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Answer:
I attached the answer as an image. I also drew in the two most acidic hydrogens.
Explanation:
This goes through the 'benzyne' intermediate, meaning it does an E2-looking reaction by expelling a leaving group (chloride) from the adjacent part of the ring using the amide as a strong base. The triple-bonded benzyne has absurd bond angle strain, and is vulnerable to a good nucleophile like an amide ion, and the resultant sp2 anion is then reprotonated by the acid. I didn't draw in the acid-base reaction in step one, or the spectator ion (sodium).
