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This question was asked in my exam

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.

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Although a mechanism was not reported in Carpino's paper of 1958, there is a more concise mechanism than the one offered by @Soumik Das. There is no need to dissemble structure 1 to a tricarbonyl compound. Vinylogous elimination of the elements of HBr from dibromide 1 leads to species 2. [Alternatively, attack of hydroxide at the carbonyl of 1 would also suffice.] Attack of hydroxide at the carbonyl group of 2 produces 3, which via a Grob-like fragmentation affords the propargylic carboxylate 4 of the acid 5. ADDENDUM: The ring opening of 3 should not be construed as concerted. Just saving space.

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Carpino, L. A., J. Org. Chem., 1958, 80, 599.

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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.

The product formation can be summarised as follows,

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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).

enter image description here

The release of nitrogen gas is the main reason for driving this kind of elimination reaction forward.

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