JEE PYQ: Hydrocarbons Question 28
Question 28 - 2019 (08 Apr Shift 2)
Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?
(1) $\text{CH}_2\text{O}-\text{CH}=\text{CH}_2$
(2) $\text{Cl}-\text{CH}=\text{CH}_2$
(3) $\text{H}_2\text{N}-\text{CH}=\text{CH}_2$
(4) $\text{F}_3\text{C}-\text{CH}=\text{CH}_2$
Show Answer
Answer: (4)
Solution
$\text{CF}_3-\text{CH}=\text{CH}_2 \xrightarrow{\text{H}^+}$ $\text{CF}_3-\text{CH}-\text{CH}_2^+$ (less stable due to very powerful $-\text{I}$ effect of $-\text{CF}_3$) + $\text{CF}_3-\text{CH}-\text{CH}_2$ (More stable). So $\text{Cl}^-$ attacks the less substituted carbon: $\text{CF}_3-\text{CH}_2-\text{CH}_2-\text{Cl}$ (anti-Markovnikov product).
BETA
