Hi Amr,

I could not figure out what is your problem. Do you have 2 isomers (?) in a mixture or only one pure compound containing 2 terminal true acetylenic moieties? In case of one compound, how could you prove the presence of these 2 acetylenic functions, since you cannot see one of them?

Then, you say that "just its own carbon was elucidated by HMBC". Are you aware of the tricky situation you are facing when recording NMR spectra of such chemical function, namely, with reverse 2D experiments such as HMBC? Indeed, for direct correlations (1J: HSQC, HMQC, ...) you apply a kind of an INEPT sequence composed of 2 delays (most often called d2), separated by a 180° pulse on the proton. This sequence is used to get rid of the unwanted much larger signals of non coupled singlet protons borne by isotope 12 carbon atoms. These d2 delays are calculated from the direct 1J coupling constant between the 13C (sensitive carbons) and the detected protons as "1/2J".

The 1J coupling constant depends on the hybridization of the carbon atom: the usual value for D2 is about 3.5-4 ms (1/2x125 Hz = 1/250 s) when "sp3" (1J ± 125-130 Hz). However, it is about 2 fold higher ( 1J ± 250 Hz) with "sp" (like it is the case for acetylenic C–H). That is to say, in order to refocus the 2 components of magnetization, if you apply indifferently the d2 delays without changing the J-value (two-fold bigger, thus d2 must be twice smaller = 1/2x250 = 1/500 = 2 ms), then, the signal for either the carbon (in a 13C spectrum) or the signal of the coupled proton in a reverse 1H-detected 2D heteronuclear spectrum like HSQC or HMQC, can simply disappear!!!

For sure, HMBC is a 2J-3J long-range correlation, but in your "triple bond attached to an acetylenic terminal residue, there is no other proton in such 2, 3 or even 4J position. Thus, I really wonder what kind of correlation you have seen on the first one. In addition, in such an experiment, you apply a "low-pass J-filter" to get rid of the "direct" correlations (the 1J = terminal proton!). Once again, this J-filter uses a d6 delay that is set with regards to the coupling constant. Therefore, it is not well adjusted to reject sp3, sp2 or sp all at once at the same time. You should record 2 sets of data using the usual value and an other one with half of this value.

Hope this helps.

Joseph Vercauteren

To answer your question, in natural compounds, you can at least expect the natural abundance of deuterium = 0.0156% (of deuterium, on each protonated position, i.e. one deuterium atom in 6,420 of hydrogen).

We consider 2 main plant metabolisms (the "C3" or the "C4") with regard to photosynthesis (carbohydration from CO2+H2O: the natural abundance in water drives the one in natural substances issued from carbohydrates). Depending on the C3 or C4 pathway, this medium value can change but very slightly. It can never enrich 100% on one specific position on the framework.

Thus, you may have not seen it for 2 reasons: because of the bad conditions used to record NMR (see previous comment), or because of the acidity of this acetylenic proton that was exchanged with a deuterium (if you exposed the compound to a protic deuterated solvent, CD3OD, for instance).

The possibility is slim.

Ya Prof. Joseph, I think more likely that this proton was exchanged with a deuterium due to the acidic character of such acetylenic proton which comes in conjugation with double bond