The acid-catalyzed hydration of propene to form 2-propanol involves three main steps: protonation of the alkene, nucleophilic attack by water, and deprotonation. Here is the detailed mechanism:1. Protonation of the alkene:In the presence of a strong acid catalyst e.g., H2SO4 or H3O+ , the alkene propene reacts with the acid to form a carbocation intermediate. The acid donates a proton H+ to the double bond, which breaks and forms a bond with the proton. This results in the formation of a more stable carbocation intermediate, which is the 2-propyl carbocation.Propene + H+ 2-propyl carbocation2. Nucleophilic attack by water:The carbocation intermediate formed in the previous step is an electrophile, meaning it is electron-deficient and can accept electron pairs. Water, which is a nucleophile, donates an electron pair to the carbocation, forming a bond with the carbocation's positively charged carbon atom. This results in the formation of an oxonium ion intermediate.2-propyl carbocation + H2O 2-propyl oxonium ion3. Deprotonation:The oxonium ion intermediate formed in the previous step is unstable due to the positive charge on the oxygen atom. A nearby water molecule acts as a base and abstracts a proton from the oxonium ion, resulting in the formation of 2-propanol and regenerating the acid catalyst.2-propyl oxonium ion + H2O 2-propanol + H+Overall, the acid-catalyzed hydration of propene to form 2-propanol involves the protonation of the alkene, nucleophilic attack by water, and deprotonation, with the acid catalyst playing a crucial role in both the protonation and deprotonation steps.