What is the mechanism for the reaction of alkenes with diborane and chloramine?

$$\ce{CH3-CH=CH2 ->[B2H6/THF][NH2Cl/H2O, Δ] Product (P)}$$

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?

• Since others already answered this question, I will not reinvent this wheel, but the OP may want to compare the result with a similar question. In both cases a carbon atom made nucleophilic by attachment to a more electropositive atom is acting as a nucleophile, displacing a chloride ion from the chloramine. May 21 '20 at 11:41

2 Answers

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

1. 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.
• For a more elaborate variation of this reaction, see: sciencedirect.com/… May 22 '20 at 22:40

I like Waylander's answer, which is 100% correct. However, I have a little concern about the question itself. The conditions given in OP's question are not exactly correct. It may mislead the viewers and people seeking answers for similar type of questions. H. C. Brown had specifically noted (Ref.1) the following condition for his groundbreaking discovery (I used OP's example):

$$\ce{CH3-CH=CH2 ->[\text{i.} B2H6/THF][\text{ii.} NH2Cl/H2O/NaOH, Room Temp] CH3-CH2-CH2-NH2}$$

After $$\pu{1 h}$$ stirring of substrate with $$\ce{B2H6}$$ under $$\ce{N2}$$, water was added to the reaction mixture to destroy residual $$\ce{B2H6}$$ and then $$\pu{3 M}$$ aq. $$\ce{NaOH}$$ has been added before adding $$\ce{NH2Cl}$$ in THF. The mixture was then stirred another hour at room temperature before acid/base workup to isolate pure amine.

The heating to reflux was done for the reaction of substrate with the other reagent, hydroxylamine-O-sulfonic acid:

$$\ce{CH3-CH=CH2 ->[\text{i.} B2H6/THF][\text{ii.} NH2OSO3H/\Delta/3 h; \text{iii. aq.} HCl] CH3-CH2-CH2-NH3+Cl-}$$

Here though, solid $$\ce{NH2OSO3H}$$ was added without any mention of water addition. Acid/base workup follows to isolate pure amine.

Reference:

1. Herbert C. Brown, Wolfgang R. Heydkamp, Eli. Breuer, William S. Murphy, “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 (https://doi.org/10.1021/ja01071a036).