In the first step of hydroboration mechanism, why is an arrow drawn from the pi bond of the alkene to the B atom of the borane reagent signifying the pi electrons are used to make a new C-B bond?

To fulfill octet of B atom in borane, nucleophilic attack by pi electrons of alkene takes place with electrophilic B center in borane.

Explanation:

In Borane (), B atom has energetically vacant 2p orbital and thereby octet is incomplete.Therefore B center in borane act as an electrophilic center.

In alkene, pi bonding electrons are loosely bound together due to side on overlap of two constituent p-orbitals. Hence this pi bond can be easily broken. Alternatively, we can say that pi bond in alkene act as a potential nucleophile.

The first step of hydroboration occurs in a concerted manner where pi electrons of alkene first attack vacant 2p orbital of B in borane to fulfill it's octet (represented by an arrow drawn from alkene to B atom) and forms a 4 memebered cyclic intermediate. Simulaneously, a B-H bond in borane is donated to alkene through 3c-2e bond (3 center-2 electron bond).

Full mechanism has been shown below.  

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