Question

Draw the structural formula of the enol formed in each alkyne hydration reaction; then draw the structural formula of the carbonyl compound with which each enol is in equilibrium. (a) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C} \equiv \mathrm{CH}+\mathrm{H}_{2} \mathrm{O} \frac{\mathrm{HgSO}_{4}}{\mathrm{H}_{2} \mathrm{SO}_{4}}\) (an enol) \(\longrightarrow\) (b) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C} \equiv \mathrm{CH} \frac{\text { 1. (sia) })_{2} \mathrm{BH}}{2 . \mathrm{NaOH}_{4} \mathrm{H}_{2}}(\) an enol \() \longrightarrow\)

Short answer

Question: Draw the structural formula of the enol formed in each alkyne hydration reaction, as well as the structural formula of the carbonyl compound with which each enol is in equilibrium. Answer: For reaction (a), the enol formed is: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(OH)=\mathrm{C}\mathrm{H}_{2}\) The carbonyl compound in equilibrium with the enol is: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(O)=\mathrm{CH}_{2}\) For reaction (b), the enol formed is: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(OH)=\mathrm{CH}_{2}\) The carbonyl compound in equilibrium with the enol is: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(O)=\mathrm{CH}_{2}\)

Step-by-step solution

Setting up the Reaction

For the given reaction (a), the first step is to set up the reaction. The starting compound is an alkyne, specifically a terminal alkyne: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C} \equiv \mathrm{CH}\)

Hydration

When this alkyne reacts with water in the presence of a strong acid catalyst like \(\mathrm{HgSO}_{4}\), hydration takes place, and the \(\mathrm{C}\equiv\mathrm{C}\) triple bond is converted into a \(\mathrm{C}=\mathrm{C}\) double bond with an OH group added to the terminal carbon: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(OH)=\mathrm{C}\mathrm{H}_{2}\) This compound is the enol form.

Tautomerization

Enols can tautomerize to their more stable keto form. In this case, a proton shift occurs, and the double bond moves to be between the carbon atoms: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(O)=\mathrm{CH}_{2}\) This is the carbonyl compound in equilibrium with the enol. #For reaction (b):#

Setting up the Reaction

For the given reaction (b), the first step is to set up the reaction, using the same starting alkyne: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C} \equiv \mathrm{CH}\)

Hydroboration

In this reaction, hydroboration-oxidation occurs. The first step of this process is to react the alkyne with \(\mathrm{BH}\) in the presence of a suitable catalyst. This introduces a hydrogen and a boron group to the alkyne: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(BH)=\mathrm{CH}_{2}\)

Oxidation

Next, the compound reacts with \(\mathrm{NaH}_{2}\mathrm{H}_{2} \mathrm{O}_{2}\), which replaces the boron group with an OH group, resulting in the formation of the enol: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(OH)=\mathrm{CH}_{2}\) This compound is the enol form.

Tautomerization

Similarly to reaction (a), this enol can tautomerize to its more stable keto form. A proton shift occurs, and the double bond moves to be between the carbon atoms: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{C}(O)=\mathrm{CH}_{2}\) This is the carbonyl compound in equilibrium with the enol.