Question

How can i dentify the nucleophile and the electrophile in `H-Br` + `HO^-)hArr Br^-`+`H_2O` acid–base reaction?

Answer 1

Nucleophiles tend to be negatively charged or have a lone pair of electrons. In this case it is easy to see that `OH^(-)` has a negative charge as a reactant, and so it is the nucleophile. Thus, `H-Br` is the electrophile.

The nucleophilic `OH^(-)` wants a proton from `HBr`, to become water.

The `pKa` of water is `15.7`, whereas the `pKa` of `HBr` is about `-8`. Since the equilibrium lies on the side of the weaker acid, this equilibrium is skewed towards the products by about `24` orders of magnitude.

(Remember that `10^(-pKa) = K_a`, thus the `K_a`'s are `10^(-15.7)` vs. `10^8`, water vs. `HBr`, and water hardly dissociates while `HBr` does quite a bit)