Question

Tetramethylsilane is used as a reference molecule for a proton NMR. It has a reference frequency of 600 MHz for its methyl protons. The chemical shift for protons in methanol is at 3.35 ppm. What is the difference in the frequency of resonance for the CH3 groups in methanol compared to the CH3 groups in tetramethulsilane?

Please explain how to get the answer of 2010 Hz

Answer 1

The chemical shift in ppm(parts per million) is defined as follows

signal Vreference x 10 spectrometer frequency

Where

5chemical shift in ppm

resonance frequency of the signal in Hz

signal

resonance frequency of the reference in Hz

reference

Here, tetramethylsilane is being used as a reference and has a frequency of 600 MHz with a ppm value of 0. Hence, the spectrometer frequency must be 600 MHz =

600X 106 H2

We have to find the difference between resonance frequency of signal and reference, i.e. the numerator in the above relation.

Hence, given that the chemical shift is 3.35 ppm

signal Vreference S spectrometer frequency 106 signalVreference 600 x 10 Hz 3.35 signal Vreference 3.35 x 600 Hz 2010 H2

Hence, the resonance frequency of CH3 from in methanol is 2010 Hz downfield from resonance frequency of tetramethylsilane methyl group.