In Friedel-Crafts alkylation of aniline, one gets:

We need to identify what happens during Friedel-Crafts alkylation of aniline. Aniline (C₆H₅NH₂) has a lone pair on nitrogen, and in Friedel-Crafts alkylation, the Lewis acid catalyst (like AlCl₃) is used to generate a carbocation. The lone pair on the nitrogen atom of aniline reacts with the Lewis acid catalyst AlCl₃, forming a coordination complex where nitrogen donates its lone pair to the Lewis acid:

$$\text{C}_6\text{H}_5\text{NH}_2 + \text{AlCl}_3 \rightarrow \text{C}_6\text{H}_5\text{NH}_2^{+}-\text{AlCl}_3^{-}$$

This effectively puts a positive charge on the nitrogen atom, and the positively charged nitrogen acts as a strong deactivating group (meta-directing). The nitrogen withdraws electron density from the ring through its positive charge, making the ring highly electron-deficient. This deactivates the benzene ring so much that Friedel-Crafts reaction does not proceed on the ring. In fact, Friedel-Crafts reactions generally fail with aniline and other amines because the amine group gets protonated/complexed, deactivating the ring.

The answer is Option D: positively charged nitrogen at benzene ring.