# Coupling in C-NMR

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)

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?

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.

• Thank you for your answer! I just have one question, why isn't the C=O group also affected by the C$^β$? Mar 10, 2020 at 9:54
• $\ce{C=O}$ is not affected by $\ce{C^\beta}$. It is affected by only $\ce{C^\alpha}$. That's why it is only a doublet. Mar 11, 2020 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.