Coupling in C-NMR


> 
> Given the structure below, sketch the $^{13}$C - spectrum in a $^{13}$C marked sample
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> The data given is the chemical shifts (left) and the coupling constants (right)
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> [![enter image description here](https://i.stack.imgur.com/EhtcD.png)](https://i.stack.imgur.com/EhtcD.png)
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> [![enter image description here](https://i.stack.imgur.com/MMPrk.png)](https://i.stack.imgur.com/MMPrk.png)
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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?


[![enter image description here](https://i.stack.imgur.com/H3M7A.png)](https://i.stack.imgur.com/H3M7A.png)




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.


 3

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. 


 2