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Chemistry · Alcohols, Phenols and Ethers

Reactions of Phenols

The reactions of phenols, set out in NCERT Class XII Chemistry §7.4.4 under "Reactions of phenols", are dominated by one structural fact: the hydroxyl group feeds electron density into the benzene ring and turns it into a highly reactive nucleophile. This makes nitration, halogenation, the Kolbe and Reimer-Tiemann reactions reliable NEET territory, where examiners reward students who can match a reagent and condition to the exact product.

Why the Ring Is So Reactive

A phenol is hydroxybenzene, $\ce{C6H5OH}$, in which the $\ce{-OH}$ group is bonded directly to an $sp^2$ carbon of the aromatic ring. The defining feature of phenol chemistry is that the oxygen lone pair is not idle: it is delocalised into the ring through resonance. NCERT (§7.4.4) states this plainly — the $\ce{-OH}$ group "activates" the ring toward electrophilic substitution and "directs the incoming group to ortho and para positions" because these positions become electron-rich.

Two distinct sites therefore exist in a phenol molecule, and NEET questions almost always test which one is involved. Reactions can occur at the aromatic ring (electrophilic substitution, Kolbe, Reimer-Tiemann, oxidation, reduction by zinc) or at the oxygen of the $\ce{-OH}$ group (ester formation, ether formation, the salt-forming acidity reactions). Keeping these two families apart is the single most useful organising idea for this subtopic.

Reaction siteWhat happensRepresentative reactions
Aromatic ringAn electrophile or carbene replaces a ring H at o/p positionsNitration, halogenation, sulphonation, Kolbe, Reimer-Tiemann
Ring (C-OH carbon)Whole ring is reduced or the ring is oxidisedZinc dust → benzene; chromic acid → benzoquinone
Oxygen of -OHThe O-H or O is modified, ring untouchedEster formation, Williamson ether formation, salt formation

Electrophilic Aromatic Substitution

Because the $\ce{-OH}$ group pushes electron density onto the ortho and para carbons, phenol is far more reactive than benzene toward electrophilic aromatic substitution (EAS), and it reacts under much milder conditions. The figure below maps where electrophiles attack: the ortho (2, 6) and para (4) positions are activated, while the meta positions are not.

Figure 1 · Directing effect OH ortho ortho para meta meta -OH donates e⁻ → o/p enriched

The hydroxyl lone pair raises electron density at positions 2, 4 and 6, so electrophiles substitute there. NCERT notes the activation is strong enough that bromine is polarised even without a Lewis acid.

Nitration: o/p-Nitrophenol to Picric Acid

Nitration is the textbook illustration of how reaction conditions decide the product. With dilute nitric acid at low temperature (298 K), phenol gives a mixture of ortho- and para-nitrophenol.

$$\ce{C6H5OH + HNO3(dilute) ->[298\ K] \underset{o\text{-nitrophenol}}{o\text{-}O2N\text{-}C6H4\text{-}OH} + \underset{p\text{-nitrophenol}}{p\text{-}O2N\text{-}C6H4\text{-}OH} + H2O}$$

The two isomers are separated by steam distillation: o-nitrophenol is steam-volatile because it forms an intramolecular hydrogen bond (a closed six-membered chelate ring), whereas p-nitrophenol associates through intermolecular hydrogen bonding and is far less volatile. With concentrated nitric acid, all three o/p positions are nitrated and phenol is converted to 2,4,6-trinitrophenol — better known as picric acid — though NCERT records the yield as poor.

$$\ce{C6H5OH + 3\,HNO3(conc.) -> \underset{\text{picric acid (2,4,6-trinitrophenol)}}{C6H2(NO2)3OH} + 3\,H2O}$$

NEET Trap

"Concentrated" does not mean simple over-nitration

NCERT adds an industrial route to picric acid: phenol is first treated with concentrated $\ce{H2SO4}$ to give phenol-2,4-disulphonic acid, which is then nitrated with concentrated $\ce{HNO3}$. The sulphonic groups are displaced by nitro groups. The point examiners test is that the major final product is still 2,4,6-trinitrophenol.

Dilute $\ce{HNO3}$ → o/p-nitrophenol (mono). Concentrated $\ce{HNO3}$ → picric acid (tri).

Halogenation and Sulphonation

Halogenation of phenol with bromine is the classic "the solvent decides the product" reaction. In a low-polarity solvent such as $\ce{CS2}$ or $\ce{CHCl3}$ at low temperature, the reaction stops at monobromination, giving mainly p-bromophenol with some o-bromophenol.

$$\ce{C6H5OH + Br2 ->[\text{CS2, low temp}] \underset{\text{(major)}}{p\text{-}BrC6H4OH} + \underset{\text{(minor)}}{o\text{-}BrC6H4OH}}$$

In bromine water — an aqueous, strongly ionising medium — all three o/p positions are brominated at once and a white precipitate of 2,4,6-tribromophenol drops out, almost quantitatively. NCERT highlights that this proceeds without any Lewis acid, because the $\ce{-OH}$ group is itself a powerful enough activator to polarise the bromine molecule.

$$\ce{C6H5OH + 3\,Br2 ->[\text{Br2 water}] \underset{\text{white ppt}}{C6H2Br3OH (v) + 3\,HBr}}$$

Sulphonation follows the same activation logic. Phenol reacts with concentrated sulphuric acid to give ortho- and para-phenolsulphonic acid; the ortho product dominates at low temperature and the para product at higher temperature, reflecting the reversible, thermodynamically controlled nature of sulphonation.

$$\ce{C6H5OH + H2SO4(conc.) <=> HO\text{-}C6H4\text{-}SO3H + H2O}$$

Build the foundation

These ring reactions all flow from how readily phenol gives up its proton. Revise Acidity of Phenols to see why the phenoxide ion is so stable.

Kolbe's Reaction

Carbon dioxide is a weak electrophile, far too feeble to attack benzene. But phenol can be pushed further: treating it with sodium hydroxide generates the phenoxide ion, which NCERT describes as "even more reactive than phenol towards electrophilic aromatic substitution". The phenoxide ring is so electron-rich that it substitutes with $\ce{CO2}$ at the ortho position. After acidification, the product is salicylic acid (ortho-hydroxybenzoic acid).

Figure 2 · Kolbe scheme Phenol C₆H₅OH NaOH Na phenoxide C₆H₅ONa CO₂, Δ then H⁺ Salicylic acid o-HO-C₆H₄-COOH

Kolbe installs a carboxyl (-COOH) group ortho to the -OH. Acetylation of the resulting salicylic acid with acetic anhydride gives aspirin.

$$\ce{C6H5OH ->[NaOH] C6H5O^- Na^+ ->[CO2] ->[H^+] \underset{\text{salicylic acid}}{o\text{-}HO\text{-}C6H4\text{-}COOH}}$$

Reimer-Tiemann Reaction

In the Reimer-Tiemann reaction, phenol is treated with chloroform in the presence of sodium hydroxide. NaOH first generates the reactive species dichlorocarbene, $\ce{:CCl2}$, which acts as the electrophile and introduces a $\ce{-CHO}$ group at the ortho position. NCERT notes that the intermediate substituted benzal chloride is hydrolysed by alkali to give salicylaldehyde (2-hydroxybenzaldehyde).

Figure 3 · Reimer-Tiemann scheme Phenol C₆H₅OH CHCl₃ NaOH o-benzal chloride via :CCl₂ carbene H₂O / OH⁻ Salicylaldehyde o-HO-C₆H₄-CHO

Reimer-Tiemann installs an aldehyde (-CHO) group ortho to the -OH. Contrast with Kolbe, which installs -COOH.

$$\ce{C6H5OH + CHCl3 + 3\,NaOH ->[\Delta] ->[H^+] \underset{\text{salicylaldehyde}}{o\text{-}HO\text{-}C6H4\text{-}CHO} + 3\,NaCl + 2\,H2O}$$

NEET Trap

Kolbe vs Reimer-Tiemann — match the group

Both reactions start from sodium phenoxide and both substitute at the ortho position, which makes them easy to confuse. The discriminator is the electrophile and the group installed: $\ce{CO2}$ gives the carboxyl group of salicylic acid, while $\ce{CHCl3}$/NaOH (via dichlorocarbene) gives the aldehyde of salicylaldehyde.

Kolbe → -COOH (salicylic acid). Reimer-Tiemann → -CHO (salicylaldehyde).

Zinc Dust and Oxidation to Benzoquinone

Two ring-level reactions complete the standard NEET set. First, NCERT records that phenol is converted to benzene on heating with zinc dust — the $\ce{-OH}$ group is reduced away and replaced by $\ce{-H}$. This is also the one reaction phenol shares with the C-O cleavage family of alcohols; phenols otherwise resist C-O bond cleavage.

$$\ce{C6H5OH + Zn ->[\Delta] C6H6 + ZnO}$$

Second, oxidation of phenol with chromic acid produces a conjugated diketone called benzoquinone (specifically 1,4-benzoquinone, or para-benzoquinone). NCERT also notes that in the presence of air, phenols are slowly oxidised to dark-coloured mixtures containing quinones — the reason old samples of phenol turn pink to brown.

$$\ce{C6H5OH ->[\text{Na2Cr2O7 / H2SO4}] \underset{\text{benzoquinone}}{C6H4O2}}$$

FeCl3 Colour Test

For qualitative identification, phenols give a characteristic colour with neutral ferric chloride. Phenol itself develops a violet colouration owing to the formation of a coloured iron(III)-phenolate complex. Because ordinary alcohols give no such colour, this is a standard bench test for a phenolic $\ce{-OH}$ group, and it appears in NEET reasoning questions on distinguishing alcohols from phenols.

$$\ce{6\,C6H5OH + FeCl3 -> [Fe(OC6H5)6]^{3-} + 3\,H+ + 3\,HCl}$$

Reactions at the -OH: Esters and Ethers

The remaining reactions modify the oxygen rather than the ring. Phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters — the introduction of an acetyl ($\ce{CH3CO}$) group being called acetylation. NCERT notes the reaction is best run with acid chlorides or anhydrides in the presence of a base such as pyridine, which mops up the HCl and drives the equilibrium forward. Acetylation of salicylic acid produces aspirin.

$$\ce{C6H5OH + (CH3CO)2O -> C6H5OCOCH3 + CH3COOH}$$

Phenols also form ethers by Williamson synthesis. Here the phenol is first converted to sodium phenoxide, which then attacks an alkyl halide by an $S_N2$ pathway to give an alkyl aryl ether such as anisole. The phenoxide must act as the nucleophile (aryl halides are unreactive toward $S_N2$), so the alkyl group must come from the halide.

$$\ce{C6H5O^- Na^+ + CH3I -> \underset{\text{anisole}}{C6H5OCH3} + NaI}$$

ReactionReagent / conditionMajor product
Nitration (mono)dilute $\ce{HNO3}$, 298 Ko- + p-nitrophenol
Nitration (tri)conc. $\ce{HNO3}$picric acid (2,4,6-trinitrophenol)
Bromination (mono)$\ce{Br2}$ in $\ce{CS2}$ / $\ce{CHCl3}$, low tempp-bromophenol (major)
Bromination (tri)$\ce{Br2}$ water2,4,6-tribromophenol (white ppt)
Kolbe$\ce{NaOH}$ then $\ce{CO2}$, then $\ce{H+}$salicylic acid
Reimer-Tiemann$\ce{CHCl3}$ + $\ce{NaOH}$, then hydrolysissalicylaldehyde
ReductionZn dust, $\Delta$benzene
Oxidationchromic acidbenzoquinone
Colour testneutral $\ce{FeCl3}$violet colouration
Quick Recap

Reactions of Phenols at a glance

  • The $\ce{-OH}$ group is strongly activating and o/p-directing; phenol undergoes EAS far more easily than benzene.
  • Nitration: dilute $\ce{HNO3}$ → o/p-nitrophenol; conc. $\ce{HNO3}$ → picric acid.
  • Bromination: $\ce{Br2}$/$\ce{CS2}$ → monobromophenol; $\ce{Br2}$ water → 2,4,6-tribromophenol (white ppt), no Lewis acid needed.
  • Kolbe installs -COOH (salicylic acid); Reimer-Tiemann installs -CHO (salicylaldehyde) via dichlorocarbene; both go ortho.
  • Zinc dust → benzene; chromic acid → benzoquinone; neutral $\ce{FeCl3}$ → violet colour test.
  • At the oxygen: ester formation (acetylation → aspirin from salicylic acid) and Williamson ether synthesis (phenoxide as nucleophile).

NEET PYQ Snapshot — Reactions of Phenols

Drawn from released NEET papers. Phenol reaction questions cluster around acidity reasoning, substituent effects and ether formation.

NEET 2022 · Q.58

Statement I: The acidic strength of monosubstituted nitrophenol is higher than phenol because of the electron-withdrawing nitro group. Statement II: o-, m- and p-nitrophenol have the same acidic strength as each has one nitro group on the ring. Choose the correct option.

  • (1) Both statements incorrect
  • (2) Statement I correct, Statement II incorrect
  • (3) Statement I incorrect, Statement II correct
  • (4) Both statements correct
Answer: (2)

The nitro group withdraws electrons by both -I and -R effects, so nitrophenols are more acidic than phenol (Statement I correct). But at the meta position only -I operates, while at o/p the strong -R effect stabilises the phenoxide — so the three isomers are not equally acidic (Statement II incorrect). This is the same resonance logic that makes the o/p positions reactive toward electrophiles.

NEET 2017 · Q.26

Which one is the most acidic compound among phenol and the nitro-substituted phenols shown?

  • (1) 2,4-dinitrophenol type (multiple o/p -NO2 groups)
  • (2)–(4) phenol and mono/methyl-substituted phenols
Answer: (1)

Acidic strength tracks the stability of the conjugate anion. The isomer bearing nitro groups at the ortho/para positions delocalises the negative charge most effectively (-M effect), giving the most stable phenoxide and hence the strongest acid.

NEET 2016 · Q.7

Phenol is treated with NaH and then with methyl iodide ($\ce{ArOH ->[NaH] ArO^- ->[CH3I] ArOCH3}$). This reaction is classified as:

  • (1) Alcohol formation reaction
  • (2) Dehydration reaction
  • (3) Williamson alcohol synthesis
  • (4) Williamson ether synthesis
Answer: (4)

NaH deprotonates phenol to the phenoxide, which then displaces iodide from $\ce{CH3I}$ by $S_N2$ to give the aryl methyl ether (anisole). This is exactly the Williamson ether synthesis applied to a phenol, $\ce{R-X + R'O^-Na^+ -> R-O-R' + NaX}$.

FAQs — Reactions of Phenols

The product-prediction and reagent-matching points that decide single-mark NEET questions.

Why does phenol undergo electrophilic substitution far more easily than benzene?

The lone pair on the oxygen of the -OH group is delocalised into the benzene ring. This raises the electron density at the ortho and para positions, making the ring a strong nucleophile toward electrophiles. The -OH group is therefore a powerful activating, ortho/para-directing group, and the ring reacts under milder conditions than benzene. The activation is so strong that bromine is polarised even without a Lewis acid catalyst.

What is the difference between bromination of phenol in CS2 and in bromine water?

In a low-polarity solvent such as CS2 or CHCl3 at low temperature, the reaction stops at monobromination, giving mainly o- and p-bromophenol. With bromine water (an aqueous, highly ionising medium) the ring is brominated three times at once, precipitating white 2,4,6-tribromophenol. So the solvent decides whether you get a mono- or tri-substituted product.

What products do dilute and concentrated nitric acid give with phenol?

Dilute nitric acid at about 298 K gives a mixture of o-nitrophenol and p-nitrophenol, which are separated by steam distillation (o-nitrophenol is steam-volatile because of intramolecular hydrogen bonding). Concentrated nitric acid converts phenol to 2,4,6-trinitrophenol, commonly called picric acid, though the yield is poor.

What is the difference between Kolbe's reaction and the Reimer-Tiemann reaction?

Both start from sodium phenoxide. In Kolbe's reaction the electrophile is carbon dioxide, which substitutes at the ortho position to give salicylic acid (o-hydroxybenzoic acid). In the Reimer-Tiemann reaction the electrophile is dichlorocarbene (generated from CHCl3 and NaOH), which introduces a -CHO group at the ortho position to give salicylaldehyde after hydrolysis. Kolbe installs -COOH; Reimer-Tiemann installs -CHO.

How does the neutral FeCl3 test identify a phenol?

Phenols form coloured iron(III) complexes with neutral ferric chloride. Phenol itself gives a characteristic violet colouration. Because most ordinary alcohols give no such colour, the reaction is used as a qualitative test to detect a phenolic -OH group.

Can phenol be converted to benzene, and is its C-O bond cleaved like in alcohols?

Phenol is reduced to benzene on heating with zinc dust, the -OH being replaced by -H. Phenols do not undergo the general C-O bond cleavage reactions that alcohols show with hydrogen halides or PCl3, because the C-O bond has partial double-bond character and the carbon is sp2. The reaction with zinc dust is the one notable exception.

Related Topics
Alcohols, Phenols and Ethers Back to the full chapter hub and all subtopics. Acidity of Phenols Why phenoxide is stable and substituents shift pKa. Preparation of Phenols From haloarenes, sulphonic acids, diazonium salts and cumene. Properties of Alcohols Contrast: C-O cleavage, oxidation and dehydration of alcohols. Preparation of Alcohols Hydration, hydroboration, reduction and Grignard routes.