Question

The natural abundance of \({ }^{13} \mathrm{C}\) is only \(1.1 \%\). Furthermore, its sensitivity in NMR spectroscopy (a measure of the energy difference between a spin aligned with or against an applied magnetic field) is only \(1.6 \%\) that of \({ }^{1} \mathrm{H}\). What are the relative signal intensities expected for the \({ }^{1} \mathrm{H}-\mathrm{NMR}\) and \({ }^{13} \mathrm{C}-\mathrm{NMR}\) spectra of the same sample of \(\mathrm{Si}\left(\mathrm{CH}_{4}\right)_{4}\) ?

Short answer

Answer: The relative signal intensities expected for the ¹H-NMR and ¹³C-NMR spectra of the same sample of Si(CH₄)₄ are 16 and ≈ 0.000704, respectively.

Step-by-step solution

Determine the number of nuclei present in the sample

First, let's determine the number of \({ }^{1}\mathrm{H}\) and \({ }^{13}\mathrm{C}\) nuclei present in a single molecule of \(\mathrm{Si}(\mathrm{CH}_{4})_{4}\). Each molecule has 4 \(\mathrm{CH}_{4}\) groups, and each \(\mathrm{CH}_{4}\) group has 1 Carbon and 4 Hydrogen atoms. So, for one molecule: Number_of_H = 4 * 4 = 16 Number_of_C = 4

Calculate proportion of \({ }^{13}\mathrm{C}\) to \({ }^{1}\mathrm{H}\) nuclei

We are given that the natural abundance of \({ }^{13}\mathrm{C}\) is only \(1.1 \%\). Abundance_of_13C = 0.011 Next, we find the proportion of \({ }^{13}\mathrm{C}\) to \({ }^{1}\mathrm{H}\) nuclei in the sample: Proportion_of_13C = Number_of_C * Abundance_of_13C Proportion_of_13C = 4 * 0.011 = 0.044 Since only \({ }^{1}\mathrm{H}\) is abundantly available in the sample, all Hydrogen atoms are \({ }^{1}\mathrm{H}\), which means: Proportion_of_1H = 1 Proportion_of_1H = 16

Calculate the sensitivity ratio

It is given that the sensitivity of \({ }^{13}\mathrm{C}\) in NMR spectroscopy is \(1.6 \%\) that of \({ }^{1}\mathrm{H}\): Sensitivity_of_13C_to_1H = 0.016 We now calculate the product of the proportion and the sensitivity ratio for both nuclei: Relative_signal_intensity_same_sample_H = Proportion_of_1H Relative_signal_intensity_same_sample_H = 16 Relative_signal_intensity_same_sample_C = Proportion_of_13C * Sensitivity_of_13C_to_1H Relative_signal_intensity_same_sample_C = 0.044 * 0.016 Relative_signal_intensity_same_sample_C = 0.000704

Present the answer

The relative signal intensities expected for the \({ }^{1}\mathrm{H}-\mathrm{NMR}\) and \({ }^{13}\mathrm{C}-\mathrm{NMR}\) spectra of the same sample of \(\mathrm{Si}(\mathrm{CH}_{4})_{4}\) are: For \({ }^{1}\mathrm{H}-\mathrm{NMR}\): 16 For \({ }^{13}\mathrm{C}-\mathrm{NMR}\): ≈ 0.000704