Question

The 1H NMR signal for bromoform (CHBr3) appears at 2065 Hz when recorded on a 300−MHz NMR spectrometer. If the spectrum was recorded on a 200−MHz instrument, what would be the chemical shift of the CHBr3 proton? Enter your answer in the provided box.

Answer 1

The chemical shift (δ) for CHBr₃ proton = 6.88 ppm

Step-by-step explanation:

In NMR spectroscopy, Chemical shift (δ) is expressed in parts per million (ppm) and is given by the equation:

`\delta (ppm)= (Observed\: frequency, \\u (Hz))/(Frequency\: of\: Spectrometer, \\u^(') (MHz))* 10^(6)` ....equation (1)

Given: Observed frequency: ν₁ = 2065 Hz,

Spectrometer frequency: ν'₁ = 300 MHz, ν'₂ = 200 MHz

To calculate the chemical shift (δ) for the given CHBr₃ proton, we use the equation (1)

`\delta = (\\u_(1) (Hz))/(\\u_(1)^(') (MHz))* 10^(6) = (2065 Hz)/(300 * 10^(6) Hz)* 10^(6) = 6.88 ppm`

Since in NMR spectroscopy, chemical shift is a field independent scaling. Thus the value of the chemical shift of a given proton, such as CHBr₃ proton, is independent of the magnetic field strength of the spectrometer.

So the value of chemical shift of a given proton remains same when measured with a 300 MHz and 200 MHz NMR spectrometer.

Therefore, the chemical shift (δ) for CHBr₃ proton = 6.88 ppm