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Diazo chemistry controlling the selectivity of olefin ketonisation by nitrous oxide
Authors:Hermans Ive  Moens Bart  Peeters Jozef  Jacobs Pierre  Sels Bert
Affiliation:Centre for Surface Chemistry and Catalysis, K.U. Leuven, Kasteelpark Arenberg 23, B-3001 Heverlee, Belgium. ive.hermans@biw.kuleuven.be
Abstract:The thermal reaction of olefins with nitrous oxide was recently put forward as a promising synthetic ketone source. The 1,3-dipolar cycloaddition of N(2)O to the C=C double bond, forming a 4,5-dihydro-[1,2,3]oxadiazole intermediate, was predicted to be the first elementary reaction step. This oxadiazole can subsequently decompose to the desired carbonyl product and N(2)via a hydrogen shift. In this contribution, Potential Energy Surfaces are constructed at the reliable G2M level of theory and used to evaluate thermal rate constants by Transition State Theory. Compelling theoretical and experimental evidence is presented that an oxadiazole intermediate not only can undergo a hydrogen shift, but eventually also a methyl- or even an alkyl-shift. Special emphasis is also given on a hitherto neglected decomposition of the oxadiazole via a concerted C-C and N-O cleavage. For some substrates, such as internal olefins, this diazo route is negligibly slow, compared to the ketone path, leaving no marks on the selectivity. For cyclopentene the diazo cleavage was however found to be nearly as fast as the desired ketone route. However, the diazo compound, viz. 5-diazopentanal, reconstitutes the oxadiazole much faster upon ring-closure than it is converted to side-products. Therefore, a pre-equilibrium between the diazoalkanal and the oxadiazole is established, explaining the high ketone yield. On the other hand, for primary alkenes, such a concerted C-C and N-O cleavage to diazomethane is identified as an important side reaction, producing aldehydes with the loss of one C-atom. For these substrates, the bimolecular back-reaction of the C(n-1) aldehyde and diazomethane is too slow to sustain an equilibrium with the oxadiazole; diazomethane rather reacts with the substrate to form cyclopropane derivatives. The overall selectivity is thus determined by a combination of H-, methyl- or alkyl-shift, and the eventual impact of a diazo cleavage in the oxadiazole intermediate.
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