Properties of Phenol

Q: Why does phenol not react with sodium bicarbonate?

Phenol (pKa 10.0) is a weaker acid than carbonic acid (pKa 6.4), so it cannot protonate the bicarbonate ion and no carbon dioxide is released. Carboxylic acids and picric acid, which are stronger than carbonic acid, do liberate carbon dioxide.

ChemistryAlcohols, Phenols And EthersFor NEET aspirants

The properties of phenol come from two reactive sites. The O-H group is far more acidic than in alcohols ( $p K_{a}$ 10), because the phenoxide ion is stabilised by resonance, so phenol dissolves in NaOH but not in NaHCO $_{3}$ . The ring is strongly activated at the ortho and para positions, so bromine water gives 2,4,6-tribromophenol at once. These two ideas explain every property of phenol, from its boiling point to the Kolbe and Reimer-Tiemann reactions. A favourite area for NEET and JEE Main.

On this page1Physical2Acidity3O-H reactions4Rearrangements5Ring substitution6Kolbe, Reimer-Tiemann7Other reactions8Examples

Key Formulas - Quick Reference

★ Must learnAcidity: RCOOH $>$ ArOH $>$ $H_{2} O$ $>$ ROH; $p K_{a}$ (phenol) = 10.0, ethanol 15.9
★ Must learnPhenol + NaOH: reacts; phenol + $NaHC O_{3}$ : no reaction (picric acid does react)
o/p $N O_{2}$ increases acidity ( $p K_{a}$ 7.1-7.2), $C H_{3}$ decreases it ( $p K_{a}$ 10.2)
★ Must learn $B r_{2}$ water: 2,4,6-tribromophenol (white ppt); $B r_{2}$ / $C S_{2}$ , 273 K: 4-bromophenol (major)
dil. $HN O_{3}$ , 298 K: o- + p-nitrophenol; conc. $HN O_{3}$ : picric acid
★ Must learnKolbe: $C_{6} H_{5} ONa$ + $C O_{2}$ (400 K, 4-7 atm), then $H^{+}$ : salicylic acid
★ Must learnReimer-Tiemann: $CHC l_{3}$ + aq. NaOH, then $H^{+}$ : salicylaldehyde (electrophile : $CC l_{2}$ )
Test: neutral $FeC l_{3}$ gives a violet colour

1. Physical Properties

Pure phenol is a colourless crystalline solid (m.p. 314 K, b.p. 455 K) that turns pink on standing because of air oxidation. The O-H group forms hydrogen bonds with other phenol molecules, which raises the melting and boiling points, and with water, which makes phenol moderately soluble (about 8 g per 100 g of water).

Figure 1: Phenol boils about 50 K above chlorobenzene, which is heavier, because its molecules are held together by O-H···O hydrogen bonds.

The nitrophenols show how the position of a group changes hydrogen bonding. In o-nitrophenol the O-H bonds to its own nitro group (intramolecular hydrogen bond), so the molecules do not associate: it has a lower melting point, is less soluble in water and is volatile in steam. p-Nitrophenol bonds to its neighbours and to water.

Figure 2: Where the O-H points decides the property. In o-nitrophenol it bonds to its own nitro group (chelation), so the molecules stay separate: low m.p., steam volatile. p-Nitrophenol bonds to its neighbours and to water.

Exam Trick

"Ortho keeps to itself." An intramolecular hydrogen bond (ortho) means lower m.p., lower solubility, steam volatile. That is how o- and p-nitrophenol are separated: steam distillation carries the ortho isomer over.

Key idea

Hydrogen bonding explains all the physical properties; intramolecular bonding (ortho) cancels the effect.

2. Acidity of Phenol

Phenol is a weak acid ( $p K_{a}$ = 10.0), about a million times stronger than ethanol ( $p K_{a}$ = 15.9). Two effects combine. The $s p^{2}$ ring carbon is more electronegative than an $s p^{3}$ carbon, which polarises the O-H bond, and, far more important, the phenoxide ion is stabilised by resonance: its negative charge spreads to the ortho and para carbons.

C_{6} H_{5} OH + H_{2} O ⇌ C_{6} H_{5} O^{-} + H_{3} O^{+}

Figure 3: The phenoxide charge is delocalised over O and the ring carbons 2, 4 and 6. No charge is separated in these contributors, so phenoxide is far more stabilised than phenol itself, which makes phenol about a million times more acidic than ethanol.

Phenol itself also has resonance structures, but they separate charge (

+

on O,

-

on the ring), which costs energy. The phenoxide contributors carry only one negative charge each, so resonance stabilises the ion much more than the acid, and the equilibrium moves towards ionisation.

2.1 Effect of substituents

Electron-withdrawing groups ( $- NO_{2}$ , $- CN$ , $- CHO$ , halogens) spread the negative charge further and increase acidity. Their effect is largest at the ortho and para positions, where they can take the charge by resonance. Electron-releasing groups (alkyl, $- OCH_{3}$ , $- NH_{2}$ ) intensify the charge and decrease acidity.

Figure 4: Longer bar, weaker acid. Phenol (

p K_{a}

10.0) sits between water and carbonic acid, so it dissolves in NaOH but not in NaHCO

_{3}

; three nitro groups make picric acid (

p K_{a}

0.5) stronger than ethanoic acid.

Compound	$p K_{a}$	Why
2,4,6-trinitrophenol (picric acid)	0.5	three o/p nitro groups; stronger than ethanoic acid
p-nitrophenol	7.1	nitro takes the charge by resonance (−R) and −I
o-nitrophenol	7.2	−R and strong −I, but intramolecular H-bond holds the H
m-nitrophenol	8.3	−I only: no resonance from meta
phenol	10.0	reference
o-, m-, p-cresol	10.2, 10.1, 10.2	+I and hyperconjugation destabilise phenoxide
ethanol	15.9	no resonance in ethoxide

Because phenol is a stronger acid than water but weaker than carbonic acid ( $p K_{a}$ 6.4), it reacts with sodium and with sodium hydroxide but does not liberate $C O_{2}$ from sodium bicarbonate.

Figure 5: Four quick tests. Phenol reacts with Na and NaOH (it is a stronger acid than water) but not with NaHCO

_{3}

(it is weaker than carbonic acid); neutral FeCl

_{3}

gives a violet complex.

Phenol,

p K_{a}

10.0reacts with Na and NaOH
no

CO_{2}

with

NaHCO_{3}

violet with neutral

FeCl_{3}

Ethanol,

p K_{a}

15.9reacts with Na only
no reaction with NaOH or

NaHCO_{3}

no colour with

FeCl_{3}

JEE Advanced

o-Nitrophenol is slightly weaker than p-nitrophenol even though the $- I$ effect is stronger from the ortho position: the intramolecular hydrogen bond must be broken before the proton can leave. The same idea explains why salicylic acid ( $p K_{a}$ 2.97) is much stronger than benzoic acid (4.20): there the hydrogen bond stabilises the salicylate anion instead.

Quick Recall: tap to check

Why does phenol dissolve in NaOH but not in

NaHC O_{3}

Water (

p K_{a}

15.7) < phenol (10.0) < carbonic acid (6.4).

Most acidic of o-, m-, p-nitrophenol?

p-Nitrophenol (

p K_{a}

7.1).

Which phenol liberates

C O_{2}

from

NaHC O_{3}

Picric acid (2,4,6-trinitrophenol,

p K_{a}

0.5).

Key idea

Acidity follows phenoxide stability: anything that spreads the negative charge (o/p nitro) raises it.

3. Reactions of the O-H Group

At the O-H group phenol reacts like an alcohol, only more readily because it is more acidic. With acid chlorides or anhydrides (in pyridine or aqueous NaOH, the Schotten-Baumann conditions) it forms esters; acetylation of salicylic acid gives aspirin. The phenoxide ion is a good nucleophile, so it gives ethers with alkyl halides or dimethyl sulphate.

C_{6} H_{5} OH + C H_{3} COCl pyridine C_{6} H_{5} OCOC H_{3} + HCl

C_{6} H_{5} ONa + C H_{3} I C_{6} H_{5} OC H_{3} + NaI

Figure 6: At the O-H group phenol behaves like an alcohol: acyl chlorides or anhydrides give esters, and its sodium salt gives ethers with alkyl halides (Williamson synthesis).

Phenol does not undergo the reactions in which alcohols lose their C-O bond: no substitution by HX,

PCl_{5}

SOCl_{2}

, and no dehydration, because the C-O bond of phenol has partial double-bond character.

4. Rearrangements: Fries and Claisen

Two rearrangements move a group from the phenolic oxygen to the ring. In the Fries rearrangement a phenyl ester heated with anhydrous $AlCl_{3}$ gives o- and p-hydroxyaryl ketones: low temperature favours para, high temperature ortho. In the Claisen rearrangement allyl phenyl ether heated to about 473 K gives 2-allylphenol; if both ortho positions are blocked, the allyl group goes para.

Figure 7: Two rearrangements that move a group from oxygen to the ring. Fries: the acyl group goes ortho (high temperature) or para (low temperature). Claisen: the allyl group goes ortho, joined through its far (γ) carbon.

The Claisen rearrangement is a concerted [3,3]-sigmatropic shift: the ring carbon bonds to the far (

γ

) carbon of the allyl group, so the allyl chain is inverted on ortho migration. A

γ

-labelled allyl ether therefore gives the label on the ring-bound carbon.

5. Electrophilic Substitution in the Ring

The $- OH$ group donates a lone pair into the ring, making the ortho and para carbons electron-rich. Phenol is therefore far more reactive than benzene: it is brominated without a Lewis acid and nitrated by dilute nitric acid.

5.1 Bromination

With bromine water, phenol gives a white precipitate of 2,4,6-tribromophenol at once. In a non-polar solvent such as $CS_{2}$ or $CHCl_{3}$ at low temperature, phenol stays un-ionised and mainly 4-bromophenol forms.

C_{6} H_{5} OH + 3B r_{2} H_{2} O C_{6} H_{2} B r_{3} OH + 3HBr

Figure 8: The solvent decides how far bromination goes. In water phenol ionises to the very reactive phenoxide and all three o/p positions are substituted at once; in CS

_{2}

at 273 K only one bromine enters, mostly para.

5.2 Nitration and sulphonation

Dilute nitric acid at 298 K gives a mixture of 2- and 4-nitrophenol, separated by steam distillation. Concentrated nitric acid gives 2,4,6-trinitrophenol (picric acid); because conc. $HNO_{3}$ also oxidises phenol, picric acid is made industrially by sulphonating phenol first and then nitrating.

Figure 9: Mild reagents are enough, because OH strongly activates the ring. Dilute HNO

_{3}

gives o- and p-nitrophenol; concentrated HNO

_{3}

gives picric acid. In sulphonation, temperature picks the isomer: ortho at 298 K, para at 373 K.

Exam Trick

"Water fills all three." Bromine water or conc. $HN O_{3}$ : 2, 4 and 6 all substituted. A non-polar solvent, low temperature or dilute acid gives just one group, para major (ortho for Kolbe and Reimer-Tiemann).

Key idea

The OH group makes the ring so reactive that the conditions, not the reagent, decide how many groups enter.

6. Kolbe and Reimer-Tiemann Reactions

Kolbe reaction. Sodium phenoxide heated with carbon dioxide at 400 K and 4-7 atm gives sodium salicylate; acid then gives salicylic acid (2-hydroxybenzoic acid), the starting material for aspirin. $CO_{2}$ is a weak electrophile, so the phenoxide ion (not phenol) is needed.

C_{6} H_{5} ONa + C O_{2} 400 K, 4-7 atm C_{6} H_{4} (OH) COONa

Figure 10: Kolbe reaction. CO

_{2}

is a weak electrophile, so only the very reactive phenoxide ion (not phenol) attacks it, at the ortho carbon, at 400 K and 4-7 atm. A proton shift restores the aromatic ring.

Reimer-Tiemann reaction. Phenol refluxed with chloroform and aqueous NaOH at about 340 K, then acidified, gives salicylaldehyde (2-hydroxybenzaldehyde) with some 4-hydroxybenzaldehyde. The electrophile is dichlorocarbene, formed from chloroform by the base. With carbon tetrachloride instead of chloroform the product is salicylic acid.

Figure 11: Reimer-Tiemann reaction. The real electrophile is dichlorocarbene, :CCl

_{2}

, made from CHCl

_{3}

and OH

^{-}

; the CHCl

_{2}

group put on the ring is hydrolysed to CHO, mostly ortho to OH.

Kolbe reactionreagent:

CO_{2}

(400 K, 4-7 atm)
electrophile:

CO_{2}

product: salicylic acid (-COOH)

Reimer-Tiemann reactionreagent:

CHCl_{3}

+ aq. NaOH
electrophile: :

CCl_{2}

product: salicylaldehyde (-CHO)

7. Other Reactions and the Test for Phenol

Phenol is oxidised by chromic acid to p-benzoquinone and turns pink-brown in air for the same reason. Distilled with zinc dust it loses its oxygen to give benzene. With ammonia at 573 K over $ZnCl_{2}$ it gives aniline. In cold alkaline solution it couples at the para position with benzenediazonium chloride to give an orange azo dye. Neutral ferric chloride gives a violet colour, the standard test.

C_{6} H_{5} OH + Zn Δ C_{6} H_{6} + ZnO

Figure 12: Four more reactions. Chromic acid oxidises phenol to p-benzoquinone; Zn dust removes O (gives benzene); NH

_{3}

/ZnCl

_{2}

replaces OH by NH

_{2}

; a diazonium salt couples at the para position to an orange azo dye.

Quick Recall: tap to check

Product of phenol with Zn dust?

Benzene (and ZnO).

Colour with neutral

FeC l_{3}

Violet.

Electrophile in the Reimer-Tiemann reaction?

Dichlorocarbene, :

CCl_{2}

Figure 13: Flowchart: first ask whether the reagent attacks O-H or the ring, then how strong the conditions are. Harsh conditions fill 2, 4 and 6; mild ones give an o/p mixture.

Figure 14: Mind map: two reactive sites. The O-H behaves like a (stronger) alcohol; the ring is strongly activated at 2, 4 and 6.

8. Solved Examples

Solved Example 1

How will you separate o-nitrophenol and p-nitrophenol?

Solution:

By steam distillation. o-Nitrophenol has an intramolecular hydrogen bond, so its molecules do not associate and it is volatile in steam; it distils over. p-Nitrophenol is held by intermolecular hydrogen bonds and stays in the flask.

Solved Example 2

Arrange in decreasing order of acidity: (a) phenol A, o-nitrophenol B, m-nitrophenol C, p-nitrophenol D; (b) phenol A, o-, m-, p-chlorophenol E, F, G; (c) phenol A, o-, m-, p-cresol H, I, J.

Solution:

(a) D > B > C > A ( $p K_{a}$ 7.1, 7.2, 8.3, 10.0): nitro withdraws by resonance only from o/p; the ortho isomer loses a little to its intramolecular H-bond. (b) E > F > G > A ( $p K_{a}$ 8.5, 9.1, 9.4, 10.0): the $- I$ effect of Cl falls with distance. (c) A > I > J ≈ H ( $p K_{a}$ 10.0, 10.1, 10.2, 10.2): methyl releases electrons; from meta only by $+ I$ .

Solved Example 3

Identify the major product when o-cresol is treated with bromine water.

Solution:

OH directs to its ortho and para positions; one ortho position carries $CH_{3}$ . Bromine enters both free positions (C-4 and C-6): the product is 4,6-dibromo-2-methylphenol.

Solved Example 4

Phenol is heated with

CCl_{4}

and NaOH, then acidified. What is formed, and what is the electrophile?

Solution:

Salicylic acid. The ring carbon ortho to $O^{-}$ attacks $CCl_{4}$ (as in the Reimer-Tiemann reaction), giving a $- CCl_{3}$ group; three hydroxides replace the three chlorines to give $- C(OH)_{3}$ , which loses water to $- COOH$ . Acid releases salicylic acid.

Solved Example 5

Acidic nature is greatest for
(A) o-aminophenol
(B) m-aminophenol
(C) p-aminophenol
(D) all are equal

Solution:

Answer: (B). $- NH_{2}$ releases electrons by resonance to the ortho and para positions, destabilising the phenoxide; from meta only its weak $- I$ effect acts.

Solved Example 6

Phenol can be distinguished from ethanol by
(A) Tollens' reagent
(B) Schiff's reagent
(C) neutral

FeC l_{3}

(D) HCl

Solution:

Answer: (C). Phenol gives a violet colour with neutral $FeCl_{3}$ ; ethanol gives none.

Solved Example 7

Phenol is less acidic than
(A) ethanol
(B) methanol
(C) o-nitrophenol
(D) p-cresol

Solution:

Answer: (C). o-Nitrophenol ( $p K_{a}$ 7.2) is stronger; alcohols and p-cresol are weaker.

Solved Example 8

Phenyl acetate is heated with anhydrous

AlCl_{3}

to give 2-hydroxyacetophenone (X) and 4-hydroxyacetophenone (Y). Which statements are correct?
(A) X forms faster at higher temperature
(B) X is more volatile than Y
(C) Y is more volatile than X
(D) Y is favoured at lower temperature

Solution:

Answer: (A), (B) and (D). High temperature favours the ortho isomer X, low temperature the para isomer Y. X has an intramolecular hydrogen bond (C=O···H-O), so it is the more volatile, steam-distillable isomer; (C) is wrong.

Solved Example 9

Why does picric acid dissolve in sodium bicarbonate solution with effervescence while phenol does not?

Solution:

Three nitro groups at 2, 4 and 6 delocalise the phenoxide charge so strongly that picric acid ( $p K_{a}$ 0.5) is a stronger acid than carbonic acid ( $p K_{a}$ 6.4); it protonates $HCO_{3}^{-}$ and $CO_{2}$ escapes. Phenol ( $p K_{a}$ 10.0) is too weak.

Solved Example 10

Write the product of the Reimer-Tiemann reaction on p-cresol.

Solution:

The CHO group enters ortho to OH: 2-hydroxy-5-methylbenzaldehyde. Some dienone (4-dichloromethyl-4-methylcyclohexa-2,5-dienone) also forms, because the para position is blocked.

Practice Questions

A compound $C_{6} H_{6} O$ gives a violet colour with $FeC l_{3}$ ; with $CO_{2}$ /NaOH (400 K) then acid it gives C, which with acetyl chloride gives a pain killer D. Identify A-D.Answer: A phenol; B sodium salicylate; C salicylic acid; D aspirin (acetylsalicylic acid).
How can benzoic acid and phenol be separated?Answer: shake with aq. $NaHCO_{3}$ : only benzoic acid dissolves.
Why does phenol not react with $NaHCO_{3}$ ?Answer: it is a weaker acid than carbonic acid.
Phenol + $K_{2} S_{2} O_{8}$ (alkaline), then hydrolysis gives B (major) and C (minor). Identify them.Answer: Elbs oxidation: B benzene-1,4-diol (quinol), C benzene-1,2-diol (catechol).
An aromatic amine A (in $CHCl_{3}$ ) and B are separated with aq. KOH. The aqueous layer with $CHCl_{3}$ and heat, then acid, gives D and E ( $C_{7} H_{6} O_{2}$ ). Identify B, D, E.Answer: B phenol; D salicylaldehyde; E 4-hydroxybenzaldehyde.
Product of phenol with acetyl chloride and $AlCl_{3}$ ?Answer: o- and p-hydroxyacetophenone (via the ester, then Fries).
Give the product of allyl phenyl ether heated to 473 K.Answer: 2-allylphenol (Claisen rearrangement).

Common Mistakes to Avoid

Watch out

Saying phenol reacts with $NaHC O_{3}$ . Only acids stronger than carbonic acid, such as picric acid, do.
Writing o-nitrophenol as more acidic than p-nitrophenol. The intramolecular H-bond makes it slightly weaker (7.2 vs 7.1).
Using $AlC l_{3}$ for the bromination of phenol. The ring is so activated that no Lewis acid is needed.
Writing monobromophenol with bromine water. Water gives 2,4,6-tribromophenol; $C S_{2}$ at 273 K gives 4-bromophenol.
Swapping Kolbe and Reimer-Tiemann products. $C O_{2}$ gives the acid (-COOH); $CHC l_{3}$ gives the aldehyde (-CHO).
Writing a substitution of phenol by HBr or $PC l_{5}$ . The phenolic C-O bond does not break.
Forgetting that the para isomer forms at low temperature in the Fries rearrangement, the ortho at high temperature.
Thinking methyl groups make phenol more acidic. Cresols ( $p K_{a}$ ≈ 10.2) are slightly weaker than phenol.

Frequently Asked Questions

Why is phenol more acidic than ethanol?

The phenoxide ion is stabilised by resonance, which spreads its negative charge over the oxygen and the ortho and para ring carbons. Ethoxide has no such delocalisation and its ethyl group even intensifies the charge. As a result phenol ( $p K_{a}$ 10.0) is about a million times stronger than ethanol (15.9).

Why does phenol not react with sodium bicarbonate?

Phenol ( $p K_{a}$ 10.0) is a weaker acid than carbonic acid ( $p K_{a}$ 6.4), so it cannot protonate the bicarbonate ion and no carbon dioxide is released. Carboxylic acids and picric acid, which are stronger than carbonic acid, do liberate carbon dioxide.

Why is p-nitrophenol more acidic than phenol?

The nitro group at the para position withdraws electron density by resonance and induction, spreading the negative charge of the phenoxide ion onto its own oxygens. This stabilises the anion, lowering the $p K_{a}$ from 10.0 to 7.1. A meta nitro group acts only inductively and lowers it to 8.3.

What is the product of phenol with bromine water?

A white precipitate of 2,4,6-tribromophenol forms immediately. In water phenol partly ionises to the very reactive phenoxide ion, so all three ortho and para positions are brominated. In carbon disulphide at 273 K only one bromine enters, mainly at the para position.

What is the difference between the Kolbe and Reimer-Tiemann reactions?

Both put a one-carbon group ortho to the OH of phenoxide. In the Kolbe reaction the electrophile is carbon dioxide and the product is salicylic acid. In the Reimer-Tiemann reaction chloroform and alkali form dichlorocarbene, and the product is salicylaldehyde.

How is phenol identified in the laboratory?

Phenol gives a violet colour with neutral ferric chloride solution, dissolves in sodium hydroxide but not in sodium bicarbonate, and decolourises bromine water with a white precipitate of 2,4,6-tribromophenol. Alcohols give none of these results.

Which properties of phenol are most asked in NEET?

NEET most often asks the acidity order of substituted phenols, the product of bromine water, the reagents of the Kolbe and Reimer-Tiemann reactions, and why phenol does not react with sodium bicarbonate. The $p K_{a}$ ladder and the o/p rule answer most of them.

How does JEE Main test the reactions of phenol?

JEE Main combines steps: phenol to salicylic acid to aspirin, Fries or Claisen rearrangement products, or identification of phenols from ferric chloride, bicarbonate and bromine-water results. Substituent effects on acidity, including the intramolecular hydrogen bond of o-nitrophenol, are a favourite.

Previous year questions on Properties of Phenol

10 questions from past papers, each with a step-by-step solution.

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