Reactions of Phenol

CHEPhenols

Phenol's $\text{-}OH$ strongly activates the benzene ring, so it undergoes easy electrophilic substitution plus its own named reactions — Kolbe and Reimer–Tiemann. This subtopic also covers the $FeCl_3$ identification test and common phenol-to-product conversions that appear as short-answer questions.

Kolbe's reaction

C_6H_5O^-Na^+ + CO_2 \xrightarrow[\text{then } H^+]{400\ K,\ \text{pressure}} \text{2-hydroxybenzoic acid (salicylic acid)}

Reimer–Tiemann reaction

C_6H_5OH \xrightarrow[\text{then } H_3O^+]{CHCl_3,\ NaOH} \text{2-hydroxybenzaldehyde (salicylaldehyde)}

Bromination of phenol

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

Williamson etherification to anisole

C_6H_5O^-Na^+ + CH_3I \rightarrow C_6H_5OCH_3 + NaI

Phenol with sodium metal

2\,C_6H_5OH + 2Na \rightarrow 2\,C_6H_5O^-Na^+ + H_2\uparrow

Esterification with acetyl chloride

C_6H_5OH + CH_3COCl \xrightarrow{\text{pyridine}} CH_3COOC_6H_5\ (\text{phenyl acetate}) + HCl

Kolbe's reaction: sodium phenoxide $+\ CO_2$ under pressure, then $H^+$ , gives salicylic acid (2-hydroxybenzoic acid) — the basis of aspirin manufacture.
Uses: phenol itself is used as an antiseptic/disinfectant (dilute solutions; e.g. chloroxylenol in Dettol is a phenol derivative) and to manufacture Bakelite (phenol–formaldehyde resin), dyes and explosives (picric acid/TNP); salicylic acid from Kolbe's reaction is the precursor for aspirin.
Reimer–Tiemann reaction: phenol $+\ CHCl_3$ /NaOH then hydrolysis gives salicylaldehyde (2-hydroxybenzaldehyde); the reactive intermediate is dichlorocarbene $:CCl_2$ .
$FeCl_3$ test: phenol gives a violet/purple colour (forming a coloured iron–phenoxide complex); ethanol gives no colour — a key distinguishing test.
With $Br_2$ water (no catalyst) phenol gives 2,4,6-tribromophenol (white ppt); with conc. $HNO_3$ it gives picric acid (2,4,6-trinitrophenol).
Phenol $\rightarrow$ anisole via Williamson: sodium phenoxide $+\ CH_3I \rightarrow C_6H_5OCH_3$ (the phenoxide $O^-$ is the nucleophile).
Distinguish phenol from ethanol with $FeCl_3$ (violet) or $Br_2$ water (white ppt); distinguish phenol from a carboxylic acid with $NaHCO_3$ (only the acid gives $CO_2$ ).
The $\text{-}OH$ group is $o,p\text{-}$ directing and ring-activating, so phenol's electrophilic substitutions occur readily under mild conditions.
Nitration: with dilute $HNO_3$ phenol gives a mixture of o- and p-nitrophenol (separable by steam distillation, as o-nitrophenol is steam-volatile); with conc. $HNO_3$ it gives 2,4,6-trinitrophenol (picric acid).
Bromination in a non-polar solvent like $CS_2$ at low temperature gives mainly p-bromophenol (monosubstitution), whereas $Br_2$ water gives the 2,4,6-tribromo product — solvent controls the extent.
Reaction with conc. $H_2SO_4$ gives o-/p-phenolsulphonic acid; the para product predominates at higher temperature (a reversible, thermodynamically controlled sulphonation).
Phenol reacts with acetic anhydride/acetyl chloride to give phenyl acetate (esterification of the $\text{-}OH$ ); with zinc dust it is reduced to benzene.
Phenol undergoes Friedel–Crafts and coupling chemistry too: with diazonium salts it gives orange/red azo dyes (coupling at the para position), another characteristic test for the activated ring.

Where the marks go

Naming the Reimer–Tiemann product as a carboxylic acid — it gives salicyl-aldehyde (an $\text{-}CHO$ group), not salicylic acid; the carbene $:CCl_2$ is the key intermediate.
Forgetting that Kolbe's reaction uses sodium phenoxide (not free phenol) and needs $CO_2$ under pressure — free phenol alone does not react.
Saying $FeCl_3$ must be acidified — the violet colour needs neutral or near-neutral $FeCl_3$ ; strongly acidic conditions can suppress the colour.
Confusing picric acid (2,4,6-trinitrophenol, from conc. $HNO_3$ ) with simple nitrophenols (from dilute $HNO_3$ ).
Treating phenol's ring as deactivated like nitrobenzene — $\text{-}OH$ is a strong activator, so substitution is faster than benzene and is $o,p\text{-}$ directing.

How the board asks it

Predict the productkolbe and reimer-tiemann named reactions
Give the structure of the major organic product, together with the conditions, when phenol is treated with (i) $CHCl_3$ and aqueous $NaOH$ followed by $H^+$ , and (ii) $CO_2$ under pressure after conversion to sodium phenoxide, then acidification.
Distinguish $FeCl_3$ test and $Br_2$ water
How will you distinguish between phenol and ethanol using a simple chemical test? State the observation for each.
Conversionphenol-to-product conversions
How will you convert phenol into (i) salicylic acid, (ii) picric acid, and (iii) anisole? Give the reagents and conditions for each step.
Give reasons $o,p$ -directing, ring-activating $\text{-}OH$
Account for the fact that phenol reacts with $Br_2$ water at room temperature without any catalyst to give a white precipitate of $2,4,6$ -tribromophenol, whereas benzene does not.
Identify / classify $FeCl_3$ violet colour and reimer-tiemann
An organic compound $A$ ( $C_6H_6O$ ) gives a violet colour with neutral $FeCl_3$ and, on treatment with $CHCl_3/NaOH$ followed by hydrolysis, yields $B$ , an aromatic compound that answers the Tollens' test. Identify $A$ and $B$ .
Predict the productsolvent control of bromination
Why does phenol give mainly $p$ -bromophenol when treated with $Br_2$ in $CS_2$ at low temperature, but $2,4,6$ -tribromophenol when treated with $Br_2$ water?

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.