Question

What is the product (Q) of the following reaction?

A.

B.

C.

D.

Answer 1

Hint: When CHFBrI will react with tertiary butoxide then, there is the formation of carbene, this carbine will react with the reactant molecule and insert itself in the reactant and then there will be the formation of carbocation leaving the most electronegative on the ring.

Complete step by step answer: Let us first see, what is the reaction between the molecules written on the arrow in the reaction, so, there are two compounds written, i.e., CHFBrI and t-BuO. CHFBrI is Bromochloroiodomethane and t-BuO is tertiary butoxide, so these react together, then there will be the formation of carbene in which the carbon atom will have a lone pair, fluorine and bromine atom. The reaction is given below:
$CHFBrI\xrightarrow{t-Bu{{O}^{-}}}\overset{-}{\mathop{C}}\,FBrI\xrightarrow{{{I}^{-}}}:CFBr$
This carbene will attack the reactant. The given reactant is a fused two pentene rings, so the carbine will attack one of the double bonds and will form a six-membered ring. After the formation of a six-membered ring, the bromine will leave and there will be the formation of a carbocation. This carbocation will be balanced by the formation of a double bond by the removal of a hydrogen atom from the ring. The series of reaction is given below:

So, the obtained product is equal to the compounds given in option (d). Therefore, the correct answer is an option (d).

Note: In the formation of carbene, the iodine was removed because the electronegativity of iodine is less than the electronegativity of bromine and fluorine. So, the atom that has the highest electronegativity will retain the aromatic ring.