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 500-MHz instrument, what would be the chemical shift of the CHBr3 proton? Enter your answer in the provided box.

Answer:The chemical shift (δ) is 6.88ppm.

Explanation:

We have the following data:

Absorption frequency of the proton in bromoform=2065Hz

Frequency of the NMR spectrometer (instrument) = 300MHz

The formula for calculating the chemical shift (δ) in PPM is:

chemical shift (δ) = {[Frequency of proton (Hz) - Frequency of reference (Hz]:Frequency of NMR spectrometer (MHz }

Using the formula for chemical shift we can calculate the value of chemical shift (δ) in ppm

chemical shift (δ) = {[2065Hz-0]:300*10⁻⁶}

chemical shift (δ) = 6.88*10⁶

chemical shift (δ) = 6.88ppm for a 300MHz NMR spectrometer

The chemical shift (δ) is a ratio of frequency absorbed by proton with that of NMR spectrometer frequency hence the chemical shift value would remain same what ever NMR spectrometer frequency we use. Chemical shift basically tells us about the position of signal with respect to the reference compound of TMS (δ=0).

chemical shift (δ) is measured in

So the value of (δ) is same for any spectrometer used.

The chemical shift (δ) for a 500MHz NMR spectrometer used would also be 6.88PPM.

Alternatively since the frequency of proton absorbed is directly related to the magnetic field applied that is the frequency of NMR spectrometer hence:

Let the frequency of proton absorbed in 300MHZ=V₁=2065

Let the frequency of proton absorbed in 500MHZ=V₂=?

frequency of proton absorbed∝Applied magnetic field (Frequency of NMR spectrometer)

So V₁/V₂=[300*10⁶]/[500*10⁶]

V₂=[2065*500]:300

V₂=3441HZ

For a 500 MHz proton the frequency of absorption would be 3441MHz

using this frequency we can calculate chemical shift (δ) using above formula:

(δ) = [3441-0]/[500*10⁻⁶]

(δ) = 6.88PPM

Hence we obtain the same value of chemical shift in both the spectrometers.