The given reaction involves a Friedel-Crafts alkylation using anhydrous \text{AlCl}_3 as a catalyst. Let's analyze the reaction step-by-step:
The presence of the methyl ether group (–OCH3) on the benzene ring makes it an activating group and directs the incoming electrophile to ortho/para positions.
The nitro group (–NO2) is a deactivating group, directing electrophiles to meta positions. However, due to the stronger activating effect of the –OCH3 group, its influence is predominant.
The chlorine on the isopropyl chloride reacts with the anhydrous \text{AlCl}_3 to generate an isopropyl cation (carbocation), which acts as the electrophile.
The methyl ether group strongly directs the electrophilic substitution to the para position due to steric hindrance at the ortho position and greater stability at para.
The major product will thus have the isopropyl group bonded to the para position relative to the –OCH3 group on the benzene ring.
The correct major product formed in this reaction is shown below:
