Question

The chemical shift of a CH₃ proton in ethanol is δ = 2.20 and the CHO aldehyde proton is δ = 9.80. What is the difference in the local magnetic field between the two regions under an applied field of 15 T

Answer 1

First calculate the resonance frequency for proton uses the following expression:

ν = γ B/2 π ; Here, γ is given for proton is 26.752 x 10⁸ /T s , and magnetic field is1.5 T.

Then; ν = γ B/2π

= [(26.752 x 10⁸/T s) x (1.5 T)] / (2 x3.14)

= 6.39 x 10⁸ /s

= 6.39 x 10⁸ Hz

Now convert it into MHz:

MHz = 10⁶ Hz

6.39 x 10⁸ Hz = 639 MHz

The resonance frequency for proton is 639 MHz.

Now calculate the difference between the local magnetic field of -CH₃ and –CHO as following:

Chemical Shift (δ) = (νH – νTMS (Hz))/operating frequency (Hz) x 10⁶ ppm

Here, δ for methyl group is 2.20,

δ for –CHO is 9.80.

(9.8 – 2.2) ppm = (ν_CH3 – ν_CHO (MHz))/639 x 10⁶ ppm

7.6 = (ν_CH3 – ν_CHO (MHz))/639 x 10⁶

(ν_CH3 – ν_CHO (MHz)) = (7.6 x 639) / 10⁶ = 4856 x 10^-6

(ν_CH3 – ν_CHO (Hz)) = 4856

The difference between the local magnetic field of –CH₃ and –CHO group in1.5 T magnetic field is 4856 Hz