[Rutherford]

Is only the tertiary nitrogen a stereocenter, or both are?

[AlphaScent]

Brain fart.

I did indeed write them out a bit ago and it is 8.

Dummy.

[AWK]

Only quaternary nitrogen atom does not interconvert between R and S configurations. In the case of tertiary one only sterically hindered N atom involved in three-fold ring (oxaziridines or aziridines) shows sufficient inversion barrier. First reports ~1976.

[Rutherford]

So there are 8 stereoisomers? The tertiary nitrogen can be regarded as a stereocenter.

[clarkstill]

All this talk of lone pair tunnelling is largely irrelevant; both nitrogens are amides, not amines, so are sp2 not tetrahedral.  The one in the ring is fixed, so definitely won't have any E/Z isomerism.  The open chain amide can potentially adopt E or Z stereochemistry, so overall there are 8 stereoisomers when the 2 stereogenic centres are considered.

[Rutherford]

I don't think that we can strictly say that they are sp2, as they are actually something between sp2 and sp3. Now I really don't know what to say about the number of the stereoisomers.

[clarkstill]

Nope, amides are most definitely sp2.  Have a look at any protein crystal structures and you'll see hundreds of them...

[Rutherford]

Okay, I read a little about it. So, there are 2 carbon stereocenters and one nitrogen atom causing E/Z isomerism.

The other nitrogen (in the rings) is still problematic for me. It can't be planar as the angle tension would be very big (30°). It has to be near sp3, and that would imply that it is a stereocenter.

[clarkstill]

Yeah I guess it's probably slightly distorted from being planar, but even so it won't add to the stereoisomer count as the ring junction could never be trans...

[Rutherford]

How do you mean trans? Why wouldn't it be possible as I draw in the picture? The six membered ring could also be below the plane.

[clarkstill]

I agree both cis stereoisomers are possible, but the two trans ones arent:

[Rutherford]

Okay, so it is 8 as you said. Thank you very much.