Reactions of phenol

Dip a strip of phenol solution in neutral $FeCl_3$ and it turns violet — a party trick, but also a clue to phenol's whole personality. The ring and the –OH oxygen share electrons through conjugation, and that single fact explains everything phenol does: it makes the O–H bond acidic enough to react with NaOH (though not with $NaHCO_3$ ), and it floods electron density into the ortho and para ring positions, making phenol far more reactive than benzene toward electrophiles. This article works through phenol's acidic reactions, its two named syntheses — Kolbe's reaction and the Reimer–Tiemann reaction — electrophilic ring substitution, and the two confirmatory tests examiners love to ask about.

The core insight

Conjugation of the –OH oxygen's lone pair with the benzene ring both stabilises the phenoxide ion (making phenol acidic) and activates the ring at the ortho/para positions (making phenol substitute far faster than benzene), and both effects drive every named reaction in this article.

The core idea

Acidic character. Phenol ( $pK_a \approx 10$ ) loses $H^+$ far more readily than an alcohol ( $pK_a \approx 16$ – $18$ ) because the resulting phenoxide ion spreads its negative charge into the ring by resonance, while an alkoxide ion has nowhere to delocalise. Phenol is still much weaker than a carboxylic acid ( $pK_a \approx 4$ – $5$ ), so it reacts with NaOH but is too weak to displace $CO_2$ from $NaHCO_3$ — a classic distinguishing test between phenols and carboxylic acids:

C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O

2C_6H_5OH + 2Na \rightarrow 2C_6H_5ONa + H_2\uparrow

2C_6H_5OH + Zn \xrightarrow{\Delta} (C_6H_5O)_2Zn + H_2\uparrow

where: $C_6H_5OH$ is phenol, $C_6H_5ONa$ is sodium phenoxide, and each equation shows the O–H proton being removed by a base ( $OH^-$ ), a metal (Na) or a metal that forms a covalent phenoxide (Zn).

Esterification and $PCl_5$ . Acetyl chloride or acetic anhydride O-acylates phenol to phenyl acetate:

C_6H_5OH + CH_3COCl \rightarrow C_6H_5OOCCH_3 + HCl

With $PCl_5$ , phenol behaves unlike an alcohol: the aromatic C–O bond resists substitution, so the main product is the phenyl ester of phosphorus oxychloride rather than chlorobenzene.

Electrophilic substitution. The –OH lone pair donates into the ring by resonance, so phenol undergoes bromination, nitration and sulphonation far more readily than benzene, usually without any catalyst, and always ortho/para to –OH:

C_6H_5OH + 3Br_2(aq) \rightarrow 2,4,6\text{-tribromophenol}\downarrow + 3HBr

Kolbe's reaction. Sodium phenoxide, heated with $CO_2$ under pressure, is attacked at the ortho position by $CO_2$ (the electrophile); acidifying the product gives salicylic acid:

C_6H_5ONa + CO_2 \xrightarrow[125^\circ C]{4\text{–}7\ atm} C_6H_4(OH)COONa \xrightarrow{H^+} C_6H_4(OH)COOH

Reimer–Tiemann reaction. Phenol with chloroform and aqueous NaOH generates dichlorocarbene ( $:CCl_2$ ), which attacks the ring ortho to –OH; hydrolysis then delivers an aldehyde:

C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{\Delta} \text{salicylaldehyde} + 3NaCl + 2H_2O

Tests. The $FeCl_3$ test gives a violet/purple colour with neutral ferric chloride; the azo dye test couples phenol with a cold diazonium salt to give a coloured (typically orange) azo compound.

Figure

Phenol ring showing the oxygen lone pair delocalising into the ortho and para positions and increasing electron density there

Why it works

Every reaction above traces back to one structural fact: the oxygen's lone pair is not stuck on oxygen — it is delocalised into the aromatic $\pi$ system. This one conjugation does two jobs at once, and understanding both is worth more than memorising six separate reactions.

Job 1 — acidity. When phenol loses $H^+$ , the resulting phenoxide ion, $C_6H_5O^-$ , is not a plain localised alkoxide. The negative charge on oxygen can be drawn as resonance structures with the charge sitting on the ortho and para carbons instead, spreading it over four positions in total. Spreading charge lowers energy, so the phenoxide ion is markedly more stable than a hypothetical localised alkoxide — and a more stable conjugate base means a stronger acid. An ordinary alcohol's alkoxide has no ring to delocalise into, so alcohols stay far less acidic ( $pK_a \approx 16$ – $18$ versus phenol's $pK_a \approx 10$ ). Carboxylate ions, by contrast, delocalise charge over two equivalent oxygens with a shorter resonance path, which stabilises them even more than phenoxide — hence carboxylic acids ( $pK_a \approx 4$ – $5$ ) beat phenol on acid strength, and phenol cannot deprotonate the weaker base $NaHCO_3$ the way a carboxylic acid can.

Job 2 — ring activation. The very same lone-pair donation that stabilises phenoxide also raises electron density at the ring carbons before any proton is even lost — the neutral phenol molecule already has extra electron density concentrated ortho and para to –OH, exactly where the resonance structures place it. An electrophile ( $Br^+$ , $NO_2^+$ , $SO_3$ , $CO_2$ , or dichlorocarbene) is attracted to the most electron-rich carbon it can find, so it always lands ortho or para. Because the ring is so strongly activated, phenol's substitutions often need no catalyst at all — a sharp contrast with benzene, which needs a Lewis acid such as $FeBr_3$ to generate enough electrophile character. This is also why bromine water alone brominates phenol at all three activated positions instantly, while benzene needs forcing conditions for just one.

Worked example

Phenol is treated first with bromine water, and separately, sodium phenoxide is treated with $CO_2$ under pressure followed by acid. Identify both products and explain why each electrophile attacks where it does.

Step 1. Bromine water reaction. The three ring carbons ortho and para to –OH (positions 2, 4 and 6) carry the extra electron density from resonance donation, so all three are attacked by the mild electrophile $Br_2$ without any catalyst:

C_6H_5OH + 3Br_2(aq) \rightarrow 2,4,6\text{-tribromophenol}\downarrow(\text{white ppt}) + 3HBr

The instant white precipitate is a standard qualitative test for phenol; benzene under the same conditions shows no reaction at all.

Step 2. Kolbe's reaction. Sodium phenoxide is even more electron-rich at the ring than neutral phenol (the full negative charge sits partly on oxygen and partly, by resonance, on the ortho/para carbons), so it reacts with the mild electrophile $CO_2$ at 125°C and 4–7 atm, attack occurring at the ortho carbon:

C_6H_5ONa + CO_2 \xrightarrow[125^\circ C]{4\text{–}7\ atm} C_6H_4(OH)COONa

Step 3. Acidify. Treating sodium salicylate with dilute acid protonates the carboxylate to give the neutral acid:

C_6H_4(OH)COONa \xrightarrow{H^+} C_6H_4(OH)COOH

The two products are 2,4,6-tribromophenol from bromine water, and salicylic acid (2-hydroxybenzoic acid) from the Kolbe sequence — both installed exactly where resonance predicts the extra electron density lies.

Competitive stretch

If the same phenoxide is instead treated with $CHCl_3$ /NaOH (Reimer–Tiemann conditions), the electrophile changes from $CO_2$ to dichlorocarbene, $:CCl_2$ , generated in situ; it still attacks ortho for the same electronic reason, but delivers a –CHO group after hydrolysis, giving salicylaldehyde instead of salicylic acid. Same position, different electrophile, different functional group — a favourite examiner twist.

Common traps

Assuming phenol is acidic enough to react with $NaHCO_3$ — it is not; only carboxylic acids liberate $CO_2$ from bicarbonate, and this non-reaction is a standard test to tell the two apart.
Forgetting that phenol's electrophilic substitutions need no Lewis-acid catalyst and readily over-substitute (all three activated sites) unless conditions like dilute bromine water and low temperature are controlled.
Mixing up Kolbe's reaction ( $CO_2$ electrophile, gives an acid) with Reimer–Tiemann (dichlorocarbene electrophile, gives an aldehyde) — both ortho-selective, but different electrophile and different product.
Dropping the word "neutral" for the $FeCl_3$ test — acidic ferric chloride solutions do not give the diagnostic violet colour.

Ask the tutor: want to work through why sulphonation or nitration of phenol also lands ortho/para, or compare Kolbe's and Reimer–Tiemann side by side for a substituted phenol?

Why doesn't phenol react with NaHCO3 if it's still an acid?

Acidity is a matter of degree, set by how stable the conjugate base is. Phenoxide delocalises its negative charge into the ring, which is a real but modest stabilisation, giving $pK_a \approx 10$ . Carboxylate ions delocalise equally over two oxygens in a shorter, more effective resonance path, giving $pK_a \approx 4$ – $5$ — a thousand-fold stronger acid. $NaHCO_3$ is only a weak enough base to be deprotonated by acids stronger than roughly $pK_a \approx 6$ – $7$ , so it reacts with carboxylic acids but is untouched by phenol. That gap is exactly what makes the "fizzes with $NaHCO_3$ " test useful for telling the two functional groups apart.

Does phenol always give tri-substitution with bromine, or can you stop at mono-substitution?

With aqueous bromine water in excess, all three activated ortho/para positions react and you get 2,4,6-tribromophenol as a white precipitate — this is the standard qualitative test. To stop at mono-bromination, chemists instead use bromine dissolved in a non-polar solvent like $CS_2$ at low temperature, which moderates the reaction enough to isolate mainly para-bromophenol. The underlying activation is the same either way; it is the reaction conditions, not the mechanism, that decide how far substitution goes.

How is dichlorocarbene generated in the Reimer-Tiemann reaction, and why does it attack the ring?

NaOH deprotonates $CHCl_3$ to give the carbanion $CCl_3^-$ , which then expels a chloride ion to form the neutral, highly reactive species dichlorocarbene, $:CCl_2$ , an electron-deficient carbon with an empty orbital. Because phenol's ring (as its phenoxide, under the basic conditions) is electron-rich exactly at the ortho and para carbons, dichlorocarbene behaves as an electrophile and attacks there, preferentially ortho. The resulting intermediate is then hydrolysed in the reaction conditions to unmask the aldehyde group, giving salicylaldehyde.