Propene on hydroboration followed by oxidation with alkaline peroxide forms.

Hydroboration-oxidation of propene proceeds through anti-Markovnikov addition of water across the double bond.

The reaction begins with hydroboration, where the boron atom adds to the less substituted carbon while hydrogen adds to the more substituted carbon. Propene reacts with borane to form tri-$n$-propylborane: $3\text{CH}_3\text{CH}=\text{CH}_2 + \text{BH}_3 \rightarrow (\text{CH}_3\text{CH}_2\text{CH}_2)_3\text{B}$.

Oxidation with alkaline hydrogen peroxide cleaves the boron-carbon bonds, replacing boron with hydroxyl groups: $(\text{CH}_3\text{CH}_2\text{CH}_2)_3\text{B} + 3\text{H}_2\text{O}_2 + \text{NaOH} \rightarrow 3\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{Na}_3\text{BO}_3$.

The product is propan-1-ol, with the hydroxyl group on the terminal carbon. Tri-$n$-propylborane is an intermediate, tri-iso-propylborane would result from Markovnikov addition, and propan-2-ol forms through acid-catalyzed hydration.

The correct choice is $\boxed{\text{C}}$.