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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)

enter image description here

enter image description here

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?

enter image description here

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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.

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  • $\begingroup$ 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. $\endgroup$ Mar 11 '20 at 3:39
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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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