Explain why phenol directs to positions 2, 4 and 6
Connect oxygen lone-pair donation into the aromatic ring with ortho and para substitution at positions 2, 4 and 6.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 32. Distinguish phenol (–OH directly on an aromatic ring) from an alcohol (–OH on an aliphatic carbon). Use the specified reagents, temperatures and products; phenol reactions are limited to those listed in the syllabus. Written practice is self-checked
Before you begin
- Recognise alcohols and phenols, aromatic substitution, acids and bases, and diazonium ions from the preceding organic topics.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If –OH is at position 1, which positions are directed toward?
What you will learn
- Identify positions 2, 4 and 6 relative to phenolic –OH
- Explain the directing effect using electron donation
- Relate the positions to phenol substitution products
Map the ring positions
When –OH is attached to ring carbon 1, positions 2 and 6 are ortho and position 4 is para. In phenol's listed reactions, electrophilic substitution is directed to positions 2, 4 and 6.
Positions 2 and 6 are equivalent by symmetry in unsubstituted phenol, so nitration gives the 2- and 4-isomers; bromination can occupy all three sites to give 2,4,6-tribromophenol.
Oxygen donates electron density into the ring
A lone pair on oxygen can overlap with the aromatic π system. Resonance structures place increased electron density at ortho and para positions, making those positions more favourable for electrophilic attack.
The rule in the syllabus is to recall that the hydroxyl group directs to 2, 4 and 6. Use the electronic explanation to understand the pattern, but keep specific product predictions within the listed reactions and available positions.
Pause and explain
What does oxygen contribute to the aromatic ring?
Worked example
Explain why –OH directs electrophilic substitution in phenol to positions 2, 4 and 6.
Show the worked solution
- The oxygen has a lone pair.
- That lone pair can delocalise into the aromatic π system.
- Resonance increases electron density at ortho and para positions.
Answer Oxygen lone-pair donation into the ring makes positions 2, 4 and 6 more electron-rich and favours electrophilic substitution there.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Label the ortho and para positions relative to phenolic –OH at carbon 1.
Give me a hint
Count around the six-carbon ring.
Compare my reasoning
- Adjacent positions 2 and 6 are ortho.
- Opposite position 4 is para.
- Together the directed positions are 2, 4 and 6.
Ortho: 2 and 6; para: 4.
Look for these in your work
- Ortho positions correct
- Para position correct
- Full set stated
How does oxygen's lone pair affect ring electron density?
Give me a hint
Connect lone-pair donation to resonance positions.
Compare my reasoning
- The lone pair overlaps with the aromatic π system.
- Resonance delocalises electron density into the ring.
- Positions 2, 4 and 6 are activated for electrophilic substitution.
Oxygen donates a lone pair by resonance, increasing electron density at ortho and para positions 2, 4 and 6.
Look for these in your work
- Orbital interaction stated
- Electron-density change stated
- Positions identified
A ring is already substituted at position 1 by –OH and at position 4 by –NO₂. A question asks only which positions –OH directs toward. State them and qualify the prediction.
Give me a hint
Answer relative to the requested group, then note the second substituent.
Compare my reasoning
- –OH directs to positions 2, 4 and 6.
- Position 4 is already occupied.
- Other substituents may influence the actual mixture; do not claim a unique ratio from –OH alone.
Relative to –OH, the directing positions are 2, 4 and 6; because position 4 is occupied and –NO₂ also directs, a full product distribution needs more context.
Look for these in your work
- OH-directed positions named
- Occupied site noted
- Competing influence qualified
Common mistakes
- Calling –OH meta directing
- Numbering from an arbitrary ring atom
- Forgetting position 1 is the carbon bonded to –OH
- Claiming the directing group supplies an electrophile
What can you explain now?
If –OH is at position 1, which positions are directed toward?
Compare with the explanation
2, 4 and 6
The hydroxyl group directs to both ortho positions and the para position.
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Make it stick
How does –OH direct phenol reactions, and which positions correspond to its ortho/para effect?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.