# How do dibromopyrazolones undergo base-promoted decomposition?

I was not sure but seeing the base as NaOH I used the fact that it will take most acidic hydrogen that is N-H and electron pair will come over it and it will resonate from two sides. I could make upto this intermediate but next I am not really sure the ring will break up and give the product as option D.

In the first question, the formation of the product D) can be justified in the following way. First the reagent $$\ce{OH-/H2O}$$ will basically perform alkaline hydrolysis of the starting compound to form 2,3-diketocarboxylate ion. But by the hydrolysis, it will also release hydrazine ($$\ce{H2N-NH2}$$) in basic medium, which will ultimately perform Wolff-Kishner reduction on the $$2,3$$-diketone part of the intermidiate compound. It is interestingly observed that 1,2-diketones under Wolff-Kishner conditions always produce alkynes.
Also the mechanism of Wolff-Kishner reduction of $$\alpha - \beta$$ diketone compound is similar to the following image (Some intermidiate steps are skipped to make it concise).