Question

Question:Using a 60-MHz spectrometer, a chemist observes the following absorption: doublet, J = 7 Hz, at 4.00

(a) What would the chemical shift () be in the 500-MHz spectrum?

(b) What would the splitting value J be in the 500-MHz spectrum?

(c) How many hertz from the TMS peak is this absorption in the 60-MHz spectrum? In the 500-MHz spectrum?

Short answer

Answer

1. The chemical shift of remains the same irrespective of the operating frequency of the spectrometer.
2. The splitting value J= 7 Hz.
3. For 60 MHz, the operating frequency is 240 Hz and for 500 MHz the operating frequency is 2000 Hz.

Step-by-step solution

Chemical shift

The chemical shift is the fraction and its formula is as shown:

$\mathbf{Chemical} \mathbf{shift}\mathbf{=}\frac{\mathbf{shift} \mathbf{downfield}\left(\mathbf{Hz}\right)}{\mathbf{spectrometer} \mathbf{frequency}\left(\mathbf{MHz}\right)}$

Subpart (a)

The given chemical shift value for the spectrometer operating frequency 60 MHz is 4.0 ppm.

The value of chemical shift does not depend on the operating frequency of the spectrometer. Hence, the value of chemical shift remains the same and it is 4.0 ppm.

Subpart (b)

The coupling constant J value does not depend on the operating frequency of the spectrometer. Hence, its value is 7 Hz.

Subpart (c)

The frequency shift is calculated as shown below:

$$\begin{gathered} frequency shift=chemical shift\times spectrometer frequency \\ =4.0 ppm\times 60 MHz \\ =240 Hz \\ frequency shift=chemical shift\times spectrometer frequency \\ =4.0 ppm\times 500 MHz \\ =2000 Hz \end{gathered}$$