Friday, January 10, 2020

Where is my INEPT signal? - Proton Exchange Issues

INEPT and DEPT sequences are routinely used to enhance the NMR signals for low γ nuclides such as 15N or 13C.  The enhancement relies on polarization transfer between the protons J-coupled and the low γ nuclide.  The pulse sequences incorporate delays based on the reciprocal of the J-coupling constant between the protons and the low γ nuclide.  In the case of 15N INEPT, the enhancement for each scan can be as much as γHN (~ 10) compared to that from a conventional one-pulse sequence with inverse gated decoupling.  Furthermore, the recycle delay for the INEPT sequence depends on the the 1H T1 relaxation time rather than that of 15N.  1H T1's are typically an order of magnitude (or more) less than those of 15N so the recycle delays required for 15N INEPT spectra are at least ten (and possibly 100 times) shorter than those required for one-pulse data collection.  These two factors mean that the true time saving for a 15N INEPT measurement compared to a one-pulse 15N measurement can be on the order of 100 - 1000 times.  There are, however cases where 15N INEPT signals are attenuated or entirely nonexistent.  Attenuated 15N INEPT signals are observed when the protons (with short T1) coupled to 15N exchange with those of water (longer T1) on a time scale of seconds.*  The problem arises because of saturation transfer during the inverse gated decoupling used during the acquisition time. The partially saturated protons are unable to transfer as much polarization to the 15N as they would were they fully polarized. The problem can be reduced if a recycle delay much greater than the T1 relaxation time of the water protons is employed.  If the protons bound to 15N undergo exchange with other labile protons at a rate fast with respect to the 1H-15N J coupling interaction, polarization transfer from 1H to 15N is not possible and a 15N INEPT signal cannot be observed.  This is demonstrated in the figure below.
Concentrated solutions of the methyl ester of anthranilic acid and anthranilic acid were prepared in DMSO-d6.  The 15N NMR data were collected on a 600 MHz instrument with a cryoprobe.  The left-hand panel of the figure compares the 15N one-pulse spectrum with inverse gated decoupling (bottom) to the INEPT spectrum (top) for the methyl ester.  The spectra were collected with the same number of scans. For the methyl ester, the 15N bound protons do not exchange with any other labile protons.  The enhancement in the 15N INEPT spectrum is clear.  Similar spectra for anthranilic acid are shown on the right-hand side of the figure.  In anthranilic acid, the 15N bound -NH2 protons undergo intramolecular exchange with the acid proton at a rate fast with respect to the one-bond 15N-1H coupling constant (~90 Hz).  As a result, polarization transfer is not possible and no INEPT signal is observed.  The same is true for the meta- and para- isomers (data not shown). 

Thank you to Jin Hong for sharing her experience with collecting 15N INEPT data for anthranilic acid and Mojmir Suchy for kindly providing the samples.

* G.D. Henry and B.D. Sykes, J. Magn. Reson. B, 102, 193 (1993).

6 comments:

Eugene said...

This is very cool. Can you please provide some implementation details for how you got this to work? (Presumably you used Bruker spectrometers? What sequences and parameters?)

Glenn Facey said...

Eugene,
For the one-pulse observations, I used the zgig30 pulse program with a 10 second recycle time. For the INEPT data, I used the ineptrd pulse program optimized for 90 Hz coupling, D1 = 2s, D3 = 1.85ms, D4 = 2.78ms. In all cases 256 scans were collected.

Glenn

Alberto said...

Dear Glenn

Thank you very much for the explanation. It saves a lot time.
I have a question. In the INEPT system (ineptrd), which nitrogen apper (NR3?) and how (Carbon coupling?)
Thank you very much for your help

tlp said...

What if the proton rapidly exchanges with other nitrogen atom (e.g. tautomer), would you see the INEPT signal for both or for none?

Also, could you please point to a good resource that explains how the optimization to J=90 Hz works? Like how did you come up with D1,D3, and D4 numbers above?

Glenn Facey said...

For truly fast exchange between tautomers, one would observe only a single averaged structure in the 1H and 15N spectra. The observation of an INEPT signal would depend on whether the observed 1H-15N coupling of the averaged structure corresponded to the delay set in the INEPT sequence.

My choice of a delay based on 90 Hz was because the one bond 1H-15N coupling constant was ~ 90 Hz.

Glenn

tlp said...

Thank you for the answer!
I've found the math in other sources. Basically delay optimization means setting
d2 = 1/(4 * J[Hz])
d3 = 1/(6 * J[Hz])