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(B) $1$. Benzene reacts with $Cl_2/AlCl_3$ (electrophilic aromatic substitution) to form chlorobenzene $(A)$.$2$. Chlorobenzene reacts with conc. $HNO

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$4-$nitrophenol

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(B) $1$. Benzene reacts with $Cl_2/AlCl_3$ (electrophilic aromatic substitution) to form chlorobenzene $(A)$.$2$. Chlorobenzene reacts with conc. $HNO_3$ and conc. $H_2SO_4$ (nitration) to form a mixture of $o$-nitrochlorobenzene and $p$-nitrochlorobenzene. The $p$-nitrochlorobenzene $(B)$ is the major product due to less steric hindrance.$3$. $p$-nitrochlorobenzene reacts with $NaOH/\Delta$ via nucleophilic aromatic substitution $(S_{NAr})$ to replace the $Cl$ atom with an $OH$ group,forming $p$-nitrophenol as the final product.

The main product of the following reaction will be:

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