12C/13C Isotope Effects on 1H T1 Relaxation Times

What is the ¹H T₁ relaxation time of chloroform?  It seems like a simple enough question, but the answer is not so simple.  The relaxation rate for any proton is the sum of relaxation rates resulting from several different mechanisms (eg. homonuclear dipolar coupling, heteronuclear dipolar coupling, chemical shielding anisotropy, spin rotation etc...).  Each of these mechanisms of relaxation depends on dynamic effects and the extent to which those processes occur at the Larmor frequency.   Often, in proton-rich organic compounds, ¹H T₁ relaxation is dominated by the homonuclear dipolar coupling interaction.  For chloroform, with only a single proton, there can be no intra-molecular homonuclear ¹H dipolar interaction and the ¹H relaxation rate must depend on other mechanisms.  One of these mechanisms is the result of the heteronuclear dipolar coupling interaction.  For the ¹³C isotopologue of chloroform, one would expect a significant heteronuclear dipolar interaction between the directly bound ¹H and ¹³C.   This interaction is absent in the ¹²C isotopologue and one would therefore expect the T₁ relaxation time of ¹³CHCl₃ to be much shorter than that of ¹²CHCl₃.  This is illustrated in the figure below.

The ¹H T₁ relaxation times for both ¹²CHCl₃ and ¹³CHCl₃ were measured with the inversion recovery method for a degassed, dilute (1%) sample of chloroform in acetone-d6.  The inversion recovery delay was varied from from  1 sec. to 300 sec.  The recycle delay was 300 sec.  Relaxation is much more efficient for ¹³CHCl₃ compared to ¹²CHCl₃.  The T₁ for ¹³CHCl₃ is only 46% that of ¹²CHCl₃, indicating the significance of the heteronuclear ¹H - ¹³C dipolar coupling interaction as a relaxation mechanism.       

Boron Isotope Effects in Fluorine NMR Spectra

In previous posts on this BLOG, examples of  ¹H/²H and ¹²C/¹³C isotope effects were discussed.  The figure below shows an example of a ¹⁰B/¹¹B isotope effect observed in the ¹⁹F NMR spectrum of tetrabutyl ammonium tetrafluorobarate.

The spectrum clearly shows two resonances separated by 0.05 ppm with an intensity ratio of approximately 20:80 corresponding to the natural abundances of  ¹⁰B and ¹¹B, respectively.  The low frequency resonance is due to ¹¹BF₄^-.  Since  ¹¹B is a spin I = 3/2 nuclide we observe a 1:1:1:1 quartet with J = 1.25 Hz corresponding to the one bond ¹⁹F - ¹¹B coupling.  The high frequency resonance is due to ¹⁰BF₄^-.  Since  ¹⁰B is a spin I = 3 nuclide we observe a very poorly resolved 1:1:1:1:1:1:1 septet with J ~ 0.4 Hz corresponding to the one bond ¹⁹F - ¹⁰B coupling.  

Isotope Effects and the 19F - 13C HMQC Spectrum of Trifluoroacetic Acid

The ¹⁹F - ¹³C HMQC spectrum of trifluoroacetic acid is shown in the figure below.

The data were collected with a delay appropriate for a ¹⁹F - ¹³C J  coupling constant between the ¹J_F-C coupling constant of 284 Hz and the ²J_F-C coupling constant of 44 Hz.  The top and side traces are the one-pulse ¹⁹F and ¹³C spectra, respectively.  Why are the HMQC responses not at the same ¹⁹F chemical shift and why aren't they correlated to the peak in the ¹⁹F spectrum?  In order to answer these questions one must take into consideration the ¹⁹F - ^{12, 13}C isotope effects.  The chemical shift of the fluorine depends on whether it is bound to a ¹²C or a ¹³C.  The effect is largest across one bond and gets smaller over multiple bonds.  The ¹⁹F NMR spectrum for trifluoroacetic acid is shown in the figure below with and without ¹³C broadband decoupling in the upper and lower traces, respectively.

Approximately 98% of the trifluoroacetic acid is the ¹²CF₃-¹²COOH isotopomer, giving rise to a large singlet plotted off-scale in the figure. Approximately 1% of the signal is from the ¹³CF₃-¹²COOH isotoponer giving rise to a doublet with ¹J_F-C = 284 Hz and approximately 1% of the signal is from the ¹²CF₃-¹³COOH isotoponer giving rise to a doublet with ²J_F-C = 44 Hz.  All of these signals are clearly present in the lower trace of the figure.  When ¹³C broadband decoupling is applied, the doublets collapse into singlets.  The singlets from each of the isotopomers are resolved in the top trace.  The one-bond ¹⁹F - ^{12, 13}C isotope effect is 0.13 ppm and the two-bond effect is 0.02 ppm.  The figure below shows the same HMQC data with the spectrum from the top trace used as a projection.

One can see that the HMQC responses are correlated to their respective isotopomers.  These effects are also present in ¹H - ¹³C HMQC spectra, but the ¹H - ^{12, 13}C isotope effect is much smaller than the ¹⁹F - ^{12, 13}C isotope effect.  

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.

Complexed Solvents

I was once asked by an inorganic chemist: why do I have two THF signals in the spectrum of my compound dissolved in THF-d8? Many inorganic compounds crystallize with complexed solvent molecules as a fundamental component of their structure. This is particularly true of tetrahydrofuran (THF). The complexed solvent molecules are released when the solid compound is re-dissolved in solution and can easily be detected by high resolution NMR. The figure below shows the 500 MHz ¹H NMR spectrum of an inorganic compound containing complexed THF which was re-dissolved in THF-d8. One can see the spectrum of the residual protons of the THF-d8 solvent and the spectrum of the complexed THF that was released when the solid was dissolved. The signals are separated due the isotope effect.

Dilute "D2O" in Benzene-d6

The ¹H NMR spectrum of a mixture of H₂O and D₂O is a single line at about 4.8 ppm. The H₂O and HDO resonances are unresolved in the spectrum due to fast chemical exchange between the isotopomers and possibly line broadening due to radiation damping. When traces of D₂O are added to benzene-d₆, which already contains traces of H₂O, the situation is different. The resonance is shifted by more than 4 ppm to lower frequency compared to the bulk and since the water is now dilute and in small quantities, chemical exchange is slow on the NMR time scale and radiation damping is no longer a problem. The figure below shows the 500 MHz ¹H NMR spectrum of dilute D₂O in benzene-d₆. The isotope shift between H₂O and HDO and the HD coupling constant can easily be measured from the spectrum.

Spin-Spin Coupling Between Equivalent Nuclei

When many chemists are asked what is the ²J_H-H coupling for compounds like methane, acetone, methylene chloride, dimethyl ether or DMSO, they will often return a look of confusion. "There is no coupling," they will say, "the proton spectrum is a singlet". Indeed the proton spectrum is a singlet for these compounds but ²J_H-H is not equal to zero. The only reason that the coupling is not observed in the spectrum is because the chemical shifts of each proton are identical. The coupling can easily be measured by observing the spectrum of a partially deuterated isotopomer. The ²J_H-H coupling constant is equal to ²J_H-D multiplied by the ratio of the gyromagnetic ratios of ¹H to ²H. This is illustrated in the figure below for methylene chloride.In fact, ²J_H-H is -7.192 Hz not +7.192 Hz however, this cannot be determined simply by observing the spectrum. Both spectra were measured for dilute solutions with CDCl₃ as solvent. The residual protons of CDCl₃ were used as the chemical shift reference (7.26 ppm). The chemical shift difference between CH₂Cl₂ and CHDCl₂ is due to an isotope effect.

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.

The Proton Decoupled 31P NMR Spectrum of Triphenyl Phosphate

I was asked about the origin of the small peaks in the 31P NMR spectrum of triphenyl phosphate posted to this blog on October 4, 2007. The small peaks are 13C satellites. The phosphorus is coupled to the 13C in the ipso (J=7.47 Hz), ortho (J=4.98 Hz), meta (J=0.93 Hz) and para (J=1.32 Hz) positions on the aromatic rings. The coupling constants were measured in a high resolution 13C NMR spectrum. The figure below shows the proton decoupled 31P NMR spectrum with the couplings color coded. The satellites due to the meta and para couplings overlap. The displacement of the doublets is due to the isotope effect (see blog entry for September 13, 2007).

Isotope Shifts for Chloroform

Yesterday, an observant student asked me why she frequently sees a small peak in the triplet for CDCl3 in the 13C spectrum.
The small peak is due to a small amount of CHCl3 in the CDCl3 solvent. The CHCl3 signal is a singlet because proton decoupling was used to collect the data. The CDCl3 signal is a 1:1:1 triplet due to the J coupling to the deuteron which is a spin I=1 nucleus having three energy levels. The chemical shift difference between the CHCl3 and CDCl3 isotopomers is called an isotope shift.

Question 1: Since chlorine has two NMR active isotopes (35Cl and 37Cl) why don't we observe J coupling between the 13C and the 35/37Cl like we do for the 13C and 2H?

Answer 1: The relaxation between the energy levels of the chlorine isotopes is very fast and the 13C "sees" each Cl at an average energy. 2H on the other hand relaxes slowly between the energy levels and the 13C "sees" all three energy states of the deuteron.

Question 2: Why don't we see an isotope shift between CD 35Cl3 and CD 37Cl3?

Answer 2: The effect is there ...... just too small to observe.