In the acid-catalyzed hydration of an asymmetric alkene, which outcome reflects Markovnikov's rule?

• A. The hydrogen adds to the carbon with more hydrogens
• B. The hydroxyl attaches to the less substituted carbon
• C. The hydrogen adds to the carbon with more hydrogens and the hydroxyl attaches to the more substituted carbon
• D. The reaction yields the same product regardless of substitution

Answer: (C)

In acid-catalyzed hydration of an asymmetric alkene, the addition follows Markovnikov's rule: the proton from the acid adds to the carbon of the double bond that has more hydrogens (the less substituted carbon), creating the more stable carbocation on the adjacent, more substituted carbon. Water then attacks that carbocation, and deprotonation yields an alcohol where the hydroxyl group ends up on the more substituted carbon. This is exactly the described outcome: hydrogen attaches to the carbon with more hydrogens and the hydroxyl attaches to the more substituted carbon. The reason this path is favored is carbocation stability—more substituted carbons form more stable carbocations (tertiary > secondary > primary), guiding the regiochemistry of the reaction. If you tried to place the hydrogen on the more substituted carbon instead, you’d generate a less stable carbocation and the reaction would not proceed as readily under acidic hydration conditions.