Sapling Chapter 7 Question #18

Sapling Chapter 7 Question #18 - Ketone-Enol Tautomerization 

In this problem, the starting material is converted from an alkyne to an enol then to a ketone. 

The first step of this mechanism occurs when the starting material (the alkyne) reacts with the reagent H2SO4. The triple bond grabs a hydrogen and the remaining electrons from the hydrogen-oxygen bond to to the oxygen that it was removed from. This produces a carbocation intermediate. Two intermediates are formed because the carbocation can occur on either carbon because the hydrogen can attach to either carbon on each side of the alkene. 

After the formation of the carbocation, the second step occurs when the water (H2O) attacks the carbon with the positive charge. This is the only place it can attach given that the octet is full for all of the other carbons in the molecule. Although the carbon is not tertiary, it will not rearrange. At the end of this attack, the water will bond to that carbon with a formal charge.

The third step is to deprotonate. This means that a proton will be removed. In this problem, it will be removed from the water that attached in the previous step. Another water molecule is responsible for this chemistry. This results in the formation of an enol and H3O as in inorganic product. 

Tautomerization occurs in the final step of this mechanism. The enol formed is not as stable as the ketone, causing it to immediately rearrange. The ketone is the predominate product because it is much more stable than the enol. 

In this sapling problem, the initial and final product is asked. Therefore, the initial product would be the enol and the final (more correct) product would be the ketone drawn above.