JEE PYQ: Haloalkanes And Haloarenes Question 28
Question 28 - 2019 (12 Jan Shift 1)
The major product in the following conversion is:
$\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}=\text{CH}-\text{CH}_3 \xrightarrow{\text{HBr (excess)}/\text{Heat}}$
(1) $\text{HO}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3$
(2) $\text{HO}-\text{C}_6\text{H}_4-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3$
(3) $\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3$
(4) $\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3$
Show Answer
Answer: (2)
Solution
$\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}=\text{CH}-\text{CH}_3 \xrightarrow{\text{HBr}}$ benzylic cation (more stable). HBr (excess) also cleaves the methyl ether. Product: $\text{HO}-\text{C}_6\text{H}_4-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3$.
BETA
