Why does enantiopure sec-butyl alcohol retain its optical activity over aqueous base but forms a racemic mixture over dilute sulphuric acid?
Although sulphuric acid is a notable dehydrating agent, it is concentrated rather than dilute sulphuric acid that would be typically used for such a purpose.

Here, it is due to the improvement of $\ce{H2O}$ as a leaving group compared to $\ce{OH-}$, due to the lack of any charge as a leaving group.

Under acidic conditions, the sec-butyl alcohol will be protonated in a small amount creating a much better leaving group.  The protonation also makes the alcoholic carbon centre more electrophilic and vulnerable to $S_N2$ attack by other water molecules in solution, leading to inversion of the chiral centre.

This in not possible under neutral conditions as the unprotonated alcohol would have $\ce{OH-}$ as a much poorer leaving group, preventing elimination or substitution reactions.

Under acidic conditions, with neutral water as a nucleophile $S_N2$ substition will be the preferred path.


A secondary alcohol on a small, un-branched carbon chain represents a relatively accessible spot for nucleophilic attack.
$E_1$ eliminiation or $S_N1$ substituion would require the formation of a secondary carbocation with relatively poor stabilisation compared to a tertiary carbocation.
$E_2$ elimination would require the removal of a hydrogen of low acidity from the carbon chain by a very weak base, water.


This allows a equilibrium converting molecules back and forth between the $R$ and $S$ forms of the alcohol to be set up and leads to racemisation in solution.