Given the structure below, sketch the $^{13}$C - spectrum in a $^{13}$C marked sample

The data given is the chemical shifts (left) and the coupling constants (right)

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The correct spectrum is given at the end of this question

I don't really understand what is meant by "in a $^{13}$C marked sample" however I assume that it means that I am supposed to look at C instead of H as my core atom. So from this I tried to determine the coupling. If looking at the α-C I assumed that its neighbouring H would split it into a doublet and then the hydrogens of the β-C (3+1 = 4) would split that doublet into an multiplet of eight. Which seems to be right if compared to the correct spectrum. Similarly, the β-C is first split into triplets and then into doublets, which makes 6 peaks.

However, why would C=O cause a peak? And how does it become a doublet?

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The compound in question is seemingly a polypeptide. If all carbons are labeled $\ce{^{13}C}$, then you can obtain very good $\ce{^{13}C}$-NMR spectrum withe a few accusations with minimum noise. That's probably what "$\ce{^{13}C}$ marked sample" means.

Since it is a polypeptide, $\ce{^{13}C=O}$ is attaced to a $\ce{-NH-}$ group from right hand side because it is a part of peptide bond (an amide). The relevant chemical shift for carbonyl carbon of an amide is around $\pu{175 ppm}$, and therefore the peak at that position is justified. However, this carbonyl carbon is adjacent to $\ce{C^\alpha}$. Since $^1J_\ce{^{13}C-^{13}C}$ coupling constant is given as $\pu{50 Hz}$, this peak is split as a doublet with $J$ value of $\pu{50 Hz}$.

Similarly, $\ce{C^\alpha}$ is attached to $\ce{C^\beta}$ and $\ce{C=O}$ groups as well as to a hydrogen. Since all $\ce{^{13}C-^{13}C}$ coupling constants are identical (according to the given values), the $\ce{^{13}C}$ peak of $\ce{C^\alpha}$ would split first to triplet with $^1J_\ce{^{13}C-^{13}C}=\pu{50 Hz}$ because of two carbons. Then, each triplet would split to doublets because of a single hydrogen on $\ce{C^\alpha}$ with $^1J_\ce{^{13}C-^1H}=\pu{125 Hz}$. Thus, the $\ce{C^\alpha}$ peak would appear as dt as shown in the answer.

Keep in mind that $\ce{C^\alpha}$ peak and $\ce{C=O}$ peaks would further split to aditional doublets with $^1J_\ce{^{13}C-^{15}N}=\pu{15 Hz}$ if present nitrogen is $\ce{^{15}N}$-labeled.

I let OP to figured out what splitting would $\ce{C^\beta}$ peak would show.

  • $\begingroup$ Thank you for your answer! I just have one question, why isn't the C=O group also affected by the C$^β$? $\endgroup$ – katara Mar 10 '20 at 9:54
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    $\begingroup$ $\ce{C=O}$ is not affected by $\ce{C^\beta}$. It is affected by only $\ce{C^\alpha}$. That's why it is only a doublet. $\endgroup$ – Mathew Mahindaratne Mar 11 '20 at 3:39

13C NMR detects only the 13 C isotope of carbon, whose natural abundance is only 1.1%, because the main carbon isotope, 12 C , is not detectable by NMR since its nucleus has zero spin. (https://en.wikipedia.org/wiki/Carbon-13_nuclear_magnetic_resonance)

You seem to have the number of peaks in the α-C and the β-C confused. For example, the α-C peak is first split into a triplet by the two neighboring carbons, and that triplet is split into a doublet because of the α-C hydrogen.

Considering the α-C and the β-C are split by neighboring carbons, I believe you can then determine why the C=O peak is a doublet.


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