Question

The major product of the following reaction is $F_3C \rightarrow CH = CH_2 + HBr$

• A. F3C-----CH2----CH2Br
• B. F3C------CH(Br)------CH3
• C.

  

• D.

  

Answer 1

The Correct Option is A

The reaction is the addition of HBr to an alkene:

$F_3C–CH=CH_2 + HBr$ 

This follows the electrophilic addition mechanism. Since there is no peroxide present, the addition follows Markovnikov’s rule. 

According to Markovnikov’s rule, the hydrogen (H) adds to the carbon of the double bond that already has more hydrogens, and the bromine (Br) adds to the other carbon. 

In the given compound:
$F_3C–CH=CH_2$
- The terminal carbon ($CH_2$) gets the H
- The middle carbon (adjacent to $CF_3$) gets the Br 

However, the $CF_3$ group is strongly electron-withdrawing, making the formation of a carbocation at the adjacent carbon (near CF₃) highly unstable. Therefore, instead of a Markovnikov product, the reaction favors anti-Markovnikov addition due to the destabilization of the carbocation by the electron-withdrawing $CF_3$ group. 

Thus, the H⁺ adds to the middle carbon and Br⁻ adds to the terminal one, forming:
$F_3C–CH_2–CH_2Br$ 

Answer: F₃C–CH₂–CH₂Br