Acetone-d6 $^{13}C$ NMR Signal Analysis
This question concerns the 13C NMR spectrum of acetone-$d_6$, specifically the signal characteristics of its methyl carbons. Acetone-$d_6$ has the chemical formula $(CD_3)_2CO$.
Deuterium Coupling Mechanism
The signal observed in the 13C NMR spectrum is from the carbon atoms located in the $CD_3$ groups. These carbon nuclei interact (couple) with the deuterium ($D$) nuclei directly bonded to them.
Deuterium ($D$) possesses a nuclear spin quantum number of $I=1$.
Calculating Signal Multiplicity
The number of lines in a split NMR signal is determined by the number of equivalent coupling nuclei ($n$) and their spin quantum number ($I$), using the formula: $ \text{Number of lines} = 2nI + 1 $
In acetone-$d_6$, each methyl carbon is bonded to $n=3$ deuterium atoms. Since $I=1$ for deuterium, the calculation is: $ 2 \times 3 \times 1 + 1 = 7 $
A signal split into 7 lines is referred to as a septet, matching the observation stated in the question.
Intensity Ratio Interpretation
The relative intensities of the lines within the septet follow a specific pattern determined by the statistical distribution of nuclear spin states during the coupling interaction.
For the coupling of a 13C nucleus to three equivalent deuterium nuclei ($n=3, I=1$), the resulting septet exhibits an intensity ratio of $1:3:6:7:6:3:1$.
Final Signal Description
Therefore, the 13C NMR spectrum of acetone-$d_6$ features a signal around 30 ppm, characterized as a septet with the intensity ratio $1:3:6:7:6:3:1$, due to coupling with the three attached deuterium atoms.
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The compound is
