[Dynamo22]

When reacting the molecule below with xs Methanol in 3eq of H2SO4 with benzene and distillation then washed with NaHCO3. It forms a product of C5H9NO (I'm sorry that's a lot of reagents). Some spectral data: IR- 3219, 2944, 2866, 1666. C13 - 173, 42, 31.5, 22.3, 20.9. H NMR 7.40 (1H, s), 3.31 (2H, m), 2.34 (2H, m), 1.75 (4H, m). I thought it may form one of the rings below, but I am not sure if you can have ring metathesis in this reaction. I'm mainly confused about the 173 C13 signal which would indicate a C=O bond but the 1666 IR stretch could be indicative of a C=C bond. Anyway, any help would be much appreciated. Thanks!

[discodermolide]

The product is the left hand ring. Now can you explain the IR and the NMR.

[Dynamo22]

I'll do my best!
The C=O is the 1666 stretch (the N makes it lower than a normal ketone), N-H is the 3219 stretch, the 2866 and 2944 stretches would be C-H stretches. I'm not sure which stretch is located where. I would imagine their significance would be because they are next door to the nitrogen and the carbonyl group. I would think the higher would be next to the carbonyl group. The C13 173 is carbonyl group, 42 would be C-C (second carbon would be carbonyl carbon) bond. The lower shifts would indicate the other C-C bonds in the ring.  Proton NMR the 7.4 shift would be the N-H bond, 3.31 would be the CH2 next to the nitrogen, 2.34 would be CH2 next to carbonyl, and 1.75 would be the two remaining CH2 hydrogens.

Not sure if those stretches and shifts are correct. From a mechanism stand point, would the acid donate a proton to the OH group (forcing the then formed HOH to leave) and then remove a hydrogen from the nitrogen after the ring formation? I'm also confused as to how the carbonyl hangs around in the process of and does not turn into a hydroxy group.

Thanks so much for your help, I really appreciate it!

[discodermolide]

The IR and NMR assignments seem ok.
Why should the carbonyl turn into a hydroxy group, this is a reduction and you do not have reducing conditions.
What happens here is this: firstly you do a Fisher esterification of the carboxylic acid which is catalysed by sulphuric acid to give the methyl ester, because of the nicely placed amino group the straight chain amino ester cyclises to give the lactam, the left hand ring system.

[Dynamo22]

Okay, we have covered Fischer esterification, but not lactam formation. So I apologize if I'm missing something. So the methyl ester forms and then does the amino group attack the carbonyl or does the ester get protonated, leave (with methanol as byproduct), then have the amino group attack? Again, I apologize for the questions we have never been shown any of this. Which makes this homework all the more frustrating.

[discodermolide]

It probably happens all at once. The methyl ester gets protonated, the NH₂ attacks the carbonyl carbon and the protonated ester leaves as methanol.

[Dynamo22]

Okay, so after I got past the steps of the acid catalyzed Fischer esterification, I had the ester pick up a proton, depart to form a positive charge on the carbonyl carbon (retaining the double bond to the oxygen), and then the NH2 attacking that positive bond, then the acid plucking off one of the hydrogens from the nitrogen. I'm not sure if those are reasonable steps or not, but I can't see any other way to form the products. I would imagine it would have to be concerted, but I have to show a step by step mechanism.

[discodermolide]

The methyl ester does not depart to give a positive charge on the carbonyl carbon. This is not favoured.
Ok we'll call it concerted,where the NH₂ attacks and the protonated ester group leaves. If you want me to draw it I will.

[Dynamo22]

That seems reasonable, thank you very much. No need to draw it. Would that be considered an SN2 step?

[discodermolide]

No I would not consider it a SN2 reaction.

[curiouscat]

The methyl ester does not depart to give a positive charge on the carbonyl carbon. This is not favoured.
Ok we'll call it concerted,where the NH₂ attacks and the protonated ester group leaves.

How feasible are such concerted steps in mechanisms? Probabistically that sounds a nightmare. Are the resultant kinetics very slow?

[discodermolide]

This reaction is not really concerted, but certainly very rapid.
But many cycloadditions are concerted where bonds are broken and formed at the same time, depends upon the orbital overlap.