Which set of reactants is better for preparation of 1-methoxy-4-nitrobenzene? I've got two set of reactants:


4-nitrobromobenzene + sodium methoxide
sodium 4-nitrophenoxide + bromomethane


I've got to tell which reaction is more preferable among these two. 

The first reaction takes place through nucleophilic addition of methoxide on the ring followed by elimination of bromine. I've been taught that such reactions take place under drastic conditions, but the presence of an electron withdrawing group at ortho and para position increases the reactivity, so the presence of $\ce{NO2}$ group should increase the reactivity . 

The second reaction takes place by SN2 mechanism, where bromomethane is attacked by the nitrophenoxide ion. This reaction "seems" to be better than the first. (Is it?)

Hence, my attempt is that the second reaction should be more favourable than the first since the first reaction should take place under more drastic conditions than the second. 

Am I right? Or does the presence of $\ce{NO2}$ group increase the reactivity so much that it is comparable to the second reaction or maybe better?
Drastic conditions for reaction 1? Not really. Reflux the reactants in $\ce{MeOH}$ and the reaction should go smoothly and in high yield.

Reaction 2 - two difficulties.
The anion of 4-nitrophenol is very stable and, because of the electron withdrawing power of the nitro group, it is not very nucleophilic. Second difficulty is that bromomethane has a boiling point of 3C so the reaction mixture has to be kept cold to keep it in the reaction mixture. This does not help you overcome the poor nucleophilicity of your phenoxide.

Without running the two reactions it is guesswork to predict which will give the better outcome, but I favour the SNAr.