^1H NMR resonance frequency of a spectrometer is 500 MHz Calculate the lower sta

Question

^1H NMR resonance frequency of a spectrometer is 500 MHz Calculate the lower state excess of^1H nuclei at 25 degree C in ppm What will be the^19 F resonance frequency of this spectrometer in MHz ? A -CH_2- group triplet in the^1H NMR appears as three lines at 1612, 1620 and 1628 Hz from TMS. Calculate the chemical shift of this -CH_2 group in ppm. What is the coupling constant of the -CH_2 signal in Hertz ? A second -CH_2- triplet at 2.62 ppm is coupled to the -CH_2- group in part (c). Calculate the positions of three lines in this triplet in Hertz Magnetogyric ratios of^1H = 2.6752 times 10^8 radian T^-1 s^-1^19F = 2.5181 times 10^8 radian T^-1s^-1, (h = 6.626 times 10^-34 Js. C = 3.0 times 10^8 ms^-1 k = 1.38 times 10^-23 JK^-1)

Explanation / Answer

a.) The Boltzmann ratio of nuclear spins is:

Nupper / Nlower = e-hv/kT

where Nupper / Nlower = ratio of nuclear spins in upper and lower levels

h = Planck's constant in Js

v = frequency in MHz

t = absolute temperature in Kelvin

Nupper / Nlower = e-6.26X10-34 Js X 500 MHz / 1.38 X 10-23 J/K X 298 K

Nupper / Nlower = e-3.13 X 10-31 / 4.112 X 10-21

Nupper / Nlower = 0.999999

If the upper energy state had 1000000 protons, then lower energy state had:

Nlower = 1000000/0.999999

Nlower = 1000001

Thus, Nlower has only 1 proton more than the upper state and hence the lower state excess of 1H nuclei is 1 ppm.

b.) We need to calculate the strength of magnetic field first using the values given for proton nmr.

v0 = magnetogyric ratio X B0 / 2 X pi

v0 in Hz = 2.6752 X 108 radian/T.s X  B0 in T / 6.28

B0 = 11.737 T

Now that we have the field strength, let's use the above mentioned formula to calculate the resonance frequency of 19F.

v0 = magnetogyric ratio X B0 / 2 X pi

v0 in MHz = 2.5181 X 108 radian/T.s X 11.737 T / 2 X 3.14

v0 in MHz = 470.62 MHz

Thus, the resonance frequency of 19F is 470.62 MHz.

c.) We have been given peak positions in Hz. To calculate the chemical shifts in ppm, divide the peak positions in Hz by the resonance frequency of the proton nmr spectrometer.

1612 Hz / 500 MHz = 3.224 ppm

1620 Hz / 500 MHz = 3.240 ppm

1628 Hz / 500 MHz = 3.256 ppm

Because we have a triplet here, coupling constant (J) can be calculated by taking a difference between the "middle" peak and an "outer" peak and multiplying it by the resonance frequency in MHz. Thus,

J = 3.256 ppm - 3.240 ppm = 0.016 ppm

J = 0.016 ppm X 500 MHz

J = 8 Hz

d.) Here, we are assuming that the coupling constant in case of the second -CH2 group is also 8 Hz like the first one. Otherwise the splitting pattern would become far too complicated. Hence, One line will appear at 2.62 ppm, second at 2.604 ppm and the theird at 2.636 ppm i.e. 1310 Hz, 1302 Hz and 1318 Hz repsectively.

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