The product of the following reaction is:

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(C) The reactant is phenyl benzoate,which contains two phenyl rings. One ring is attached to an oxygen atom $(-O-)$,and the other is attached to a car

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(C) The reactant is phenyl benzoate,which contains two phenyl rings. One ring is attached to an oxygen atom $(-O-)$,and the other is attached to a carbonyl group $(-C(=O)-)$. The oxygen atom is electron-donating by resonance ($+M$ effect),which activates the attached phenyl ring towards electrophilic aromatic substitution. The carbonyl group is electron-withdrawing by resonance ($-M$ effect),which deactivates the other phenyl ring. Therefore,the electrophile $(Br^+)$ generated by $Br_2/FeBr_3$ will attack the more activated ring,which is the one attached to the oxygen atom. In the ring attached to the oxygen atom,the $-O-$ group is ortho/para-directing. The para position is sterically less hindered than the ortho position,making the para-substituted product the major one. Thus,the product is phenyl $4$-bromobenzoate.

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