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(B) $(I)$ Reaction with $HBr$ in the presence of peroxide follows anti-Markovnikov addition to form $1-$bromopropane.$(II)$ Reaction with $HBr$ in $CC

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$II$ and $III$

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(B) $(I)$ Reaction with $HBr$ in the presence of peroxide follows anti-Markovnikov addition to form $1-$bromopropane.$(II)$ Reaction with $HBr$ in $CCl_4$ follows Markovnikov's addition to form $2-$bromopropane.$(III)$ Free radical bromination of propane $(CH_3-CH_2-CH_3)$ with $Br_2$ under $hv$ (light) preferentially forms $2-$bromopropane due to the higher stability of the secondary free radical intermediate.$(IV)$ Reaction with $Br_2$ in $CCl_4$ results in the formation of $1,2-$dibromopropane.Therefore,reactions $(II)$ and $(III)$ result in the formation of $2-$bromopropane.

Select the reaction$(s)$ that would result in the formation of $2-$bromopropane.
$(I) \ CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}}$
$(II) \ CH_3-CH=CH_2 + HBr \xrightarrow{CCl_4}$
$(III) \ CH_3-CH_2-CH_3 + Br_2 \xrightarrow{hv}$
$(IV) \ CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4}$

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Select the reaction$(s)$ that would result in the formation of $2-$bromopropane.$(I) \ CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}}$$(II) \ CH_3-CH=CH_2 + HBr \xrightarrow{CCl_4}$$(III) \ CH_3-CH_2-CH_3 + Br_2 \xrightarrow{hv}$$(IV) \ CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4}$