Which of these will produce the highest yield Friedel Crafts reaction?

In a Friedel-Crafts (alkylation or acylation) reaction the aromatic ring behaves as a nucleophile while the electrophile is generated with the help of a Lewis‐acid catalyst such as $$AlCl_3$$. Hence, the rate and overall yield of the reaction depend mainly on two factors:

1. The electron density of the aromatic ring (controlled by the substituents already present).
2. Possible inhibition caused by the substituent forming a tight complex with the Lewis acid, thereby tying up the catalyst.

Step 1 - Classify the substituents in the four compounds
• Electron-withdrawing, $$-I$$ and $$-M$$ (deactivating) groups such as $$-NO_2$$ decrease the ring’s nucleophilicity and give very poor, often zero, Friedel-Crafts yield.
• Electron-withdrawing but weakly deactivating halogens ($$-I$$, weak $$+M$$) also lower the yield.
• Alkyl groups ($$+I$$, hyperconjugation) activate the ring moderately and usually provide good yields.
• Strongly electron-donating groups having lone pairs in conjugation with the ring (e.g. $$-OCH_3$$, $$-NH_2$$) possess a powerful $$+M$$ effect and give the fastest reactions and the best isolated yields.

Step 2 - Examine each option
Option A and Option B contain deactivating substituents (for example $$NO_2$$ or a halogen), so they will suffer from very low conversion.
Option C contains only an alkyl group. The ring is activated, but not as strongly as it could be.
Option D is anisole (methoxybenzene). The $$-OCH_3$$ group donates electron density through resonance ($$+M$$ effect), creating a highly nucleophilic ortho-/para-rich ring. Although oxygen can coordinate weakly with $$AlCl_3$$, this complex is reversible, and the strong activation dominates, giving the highest overall yield among the four compounds.

Therefore, the substrate that delivers the highest yield in a Friedel-Crafts reaction is the one bearing the methoxy group.

Option D which is: anisole (methoxybenzene)