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(A) The reaction proceeds in two steps:$1$. The first step is the anti-Markovnikov addition of $HBr$ to the alkene in the presence of a peroxide $((C_

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$4-$($3$-fluoropropyl)bromobenzene

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(A) The reaction proceeds in two steps:$1$. The first step is the anti-Markovnikov addition of $HBr$ to the alkene in the presence of a peroxide $((C_6H_5CO)_2O_2)$. This results in the formation of $1-bromo-4-(3-bromopropyl)benzene$.$2$. The second step involves the reaction with $CoF_2$,which is a Swarts reaction. This reagent replaces the terminal bromine atom with a fluorine atom,yielding $1-bromo-4-(3-fluoropropyl)benzene$ as the major product $P$.

The major product $P$ of the above reaction is:

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