The compound (X) will be: | vedclass

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(A) The reaction involves the removal of two bromine atoms from the starting material using $2AgBF_4$. $Ag^+$ ions react with $Br^-$ to form $AgBr$ pr

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(A) The reaction involves the removal of two bromine atoms from the starting material using $2AgBF_4$. $Ag^+$ ions react with $Br^-$ to form $AgBr$ precipitate,leaving behind a carbocation intermediate. Since two $Br$ atoms are removed,a dication is formed. The resulting species is a cyclobutadiene dication,which is aromatic ($2\pi$ electrons,$H$ückel's rule $4n+2$ where $n=0$). Thus,the compound $(X)$ is the cyclobutadiene dication with four phenyl groups attached,represented by a $+2$ charge in the center of the ring.

The compound $(X)$ will be:

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