4
$\begingroup$

Passage-II

1,3‐dibromocyclobutane, reaction scheme

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

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).

$\endgroup$
5
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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, $\ce{HC#C-CH=CH2})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.