

The expression of the intensity of NMR is shown below.

    $\text{Intensity}\propto \frac{N{\gamma }_{\text{N}}^2{B}_0^2}{T}$

Where,

• $N$ is the total number of nuclei.
• $B_0$ is the magnetic field.
• $T$ is the temperature.
• ${\gamma }_{\text{N}}$ is the magnetogyric ratio.

The number of nuclei can be written in term of level of enrichment.

When the intensity of NMR spectra of ${}^{15}\text{N}$ and ${}^{13}\text{C}$ is the same, then the relationship number of total spin is given by the expression as shown below.

    $\begin{array}{c}\frac{N\left({}^{15}\text{N}\right){\left({\gamma }_{\text{N}}\left({}^{15}\text{N}\right)\right)}^2{B}_0^2}{T}=\frac{N\left({}^{13}\text{C}\right){\left({\gamma }_{\text{N}}\left({}^{13}\text{C}\right)\right)}^2{B}_0^2}{T}\\ N\left({}^{15}\text{N}\right){\left({\gamma }_{\text{N}}\left({}^{15}\text{N}\right)\right)}^2=N\left({}^{13}\text{C}\right){\left({\gamma }_{\text{N}}\left({}^{13}\text{C}\right)\right)}^2\\ N\left({}^{15}\text{N}\right)=\frac{N\left({}^{13}\text{C}\right){\left({\gamma }_{\text{N}}\left({}^{13}\text{C}\right)\right)}^2}{{\left({\gamma }_{\text{N}}\left({}^{15}\text{N}\right)\right)}^2}\end{array}$

Substitute the value of $N\left({}^{13}\text{C}\right)$, $\left({\gamma }_{\text{N}}\left({}^{13}\text{C}\right)\right)$, and $\left({\gamma }_{\text{N}}\left({}^{15}\text{N}\right)\right)$ in the above expression.

  $\begin{array}{c}N\left({}^{15}\text{N}\right)=\frac{1.108\%{\left(6.73\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}\right)}^2}{{\left(2.7126\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}\right)}^2}\\ =\frac{1.108\%\times 4.5293\times {10}^{15}}{\left(7.3582\times {10}^{14}\right)}\\ =\underset{\_}{6.8\%}\end{array}$

The level of enrichment required for ${}^{15}\text{N}$ signal to have same intensity as that from ${}^{13}\text{C}$ with its natural abundance is $\underset{\_}{6.8\%}$.

(b)

Interpretation Introduction

Interpretation:

The intensity of achievable by natural abundance of ${}^{13}\text{C}$ has to be calculated.  The intensity of achievable by $100\%$ of ${}^{17}\text{O}$ has to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 12A.3P

The intensity of achievable by the natural abundance of ${}^{13}\text{C}$ is proportional to $\frac{5.018\times {10}^{13}\times B_0^2}{T}{\text{T}}^{-1}{\text{ s}}^{-1}$ at given temperature and magnetic field.  The intensity of achievable by $100\%$ of ${}^{17}\text{O}$ is proportional to $\frac{3.58644\times {10}^{14}\times B_0^2}{T}{\text{T}}^{-1}{\text{ s}}^{-1}$ at given temperature and magnetic field.

Explanation of Solution

The expression of the intensity of NMR is shown below.

    $\text{Intensity}\propto \frac{N{\gamma }_{\text{N}}^2{B}_0^2}{T}$        (1)

Where,

• $N$ is the total number of spin.
• $B_0$ is the magnetic field.
• $T$ is the temperature.
• ${\gamma }_{\text{N}}$ is the magnetogyric ratio.

The natural abundance of ${}^{13}\text{C}$ is $1.108\%$.

The value of ${\gamma }_{\text{N}}$ for ${}^{13}\text{C}$ is $6.73\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}$.

The value of ${\gamma }_{\text{N}}$ for ${}^{17}\text{O}$ is $-1.89379\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}$.

The percentage enrichment of ${}^{17}\text{O}$ is $100\%$.

The number of nuclei can be written in term of level of enrichment.

Substitute the value of $N$ and ${\gamma }_{\text{N}}$ for ${}^{13}\text{C}$ in the equation (1).

    $\begin{array}{c}\text{Intensity}\propto \frac{\left(1.108\%\right){\left(6.73\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}\right)}^2{B}_0^2}{T}\\ \propto \frac{1.108\times 4.5293\times {10}^{15}\times B_0^2}{100\times T}\\ \propto \frac{5.018\times {10}^{13}\times B_0^2}{T}{\text{T}}^{-1}{\text{ s}}^{-1}\end{array}$

The intensity of achievable by natural abundance of ${}^{13}\text{C}$ is proportional to $\frac{5.018\times {10}^{13}\times B_0^2}{T}{\text{T}}^{-1}{\text{ s}}^{-1}$ at given temperature and magnetic field.

Substitute the value of $N$ and ${\gamma }_{\text{N}}$ for ${}^{17}\text{O}$ in the equation (1).

    $\begin{array}{c}\text{Intensity}\propto \frac{\left(100\%\right){\left(1.89379\times {10}^7{\text{T}}^{-1}{\text{ s}}^{-1}\right)}^2{B}_0^2}{T}\\ \propto \frac{100\times 3.58644\times {10}^{14}\times B_0^2}{100\times T}\\ \propto \frac{3.58644\times {10}^{14}\times B_0^2}{T}{\text{T}}^{-1}{\text{ s}}^{-1}\end{array}$

The intensity of achievable by $100\%$ of ${}^{17}\text{O}$ is proportional to $\frac{3.58644\times {10}^{14}\times B_0^2}{T}{\text{T}}^{-1}{\text{ s}}^{-1}$ at given temperature and magnetic field.