An organic compound 'A' with emperical formula C₆ H₆ O gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is

Solution

Chemistry ma’am · 4 steps

Sooty flame plus C₆H₆O means phenol, and a non-polar solvent lets bromine in only once, at para.

1. Identify A. A sooty flame means a high carbon-to-hydrogen ratio, which points to an aromatic ring. C₆H₆O has 4 degrees of unsaturation, exactly what a benzene ring needs. So A is phenol, C₆H₅OH.
2. Why only monobromination. In water, phenol ionises partly and the ring becomes very strongly activated, so you get 2,4,6-tribromophenol. In a low-polarity solvent such as CS₂ or CHCl₃, bromine is not polarised much and phenol does not ionise. Only one bromine goes in.

   C₆H₅OH + Br₂ →(CS₂) BrC₆H₄OH + HBr

3. Para over ortho. The -OH group directs to ortho and para. Both isomers form, but the para position is far from the -OH group and less crowded. So 4-bromophenol is the major, high-yield product.
4. So. B is p-bromophenol, which is option (d).

The idea to keep. Phenol with bromine water gives 2,4,6-tribromophenol. With Br₂ in CS₂ or CHCl₃ you get monobromophenols, and para is the major one.

Answer from the NTA final answer key. Solution by PrepWiser.