Product of treating 1,3‐dibromocyclobutane with alcoholic KOH at elevated temperatures

Question

Passage-II

$$ \ce{ A ->[\text{1.}~\ce{NH2-}(\pu{2 mol})][\text{2.}~\ce{CH3-I}(\pu{2 mol})] B ->[\ce{HgSO4}, \ce{H2SO4}][\ce{H2O}] C ->[\text{dil.}~\ce{OH-}][Δ] D [Mixture] } $$

31. "A" gives

A) Lucas test
B) Silver mirror test
C) White precipitate with ammoniacal $\ce{AgNO3}$
D) Yellow precipitate with ammoniacal cuprous chloride

32. How many isomeric products are existed in the mixture "D"?

A) 2
B) 4
C) 6
D) 8

What would be 'A' in the reaction? Any help would be deeply appreciated.

I know that alc. KOH does elimination, but I was confused whether it would eliminate one bromine each from two rings and create a bond between both the rings or it would simply eliminate both bromines from same ring and create two conjugated double bonds (cyclobut-1,3-diene).

EDIT: There is one more possibility: both the bromines get eliminated and a diagonal bond is formed in the ring (cyclobutane with any two diagonal carbons joined).

Answer 1

The two equivalents $\ce{HBr}$ are eliminated from two different molecules, one from the starting material and one from an intermediate. In the first step, $\ce{HBr}$ is eliminated under cleavage of the cyclobutane ring, which is favored by the release of ring strain. Elimination of $\ce{HBr}$ from 1-bromo-1,3-butadiene finally yields but-1-en-3-yne.

Answer 2

When 1,2-dibromocyclobutane is heated with KOH, 2 moles of acetylene are formed, probably via a cyclobutadiene intermediate. 1,3-dibromocyclobutane reacts differently.

Here, the final product is vinylacetylene (but-1-en-3-yne, $\ce{HC#C-CH=CH2})$.