Q.When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the product.

Concept: Acidity Of Phenol — but here the key reaction is electrophilic aromatic substitution; the –OH group activates the ring so strongly that bromination occurs at all three available positions without a catalyst.

Reasoning:

1. Bromine water ($\text{Br}_2/\text{H}_2\text{O}$) provides $\text{Br}^+$ electrophile.
2. The –OH group is a powerful ortho/para director. In phenol, all ortho and para positions are free.
3. Substitution happens at both ortho positions and the para position — a tribromination. …

The key to this reaction lies in understanding how the –OH group activates the benzene ring. The oxygen’s lone pairs donate electron density into the ring through resonance, making the ortho and para positions strongly nucleophilic. Bromine water ($\text{Br}_2$ in water) provides a source of electrophilic bromine ($\text{Br}^+$), which attacks these activated sites.

Unlike bromination of benzene (which requires a Lewis acid catalyst like $\text{FeBr}_3$), phenol is so reactive that it undergoes triple substitution even in mild conditions — no catalyst needed. The water in bromine water also helps by polarising the $\text{Br}_2$ molecule, generating $\text{Br}^+$ more readily.

1. Electrophile generation: In bromine water, $\text{Br}_2$ is polarised by water molecules, producing a small concentration of $\text{Br}^+$ ions:

2. First substitution: The $\text{Br}^+$ attacks the para position (most activated, least sterically hindered) of phenol. A $\sigma$-complex forms, then loses $\text{H}^+$ to restore aromaticity, giving 4-bromophenol.

3. Second substitution: The remaining ortho positions are still activated by the –OH group. Another $\text{Br}^+$ attacks one ortho position, yielding 2,4-dibromophenol.

4. Third substitution: The last ortho position (position 6) is still sufficiently activated. A third $\text{Br}^+$ attacks here, producing 2,4,6-tribromophenol.

5. Precipitation: 2,4,6-Tribromophenol is sparingly soluble in water due to its large nonpolar bromine atoms and the heavy molecular mass. It forms a white precipitate — often described as a "creamy white" or "pale yellow" solid in practice, but exam answers specify white. …

Method: Electrophilic Aromatic Substitution (Bromination of Phenol)

Why This Happens (Concept First)

Phenol ($C_6H_5OH$) is highly activated towards electrophilic substitution because the –OH group donates electrons via resonance into the benzene ring. This makes the ortho and para positions extremely electron-rich. Bromine water ($Br_2/H_2O$) provides the electrophile ($Br^+$), which attacks these positions.

However, because phenol is so reactive, tribromination occurs instantly — even at room temperature — without needing a catalyst. The reaction is so fast that all three available positions (two ortho, one para) get substituted.

Steps of the Method

1. Identify the activating group

   The –OH group on phenol donates electrons, making the ring ortho/para directing.

2. Recognize the reagent

   Bromine water ($Br_2/H_2O$) provides $Br^+$ ions. No catalyst (like $FeBr_3$) is needed because phenol is reactive enough.

3. Determine the substitution pattern

   The –OH group directs bromine to the 2, 4, and 6 positions (ortho and para). All three positions get substituted.

4. Write the reaction

5. Name and structure of the product …

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Writing the wrong product structure (mono-brominated instead of tri-brominated)

Why it happens:

Students often think bromination of phenol is similar to benzene — one substitution at a time. They forget that the –OH group is a powerful activating group, making the ring highly reactive toward electrophilic substitution.

How to avoid:

• Remember: Phenol + bromine water (excess bromine) gives 2,4,6-tribromophenol, not just monobromophenol.
• The –OH group directs bromine to the ortho and para positions. With excess bromine, all three positions get substituted.

Correct structure:

(Image: A benzene ring with –OH at C1, Br at C2, C4, C6)

Mistake 2: Writing the wrong IUPAC name

Why it happens:

Students confuse common names with IUPAC names, or they number the ring incorrectly.

How to avoid:

• The IUPAC name is 2,4,6-tribromophenol.
• Numbering starts from the carbon attached to –OH (C1). Bromines go to positions 2, 4, and 6.
• Common name: Tribromophenol (acceptable in many exam contexts, but IUPAC is safer).

Correct name:

2,4,6-Tribromophenol

Mistake 3: Forgetting the white precipitate observation

Why it happens:

Students focus only on the product structure and miss the physical state — which is often asked in exams.

How to avoid:

• Always note: 2,4,6-tribromophenol is a white precipitate (insoluble in water).
• This is a test for phenol — the formation of a white precipitate with bromine water confirms the presence of a phenolic –OH group.

Key point:

White precipitate = 2,4,6-tribromophenol

Mistake 4: Writing the reaction without balancing or showing bromine water correctly

Why it happens:

Students write “Br₂” without specifying “bromine water” or forget that 3 moles of Br₂ are needed.

How to avoid:

• Write the balanced equation:

• Bromine water is Br₂ in water — it provides excess bromine for complete substitution.

Mistake 5: Confusing this with Friedel-Crafts bromination or catalytic bromination

Why it happens: …