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What happens during this reductive amination?

I got the first product; the lithium aluminium hydride will reduce the nitrile to an amine. But what about the reductive animation step? As far as I can see the C=O will be protonated by the strong tosic acid, and then there will be a competition between the methanol and the internal amine to attack C=O. Actually, the amine is able to attack the C=O without the C=O being protonated, but still, this intramolecular seem too favorable given that we are forming a 4-membered ring. So will the methanol outcompete the amine? In that case what's up with the reductive animation step?

enter image description here

This is what I think happens after the second step ... I am going through the intramolecular route ... since that's the only way we get an iminium ion, on which the sodium cyanoborohydride can act.

enter image description here

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    $\begingroup$ In the condensation to make the iminium intermediate, the hydroxyl should have been eliminated. $\endgroup$ – jerepierre Oct 29 '14 at 21:11
  • $\begingroup$ Just curious, did the instructor ever say what he thought "product 2" was? $\endgroup$ – ron Nov 4 '14 at 16:21
  • $\begingroup$ @ron I'll let you know when I find out $\endgroup$ – Dissenter Nov 4 '14 at 16:22
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Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.

But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:

The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.

In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):

enter image description here

But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.

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I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.

In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).

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