Friedel-Craft reaction with amide substituent In this question, I know that aniline will not undergo FC reaction as it forms a complex with $\ce {AlCl3}$ and precipitates out. The answer given is (c) and I agree with it. However, the benzene ring with the amide substituent in (d) will be deactivated as the amide substituent will put a positive charge into resonance. So, will it give FC reaction? Should the answer be both c and d?
-NHCOCH3 is a slightly activating group due to the lone pair of N which shows +M effect. Hence it is Ortho para directing.