[camptzak]

The sodium borohydride reduces the imines to sodium amides

The sodium amides take a proton from methanol?

[Dan]

There is no amide functionality present, we have two oximes. Can you draw what you mean?

[camptzak]

I am guessing that the oxime reacts with the sodium borohydride to form a primary amine. I cant seem to make the arrow pushing work though.

I have the sodium borohydride attacking the ketone, then the negative charge ends up on the nitrogen. It stands to reason (at least in my head it does ) that the negative charge would move over to the oxygen to form sodium hydroxide.

Does the boron accept the electron pair from nitrogen, charging the boron?
Then perhaps the nitrogens lumo could be attacked by a hydride again..


heres a really bad drawing to accompany my idea

[Dan]

I am guessing that the oxime reacts with the sodium borohydride to form a primary amine. I cant seem to make the arrow pushing work though.

So you are suggesting: oxime  hydroxylamine  amine by the action of sodium borohydride?

No, as far as I know boron hydrides do not generally reduce hydroxylamines to amines, and do not in this case either.

I have the sodium borohydride attacking the ketone, then the negative charge ends up on the nitrogen.

Not ketone, oxime. You need to use correct terminology, nomenclature exists to help chemists communicate.

Right idea to start with. Now you have generated a nucleophile, what might that do? (also don't ignore the ZrCl₄).

[yesway]

Hello,

after nucleophile formation due to hydride attack, I would suggest formation of a N-hydroxyl piperidine moiety by loss of hydroxylamine (which is followed by another hydride attack on activated cyclic "oxime" that formed after expelling the hydroxylamine). I only can guess what ZrCl4 does: 1. Lewis acid? 2. Chelating Oximes via their hydroxyl group? 3. Both?  When there is activation by ZrCl4 I suppose overreduction to primary amines could be possible as a side product. Can that be? I also wonder if there is any stereochemistry involved in the first/second hydride attack?

For the second step, I'm guessing Zn/AcOH reduces the hydroxylamine to an amine (although I have never seen this reduction but I believe so simply because I dont see any other unprot. fct. group).

Best

[Dan]

after nucleophile formation due to hydride attack, I would suggest formation of a N-hydroxyl piperidine moiety by loss of hydroxylamine (which is followed by another hydride attack on activated cyclic "oxime" that formed after expelling the hydroxylamine).

Well done, that is correct.

I only can guess what ZrCl4 does: 1. Lewis acid? 2. Chelating Oximes via their hydroxyl group? 3. Both? 

Yes. Following reduction of one of the oximes to a hydroxylamine, the presence of Lewis acidic Zr(IV) species facilitates condensation of the hydroxylamine with the remaining oxime, forming a nitrone, which is reduced by hydride to the corresponding N-hydroxypiperidine.

For the second step, I'm guessing Zn/AcOH reduces the hydroxylamine to an amine (although I have never seen this reduction but I believe so simply because I dont see any other unprot. fct. group).

Yes, Zn/AcOH reduces the N-O bond.

I also wonder if there is any stereochemistry involved in the first/second hydride attack?

Me too. The yield for this sequence (borohydride/ZrCl₄, then Zn/AcOH, followed by [not shown here] TFAA protection) is 32% for the desired stereoisomer. It is claimed that this is the major product, but product ratios are not given in the paper.

Chemistry is from Mander's synthesis of Galbulimima alkaloid GB13: J. Am. Chem. Soc. 2003, 125, 2400. DOI: 10.1021/ja029725o