What is the mechanism for the acid-catalyzed hydration of but-2-yne?
Answer:
See below.
Explanation:
The reaction has two parts:
- Addition of water to the alkyne to form an enol.
- Tautomerization of the enol to form a ketone.
Part 1. Formation of the enol
Step 1. Protonation of the alkyne to form a carbocation.
\(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}+\mathrm{H}_{3} \stackrel{+}{\mathrm{O}} \rightarrow \mathrm{CH}_{3}{ }^{+} \mathrm{C}=\mathrm{CHCH}_{3}+\mathrm{H}_{2} \mathrm{O}\)
Step2. Water attacks the carbocation and forms an oxonium ion.
\(\mathrm{CH}_{3}{ }^{+} \mathrm{C}=\mathrm{CHCH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{C}\left(\stackrel{+}{\mathrm{OH}}_{2}\right)-\mathrm{CHCH}_{3}\)
Step 3. Deprotonation of the oxonium to form an enol.
\(\mathrm{CH}_{3} \mathrm{C}\left(\stackrel{+}{\mathrm{OH}}_{2}\right)-\mathrm{CHCH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{OH})=\mathrm{CHCH}_{3}+\mathrm{H}_{3} \mathrm{O}\)
Part 2. Tautomerization of the enol to a ketone
Step 4. Protonation of the alkene carbon atom to form a carbocation.
\(\mathrm{CH}_{3} \mathrm{C}(\mathrm{OH})=\mathrm{CHCH}_{3}+\mathrm{H}_{3} \stackrel{+}{\mathrm{O}} \rightarrow \mathrm{CH}_{3} \mathrm{C}(=\stackrel{+}{\mathrm{OH}})-\mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O}\)
Step 5. Deprotonation of the oxygen to form the ketone.
\(\mathrm{CH}_{3} \mathrm{C}(=\stackrel{+}{\mathrm{OH}})-\mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{C}(=\mathrm{O})-\mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{3} \stackrel{+}{\mathrm{O}}\)
If you have trouble following the steps above with condensed structures, here’s an image of the same mechanism for the hydration of propyne.
