Question

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

• A. $C{H}_{3}O-CH=C{H}_2$
• B. $H_{2}N-CH=C{H}_2$
• C. $F_{3}C-CH=C{H}_2$
• D. $Cl-CH=C{H}_2$

Answer 1

The correct option is C $F_{3}C-CH=C{H}_2$

Markovnikov's rule: When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part of the reagent is attached to the unsaturated C atom having less number of hydrogen atoms.

In (d), we getting exclusive anti-Markovnikov product. This is due to the strong electron withdrawing nature of $-C{F}_3$. It will destabilise the carbocation formed at carbon next to it.

Hence, option (d) is correct.