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

Step 1: Understand when anti-Markovnikov addition occurs

For HCl addition to alkenes:

• Markovnikov: $\ce{H^+}$ adds to the carbon with more H atoms; carbocation forms at the more substituted carbon
• Anti-Markovnikov: occurs when the more substituted carbon is destabilized (cannot support positive charge)

Step 2: Analyze the effect of substituents

(a) $\ce{F3C-CH=CH2}$: The $\ce{-CF3}$ group is strongly electron-withdrawing (by induction and field effect):

• $\ce{H^+}$ adds to C2 ($\ce{=CH2}$) would give primary carbocation at C2
• $\ce{H^+}$ adds to C1 ($\ce{=CH-CF3}$) would give secondary carbocation at C2
• BUT: $\ce{-CF3}$ destabilizes the adjacent carbocation by its strong $-I$ effect
• Therefore, $\ce{H+}$ adds to C1 (adjacent to $\ce{CF3}$) → carbocation at C2 (away from $\ce{CF3}$)
• Product: $\ce{Cl}$ at C1 → anti-Markovnikov ✓

(b)–(d) Substituents like $\ce{-Cl}$, $\ce{-OCH3}$, $\ce{-NH2}$ donate electrons via resonance (+M effect), stabilizing adjacent carbocations → favor Markovnikov addition.

Step 3: Conclusion

$\ce{F3C-CH=CH2}$ (trifluoropropene) gives anti-Markovnikov addition with HCl due to the powerful electron-withdrawing $\ce{-CF3}$ group destabilizing the adjacent carbocation.

Answer: Option (a) — $\ce{F3C-CH=CH2}$

Key Points to Remember:

• Strong EWG (like $\ce{-CF3}$, $\ce{-NO2}$) adjacent to $\ce{C=C}$ reverses Markovnikov selectivity
• The carbocation forms at the carbon farther from the EWG
• EDG ($\ce{-OR}$, $\ce{-NR2}$) = Markovnikov addition; EWG ($\ce{-CF3}$) = anti-Markovnikov