# Photolysis of N-bromo compound in acidic conditions (radical cyclisation)

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.

-

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.