Question:
An organic compound (X) dissolves in dilute NaOH but not in dilute \(NaHCO_3\) solution. X with \(Br_2/H_2O\) gives tribromo derivative. What is X?
An organic compound (X) dissolves in dilute NaOH but not in dilute \(NaHCO_3\) solution. X with \(Br_2/H_2O\) gives tribromo derivative. What is X?
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Phenol + bromine water gives a white precipitate of 2,4,6-tribromophenol without any catalyst.
Updated On: Jun 18, 2026
- Option 1
- Option 2
- Option 3
- Option 4
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The Correct Option is B
Solution and Explanation
Concept:
Phenol is acidic enough to react with NaOH but not strong enough to react with sodium bicarbonate.
Step 1: Analyze the NaOH test.
Phenol reacts with sodium hydroxide. \[ C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O \] Therefore it dissolves in NaOH.
Step 2: Analyze the \(NaHCO_3\) test.
Phenol is weaker than carbonic acid. Hence it cannot liberate \(CO_2\) from sodium bicarbonate. Therefore it does not react with \[ NaHCO_3 \]
Step 3: Reaction with bromine water.
Phenol strongly activates the benzene ring. \[ C_6H_5OH + 3Br_2 \rightarrow 2,4,6\text{-Tribromophenol} +3HBr \] A white precipitate is obtained. Hence \[ \boxed{X=\text{Phenol}} \]
Step 1: Analyze the NaOH test.
Phenol reacts with sodium hydroxide. \[ C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O \] Therefore it dissolves in NaOH.
Step 2: Analyze the \(NaHCO_3\) test.
Phenol is weaker than carbonic acid. Hence it cannot liberate \(CO_2\) from sodium bicarbonate. Therefore it does not react with \[ NaHCO_3 \]
Step 3: Reaction with bromine water.
Phenol strongly activates the benzene ring. \[ C_6H_5OH + 3Br_2 \rightarrow 2,4,6\text{-Tribromophenol} +3HBr \] A white precipitate is obtained. Hence \[ \boxed{X=\text{Phenol}} \]
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