# Sensitivity of Carbon NMR Compared to Proton NMR

I am trying to figure out exactly how much less sensitive carbon NMR is from proton NMR. In a lecture, my professor said that carbon NMR is 1/5800th as sensitive as proton NMR for the following reasons:

• Abundance of Carbon-13 is 1.1% so sensitivity is decreased by a factor of (0.011)
• The precession frequency (difference in energy between the alpha and beta states) is about 1/4 of the precession frequency of a proton. Because this is equivalent to electrical output on the oscillator coil, this will be 1/4 as powerful (0.25)
• The magnetogyric ratio of Carbon-13 is 1/4 that of Hydrogen-1 (0.25)
• The dipole generated by Carbon-13 is 1/4 that of Hydrogen-1 (0.25)

Thus: 0.011 * 0.25 * 0.25 * .25 = 1/5800 (roughly)

My only objection is that the magnetogyric ratio is proportional to the magnetic moment and the spin number. Thus, I would not expect the 1/4 factor to be counted twice since it is assumed based on the definition of magnetogyric ratio. For this reason, I would expect the sensitivity of carbon NMR to only by 1/1600th of proton NMR.

i.e. 0.011 (abundance) * 0.25 (lower precession frequency) * 0.25 (magnetogyric ratio / dipole moment) = 1/1600

I am not sure who is right. I cannot find any sources that will detail the exact sensitivy differences between proton and carbon NMR and why it is so

• – Buttonwood Apr 12 '20 at 22:42
• I've already read that article; however, it does not specify the effect of the precession frequency or the relationship between the dipole moment and the gyromagnetic ratio – Eli Jones Apr 12 '20 at 22:58

The receptivity ratio is ca. $$5800:1$$, as listed in the Bruker tables (under the column Receptivity > Natural rel. $$\ce{^13C}$$). Receptivity is a product of the natural abundance and an "inherent NMR sensitivity", which scales as $$\gamma^3 I(I+1)$$. For proton and carbon, $$I = 1/2$$, so the $$I(I+1)$$ term cancels out and we are left with

\begin{align} \frac{\text{receptivity }\ce{^1H}}{\text{receptivity }\ce{^13C}} &= \frac{\text{abundance }\ce{^1H}}{\text{abundance }\ce{^13C}}\cdot \left(\frac{\gamma_\ce{H}}{\gamma_{C}}\right)^3 \\ &= \frac{100\%}{1.1\%} \cdot (4)^3 \\ &\approx 5800. \end{align}

Why $$\gamma^3$$? One must bear in mind that at the end of the day, the signal that is detected is not the magnetisation directly. The precessing magnetisation that is excited in an NMR experiment induces a voltage in the coil that surrounds the sample. This voltage is the actual quantity that is detected.

By Faraday's Law, the voltage generated is proportional to the rate of precession, i.e. the Larmor frequency $$\omega_0$$, as well as the actual magnitude of the magnetisation $$M$$. The Larmor frequency is itself proportional to $$\gamma$$ (it is defined by $$\omega_0 = -\gamma B_0$$), so that accounts for one out of three.

Any magnetisation that we can measure must be derived from the equilibrium magnetisation $$M_0$$. Since the equilibrium magnetisation points along the $$z$$-axis, $$M_0$$ is the expectation value of $$z$$-magnetisation $$\langle M_z \rangle$$ at equilibrium. $$\langle M_z \rangle$$ itself can be related to the populations of the spin-up and spin-down states, conventionally denoted as $$n_\alpha$$ and $$n_\beta$$:*

\begin{align} \langle M_z \rangle &= \gamma \langle I_z \rangle \\ &= \gamma \left[ \left(\frac{\hbar}{2}\right)n_\alpha + \left(-\frac{\hbar}{2}\right)n_\beta \right] \\ &= \frac{\hbar\gamma}{2}(n_\alpha - n_\beta) \end{align}

So that accounts for one more occurrence of $$\gamma$$. The third and final one comes from the population difference $$(n_\alpha - n_\beta)$$. According to the Boltzmann distribution, we have

\begin{align} \frac{n_\alpha}{n_\beta} &= \exp\left(\frac{\Delta E}{k_\mathrm{B}T}\right) \\ &\approx 1 + \frac{\Delta E}{k_\mathrm{B}T} \end{align}

where $$\Delta E = E_\beta - E_\alpha$$ is the (positive) energy difference between the $$\alpha$$ and $$\beta$$ states. In the second step we have expanded the exponential as a Taylor series and truncated at first-order, making the (in this context, very valid) assumption that $$\Delta E \ll k_\mathrm{B}T$$. So now:

\begin{align} n_\alpha - n_\beta &= \left(1 + \frac{\Delta E}{k_\mathrm{B}T}\right) n_\beta - n_\beta \\ &= \frac{\Delta E}{k_\mathrm{B}T} \cdot n_\beta \end{align}

The energy of the spin states in an external magnetic field (the Zeeman splitting) will depend on the gyromagnetic ratio:

$$\Delta E = \gamma\hbar B$$

So, working backwards: a nuclide with a smaller gyromagnetic ratio, when placed in an external magnetic field,

• has a smaller difference between the population of spin-up and spin-down states (by a factor of $$\gamma$$);
• which leads to a smaller equilibrium magnetisation (by a factor of $$\gamma$$);
• which, when excited, generates a smaller voltage in the detection coil (by a factor of $$\gamma$$).

Finally, it should be noted that this does not take into account any noise in the spectrum; it is just a measure of the signal that is available. As far as I can tell, the (root-mean-square) noise also scales with the frequency, and this leads to a theoretical signal-to-noise ratio that scales as either $$\gamma^{3/2}$$ or $$\gamma^{11/4}$$ depending on the analysis made (essentially, the exponent is slightly reduced from the nominal value of 3). I don't have the ability to explain this properly, but I can provide references that are probably helpful:

* Populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $$|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$$, then we have $$n_\alpha = \overline{c_\alpha^* c_\alpha}$$ where the bar indicates an average over all spins in the ensemble. For more information please consult a suitable resource on density matrices.