Question

A resonance is displaced $90\mathrm{Hz}$from TMS at a magnetic field strength of 1.41 T. What will be the frequency difference at $\left(\mathrm{a}\right)4.69\mathrm{T},\left(\mathrm{b}\right)7.05\mathrm{T},\left(\mathrm{c}\right)\mathrm{and}18.8\mathrm{T}$? What will be the chemical shifts d at these same magnetic field strengths?

Short answer

(a) The frequency difference at $4.69\mathrm{T}$is $300\mathrm{Hz}$.

(b) The frequency difference at $7.05\mathrm{T}$is $450\mathrm{Hz}$.

(c) The frequency difference at $18.8\mathrm{T}$is $1200\mathrm{Hz}$.

The chemical shift is $1.5\mathrm{ppm}$.

Step-by-step solution

Part (a) Step 1. Given information

The resonance is displaced by $90\mathrm{Hz}$from TMS.

The magnetic field strength is$1.41\mathrm{T}$.

The external field is $4.69\mathrm{T}$.

Part (a) Step 2. Calculate frequency difference

The frequency difference is calculated as follows:

$∆v=\frac{\mathrm{Displaced}\mathrm{frequency}\mathrm{of}\mathrm{sample}\times \mathrm{External}\mathrm{field}}{\mathrm{Magnetic}\mathrm{field}\mathrm{strength}}........\left(1\right)$

Subsitute displaced frequency, external field, and field strength in the equation (1) as follows:

$$\begin{gathered} ∆v=\frac{90\mathrm{Hz}\times 4.69\mathrm{T}}{1.41\mathrm{T}} \\ =300\mathrm{Hz} \end{gathered}$$

Part (b) Step 1. Given information

The external field is $7.05\mathrm{T}$.

Part (b) Step 2. Calculate frequency difference 

Substitute displaced frequency, external field, and field strength in the equation (1) as follows:

$$\begin{gathered} ∆v=\frac{90\mathrm{Hz}\times 7.05\mathrm{T}}{1.41\mathrm{T}} \\ =450\mathrm{Hz} \end{gathered}$$

Part (c) Step 1. Given information

The external field is$18.8\mathrm{T}$.

Part (c) Step 2. Calculate frequency difference 

Substitute displaced frequency, external field, and field strength in the equation (1) as follows:

$$\begin{gathered} ∆v=\frac{90\mathrm{Hz}\times 18.8\mathrm{T}}{1.41\mathrm{T}} \\ =1200\mathrm{Hz} \end{gathered}$$

Step 7. Calculate the chemical shift

The chemical shift does not depend on the external magnetic field. Therefore, the chemical shift will be the same in all the subparts.

Calculate the chemical shift as follows:

$\delta =\frac{\mathrm{Observed}\mathrm{shift}\left(\mathrm{Hz}\right)\times {10}^6}{\mathrm{Operating}\mathrm{frequency}}$

It is considered that the proton is absorbed in data-custom-editor="chemistry" $60\mathrm{Hz}$in a data-custom-editor="chemistry" $60\mathrm{MHz}$instrument.

Substitute the observed shift and operating frequency in the above equation as follows:

$$\begin{gathered} \delta =\frac{90\mathrm{Hz}\times {10}^6}{60\mathrm{Hz}} \\ =1.5\mathrm{ppm} \end{gathered}$$

Thus, the chemical shift is data-custom-editor="chemistry" $1.5\mathrm{ppm}$.