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(A) $1$. The reaction of $1$-bromo-$3$-chlorocyclobutane with $Na/Et_2O$ is a Wurtz reaction. Since the $Br$ atom is a better leaving group than $Cl$,

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$D$-cyclobutylcyclobutane-$D$ and $F$-cyclobutylcyclobutane-$F$

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(A) $1$. The reaction of $1$-bromo-$3$-chlorocyclobutane with $Na/Et_2O$ is a Wurtz reaction. Since the $Br$ atom is a better leaving group than $Cl$,the coupling occurs at the $Br$ position,forming $1,1'$-dichloro-$3,3'$-bicyclobutane (compound $A$).$2$. Compound $A$ reacts with $(i) Mg/Et_2O$ followed by $(ii) D_2O$. The $Mg$ forms a Grignard reagent with the $Cl$ atoms,which is then quenched by $D_2O$ to replace $Cl$ with $D$,yielding $B$ as $1,1'$-dideuterio-$3,3'$-bicyclobutane.$3$. Compound $A$ reacts with $CoF_2$ (a fluorinating agent used in the Swarts reaction) to replace the $Cl$ atoms with $F$ atoms,yielding $C$ as $1,1'$-difluoro-$3,3'$-bicyclobutane.$4$. Therefore,the products $B$ and $C$ are $1,1'$-dideuterio-$3,3'$-bicyclobutane and $1,1'$-difluoro-$3,3'$-bicyclobutane respectively.

In the following sequence of reaction,the major products $B$ and $C$ respectively are :

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