Match the following reactions in List-I with | vedclass

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(B) The addition of $HI$ to propene follows Markownikoff's rule,proceeding via the formation of a $2^{\circ}$ carbocation,which is more stable tha

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$A-III, B-IV, C-IV$

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(B) The addition of $HI$ to propene follows Markownikoff's rule,proceeding via the formation of a $2^{\circ}$ carbocation,which is more stable than a $1^{\circ}$ carbocation. The major product is $2	ext{-iodopropane}$ $(III)$.
$(B)$ The reaction of ethylbenzene $(C_6H_5CH_2CH_3)$ with $Br_2$ under heating $(\Delta)$ is a free radical substitution at the benzylic position. The $1	ext{-phenylethyl}$ radical is stabilized by resonance,leading to $1	ext{-bromo-}1	ext{-phenylethane}$ $(IV)$.
$(C)$ The addition of $HBr$ to styrene $(C_6H_5CH=CH_2)$ follows Markownikoff's rule. The electrophilic addition of $H $ forms a stable benzylic carbocation $(C_6H_5CH^ CH_3)$,which then reacts with $Br^-$ to form $1	ext{-bromo-}1	ext{-phenylethane}$ $(IV)$.
Therefore,the correct match is $A-III, B-IV, C-IV$.

List-$I$ (Reaction)	List-$II$ (Major product)
$A. \ CH_3CH=CH_2 \xrightarrow{HI}$	$I. \ 1\text{-bromo-}2\text{-phenylethane}$
$B. \ C_6H_5CH_2CH_3 \xrightarrow{Br_2/\Delta}$	$II. \ 1\text{-iodopropane}$
$C. \ C_6H_5CH=CH_2 \xrightarrow{HBr}$	$III. \ 2\text{-iodopropane}$
	$IV. \ 1\text{-bromo-}1\text{-phenylethane}$

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