Question: Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?
$\mathrm{F}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}$
$\mathrm{Cl}-\mathrm{CH}=\mathrm{CH}_{2}$
$\mathrm{CH}_{3} \mathrm{O}-\mathrm{CH}=\mathrm{CH}_{2}$
$\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}=\mathrm{CH}_{2}$
Correct Option: 1
Solution:
Due to higher $\mathrm{e}^{-}$withdrawing nature of $\mathrm{CF}_{3}$ group.
It follow anti markovnikoff product
