Chapter 12, Problem 12D.3BE

Interpretation Introduction

Interpretation:

The magnetic fields of hyperfine lines and their relative intensities have to be stated for a radical that has three inequivalent protons.

Concept introduction:

EPR stands for electronic paramagnetic resonance.  It is used to study species that contain unpaired electrons.  EPR can be used to study both solids and liquids.  However, the study of the gas phase samples is difficult due to the free rotation of the molecules.  The $g-$ value represents the ease with which the applied field can generate currents through the molecule.

Answer to Problem 12D.3BE

The magnetic fields of hyperfine lines are $\underset{\_}{328.865mT}$, $\underset{\_}{330.975mT}$, $\underset{\_}{331.735mT}$, $\underset{\_}{331.755mT}$, $\underset{\_}{333.845mT}$, $\underset{\_}{333.865mT}$, $\underset{\_}{334.625mT}$, and $\underset{\_}{336.735mT}$.  The relative intensities of these lines are $1:1:2:2:1:1$.

Explanation of Solution

The coupling constants of the ESR spectrum of a radical that has two inequivalent protons are $2.11\text{ mT}$ , $2.87\text{ mT}$ and $2.89\text{ mT}$.

The center of the ESR spectrum is $332.8\text{ mT}$.

When two different nuclei are coupled, then the splitting of two adjacent ESR lines is given by $a$.

The hyperfine lines form by first splitting is separated by $2.11\text{ mT}$.  The magnetic field of the hyperfine lines can be calculated by adding or subtracting $a/2$ in the value of $332.8\text{ mT}$.

The hyperfine line on the left side of line with magnetic field equal to $332.8\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=332.8\text{ mT}-\frac{2.11\text{ mT}}{2}\\ =332.8\text{ mT}-1.055\text{ mT}\\ =331.745\text{ mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $332.8\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=332.8\text{ mT}+\frac{2.11\text{ mT}}{2}\\ =332.8\text{ mT}+1.055\text{ mT}\\ =333.855\text{ mT}\end{array}$

The hyperfine lines form by second splitting is separated by $2.87\text{ mT}$.  The magnetic field of the hyperfine lines can be calculated by adding or subtracting $a/2$ in the value of magnetic field of central line.

The hyperfine line on the left side of line with magnetic field equal to $331.745\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=331.745\text{ mT}-\frac{2.87\text{ mT}}{2}\\ =331.745\text{ mT}-1.435\text{ mT}\\ =330.31\text{ mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $331.745\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=331.745\text{ mT}+\frac{2.87\text{ mT}}{2}\\ =331.745\text{ mT}+1.435\text{ mT}\\ =333.18\text{ mT}\end{array}$

The hyperfine line on the left side of line with magnetic field equal to $333.855\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=333.855\text{ mT}-\frac{2.87\text{ mT}}{2}\\ =333.855\text{ mT}-1.435\text{ mT}\\ =332.42\text{ mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $333.855\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=333.855\text{ mT}+\frac{2.87\text{ mT}}{2}\\ =333.855\text{ mT}+1.435\text{ mT}\\ =335.29\text{ mT}\end{array}$

The hyperfine lines form by second splitting is separated by $2.89\text{ mT}$.  The magnetic field of the hyperfine lines can be calculated by adding or subtracting $a/2$ in the value of magnetic field of central line.

The hyperfine line on the left side of line with magnetic field equal to $330.31\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=330.31\text{ mT}-\frac{2.89\text{ mT}}{2}\\ =330.31\text{ mT}-1.445\text{ mT}\\ =\underset{\_}{328.865mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $330.31\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=330.31\text{ mT}+\frac{2.89\text{ mT}}{2}\\ =330.31\text{ mT}+1.445\text{ mT}\\ =\underset{\_}{331.755mT}\end{array}$

The hyperfine line on the left side of line with magnetic field equal to $333.18\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=333.18\text{ mT}-\frac{2.89\text{ mT}}{2}\\ =333.18\text{ mT}-1.445\text{ mT}\\ =\underset{\_}{331.735mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $333.18\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=333.18\text{ mT}+\frac{2.89\text{ mT}}{2}\\ =333.18\text{ mT}+1.445\text{ mT}\\ =\underset{\_}{334.625mT}\end{array}$

The hyperfine line on the left side of line with magnetic field equal to $332.42\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=332.42\text{ mT}-\frac{2.89\text{ mT}}{2}\\ =332.42\text{ mT}-1.445\text{ mT}\\ =\underset{\_}{330.975mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $332.42\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=332.42\text{ mT}+\frac{2.89\text{ mT}}{2}\\ =332.42\text{ mT}+1.445\text{ mT}\\ =\underset{\_}{333.865mT}\end{array}$

The hyperfine line on the left side of line with magnetic field equal to $335.29\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=335.29\text{ mT}-\frac{2.89\text{ mT}}{2}\\ =335.29\text{ mT}-1.445\text{ mT}\\ =\underset{\_}{333.845mT}\end{array}$

The hyperfine line on the right side of line with magnetic field equal to $335.29\text{ mT}$ is calculated as shown below.

    $\begin{array}{c}B=335.29\text{ mT}+\frac{2.89\text{ mT}}{2}\\ =335.29\text{ mT}+1.445\text{ mT}\\ =\underset{\_}{336.735mT}\end{array}$

Therefore, the magnetic fields of hyperfine lines are $\underset{\_}{328.865mT}$, $\underset{\_}{330.975mT}$, $\underset{\_}{331.735mT}$, $\underset{\_}{331.755mT}$, $\underset{\_}{333.845mT}$, $\underset{\_}{333.865mT}$, $\underset{\_}{334.625mT}$, and $\underset{\_}{336.735mT}$.

The difference in the peaks $\underset{\_}{331.735mT}$ and $\underset{\_}{331.755mT}$ is very less.  Similarly, the difference in the peaks $\underset{\_}{333.865mT}$ and $\underset{\_}{333.845mT}$ is very less.  The ESR spectrometer will not detect them as two different peak.  Therefore, the relative intensities of these lines is $1:1:2:2:1:1$.