(i) chlorobenzene is mono-nitrated to M.(ii) | vedclass

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(D) $(i)$ Chlorobenzene on mono-nitration gives $1-$chloro$-2-$nitrobenzene and $1-$chloro$-4-$nitrobenzene $(M)$.$(ii)$ Nitrobenzene on mono-chlorina

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(D) $(i)$ Chlorobenzene on mono-nitration gives $1-$chloro$-2-$nitrobenzene and $1-$chloro$-4-$nitrobenzene $(M)$.$(ii)$ Nitrobenzene on mono-chlorination gives $1-$chloro$-3-$nitrobenzene $(N)$.$(iii)$ Anisole on mono-nitration gives $1-$methoxy$-2-$nitrobenzene and $1-$methoxy$-4-$nitrobenzene $(P)$.$(iv)$ $2-$nitrochlorobenzene on mono-nitration gives $1-$chloro$-2,4-$dinitrobenzene $(Q)$.The reaction with $aq. NaOH$ is a nucleophilic aromatic substitution reaction $(S_NAr)$. This reaction is facilitated by the presence of electron-withdrawing groups (like $-NO_2$) at ortho and para positions relative to the leaving group $(-Cl)$.Compound $Q$ ($1-$chloro$-2,4-$dinitrobenzene) has two strong electron-withdrawing $-NO_2$ groups at ortho and para positions relative to the chlorine atom,which makes the carbon atom attached to chlorine highly electrophilic. Therefore,$Q$ undergoes nucleophilic substitution with $aq. NaOH$ fastest.

$(i)$ chlorobenzene is mono-nitrated to $M$.
$(ii)$ nitrobenzene is mono-chlorinated to $N$.
$(iii)$ anisole is mono-nitrated to $P$.
$(iv)$ $2-$nitrochlorobenzene is mono-nitrated to $Q$.
Out of $M$,$N$,$P$ and $Q$,the compound that undergoes reaction with $aq. NaOH$ fastest is:

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$(i)$ chlorobenzene is mono-nitrated to $M$.$(ii)$ nitrobenzene is mono-chlorinated to $N$.$(iii)$ anisole is mono-nitrated to $P$.$(iv)$ $2-$nitrochlorobenzene is mono-nitrated to $Q$.Out of $M$,$N$,$P$ and $Q$,the compound that undergoes reaction with $aq. NaOH$ fastest is: