Q56 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry
When 2-butyne is treated with dilute $\text{H}_2\text{SO}_4/\text{HgSO}_4$, the product formed is:
- Correct answerAButanone
- BButanal
- C2-butanol
- D1-butanol
Solution
The reaction of an alkyne with dilute $\text{H}_2\text{SO}_4$ in the presence of $\text{HgSO}_4$ is known as Kucherov's reaction, which results in the hydration of the alkyne.
For 2-butyne ($\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$), the addition of a water molecule across the triple bond initially forms an unstable enol intermediate.
$\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}, \text{H}^+} \text{CH}_3-\text{C}(\text{OH})=\text{CH}-\text{CH}_3$
The enol intermediate undergoes tautomerization to form the more stable keto form, which is butanone.
$\text{CH}_3-\text{C}(\text{OH})=\text{CH}-\text{CH}_3 \rightleftharpoons \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_3$
Thus, the final product is butanone.
Answer: Butanone
For 2-butyne ($\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3$), the addition of a water molecule across the triple bond initially forms an unstable enol intermediate.
$\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}, \text{H}^+} \text{CH}_3-\text{C}(\text{OH})=\text{CH}-\text{CH}_3$
The enol intermediate undergoes tautomerization to form the more stable keto form, which is butanone.
$\text{CH}_3-\text{C}(\text{OH})=\text{CH}-\text{CH}_3 \rightleftharpoons \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_3$
Thus, the final product is butanone.
Answer: Butanone
Hence, the correct answer is option: 'A'.