Question: > 
> ### Passage-II
> 
> 
> [![1,3‐dibromocyclobutane, reaction scheme](https://i.stack.imgur.com/3fdJR.png)](https://i.stack.imgur.com/3fdJR.png)
> 
> 
> $$
> \ce{
> A ->[\text{1.}~\ce{NH2-}(\pu{2 mol})][\text{2.}~\ce{CH3-I}(\pu{2 mol})]
> B ->[\ce{HgSO4}, \ce{H2SO4}][\ce{H2O}]
> C ->[\text{dil.}~\ce{OH-}][Δ]
> D [Mixture]
> }
> $$
> 
> 
> **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: 
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.

![enter image description here](https://i.stack.imgur.com/9Csnn.png)
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, $\ce{HC#C-CH=CH2})$.