Ask Question
22 October, 18:20

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.

+5
Answers (1)
  1. 22 October, 21:16
    0
    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.
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers