1-butanol reacts with NaBr in the presence of conc. `H_(2)SO_(4)` to form 1-bromobutane. However, the reaction does not take place in the absence of `H_(2)SO_(4)`. Explain.

Explanation

`OH^(-)` is a very poor leaving group, so it cannot be displaced by `Br^(Theta)`, in the presence of `H_(2)SO_(4)` the -OH group becomes converted into -`overset(o+)(O)H_(2)` which is a much better leaving group hence, substitution occurs