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(A) The reaction of $F_3C-CH=CH_2$ with $HBr$ follows electrophilic addition. The $F_3C-$ group is a strong electron-withdrawing group due to the indu

Vedclass · Accepted Answer

$F_3C-CH_2-CH_2Br$

Vedclass · Answer

(A) The reaction of $F_3C-CH=CH_2$ with $HBr$ follows electrophilic addition. The $F_3C-$ group is a strong electron-withdrawing group due to the inductive effect ($-I$ effect). When the proton $(H^+)$ attacks the double bond, it can form two possible carbocations: $1$. $F_3C-CH^+-CH_3$ (secondary carbocation, destabilized by the strong $-I$ effect of the $F_3C$ group). $2$. $F_3C-CH_2-CH_2^+$ (primary carbocation, less destabilized by the $-I$ effect as the positive charge is further away). Since the primary carbocation $F_3C-CH_2-CH_2^+$ is more stable than the secondary carbocation $F_3C-CH^+-CH_3$ in this specific case due to the proximity of the electron-withdrawing group, the bromide ion $(Br^-)$ attacks the primary carbocation to form the major product $F_3C-CH_2-CH_2Br$.

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

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