Ketone/aldehyde synthesis from N-acylazetidines or aziridines

Question

A common principle in organic chemistry is to make sure the product is less reactive than the starting material, to prevent overreaction. So, for example, the treatment of an ester with a Grignard reagent will afford not the corresponding ketone, but the alcohol.

However there are some cases where the contrary is true. In particular, when the tetrahedral intermediate is stable, then the reaction can be worked up to 1) produce the desired ketone and 2) destroy any unreacted organometallic reagent, thus preventing overreaction. The classic example is a Weinreb amide, where the tetrahedral intermediate is stabilised by chelation, AFAIK.

Somewhat recently the synthesis of ketones from N-acylazetidines was described:¹

and also the reduction of N-acylaziridines is known to produce aldehydes:²

Presumably the idea is the same: the tetrahedral intermediate is stable under the conditions used. Why is this so? Is it connected to the small size of the ring?

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References

1. Liu, C.; Achtenhagen, M.; Szostak, M. Chemoselective Ketone Synthesis by the Addition of Organometallics to N-Acylazetidines. Org. Lett. 2016, 18 (10), 2375–2378. DOI: 10.1021/acs.orglett.6b00842.

2. Brown, H. C.; Tsukamoto, A. Selective Reductions. I. The Partial Reduction of Tertiary Amides with Lithium Aluminum Hydride. A New Aldehyde Synthesis via the 1-Acylaziridines. J. Am. Chem. Soc. 1961, 83 (22), 4549–4552. DOI: 10.1021/ja01483a013.

Answer 1

I think your suggestion about the stability of the tetrahedral intermediate is correct, but it's not that the tetrahedral intermediate itself has stability. I think it's because the collapse of the tetrahedral intermediate leads to a less stable intermediate.

My key insight for rationalizing this is that lithium hydride reduction of amides generally affords amines. We generally prefer to eliminate the lithium-oxygen adduct from the tetrahedral intermediate, forming an iminium ion.

With the two substrates you've provided, the iminium ion would be very highly strained because of the geometry of the small ring. I think collapse of the intermediate is thus disfavored. On work up, we actually kick out the amine to collapse the intermediate, providing an aldehyde or ketone.