# Mechanism of acid-catalyzed addition of water across an alkene

My book tells me that when 1-propene reacts with water and sulfuric acid, I get 2-propanol. The proposed mechanism is that one of the electron pairs from the double bond of 1-propene travels to sulfuric acid, de-protonating it. At the next step, water comes in as a nucleophile and attacks the carbocation intermediate - the terminal carbon gets hydrogen added on it because its the carbon with the most hydrogens. We get CH3CH(H2O)CH3. Now water comes in as a nucleophile again and de-protonate a hydrogen to form 2-propanol.

Now when I asked someone on my ship who used to study organic chemistry at a university, she said that she remembered that it was one of the electron pairs from the double bond of 1-propene that travels to H3O+ instead.

Is either proposed mechanism acceptable?

• Do you mean that propene attacks the oxygen or the hydrogen of H3O+? – jerepierre Apr 23 '17 at 17:28

An electron pair can only be transferred to electrophiles, which are species having a vacant orbital present on them. $\ce{H3O+}$ doesn't have a vacant orbital, so that can't happen. Here is a description as to why:
However, $\ce{H+}$ ions, which have vacant orbitals can accept the pair and have it shift, while the other carbon sharing the double bond loses its electron, making it electrophilic, and can be approached by a nucleophile, in this case, $\ce{H2O}$: