Question

In a \(^1\mathrm{H}\) NMR spectrum obtained from a spectrometer operating at a magnetic field of \(14.1\ \mathrm{T}\), two resonances are observed at \(1.25\) ppm and \(5.75\) ppm. The separation between the two resonances (in Hz) is (rounded off to one decimal place).

(Given: gyromagnetic ratio (\(\gamma\)) of \(\mathrm{H} = 2.675\times10^{8}\ \mathrm{T^{-1}\,s^{-1}}\))

Hint:

Find the spectrometer operating (Larmor) frequency from \(\nu_0=\gamma B_0/2\pi\), then convert the ppm gap to Hz using \(\Delta\nu = \delta_{ppm}\times\nu_0(\mathrm{MHz})\).

Answer 1

Correct Answer: 2701

Step 1: Find the spectrometer operating frequency.
The Larmor (resonance) frequency of a nucleus in a field \(B_0\) is
\[ \nu_0 = \frac{\gamma B_0}{2\pi} \]
Put in \(\gamma = 2.675\times10^{8}\ \mathrm{T^{-1}\,s^{-1}}\) and \(B_0 = 14.1\ \mathrm{T}\):
\[ \nu_0 = \frac{(2.675\times10^{8})(14.1)}{2\pi} = \frac{3.7718\times10^{9}}{6.2832} = 6.0028\times10^{8}\ \mathrm{Hz} \]
So the spectrometer runs at \(\nu_0 = 600.28\ \mathrm{MHz}\). This is what turns a ppm value into an actual Hz value, since ppm is defined relative to the operating frequency.

Step 2: Find the ppm separation between the two peaks.
\[ \Delta\delta = 5.75 - 1.25 = 4.50\ \mathrm{ppm} \]

Step 3: Convert the ppm gap into a frequency gap.
By definition, \(\delta(\mathrm{ppm}) = \dfrac{\Delta\nu(\mathrm{Hz})}{\nu_0(\mathrm{MHz})}\), so
\[ \Delta\nu = \Delta\delta\times\nu_0(\mathrm{MHz}) = 4.50\times600.28 = 2701.3\ \mathrm{Hz} \]

Final Answer:
The two resonances are separated by about \(2701.3\ \mathrm{Hz}\) on this 600 MHz-class spectrometer.
\[ \boxed{\Delta\nu \approx 2701\ \mathrm{Hz}} \]