Which one of the following alkenes when treated with HCl yields majorly an anti-Markovnikov product?<p></p>

Correct option is A. F_2C - CH = CH_2
Solution:- (A) F_2C - CH = CH_2
CF_3 - CH = CH_2
CF_3 - H|CH - ⊕CH_2
↓
CF_3 - H|CH - Cl|CH_2
Due to higher e^- withdrawing nature of CF_3 group. It follows the anti-markonikoff product. Hence, the correct option is A

Eduprobe

Question:

Which one of the following alkenes when treated with HCl yields majorly an anti-Markovnikov product?

Solution: