Phenol reactions — electrophilic substitution, Kolbe, Reimer-Tiemann, coupling

Question

Why is phenol highly reactive towards electrophilic substitution, and what are the key named reactions of phenol — Kolbe, Reimer-Tiemann, and diazo coupling?

Solution — Step by Step

The -OH group is a powerful activating group and ortho-para director. The lone pair on oxygen delocalises into the benzene ring through resonance, increasing electron density at ortho and para positions.

This is why phenol undergoes bromination with just $\text{Br}_2$ (aq) — no Lewis acid catalyst needed. Compare this with benzene, which requires $\text{Br}_2$ + $\text{FeBr}_3$ .

\text{Phenol} + 3\text{Br}_2(\text{aq}) \rightarrow \text{2,4,6-tribromophenol} + 3\text{HBr}

Sodium phenoxide is treated with CO $_2$ under high pressure (4-7 atm) and temperature (125 degrees C), followed by acidification:

\text{C}_6\text{H}_5\text{ONa} + \text{CO}_2 \xrightarrow{125°\text{C, 4-7 atm}} \text{sodium salicylate} \xrightarrow{\text{H}^+} \text{salicylic acid}

The carboxyl group enters at the ortho position. Salicylic acid is the precursor for aspirin — a fact NEET loves to test.

Phenol is heated with chloroform (CHCl $_3$ ) in alkaline medium. The reactive species is dichlorocarbene (:CCl $_2$ ), generated in situ:

\text{CHCl}_3 + \text{OH}^- \rightarrow \text{:CCl}_2 + \text{H}_2\text{O} + \text{Cl}^-

The dichlorocarbene attacks the ortho position of the phenoxide ring, and after hydrolysis, gives salicylaldehyde (ortho-hydroxybenzaldehyde).

If CCl $_4$ is used instead of CHCl $_3$ , the product is salicylic acid (not aldehyde).

Phenol reacts with benzene diazonium chloride ( $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ ) in mildly alkaline medium to form an azo dye (orange coloured):

\text{C}_6\text{H}_5\text{OH} + \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{NaOH, 0-5°C}} \text{p-hydroxyazobenzene}

The coupling occurs at the para position preferentially. The reaction must be done at 0-5 degrees C to prevent decomposition of the diazonium salt.

flowchart TD
    A["Phenol C6H5OH"] --> B["Electrophilic Substitution"]
    A --> C["Kolbe Reaction"]
    A --> D["Reimer-Tiemann"]
    A --> E["Diazo Coupling"]
    B --> F["Br2 aq: 2,4,6-tribromophenol"]
    B --> G["Dilute HNO3: ortho + para nitrophenol"]
    C --> H["CO2, NaOH, pressure then H+: Salicylic acid"]
    D --> I["CHCl3/NaOH: Salicylaldehyde"]
    D --> J["CCl4/NaOH: Salicylic acid"]
    E --> K["ArN2+Cl-, mild base: Azo dye"]

Why This Works

All these reactions trace back to the high electron density in phenol’s ring. The -OH group donates electrons via resonance, making the ring nucleophilic enough to attack even weak electrophiles like CO $_2$ (Kolbe) or dichlorocarbene (Reimer-Tiemann). Ordinary benzene would not react with these mild electrophiles.

Alternative Method

For remembering Kolbe vs Reimer-Tiemann products: Kolbe gives -COOH (carbon dioxide adds a carboxyl group), Reimer-Tiemann gives -CHO (chloroform is the carbon source for an aldehyde). The reagent itself hints at the oxidation level of the product — CO $_2$ is fully oxidised carbon (gives acid), CHCl $_3$ has H on carbon (gives aldehyde).

Common Mistake

The most tested trap: “What happens when CHCl $_3$ is replaced by CCl $_4$ in the Reimer-Tiemann reaction?” Students write salicylaldehyde again. Wrong — CCl $_4$ gives salicylic acid, not salicylaldehyde. The extra chlorine means the intermediate hydrolyses to -COOH instead of -CHO. JEE Main 2022 Shift 2 had exactly this question.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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