[sh20008]

Hi,
i have p-aminobenzenesulfonamide with a stereocenter next to the nitrogen of the sulfonamide bond.
The h-nmr (CD3OD) shows a dt for all aromatic hydrogens. The XJ coupling constants vary on both sides of the triplett.

7.60 (dt, 2H, 3J = 8.7 Hz, XJ=1.8-2.8 Hz)
6.70 (dt, 2H, 3J= 8.7 Hz, XJ=2.0-2.8 Hz)

What kind of coupling is XJ and why is it a triplett?
Thanks a lot for your *delete me*
Chris

[chiralic]

The aromatic protons of p-aminobenzenesulfonamide will be shifted more downfield due to the presence of de-shielding groups.  These aromatic protons will appear as two doublets.

p-aminobenzenesulfonamide have 4 signals on NMR (2 dt from aromatic protons and two st signal from NH2 proton)...

[sh20008]

Thank you for your answer.

I do not see any H-NMR signal for the NH2-hydrogen atoms. However, that is no problem.
Am I right, that you think, that all aromatic hydrogens also couple with the NH2 hydrogens to give the triplett component of the dt?

[movies]

More likely that you are seeing hyperfine coupling from the protons on the other side of the ring.  So it's not really a dt, but a ddd.

[Ψ×Ψ]

Aromatics are fun!  In addition to the expected ortho-coupling on a benzene ring (usually somewhere around 7.6 Hz), you usually see meta-coupling (with a smaller constant) also.  Para-coupling is pretty rare.  IIRC, protons on N usually don't couple to anything.