An organic compound "A" with empirical formula CnH2nO gives sooty flame on burning. Its reaction with bromine solution…

Step 1: Identify compound A from clues

• Empirical formula $\ce{C_nH_{2n}O}$: degree of unsaturation = 1 (one ring or one double bond).
• Sooty flame: indicates high carbon-to-hydrogen ratio → aromatic compound (benzene ring gives sooty flame due to incomplete combustion).
• Empirical formula $\ce{CnH2nO}$ for an aromatic compound: $\ce{C6H6O}$ fits if n=6 → phenol ($\ce{C6H5OH}$, MW=94, empirical formula $\ce{C6H6O}$).

A = Phenol ✓

Step 2: Phenol + Br2 in low polarity solvent (CS2 or CCl4)

In a low polarity solvent (no water), bromination of phenol is controlled:

• Without water, only one or two Br enter (water promotes complete tribromination).
• In CS2/CCl4 at low temperature, monobromination occurs preferentially at the para position (less steric than ortho).
• Major product: p-bromophenol (4-bromophenol).

B = p-Bromophenol ✓

Step 3: Why not ortho?

Para position is preferred because:

• Less steric hindrance than ortho.
• Para product is more stable.
• Intramolecular H-bond possible in ortho product but steric factor dominates.

Key Points to Remember:

• Sooty flame + CnH2nO formula → phenol (aromatic).
• Phenol + Br2 in CS2/CCl4 (low polarity): monobromination → p-bromophenol (major).
• Phenol + Br2(aq) (water): tribromination → 2,4,6-tribromophenol.
• Water increases reactivity of Br2 and promotes multiple substitutions.