Why is Aniline a poor substrate for Nitration reaction? I have been asked to find out which one of the following is a poor substrate for Nitration reaction:



Upon encountering this question, my first reaction was the option c, as I know that the nitration mechanism involves the substitution of a positively charged Nitronium ion ($\ce{NO2+}$) and hence a $\ce{-OCOCH_3}$ group that shows $\mathrm{-M}$ effect will not be helpful for that process as it reduces the electron density in the benzene ring.
But the solution states that the answer is aniline (given in option a). I do not understand how is that possible. Even in the case when we use mixed acid as the reagent, it will form para- and meta-substituted nitro compounds. But, I do not know why and how does it behave as a 'poor substrate' in any case of nitration.
Please explain.
In the case of option c, the $\ce{-OCOCH_3}$ group will not be showing $\mathrm{-M}$ effect as the ring is directly attached to the oxygen atom. The group rather shows $\mathrm{+M}$ effect because of the two lone pairs present on the oxygen atom, increasing the electron density in the benzene ring and thus making it a good substrate for a nitration reaction.
On the other hand, as nitration is generally done with mixed acid as the reagent, the $\ce{-NH_2}$ group in aniline takes up a hydrogen atom to become $\ce{-NH_3^+}$,  which is a ring deactivating group.
Therefore, among the options given, aniline is the poorest substrate for nitration reactions.