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

1 Answer
Truong-Son N.
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)

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