Understanding Tribromo Derivative Formation in Phenol

Phenol undergoes electrophilic aromatic substitution reactions readily due to the activating effect of the -OH group. When treated with bromine water, phenol forms a white precipitate of 2,4,6-tribromophenol. To form a tribromo derivative, the compound must have three positions ortho and para to the -OH group that are available for substitution and not already occupied.

Step 1: Identify the Key Requirement

For a phenol derivative to form a tribromo compound, it must have at least three hydrogen atoms on the aromatic ring that are ortho or para to the -OH group. These positions are activated for electrophilic substitution.

Step 2: Analyze Each Option

We need to check which structure has three available ortho/para positions relative to the phenolic -OH group.

Option 1: $C H_{3} - C_{6} H_{4} - O H$ (Methyl phenyl ether, but drawn as cresol derivative? The image suggests a methyl group ortho to OH). If it is o-cresol (methyl group ortho to OH), it has only two available ortho/para positions (one ortho and two para, but one ortho is blocked by CH3). So it forms dibromo derivative.

Option 2: $H O - C_{6} H_{4} - C H_{3}$ (p-cresol, methyl group para to OH). It has two ortho positions to OH available, but the para position is blocked by CH3. So it forms dibromo derivative.

Option 3: $H O - C_{6} H_{4} - C H_{2} - C H_{3}$ (Phenylethanol, but the -CH2CH3 group is attached to the ring). The ethyl group is an ortho/para director, but it is a substituent on the ring. If the -OH is directly on the ring (phenol with an ethyl substituent), we need to see its position. If it is p-ethylphenol, the para position is blocked. However, the image shows a structure with the -OH on the ring and an ethyl group attached. Typically, if the ethyl group is at meta position to OH, it doesn't block ortho/para positions, but it is not a strong activator. But most importantly, if the ethyl group is at any position, it doesn't block all three positions. However, the key point is that the molecule has three available H atoms ortho/para to OH? Actually, if the ethyl group is at meta position, it leaves two ortho and one para position free, so it can form tribromo derivative. This is a likely candidate.

Option 4: $H O - C_{6} H_{4} - C O O H$ (Salicylic acid, -COOH group ortho to OH). The ortho position is blocked by -COOH, which is meta-directing. So only one ortho and one para position are available (since meta director deactivates ortho/para positions). It forms dibromo derivative.

Step 3: Final Answer

Option 3 will form the tribromo derivative because it has three available positions ortho and para to the -OH group for bromination. The ethyl group is attached to the ring, and if it is at a meta position (as commonly drawn), it does not block the ortho or para positions relative to the -OH group. Thus, it can undergo substitution at the two ortho and one para positions to give 2,4,6-tribromophenol derivative.

Related Topics

Electrophilic Aromatic Substitution
Directing Effects of Substituents
Reactivity of Phenols
Bromination of Aromatic Compounds

Key Formulae and Theory

Phenol bromination: $C_{6} H_{5} O H + 3 {Br}_{2} \to C_{6} H_{2} B r_{3} O H + 3 H B r$

Ortho/para directing groups (e.g., -OH, -OR, -alkyl) activate the ring for electrophilic substitution. Meta directing groups (e.g., -COOH, -NO2) deactivate the ring.