Question
Phenol undergoes electrophilic substitution much faster than benzene. Explain why, and write the products of: (a) Kolbe reaction, (b) Reimer-Tiemann reaction, and (c) coupling with diazonium salt.
(JEE Main / NEET pattern)
Solution — Step by Step
The -OH group has a lone pair on oxygen that delocalises into the benzene ring through resonance. This increases electron density at the ortho and para positions, making electrophilic attack easier. Phenol reacts about 1000 times faster than benzene in electrophilic substitution.
The -OH group is an activating, ortho-para directing group.
flowchart TD
A["Phenol\nC₆H₅OH"] -->|"CO₂ + NaOH\n125°C, 4-7 atm"| B["Salicylic acid\n(Kolbe reaction)"]
A -->|"CHCl₃ + NaOH"| C["Salicylaldehyde\n(Reimer-Tiemann)"]
A -->|"Br₂(aq)"| D["2,4,6-tribromophenol\n(white ppt)"]
A -->|"dil. HNO₃"| E["o-nitrophenol +\np-nitrophenol"]
A -->|"ArN₂⁺Cl⁻ / NaOH"| F["p-hydroxyazobenzene\n(Azo dye coupling)"]
B -->|"CH₃COCl"| G["Aspirin\n(acetylsalicylic acid)"]Sodium phenoxide is treated with CO₂ at high temperature and pressure. The CO₂ acts as a weak electrophile and attacks the ortho position. Salicylic acid is the precursor to aspirin — a real-world application.
CHCl₃ in NaOH generates the electrophile dichlorocarbene (:CCl₂), which attacks the ortho position of phenoxide. After hydrolysis, we get salicylaldehyde (2-hydroxybenzaldehyde).
If CCl₄ is used instead of CHCl₃, we get a carboxylic acid (salicylic acid) — but that is less commonly asked.
The diazonium salt is a weak electrophile that attacks at the para position of phenoxide ion. The resulting azo compound is a brightly coloured dye. This reaction needs alkaline medium to convert phenol to the more reactive phenoxide ion.
Why This Works
Phenol’s enhanced reactivity comes from the +M (mesomeric) effect of -OH, which donates electron density into the ring. In each named reaction, the electrophile is relatively weak (CO₂, :CCl₂, ArN₂⁺), but phenol’s activated ring is nucleophilic enough to react with them — benzene would not react with such weak electrophiles.
The ortho preference in Kolbe and Reimer-Tiemann reactions is due to the chelation of the sodium ion with both the phenoxide oxygen and the incoming group, stabilising the ortho transition state.
Alternative Method — Predicting Products via Directing Effects
For any electrophilic substitution on phenol: -OH is ortho-para directing. With aqueous bromine, all three positions get substituted (2,4,6-tribromophenol). With bromine in CS₂ (non-polar solvent, less reactive conditions), you get only mono-substitution at the para position.
The Kolbe reaction and Reimer-Tiemann reaction are the two most tested named reactions from phenol chemistry in both JEE and NEET. Remember: Kolbe gives an acid (CO₂ is the reagent), Reimer-Tiemann gives an aldehyde (CHCl₃ is the reagent). If you confuse the reagents, you get zero marks.
Common Mistake
Students write phenol directly instead of sodium phenoxide in the Kolbe reaction. The reaction requires sodium phenoxide (C₆H₅ONa), not free phenol. Phenol itself does not react with CO₂ under these conditions — the phenoxide ion is the actual nucleophile. Similarly, Reimer-Tiemann needs phenol in NaOH, which converts it to phenoxide in situ.