Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure?

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](https://i.stack.imgur.com/SpR5z.jpg)


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](https://i.stack.imgur.com/xqPmD.jpg)




Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.


![](https://i.stack.imgur.com/8YVgb.png)


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](http://en.wikipedia.org/wiki/Baldwin%27s_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:


![](https://i.stack.imgur.com/uVxpb.png)


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](https://i.stack.imgur.com/9Yp3k.png)


![](https://i.stack.imgur.com/6RWtq.png)


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.


 3

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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