[Rutherford]

Why is then the coupling constant so small that the signals don't split? One N-H is coupled also with a C-H proton, so why is it a singlet?

[discodermolide]

None of the NH protons are coupled with a CH proton.

[orgopete]

If the CH is split by the NH, it too should be split. Whether the NH is split or not depends. In CDCl3, there is enough HCl that speeds the exchange rate and the NH coupling is absent. If the HCl were removed or the spectrum were obtained in DMSO-d6, you would see the splitting.

[Rutherford]

Good explanation. Thanks.

[SinkingTako]

Sorry for thread crashing, (please don't down vote me...)

My teacher has gone through this question, and the answer given was D3. A few reasons: firstly it's reflux for 12 hours in EtOH, so the highly strained ring in D2 and D4 would have been broken in the process. Furthermore D4 cannot be formed in the reaction. (But I don't know how, neither do I know how to form D3.)

Judging from the spectra, D3 also matches the best. There's no amide H, which solves the problem of assigning it in the discussion above. Yeah.

[discodermolide]

The beta-lactam ring is certainly strained, but actually quite stable when you don't want it to be!
Anyway nature makes these 4 membered lactam rings all the time.

[Rutherford]

The production of D4 looks like this. I don't see a way to produce D3.

[SinkingTako]

Isn't D3 just the first structure after C?

[Rutherford]

I see a negative charge on the nitrogen.

[SinkingTako]

Yeah I suppose it can be deprotonated. The product will be a negative charge on the amide, which I think will be relatively stabilised. Even more so than the beta lactam ring which is more susceptible to nucleophilic attack.

Dunno,let me go find the paper.