Phenols - Electrophilic Substitution
Phenol reacts with bromine water at room temperature without any catalyst, so which product is obtained as a white precipitate in this reaction?
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Solution
2,4,6-tribromophenol
Phenol is a strongly activated aromatic compound because the hydroxyl oxygen donates electron density into the ring by resonance, greatly increasing reactivity toward electrophilic substitution at the ortho and para positions. In bromine water the ring is so activated that no Lewis acid catalyst is needed, and substitution proceeds at all the available activated positions rather than stopping at one. As a result phenol gives 2,4,6-tribromophenol, which separates as a white precipitate, making it the correct answer. The monobromo products o-bromophenol and p-bromophenol form only under controlled, less polar conditions such as bromine in carbon disulphide at low temperature, not in aqueous bromine, so they are wrong here. m-bromophenol is impossible because the hydroxyl group is ortho-para directing, never meta directing. This contrast between aqueous and non-aqueous bromination is highlighted in NCERT and is a frequent JEE Advanced product question. The reason a polar aqueous medium pushes the reaction to completion is that water promotes ionisation of bromine and stabilises the charged intermediates, raising effective electrophile concentration, whereas a non-polar solvent at low temperature keeps the bromine sluggish enough to allow clean monosubstitution. The same activation explains why phenol also reacts readily with dilute nitric acid where benzene would not. As a plausibility check, the highly activating OH group drives complete substitution at the two ortho and one para sites, matching the tribromo product.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing oxygen
- Topic
- phenols - electrophilic substitution
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
2,4,6-tribromophenol
Phenol is a strongly activated aromatic compound because the hydroxyl oxygen donates electron density into the ring by resonance, greatly increasing reactivity toward electrophilic substitution at the ortho and para positions. In bromine water the ring is so activated that no Lewis acid catalyst is needed, and substitution proceeds at all the available activated positions rather than stopping at one. As a result phenol gives 2,4,6-tribromophenol, which separates as a white precipitate, making it the correct answer. The monobromo products o-bromophenol and p-bromophenol form only under controlled, less polar conditions such as bromine in carbon disulphide at low temperature, not in aqueous bromine, so they are wrong here. m-bromophenol is impossible because the hydroxyl group is ortho-para directing, never meta directing. This contrast between aqueous and non-aqueous bromination is highlighted in NCERT and is a frequent JEE Advanced product question. The reason a polar aqueous medium pushes the reaction to completion is that water promotes ionisation of bromine and stabilises the charged intermediates, raising effective electrophile concentration, whereas a non-polar solvent at low temperature keeps the bromine sluggish enough to allow clean monosubstitution. The same activation explains why phenol also reacts readily with dilute nitric acid where benzene would not. As a plausibility check, the highly activating OH group drives complete substitution at the two ortho and one para sites, matching the tribromo product.
This medium difficulty chemistry question is from the chapter organic compounds containing oxygen, covering the topic of phenols - electrophilic substitution. It appeared in the 2025 exam.
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