# Which base accepts the proton abstracted from water in propyne hydration?

I am enrolled in Organic Chemistry 1 at a junior college and I need some assistance in understanding this mechanism. I have previously taken chemistry in high school with no problems, this would be my first time not comprehending the subject. We are covering the hydration of alkynes.

In the general mechanism, the triple bond of propyne attacks $\ce{H2SO4}$. The hydrogen, the electrophile, goes to the least substituted carbon. We now have a carbocation so water comes in as a nucleophile and attacks. But, now we have a positive charge on the oxygen, so a base de-protonates the molecule. My question is, what is this base?

After the base attacks the lone pairs from oxygen travels down to the carbon and the pi bond then attacks $\ce{H3O+}$. This is the keto-enol tautomerizatiom. My second question is, where did we get this $\ce{H3O+}$? Our list of reagents is $\ce{H2O, H2SO4}$ and $\ce{HgSO4}$. We get a product called acetone. I'm just very fuzzy on the in-between steps.

• You're in water. How about just water? – Zhe Oct 21 '17 at 13:19

Protons can basically jump on and off at any time if there is sufficient incentive for them to do so. In your reaction, you not only have the hydrogensulphate anion $\ce{HSO4-}$ that can accept a proton to go back to sulphuric acid but also water itself which can be protonated to $\ce{H3O+}$. The only important aspect is to identify the least weak base and use that in your mechanism.
Furthermore, general chemistry should have taught you that a solution containing sulphuric acid and water will contain $\ce{H3O+}$ ions by deprotonation immediately.