JEE PYQ: Hydrocarbons Question 24
Question 24 - 2020 (06 Sep Shift 1)
The major product obtained from the following reaction is:
$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{C}\equiv\text{C}-\text{C}_6\text{H}_4-\text{OCH}_3 \xrightarrow{\text{Hg}^{2+}/\text{H}^+/\text{H}_2\text{O}}$
(1) $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CO}-\text{C}_6\text{H}_4-\text{OCH}_3$
(2) $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{CO}-\text{CH}_2-\text{C}_6\text{H}_4-\text{OH}$
(3) $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CO}-\text{C}_6\text{H}_4-\text{OH}$
(4) $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{CO}-\text{CH}_2-\text{C}_6\text{H}_4-\text{OCH}_3$
Show Answer
Answer: (1)
Solution
Hydration of alkyne with $\text{Hg}^{2+}/\text{H}^+/\text{H}_2\text{O}$ follows Markovnikov’s rule. The $\text{OH}$ attaches to the carbon bearing the $p$-$\text{OCH}_3$ phenyl group (more electron-rich side), which tautomerizes to give the ketone $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CO}-\text{C}_6\text{H}_4-\text{OCH}_3$.
BETA
