Say im reacting a base with a halobenzene with strong electron withdrawing groups at o- and p- positions. My textbook says this reaction will take place, albeit slowly, as the e- withdrawing groups will delocalize the charge on the intermediate carbanion. But my doubt is, won't the charge be on the carbon meta to these groups, so that the -M effect won't affect the charge at all?

ok we need to develop this, when dealing with nucleophilic aromatic substituion, the electron withdrawing group will infact activate the ring towards the nucleophilic substituion but they still direct the substituent towards the meta position.

 

is this what you wanted to ask?

The -ve charge is delocalised ortho and para to the halide carrying carbon and so the presence of eg NO2 groups ortho and para can delocalise the charge by a negative conjugative effect. The intermediates are known as Meisenheimer complexes