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

1 Answer
Jul 24, 2015

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)