1-Chloro-2,4-dinitrobenzene reacts with sodium methoxide in methanol to give 2,4-dinitroanisole. Which statements are correct?

AThe slow step is attack of $CH X_{3} O X^{-}$ on the carbon bearing $Cl$ giving an anionic sigma (Meisenheimer) complex
BThe negative charge of the intermediate is delocalised onto the oxygen atoms of the ortho and para nitro groups
C1-Chloro-3,5-dinitrobenzene would react at the same rate under identical conditions
DThe reaction goes through a benzyne intermediate

Answer: A, B

This is addition-elimination (SNAr). The nucleophile adds to

C - Cl

, and charge is delocalised onto the ortho and para

NO X_{2}

groups; then

Cl X^{-}

is lost. No benzyne forms because the reagent is not a very strong base like

NH X_{2} X^{-}

. Nitro groups meta to

Cl

cannot delocalise the charge by resonance, so the 3,5-dinitro isomer is far less reactive.