13C NMR of "Perdeuterated" Solvents

When one acquires a ¹³C NMR spectrum of a sample, the deuterated solvent is observed with resonances characterized by the J coupling pattern of the deuterons attached to the carbon atoms. (CD = 1:1:1 triplet, CD₂ = 1:2:3:2:1 quintet, CD₃ = 1:3:6:7:6:3:1 septet). Often, however; small peaks are observed near the main solvent resonances. These are due to other isotopomers of the solvent. An example of this is shown in the bottom spectrum of the figure below for the high frequency resonance of "THF-d₈". The spectrum consists mainly of the expected 1:2:3:2:1 quintet however, there is a small peak present on the high frequency side of the quintet marked by the arrow. It is due to one of the components of the C1 resonance from THF-1,2,2,3,3,4,4-d₇. The supplier of the solvent claims that the isotopic purity is 99.5 atom % D. If the deuteration is uniform, approximately 0.5 % of the molecules will contain a single proton. Half of these molecules will be mono-protonated on the high frequency carbons. If a DEPT spectrum is acquired on the same sample, the non-protonated carbons are suppressed leaving only the mono-protonated carbons. This is shown in the top spectrum of the figure, which shows a 1:1:1 triplet from CHD. The isotope shift between CHD and CD₂ is 0.354 ppm. Other examples can be found here and here.

The 13C and 13C DEPT Spectrum of "Acetone-d6"

A student recently asked me why her solvent resonance for acetone-d₆ was showing up in her ¹³C DEPT spectrum. Since the methyls of acetone-d₆ have no protons they will not show up in a DEPT spectrum however, the solvent is typically bought at 99.9 atom % deuterium which means that there is a very small amount of acetone-d₅ (CD₃ - CO - CD₂H). The - CD₂H group will show up as a positive signal. Both ¹³C and ¹³C DEPT spectra are shown below for "acetone-d₆". In the ¹³C spectrum in the bottom trace, one can see the expected 1:3:6:7:6:3:1 septet for a spin I = 1/2 nucleus coupled to three equivalent spin I = 1 nuclei. In the DEPT spectrum in the upper trace one sees only the small fraction of protonated carbons and the spectrum is a 1:2:3:2:1 quintet resulting from a spin I = 1/2 nucleus coupled to two equivalent spin I = 1 nuclei. One can also see the isotope effect of 0.254 ppm between the two isotopomers. Although the quintet is also present in the ¹³C spectrum, it is not seen simply because it is too small in comparison to the septet.

Finding "Lost" Deuterated 13C Signals

Several times a year I will get a question like, "I just deuterated my compound and now I can't see the ¹³C NMR signals for the deuterated carbons - why not?" There are at least three reasons why seeing these signals may be difficult. First of all, the principal relaxation mechanism for protonated carbons is the dipolar interaction between the ¹³C and the attached protons. When the protons are replaced by deuterium the dipolar mechanism for relaxation is much less efficient and the T₁ for the deuterated carbon may be more than an order of magnitude greater than that of the comparable protonated carbon. The long T₁ for the deuterated carbon may mean that it will be saturated when using the "standard parameters" and either not show up or be of much reduced intensity. Secondly, the intensity of the carbon signal will be split into a multiplet due to ¹³C - ²H J coupling depending on how many deuterons are attached to the carbon (1:1:1 triplet for CD, 1:2:3:2:1 pentet for CD₂ and 1:3:6:7:6:3:1 septet for CD₃). This means that the intensity of the proton decoupled CH_n singlet will be spread over all of the lines of the ¹³C - ²H J multiplet with a corresponding large decrease in signal-to-noise ratio for any one of the lines of the multiplet compared to the proton decoupled singlet for the CH_n carbon. Thirdly, there is almost no nuclear Overhauser enhacement for deuterated carbons as there are no directly bound protons. This means that if proton decoupling is applied both during the acquisition and the recycle delay, the protonated carbon signals will grow at a much faster rate than the deuterated carbon signals as the number of scans is increased. In the figure below is an example. The lower panel shows the partial spectrum of a protonated organic molecule using proton decoupling during both the acquisition time and the 2 second recycle delay. The middle panel shows the similar spectrum of a partially deuterated molecule. Note that the deuterated carbon signals are "lost". The top panel shows a spectrum acquired with the same number of scans as that in the middle panel but with a 60 second recycle delay and decoupling only during the acquisition time (i.e. not during the recycle delay). Acquirng the data in this way eliminates the problems of long relaxation times and differential nOe's. In this spectrum one can see the deuterated carbon signals however the signal-to-noise ratio is lower than those for the protonated carbons due to the splitting from ¹³C - ²H J coupling. In this particular case also, a further loss in signal-to-noise ratio results because the nitrogen bearing carbon is also broadened due to partial self decoupling of the ¹⁴N from the ¹³C.

Bottom line - collect your data over a long period of time with a long recycle delay and inverse gated proton decoupling.

Thank you to David Lapointe for providing the samples for the figure.