Identify the substrate

$p$-Hydroxybenzophenone is benzophenone with an $-OH$ group para to the carbonyl-linked phenyl bond on one ring.

Its structure can be viewed as: $\mathrm{HO{-}C_6H_4{-}CO{-}C_6H_5}$ where the $-OH$ is para on the left ring.

Nature of the reagent

Bromine in carbon tetrachloride, $\mathrm{Br_2/CCl_4}$, is a non-polar brominating reagent. It typically causes electrophilic bromination on activated aromatic rings.

Since no Lewis acid like $\mathrm{FeBr_3}$ is mentioned, bromination will occur preferentially on the more activated ring.

Effect of substituents on the aromatic ring

On the hydroxy-substituted ring, there are two substituents:

• $-OH$: strongly activating, ortho/para directing
• $-COPh$: deactivating, meta directing

The $-OH$ group dominates activation and directs bromination to its ortho/para positions.

Available positions for substitution

In $p$-hydroxybenzophenone, the para position to $-OH$ is already occupied by the $-COPh$ group. So bromination can occur only at the two ortho positions to $-OH$.

These are positions 2 and 6 of the ring, and due to symmetry they are equivalent.

Product formed

Therefore, monobromination gives: $\mathrm{2\text{-}bromo\text{-}4\text{-}hydroxybenzophenone}$

Since positions 2 and 6 are equivalent, only one monobromo product is formed.

Conclusion

So the correct option is the one showing bromination ortho to the phenolic $-OH$ group on the hydroxy-containing ring.

Hence, the answer is Option B.