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

Answer:

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

Explanation:

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

$\delta (ppm)= \frac{Observed\: frequency, \nu (Hz)}{Frequency\: of\: Spectrometer, \nu^{'} (MHz)}\times 10^{6}$                     ....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)

$\delta = \frac{\nu_{1} (Hz)}{\nu_{1}^{'} (MHz)}\times 10^{6} = \frac{2065 Hz}{300 \times 10^{6} Hz}\times 10^{6} = 6.88 ppm$

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