Question

How many transition states are there in the mechanism for the acid-catalyzed hydration of an alkene?

Answer 1

Hint: When the alkene is hydrated or treated with water in the presence of an acid, then there is the formation of alcohol, in this reaction, there are three steps, i.e., the addition of a proton to the alkene from hydronium ion, water molecule attacks the carbocation, and removal of a hydrogen ion from the oxonium.

Complete answer:
When the alkene is hydrated or treated with water in the presence of an acid, then there is the formation of alcohol, in this reaction, there are three steps, i.e., the addition of a proton to the alkene from hydronium ion, water molecule attacks the carbocation, and removal of a hydrogen ion from the oxonium ion.
All the steps given above have a transition state.
The hydration of alkene involves strong acids like hydrochloric acid (HCl), sulfuric acid (${{H}_{2}}S{{O}_{4}}$), etc. The proton from reacts with water to form hydronium ion (${{H}_{3}}{{O}^{+}}$), as shown below:

The mechanism takes place as, first, in the alkene the double breaks, and at one carbon atom there will be the addition of hydrogen ion from the hydronium ion and the other carbon atom will have a positive charge. Now, at this positive charge, the water molecule will attack. This will create a positive charge on the oxygen atom, so another water molecule will take a hydrogen ion from the oxonium ion. In all three steps transition states are formed and alcohol is formed. The reactions are given below:

So, there are three transition states.

Note:
When symmetrical alkene is treated with water, then only one alcohol is formed, but when the alkene is unsymmetrical then the hydroxyl ion will be attached to the carbon atom having a lesser number of hydrogen atoms.