Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Experimental reaction of $$CH_3Cl$$ with aniline and anhydrous $$AlCl_3$$ does not give $$o$$ and $$p$$-methylaniline.
Reason (R): The $$-NH_2$$ group of aniline becomes deactivating because of salt formation with anhydrous $$AlCl_3$$ and hence yields $$m$$-methyl aniline as the product.
In the light of the above statements, choose the most appropriate answer from the options given below

We evaluate the assertion that the reaction of $$CH_3Cl$$ with aniline and anhydrous $$AlCl_3$$ does not give ortho- and para-methylaniline. This is true because when aniline is treated with $$CH_3Cl$$ in the presence of anhydrous $$AlCl_3$$ (Friedel-Crafts alkylation), the expected ortho- and para-methylanilines are not obtained.

The $$-NH_2$$ group reacts with $$AlCl_3$$ to form a salt:

$$C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5\overset{+}{N}H_2\text{-}AlCl_3^-$$

This salt formation converts the activating, ortho/para-directing $$-NH_2$$ group into the deactivating, meta-directing $$-\overset{+}{N}H_2AlCl_3^-$$ group, so the Friedel-Crafts reaction either does not proceed or gives poor yields.

The stated reason is that the $$-NH_2$$ group becomes deactivating because of salt formation with anhydrous $$AlCl_3$$ and hence yields meta-methylaniline. While the first part is true, the claim that it yields meta-methylaniline is false: the ring is so deactivated that the Friedel-Crafts reaction essentially does not occur, rather than giving the meta product, since Friedel-Crafts reactions do not work on strongly deactivated rings.

The assertion is true and the reason is false.

Hence, the correct answer is Option C: A is true, but R is false.