JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Hydrocarbons

An anti-Markovnikov product appears when the normally preferred carbocation cannot form. In $\mathrm{F}{}_3\mathrm{C}-\mathrm{CH}=\mathrm{CH}{}_2$ the trifluoromethyl group is strongly electron-withdrawing, so a positive charge on the carbon next to it would be badly destabilised. The proton therefore adds to avoid that, placing the positive charge on the terminal carbon and sending chloride to the carbon bearing the $\mathrm{CF}{}_3$ group, which is the anti-Markovnikov outcome. So $\mathrm{F}{}_3\mathrm{C}-\mathrm{CH}=\mathrm{CH}{}_2$ is the alkene that gives mainly the anti-Markovnikov product with HCl.