An organic compound 'A' with empirical formula C$$_6$$H$$_6$$O gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is

The empirical formula of compound $$A$$ is $$C_6H_6O$$. A molecular formula with six carbons and a single oxygen that burns with a sooty flame suggests an aromatic ring containing one oxygen-bearing group. The two most common possibilities are phenol $$(C_6H_5OH)$$ and benzaldehyde $$(C_6H_5CHO)$$.

Benzaldehyde is deactivating and meta-directing; it hardly undergoes bromination in the absence of a Lewis acid. By contrast, phenol has a strongly activating $$-OH$$ group that makes bromination easy even without a catalyst. Hence $$A$$ is phenol.

Next, bromination is carried out “in a low-polarity solvent” (e.g. $$CS_2$$, $$CCl_4$$). In such solvents the bromine remains molecular $$Br_2$$ and the medium contains no water. Therefore:

• Only one bromine atom is introduced (no poly-bromination as in aqueous medium).
• Because the $$-OH$$ group is ortho-/para-directing, the substitution can occur at the ortho or para positions.
• Steric hindrance near the $$-OH$$ group suppresses the ortho product, so the para product is formed in the highest yield.

Thus compound $$B$$ is para-bromophenol (4-bromophenol).

According to the given options, para-bromophenol corresponds to Option A.

Option A which is: para-bromophenol