Questions about propyne reacting with H2O, H2SO4 and HgSO4?

Question

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 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 H3O+. This is the keto-enol tautomerizatiom. My second question is, where did we get this H3O+? Our list of reagents is H2O, H2SO4 and HgSO4. We get a product called acetone. I'm just very fuzzy on the in-between steps.

Answer 1

Hello Shaniqua

 

The base that deprotonates the positive oxygen to form the enol is water.  Water is amphoteric, which means it can behave as both an acid (proton donor) or a base (proton acceptor).

 

When water deprotonates the positively charged oxygen atom, it acts as a base by accepting the proton to form its conjugate acid, H₃O⁺  (H₂O + H⁺ --> H₃O⁺).

 

During the keto-enol tautomerism, this conjugate acid (H₃O⁺) then donates the proton to the C=C pi bond resulting in the formation of acetone as your final product.

 

Hope the above helps!