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(B) The reaction involves two steps when $HBr$ is in excess and heat is applied:$1$. The ether group $(-OCH_3)$ undergoes cleavage by $HBr$ to form a

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(B) The reaction involves two steps when $HBr$ is in excess and heat is applied:$1$. The ether group $(-OCH_3)$ undergoes cleavage by $HBr$ to form a phenol $(-OH)$ and methyl bromide $(CH_3Br)$.$2$. The alkene group $(-CH=CH-CH_3)$ undergoes electrophilic addition of $HBr$ following Markownikoff's rule.According to Markownikoff's rule,the electrophile $H^+$ adds to the carbon with more hydrogens,and the nucleophile $Br^-$ adds to the more substituted carbon,resulting in the formation of a stable carbocation intermediate. Thus,the product is $4-(1-bromopropyl)phenol$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A) \ CH_3CH_2CH_2-O-CH_3 + HI \rightarrow$	$(i) \ (CH_3)_3C-I + CH_3CH_2OH$
$(B) \ C_6H_5OCH_2CH_3 + HBr \rightarrow$	$(ii) \ CH_3CH_2CH_2OH + CH_3I$
$(C) \ (CH_3)_3C-OCH_2CH_3 + HI \rightarrow$	$(iii) \ CH_3CH_2I + C_6H_5OH$
$(D) \ C_6H_5OCH_2CH_3 + \text{conc. } H_2SO_4 + \text{conc. } HNO_3 \rightarrow$	$(iv) \ p-NO_2-C_6H_4-OCH_2CH_3$

The major product in the following conversion is

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