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Photolysis of a N-bromo compound to an alkaloid (delta coneceine)

For the proposed mechanism, I believe that the N-Br bond is firstly broken by the absorbed light hv, then the radical formed on the N will perform an intra-molecular attack on the end of the carbon chain to form a 5 membered ring. Then the sulphuric acid becomes deprotonated by the nitrogen lone pair? Just want to check whether this is correct.

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up vote 6 down vote accepted

This is the Hofmann-Löffler-Freytag reaction, which was later examined by E. J. Corey (J. Am. Chem. Soc., 1960, 82, 1657-1668, DOI).

When the reaction is performed in sulfuric acid, it is conceivable that the nitrogen atom of the N-Bromoalkane is protonated. Homolytic cleavage of the $\ce{N-Br}$ thus probably leads to a radical cation.

The latter then abstracts a $\delta$-hydrogen atom, resulting in a radical centre in the side chain. It has been suggested that this centre might recombine with the bromo radical previously released. With other words: $\ce{H}$ and $\ce{Br}$ have changed their position. Deprotonation of the ammonium cation and a subsequent nucleophilic substitution, releasing $\ce{Br-}$ would then result in the formation of the indolizidinium skeleton.

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Thank you so much! I do however have a question. How can one determine the decomposition pathways in this reaction? – Michael Nguyen Feb 14 at 23:39
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Interesting and difficult question! If there would be any pi-system at all, I'd suggest time-resolved uv-vis spectroscopy. All the radicals and ions involved here are spectroscopically invisible, just like a DABCO radical cation. Is there any chance to trap an intermediate? Intermolecular vs intramolecular often isn't very promising either and sulfuric acid as a solvent limits the options too. Maybe you can find some more recent mechanistical papers that cite the old Corey paper? Did anybody perform CIDNP or ESR spectroscopy on this reaction? Honestly, I don't know. – Klaus Warzecha Feb 14 at 23:51

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