What is the mechanism for the reaction of alkenes with diborane and chloramine? $$\ce{CH3-CH=CH2 -&gt;[B2H6/THF][NH2Cl/H2O, Δ] Product (P)}$$
What is $\ce{Product (P)}?$

First of all, in this reaction I performed hydroboration to form an alkylborane. But I don't have any ideas what to do next. I tried a lone pair attack of nitrogen on the boron atom of the alkylborane, but what should I do next? What is the product and by what mechanism is it formed?
Organoboranes react with chloramine to give the corresponding amine as reported by Brown et al. [1], so P is n-propylamine (propan‐1‐amine).

The mechanism is similar to the oxidation of boranes using $\ce{H2O2}/\ce{NaOH}$, but instead of migration to an oxygen atom, the alkyl chain migrates to a nitrogen atom. The $\ce{N-B}$ bond is cleaved in the same way.



Reference


Brown, H. C.; Heydkamp, W. R.; Breuer, Eli.; Murphy, W. S. The Reaction of Organoboranes with Chloramine and with Hydroxylamine-O-Sulfonic Acid. A Convenient Synthesis of Amines from Olefins via Hydroboration. J. Am. Chem. Soc. 1964, 86 (17), 3565–3566. DOI: 10.1021/ja01071a036.