1. **Stating the problem:** We need to find the product formed when propene undergoes hydration. 2. **Formula and rules:** Hydration of an alkene involves the addition of water ($H_2O$) across the double bond. The general reaction is: $$\text{Alkene} + H_2O \xrightarrow{H^+} \text{Alcohol}$$ This reaction follows Markovnikov's rule, which states that the hydrogen atom from water adds to the carbon with more hydrogen atoms, and the hydroxyl group ($-OH$) adds to the carbon with fewer hydrogen atoms. 3. **Applying to propene:** Propene is $CH_3-CH=CH_2$. The double bond is between the second and third carbon atoms. 4. **Markovnikov addition:** The $H$ from $H_2O$ adds to the $CH_2$ end (carbon 3), and the $OH$ adds to carbon 2. 5. **Product formed:** The product is 2-propanol (isopropanol), with the structure $CH_3-CHOH-CH_3$. 6. **Summary:** Hydration of propene yields 2-propanol following Markovnikov's rule. **Final answer:** 2-propanol ($CH_3-CHOH-CH_3$)