Key fact about aniline in Friedel–Crafts reaction

Aniline is $\ce{C6H5NH2}$. In Friedel–Crafts alkylation, a Lewis acid catalyst such as $\ce{AlCl3}$ is used.

The amino group of aniline has a lone pair on nitrogen. This lone pair strongly coordinates with $\ce{AlCl3}$:

$\ce{C6H5NH2 + AlCl3 -> C6H5NH2->AlCl3}$

Effect of coordination/protonation on the ring

When nitrogen donates its lone pair, it acquires a positive character. The species behaves like an anilinium-type complex, and the benzene ring becomes strongly deactivated toward Friedel–Crafts alkylation.

This may be represented by resonance/complex forms in which nitrogen bears a positive charge:

$\ce{C6H5-NH2->AlCl3} \quad \text{or} \quad \ce{C6H5-NH2^{+}}$

Thus, instead of normal ortho/para alkylation, aniline forms a complex where the nitrogen is effectively positively charged and the ring is not activated for Friedel–Crafts reaction.

Check each option

A: alkylated product with ortho and para substitution
Incorrect. Although $\ce{-NH2}$ is ordinarily an ortho/para-directing group, under Friedel–Crafts conditions it coordinates with $\ce{AlCl3}$ and deactivates the ring.

B: secondary amine after acidic treatment
Incorrect. Friedel–Crafts alkylation of aniline does not give a secondary amine after acidic treatment.

C: an amide product
Incorrect. No amide is formed in Friedel–Crafts alkylation.

D: positively charged nitrogen at benzene ring
Correct. Due to interaction of the amino group with the Lewis acid catalyst, nitrogen develops positive charge, preventing normal Friedel–Crafts alkylation.

Final answer

Therefore, the correct option is:

$\boxed{\text{D}}$