Hydroxy compounds

Class 12 Chemistry, unit 10, book pages 183–192

Partial. Printed pages 183, 186, 187, 190 and 191 of the textbook are not yet covered from the book itself: the unit's opening page with the learning outcomes, the end of 10.3 and the start of 10.4 (reactions of phenol), the key points and the start of the exercise. The learning outcomes here follow the National Curriculum, and the reactions of phenol with NaOH, Na, bromine water and dilute nitric acid are written from the curriculum and the exercise. The exercise MCQs and short questions i to viii are still to be added; short questions ix to xi and all ten long questions are solved.

Practise 37 MCQs Take the unit quiz Review 21 cards

Hydroxy compounds contain an $- OH$ group on carbon: on an $s p^{3}$ carbon in alcohols, and directly on an aromatic ring in phenols and naphthols. This unit shows how alcohols form esters with acyl chlorides, how phenol is made from phenylamine through a diazonium salt, and why phenol is a stronger acid than water and ethanol. The lone pair of the $- OH$ oxygen also activates the ring, so phenol and the naphthols are brominated and nitrated under mild conditions and couple with diazonium salts to give azo dyes.

Learning outcomes (8)

Describe the preparation of esters by the reaction of alcohols with acyl chlorides, for example ethanoyl chloride with ethanol.
Describe the preparation of phenol from phenylamine: reaction with nitrous acid ( $NaNO X_{2}$ and dilute acid) below $10^{\circ} C$ to give the diazonium salt, followed by warming with water.
Describe and explain the relative acidities of water, phenol and ethanol in terms of the stability of their conjugate bases.
Describe the reactions of phenol with bases such as $NaOH (aq)$ and with sodium metal to form sodium phenoxide.
Describe the reaction of phenol with diazonium salts in $NaOH (aq)$ to form azo compounds.
Describe the nitration of phenol with dilute $HNO X_{3}$ at room temperature and the bromination of phenol with $Br X_{2} (aq)$ to form 2,4,6-tribromophenol, and explain why the conditions differ from those for benzene.
Explain that the $- OH$ group of phenol is activating and 2,4-directing.
Apply knowledge of the reactions of phenol to predict the reactions of other phenolic compounds, such as the naphthols.

IntroOrganic hydroxy compounds: alcohols, phenols and naphthols

You already know water, $H - O - H$ . Replace one hydrogen atom of water by a carbon group and you get an organic hydroxy compound: a compound with one or more hydroxyl groups ( $- OH$ ) attached to a carbon atom. The $- OH$ group decides most of the properties of these compounds, and the kind of carbon it is joined to decides how the $- OH$ group behaves.

Alcohol: A hydroxy compound in which the $- OH$ group is attached to a saturated ( $s p^{3}$ hybridised) carbon atom. General formula $R - OH$ , where R is an alkyl group.
Phenol: A hydroxy compound in which the $- OH$ group is attached directly to a carbon atom of a benzene ring ( $s p^{2}$ carbon). The simplest member, $C X_{6} H X_{5} OH$ , is also called phenol.
Naphthol: A hydroxy derivative of naphthalene, $C X_{10} H X_{7} OH$ : a naphthalene ring system (two fused benzene rings) bonded to one $- OH$ group.

1. Alcohols

Alcohol	Formula	Use given in the book
Methanol	$CH X_{3} OH$	the simplest alcohol; solvent and antifreeze
Ethanol	$C X_{2} H X_{5} OH$	found in alcoholic drinks; solvent and fuel
Isopropanol (propan-2-ol, isopropyl alcohol)	$C X_{3} H X_{7} OH$ , that is $CH X_{3} CH (OH) CH X_{3}$	disinfectant and solvent

methanol

ethanol

propan-2-ol

Alcohols are classified by the number of carbon atoms joined to the carbon that carries the $- OH$ group, in the same way as halogenoalkanes: primary ( $RCH X_{2} OH$ , for example ethanol), secondary ( $R X_{2} CHOH$ , for example propan-2-ol) and tertiary ( $R X_{3} COH$ , for example 2-methylpropan-2-ol).

2. Phenols

In phenols the $- OH$ group is joined directly to the benzene ring. Phenols are used to make plastics (for example phenol–methanal resins such as Bakelite), medicines, antiseptics and disinfectants.

phenol

3. Naphthols

Naphthols have the formula $C X_{10} H X_{7} OH$ and exist as two isomers, depending on which carbon of naphthalene carries the $- OH$ group.

1-Naphthol ( $α$ -naphthol): $- OH$ on carbon 1, the carbon next to the ring junction.
2-Naphthol ( $β$ -naphthol): $- OH$ on carbon 2.

1-naphthol

2-naphthol

Naphthols are used to make dyes and pigments, and as intermediates in the manufacture of medicines and agrochemicals.

Alcohols and phenols compared
Feature	Alcohol (ethanol)	Phenol
Carbon joined to $- OH$	$s p^{3}$ carbon of an alkyl group	$s p^{2}$ carbon of a benzene ring
Lone pair on oxygen	stays on oxygen	partly delocalised into the ring
Acidity	weaker acid than water ( $K_{a} \approx 1 0^{- 16}$ )	weak acid, stronger than water ( $K_{a} = 1.3 \times 1 0^{- 10}$ )
Reacts with $NaOH (aq)$	no	yes, forms sodium phenoxide
Reacts with sodium metal	yes, $H X_{2}$ given off	yes, $H X_{2}$ given off
Bromine water	no reaction	decolourised at once, white precipitate

10.1Preparation of esters from acyl chloridesOften examined

An ester ( $RCOOR X^{'}$ ) is formed when an acid combines with an alcohol and a small molecule is lost. Esters are used in fragrances (perfumes), flavourings and pharmaceuticals. You met the slow reaction of a carboxylic acid with an alcohol before; a much faster method uses an acyl chloride (acid chloride, $RCOCl$ ).

Ethanoyl chloride reacts immediately with cold ethanol to form ethyl ethanoate. The reaction is very exothermic and steamy fumes of hydrogen chloride gas are given off.

CH X_{3} COCl + CH X_{3} CH X_{2} OH CH X_{3} COOCH X_{2} CH X_{3} + HCl

ethanoyl chloride

ethanol

ethyl ethanoate

Mechanism: nucleophilic addition followed by elimination

In an acyl chloride the carbonyl carbon is joined to two electronegative atoms, O and Cl, so it carries a large $δ +$ charge and is strongly electrophilic. The oxygen atom of the alcohol has lone pairs, so the alcohol is the nucleophile.

Ethanoyl chloride + ethanol

Addition. A lone pair on the oxygen of ethanol attacks the $δ +$ carbonyl carbon. The π electrons of $C = O$ move onto the carbonyl oxygen. A tetrahedral intermediate forms, with $O X^{-}$ on one side and a positive oxygen ( $O X^{+}$ , still carrying its H) on the other.
$CH X_{3} COCl + C X_{2} H X_{5} OH CH X_{3} C (O X^{-}) (Cl) (O X^{+} HC X_{2} H X_{5})$
Elimination of chloride. The lone pair on $O X^{-}$ moves back to re-form the $C = O$ bond, and the $C - Cl$ bond breaks. Chloride ( $Cl X^{-}$ ) is a good leaving group.
$CH X_{3} C (= O) O X^{+} HC X_{2} H X_{5} + Cl X^{-}$
Loss of a proton. The chloride ion removes $H X^{+}$ from the positive oxygen, giving the ester and hydrogen chloride.
$CH X_{3} C (= O) O X^{+} HC X_{2} H X_{5} + Cl X^{-} CH X_{3} COOC X_{2} H X_{5} + HCl$

tetrahedral intermediate

ethyl ethanoate

Two ways to make an ester
	Carboxylic acid + alcohol	Acyl chloride + alcohol
Speed	slow	fast, even in the cold
Conditions	heat with conc. $H X_{2} SO X_{4}$ catalyst	room temperature, no catalyst
Reversible?	yes, an equilibrium	no, goes to completion ( $HCl$ escapes as a gas)
By-product	$H X_{2} O$	$HCl$ (steamy fumes)
Yield	limited by equilibrium	high

Methanol instead of ethanol

Write the equation for the reaction of ethanoyl chloride with methanol and name the ester.

The alkyl part of the alcohol becomes the alkyl part of the ester; the acyl part comes from the acyl chloride.

CH X_{3} COCl + CH X_{3} OH CH X_{3} COOCH X_{3} + HCl

methyl ethanoate, CH X_{3} COOCH X_{3}

10.2Preparation of phenol from phenylamine (aniline)Often examined

Phenol is made in the laboratory from phenylamine (aniline, $C X_{6} H X_{5} NH X_{2}$ ) in two stages: first the amine is turned into a diazonium salt, then the diazonium salt is warmed with water.

Stage 1: diazotisation (below 10 °C)

Nitrous acid ( $HNO X_{2}$ ) is unstable, so it is made in the flask from sodium nitrite and dilute hydrochloric acid:

NaNO X_{2} + HCl HNO X_{2} + NaCl

Phenylamine dissolved in excess dilute $HCl$ is cooled in ice to $0$ to $5^{\circ} C$ (always below $10^{\circ} C$ ) and the nitrous acid reacts with it to give benzenediazonium chloride:

C X_{6} H X_{5} NH X_{2} + HNO X_{2} + HCl 0 - 5 X_{\circ} X_{2}^{2 \circ} C C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} + 2 H X_{2} O

phenylamine (aniline)

benzenediazonium chloride

Stage 2: hydrolysis (warm, above 10 °C)

When the solution of the diazonium salt is warmed above $10^{\circ} C$ , water replaces the diazonium group. Nitrogen gas bubbles off and phenol is formed:

C X_{6} H X_{5} N X_{2} X^{+} Cl X^{-} + H X_{2} O > 10 X_{\circ} X_{2}^{2 \circ} C C X_{6} H X_{5} OH + N X_{2} ↑ + HCl

phenol

10.3Acidity of water, phenol and ethanolOften examined

All three compounds contain an \ce{O-H} bond and can give away a proton (\ce{H+}). They differ in how easily they do so. The key idea is: an acid is stronger when the conjugate base (the anion left behind after the proton is lost) is more stable. A stable anion does not easily take the proton back, so the equilibrium lies further to the right.

\ce{H2O + H2O <=> H3O+ + OH-}

water: conjugate base hydroxide ion

\ce{C6H5OH + H2O <=> C6H5O- + H3O+}

phenol: conjugate base phenoxide ion

\ce{C2H5OH + H2O <=> C2H5O- + H3O+}

ethanol: conjugate base ethoxide ion

Acid dissociation constant, K_a: For \ce{HA + H2O <=> A- + H3O+}, K_a=\dfrac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}. A larger K_a (smaller \mathrm{p}K_a=-\log K_a) means a stronger acid.
Conjugate base: The species left when an acid loses a proton, for example \ce{C6H5O-} from \ce{C6H5OH}.

K_a=\dfrac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}\qquad \mathrm{p}K_a=-\log_{10}K_a

Order of acid strength: ethanoic acid > phenol > water > ethanol
Compound	K_a at 25\,^\circ\mathrm{C}	\mathrm{p}K_a	Conjugate base
Ethanoic acid \ce{CH3COOH}	1.8\times10^{-5}	4.7	ethanoate \ce{CH3COO-}, charge shared equally by two O atoms
Phenol \ce{C6H5OH}	1.3\times10^{-10}	9.9	phenoxide \ce{C6H5O-}, charge delocalised into the ring
Water \ce{H2O}	1.8\times10^{-16}	15.7	hydroxide \ce{OH-}, charge on one O atom
Ethanol \ce{C2H5OH}	about 1\times10^{-16}	about 16	ethoxide \ce{C2H5O-}, charge on one O atom and made worse by the ethyl group

Why phenol is more acidic than water

In the phenoxide ion, oxygen carries the negative charge but one of its lone pairs overlaps with the π electron cloud of the benzene ring. The negative charge is delocalised (spread) over the oxygen atom and the carbon atoms at positions 2, 4 and 6 of the ring. The resonance structures show this:

Resonance structures of the phenoxide ion. In I the charge is on oxygen; in II, III and IV it is on ring carbons 2, 4 and 6 (ortho and para).

Spreading a charge over several atoms lowers its energy, so phenoxide is resonance stabilised. In the hydroxide ion the charge stays on one oxygen atom; no delocalisation is possible. So phenoxide is more stable than hydroxide, and phenol gives up its proton more readily than water: K_a(\text{phenol})=1.3\times10^{-10} > K_a(\text{water})=1.8\times10^{-16}.

Why ethanol is less acidic than water

In the ethoxide ion \ce{CH3CH2-O-} the charge is on one oxygen atom and cannot be delocalised. Worse, the ethyl group is electron-releasing (+I inductive effect). It pushes electron density towards the oxygen, which intensifies (concentrates) the negative charge. Ethoxide is therefore less stable than hydroxide, and it takes a proton back readily (it is a stronger base than \ce{OH-}).

\ce{CH3-CH2}\!\rightarrow\!\ce{O^-}\qquad(+I\ \text{effect of the ethyl group intensifies the charge on O})

Why ethanoic acid is more acidic than phenol

In the ethanoate ion the negative charge is shared by two oxygen atoms. The two resonance structures are identical, so they contribute equally and the charge is spread evenly over two highly electronegative atoms. In phenoxide, structures II to IV put the charge on carbon, which is less electronegative than oxygen, so they are less stable, contribute less, and most of the charge stays on the single oxygen. Ethanoate is therefore more stable than phenoxide.

ethanoate (form A)

ethanoate (form B)

\text{ethanoic acid} > \text{phenol} > \text{water} > \text{ethanol}

acid strength

Evidence from simple tests

Test	Ethanoic acid	Phenol	Ethanol
Blue litmus	turns red	hardly changes (very weak acid)	no change
\ce{NaOH(aq)}	neutralised	dissolves, forms sodium phenoxide	no reaction
\ce{Na2CO3} or \ce{NaHCO3}	effervescence of \ce{CO2}	no \ce{CO2} (phenol is weaker than carbonic acid)	no reaction
Sodium metal	\ce{H2} given off	\ce{H2} given off	\ce{H2} given off, slowly

10.4Reactions of the O–H group: sodium hydroxide and sodium

Phenol with sodium hydroxide

Phenol is acidic enough to be neutralised by sodium hydroxide solution. It dissolves, forming a colourless solution of sodium phenoxide and water:

\ce{C6H5OH + NaOH(aq) -> C6H5O^-Na+(aq) + H2O}

phenol

sodium phenoxide

Ethanol does not react with \ce{NaOH(aq)}, because ethanol is a weaker acid than water. Adding a stronger acid (for example \ce{HCl}) to sodium phenoxide gives phenol back.

Phenol and alcohols with sodium metal

Sodium reacts with any compound that has an acidic \ce{O-H} hydrogen. Hydrogen gas is given off (effervescence) and a sodium salt is formed:

\ce{2C6H5OH + 2Na -> 2C6H5O^-Na+ + H2 ^}

sodium phenoxide

\ce{2C2H5OH + 2Na -> 2C2H5O^-Na+ + H2 ^}

sodium ethoxide

The reaction with ethanol is gentler than that of sodium with water, because ethanol is a weaker acid than water. The reaction with phenol is faster than with ethanol.

10.4Electrophilic substitution in phenol: bromination and nitrationOften examined

Benzene reacts with electrophiles only with a catalyst, because its π electrons are held quite firmly. In phenol, one lone pair on the oxygen atom of \ce{-OH} overlaps with the π system and is donated into the ring. The ring becomes much richer in electrons, especially at the carbons 2, 4 and 6 (the two ortho and the para positions). The \ce{-OH} group is therefore a strongly activating, 2,4-directing (ortho, para-directing) group.

Resonance structures of phenol. The oxygen lone pair moves into the ring, putting extra electron density (negative charge) on carbons 2, 4 and 6.

Bromination with bromine water

When bromine water is added to aqueous phenol at room temperature, the orange-brown colour of bromine disappears at once and a white precipitate of 2,4,6-tribromophenol forms. No catalyst is needed and all three activated positions are substituted:

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

phenol

2,4,6-tribromophenol (white ppt)

Why no catalyst is needed. The electron-rich ring of phenol repels the electrons of an approaching \ce{Br2} molecule and polarises it (\ce{Br^{\delta+}-Br^{\delta-}}). The \delta+ end is attacked by the ring directly. In addition, the positive intermediate (arenium ion) formed by attack at positions 2, 4 or 6 is stabilised by the oxygen lone pair. In benzene there is no such help, so a Lewis acid (\ce{FeBr3}) is needed to make the electrophile \ce{Br+}.

Comparison with the bromination of benzene

Bromination of phenol and benzene
	Phenol	Benzene
Reagent	bromine water, \ce{Br2(aq)}	liquid (pure) bromine, \ce{Br2(l)}
Catalyst	none	\ce{FeBr3} or iron filings
Conditions	room temperature	warm, dry
Product	2,4,6-tribromophenol (three Br)	bromobenzene (one Br)
Observation	decolourised at once; white precipitate	slow; misty fumes of HBr
Reason	\ce{-OH} lone pair activates ring	ring not activated

\ce{C6H6 + Br2 ->[FeBr3][\text{warm}] C6H5Br + HBr}

Nitration with dilute nitric acid

Phenol reacts with dilute nitric acid at room temperature, without sulfuric acid. A mixture of 2-nitrophenol and 4-nitrophenol forms, because the \ce{-OH} group directs to positions 2 and 4:

\ce{2C6H5OH + 2HNO3 ->[\text{dilute}][\text{room temp.}] $\underset{\text{2-nitrophenol}}{\ce{C6H4(OH)NO2}}$ + $\underset{\text{4-nitrophenol}}{\ce{C6H4(OH)NO2}}$ + 2H2O}

2-nitrophenol

4-nitrophenol

Benzene needs a nitrating mixture of concentrated \ce{HNO3} and concentrated \ce{H2SO4} at about 50\,^\circ\mathrm{C} to form the nitronium ion \ce{NO2+}, and gives only nitrobenzene. With concentrated nitric acid, phenol is nitrated at all three positions to give 2,4,6-trinitrophenol (picric acid).

10.4 (c)Reaction of phenol with a diazonium salt (azo coupling)

Phenol is dissolved in sodium hydroxide solution to give sodium phenoxide. This solution is cooled in ice, and cold benzenediazonium chloride solution is added. A yellow-orange solution or precipitate forms. The product is one of the simplest azo compounds, 4-hydroxyazobenzene (4-(phenylazo)phenol).

\ce{C6H5OH + NaOH(aq) -> C6H5O^-Na+ + H2O}

\ce{C6H5N2+Cl- + C6H5O^-Na+ ->[H2O][0\text{–}5\,^\circ C] C6H5-N=N-C6H4-OH + NaCl}

benzenediazonium chloride

sodium phenoxide

4-hydroxyazobenzene (yellow-orange)

The diazonium ion is a weak electrophile. It attacks the activated ring of the phenoxide ion at the 4-position (para to \ce{-O-}). This is called coupling. In azo dyes two aromatic rings are linked by a nitrogen bridge, \ce{-N=N-} (the azo group). Azo dyes are synthetic; they do not occur naturally.

10.5Reactions of naphthols

Naphthols behave like simple phenols: their \ce{-OH} group is acidic and it activates the ring it is attached to. So each reaction of phenol has a naphthol version. The only new point is where the ring is substituted.

1. With sodium hydroxide

Naphthols dissolve in sodium hydroxide solution, forming sodium naphthoxides:

\ce{C10H7OH + NaOH(aq) -> C10H7O^-Na+ + H2O}

1-naphthol

sodium 1-naphthoxide

2-naphthol

sodium 2-naphthoxide

2. With bromine water

1-Naphthol decolourises bromine water. Bromine enters at positions 2 and 4 (ortho and para to \ce{-OH}, both in the same ring):

\ce{2C10H7OH + 2Br2(aq) -> $\underset{\text{2-bromo-1-naphthol}}{\ce{C10H6Br(OH)}}$ + $\underset{\text{4-bromo-1-naphthol}}{\ce{C10H6Br(OH)}}$ + 2HBr}

2-bromo-1-naphthol

4-bromo-1-naphthol

2-Naphthol decolourises bromine water with substitution at position 1, the carbon between the \ce{-OH} group and the other ring:

\ce{C10H7OH + Br2(aq) -> $\underset{\text{1-bromo-2-naphthol}}{\ce{C10H6Br(OH)}}$ + HBr}

1-bromo-2-naphthol

3. With nitric acid

Nitration follows the same pattern as bromination. 1-Naphthol gives 2-nitro-1-naphthol and 4-nitro-1-naphthol; 2-naphthol gives 1-nitro-2-naphthol:

\ce{2C10H7OH + 2HNO3 -> $\underset{\text{2-nitro-1-naphthol}}{\ce{C10H6(NO2)OH}}$ + $\underset{\text{4-nitro-1-naphthol}}{\ce{C10H6(NO2)OH}}$ + 2H2O}

\ce{C10H7OH + HNO3 -> $\underset{\text{1-nitro-2-naphthol}}{\ce{C10H6(NO2)OH}}$ + H2O}

2-nitro-1-naphthol

4-nitro-1-naphthol

1-nitro-2-naphthol

4. With a diazonium salt

Naphthols couple with benzenediazonium chloride under the same conditions as phenol: the naphthol is dissolved in \ce{NaOH} solution, cooled in ice and mixed with the diazonium salt solution. An intense orange-red precipitate of an azo dye forms. 1-Naphthol couples at position 4; 2-naphthol couples at position 1.

\ce{C6H5N2+Cl- + C10H7O^-Na+ ->[H2O] C6H5-N=N-C10H6-OH + NaCl}

4-(phenylazo)-1-naphthol

1-(phenylazo)-2-naphthol

Summary of naphthol reactions
Reaction	1-Naphthol	2-Naphthol
\ce{NaOH(aq)}	sodium 1-naphthoxide	sodium 2-naphthoxide
Bromine water	substitution at 2 and 4	substitution at 1
Nitric acid	substitution at 2 and 4	substitution at 1
Diazonium salt (in NaOH, cold)	coupling at 4; orange-red ppt	coupling at 1; orange-red ppt

Alcohols: reactions to remember

The book's readable pages treat alcohols mainly as the partners of acyl chlorides and as weak acids. Three more facts are needed for paper questions: the functional group, the reaction with sodium, and oxidation.

Functional group. The functional group of alcohols is the hydroxyl group, \ce{-OH} (not \ce{-COOH}, which is the carboxyl group of acids, and not \ce{-O-}, which is the ether link).
With sodium. \ce{2ROH + 2Na -> 2RO^-Na+ + H2}: the alcohol behaves as a very weak acid (see the section on the O–H group).
Oxidation. Primary alcohols are oxidised by acidified potassium dichromate(VI): first to an aldehyde, then to a carboxylic acid. The orange dichromate turns green (\ce{Cr^3+}).

\ce{CH3CH2OH ->[K2Cr2O7/H+][\text{heat}] CH3CHO ->[K2Cr2O7/H+][\text{heat under reflux}] CH3COOH}

This is a redox reaction: ethanol is oxidised (it gains oxygen and loses hydrogen; the oxidation number of the carbon rises from -1 to +3) while dichromate is reduced (\ce{Cr}: +6 \to +3). Secondary alcohols give ketones; tertiary alcohols are not oxidised under these conditions.

Formula sheet

Acid dissociation constant

K_a=\dfrac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}

for \ce{HA + H2O <=> A- + H3O+}; square brackets are equilibrium concentrations

Unit: \mathrm{mol\,dm^{-3}}

pKa

\mathrm{p}K_a=-\log_{10}K_a

smaller \mathrm{p}K_a means a stronger acid

no unit

K_a of water

K_a(\ce{H2O})=\dfrac{K_w}{[\ce{H2O}]}=\dfrac{1.0\times10^{-14}}{55.5}=1.8\times10^{-16}

K_w is the ionic product of water; [\ce{H2O}]=55.5\,\mathrm{mol\,dm^{-3}} in pure water

Unit: \mathrm{mol\,dm^{-3}}

Key terms

Hydroxy compound: An organic compound containing one or more \ce{-OH} groups attached to carbon.
Alcohol: A hydroxy compound with the \ce{-OH} group on a saturated (sp^3) carbon, \ce{R-OH}.
Phenol: A compound with an \ce{-OH} group attached directly to a benzene ring; also the name of \ce{C6H5OH}.
Naphthol: \ce{C10H7OH}, naphthalene with one \ce{-OH} group; 1-naphthol (\alpha) and 2-naphthol (\beta).
Ester: A compound \ce{RCOOR'} formed from an acid (or acyl chloride) and an alcohol with loss of a small molecule.
Acyl chloride: A compound \ce{RCOCl}; reacts rapidly and irreversibly with alcohols to give esters and \ce{HCl}.
Diazotisation: Reaction of an aromatic amine with nitrous acid (\ce{NaNO2 + HCl}) below 10\,^\circ\mathrm{C} to form a diazonium salt.
Diazonium salt: A salt containing the \ce{Ar-{N+}#N} ion, for example benzenediazonium chloride \ce{C6H5N2+Cl-}; stable only when cold.
Azo compound (azo dye): A compound in which two aromatic rings are linked by the azo group \ce{-N=N-}; coloured, used as synthetic dyes.
Coupling reaction: The reaction of a diazonium salt with a phenol or naphthol in cold alkaline solution to give an azo dye.
Conjugate base: The species formed when an acid loses a proton, for example phenoxide \ce{C6H5O-} from phenol.
Phenoxide ion: \ce{C6H5O-}, the conjugate base of phenol; stabilised by delocalisation of its charge into the ring.
Resonance stabilisation: The extra stability of a species whose electrons (or charge) are delocalised over several atoms, shown by drawing resonance structures.
Acid dissociation constant (K_a): The equilibrium constant for the ionisation of a weak acid in water; a larger K_a means a stronger acid.
Inductive effect (+I): Electron release through σ bonds by alkyl groups; it destabilises an anion such as ethoxide.
Activating group: A substituent, such as \ce{-OH}, that makes an aromatic ring more reactive towards electrophiles than benzene.
2,4-directing (ortho, para-directing): A group that sends an incoming electrophile to the positions next to it (2, 6) and opposite it (4).
Electrophilic substitution: Replacement of a hydrogen atom of an aromatic ring by an electrophile, the ring keeping its aromatic π system.

Short answer questionsp. 190

Printed page 192. Questions i to viii (printed page 191) are still to be added; ix to xi are given here.

ix3 marks

Explain the resonance stabilization of the phenoxide ion and how it contributes to the acidity of phenol.

Solution

When phenol loses a proton it forms the phenoxide ion, \ce{C6H5O-}: \ce{C6H5OH + H2O <=> C6H5O- + H3O+}.

One lone pair on the negatively charged oxygen overlaps with the π electron cloud of the benzene ring. The negative charge is therefore delocalised over the oxygen atom and the ring carbons at positions 2, 4 and 6, as the resonance structures I to IV show.

Resonance structures of the phenoxide ion: the negative charge is spread over O and ring carbons 2, 4 and 6.

A charge spread over several atoms is lower in energy, so the phenoxide ion is stabilised by resonance. A stable conjugate base does not readily recombine with \ce{H+}, so the ionisation equilibrium lies further to the right than for water or ethanol. This is why phenol is weakly acidic (K_a=1.3\times10^{-10}) and reacts with \ce{NaOH}, while ethanol does not.

x2026 Annual I, Set D, Q.2 (ix) OR3 marks

Compare the acidities of phenol, ethanol, and water. Provide a brief explanation for the differences in their acid strengths.

Solution

\text{phenol }(K_a=1.3\times10^{-10}) \;>\; \text{water }(K_a=1.8\times10^{-16}) \;>\; \text{ethanol }(K_a\approx10^{-16})

Phenol is the strongest of the three because the negative charge of its conjugate base, the phenoxide ion, is delocalised into the benzene ring by resonance (over O and carbons 2, 4 and 6). This stabilises the ion.
Water is weaker: in \ce{OH-} the charge is localised on one oxygen atom, with no resonance.
Ethanol is the weakest: in \ce{C2H5O-} the charge is localised on oxygen and the ethyl group releases electrons (+I effect), which intensifies the charge and destabilises the ion.

The more stable the conjugate base, the stronger the acid. So phenol reacts with \ce{NaOH(aq)}, while ethanol reacts only with a very reactive metal such as sodium.

xi3 marks

Predict the products of the reaction of 1-naphthol with NaOH and bromine water. Provide the balanced chemical equations and indicate the positions of the substituents.

Solution

With sodium hydroxide. 1-Naphthol is acidic like phenol and dissolves in \ce{NaOH(aq)}, forming sodium 1-naphthoxide and water. The \ce{-ONa} group stays at position 1.

\ce{C10H7OH + NaOH(aq) -> C10H7O^-Na+ + H2O}

With bromine water. 1-Naphthol decolourises bromine water. The \ce{-OH} group activates its own ring and directs bromine to the positions ortho and para to it, that is carbons 2 and 4. The book gives the two monobromo products:

\ce{2C10H7OH + 2Br2(aq) -> $\underset{\text{2-bromo-1-naphthol}}{\ce{C10H6Br(OH)}}$ + $\underset{\text{4-bromo-1-naphthol}}{\ce{C10H6Br(OH)}}$ + 2HBr}

sodium 1-naphthoxide

2-bromo-1-naphthol

4-bromo-1-naphthol

Long answer questionsp. 190

18 marks

Compare and contrast the reactivity of phenol, benzene, and ethanol in electrophilic aromatic substitution reactions. Explain the role of the hydroxyl group in phenol and its influence on these reactions.

Solution

Electrophilic aromatic substitution

In electrophilic aromatic substitution an electron-seeking species (electrophile, \ce{E+}) attacks the π electron cloud of an aromatic ring and replaces a hydrogen atom. So only compounds with an aromatic ring can take part, and the richer the ring is in electrons, the faster the reaction.

Order of reactivity

\text{phenol} \;\gg\; \text{benzene};\qquad \text{ethanol: no reaction}

Ethanol has no benzene ring. It has no delocalised π electrons, so it cannot undergo electrophilic aromatic substitution at all. Its reactions are those of the \ce{-OH} group (with Na, acyl chlorides, oxidation).
Benzene has six delocalised π electrons, which are held quite firmly because the delocalisation makes the ring stable. It reacts with electrophiles only when a catalyst makes a strong electrophile: bromination needs liquid \ce{Br2} with \ce{FeBr3}; nitration needs concentrated \ce{HNO3} and \ce{H2SO4} at about 50\,^\circ\mathrm{C}. One H is substituted.
Phenol reacts very much faster, under mild conditions and without a catalyst: bromine water at room temperature gives 2,4,6-tribromophenol, and dilute \ce{HNO3} at room temperature gives 2- and 4-nitrophenol.

Role of the hydroxyl group

The oxygen of \ce{-OH} has lone pairs. One lone pair overlaps with the π system of the ring and is delocalised into it, as shown below. This raises the electron density of the ring, particularly at carbons 2, 4 and 6.

Resonance in phenol: the O lone pair is donated into the ring, making carbons 2, 4 and 6 electron-rich.

Activating effect. The electron-rich ring attracts electrophiles strongly and can even polarise a neutral \ce{Br2} molecule, so no catalyst is needed.
Stabilised intermediate. The positive arenium ion formed after attack at C2, C4 or C6 is stabilised by the oxygen lone pair, so the activation energy is lower.
Directing effect. Because extra electron density sits at C2, C4 and C6, substituents enter at the ortho and para positions (2,4-directing), and with bromine water all three positions are substituted.

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

\ce{C6H6 + Br2 ->[FeBr3] C6H5Br + HBr}

	Phenol	Benzene	Ethanol
Aromatic ring	yes, activated by OH	yes	no
Bromination	bromine water, room temp., 3 Br	liquid \ce{Br2} + \ce{FeBr3}, 1 Br	no substitution
Nitration	dilute \ce{HNO3}, room temp.	conc. \ce{HNO3} + conc. \ce{H2SO4}, 50\,^\circ\mathrm{C}	no substitution

22026 Annual I, Set B, Q.68 marks

Describe the acidity of phenol in terms of its ability to donate protons. Compare its acidity to that of water and ethanol and explain the molecular basis for the differences observed.

Solution

Phenol as a proton donor

Phenol is a weak Brønsted acid: the hydrogen of its \ce{O-H} group can be donated to water or to a base.

\begin{aligned}\ce{C6H5OH + H2O &<=> C6H5O- + H3O+}\\K_a&=\dfrac{[\ce{C6H5O-}][\ce{H3O+}]}{[\ce{C6H5OH}]}=1.3\times10^{-10}\end{aligned}

Evidence of its acidity: phenol dissolves in \ce{NaOH(aq)} to form sodium phenoxide, and it gives hydrogen with sodium. It is too weak to liberate \ce{CO2} from carbonates or to turn blue litmus clearly red.

\ce{C6H5OH + NaOH(aq) -> C6H5O^-Na+ + H2O}

Comparison with water and ethanol

\text{phenol }(1.3\times10^{-10}) > \text{water }(1.8\times10^{-16}) > \text{ethanol }(\approx10^{-16})

Molecular basis

Rule. An acid is stronger when its conjugate base (the anion left after losing \ce{H+}) is more stable, because a stable anion has little tendency to take the proton back.

\begin{aligned}\ce{C6H5OH &<=> C6H5O- + H+}\\\ce{H2O &<=> OH- + H+}\\\ce{C2H5OH &<=> C2H5O- + H+}\end{aligned}

Phenoxide ion. One lone pair on the negative oxygen overlaps with the π electrons of the ring, so the charge is delocalised over O and carbons 2, 4 and 6:

Resonance structures of the phenoxide ion: the negative charge is spread over O and ring carbons 2, 4 and 6.

Hydroxide ion. The charge stays on one oxygen atom; no resonance is possible. So \ce{OH-} is less stable than phenoxide and water is a weaker acid than phenol.

Ethoxide ion. The charge also stays on one oxygen atom, and the ethyl group is electron-releasing (+I effect). It pushes more electron density onto the oxygen and concentrates the charge: \ce{CH3-CH2}\!\rightarrow\!\ce{O^-}. Ethoxide is the least stable of the three, so ethanol is the weakest acid.

36 marks

Describe how 2-naphthol will undergo a reaction with bromine water and nitric acid.

Solution

Why 2-naphthol reacts easily

2-Naphthol behaves like phenol. The lone pair on the oxygen of its \ce{-OH} group is delocalised into the ring system, so the ring is activated towards electrophiles and substitution happens under mild conditions, without a catalyst. The \ce{-OH} group directs the electrophile to an ortho or para position. In 2-naphthol the para position is a ring-junction carbon with no hydrogen, and of the two ortho positions (C1 and C3) substitution takes place at C1, the carbon between the \ce{-OH} group and the other ring. Attack at C1 gives an intermediate in which the second ring stays fully aromatic, so it is more stable.

2-naphthol

With bromine water

2-Naphthol decolourises bromine water at room temperature. One bromine atom enters at position 1, giving 1-bromo-2-naphthol and hydrogen bromide:

\ce{C10H7OH + Br2(aq) -> C10H6Br(OH) + HBr}

1-bromo-2-naphthol

With nitric acid

Nitration follows the same pattern. With nitric acid (dilute, no sulfuric acid needed) the nitro group enters at position 1, giving 1-nitro-2-naphthol and water:

\ce{C10H7OH + HNO3 -> C10H6(NO2)OH + H2O}

1-nitro-2-naphthol

46 marks

Predict the products if methanol is used instead of ethanol in the reaction with ethanoyl chloride. Provide the balanced chemical equation.

Solution

Products

Acyl chlorides react with any alcohol in the same way: the \ce{-OR} group of the alcohol replaces the \ce{-Cl} of the acyl chloride. With methanol the products are the ester methyl ethanoate and hydrogen chloride. The reaction is fast and exothermic at room temperature and gives steamy fumes of HCl.

\ce{CH3COCl + CH3OH -> CH3COOCH3 + HCl}

ethanoyl chloride

methanol

methyl ethanoate

Mechanism (nucleophilic addition–elimination)

A lone pair on the oxygen of methanol attacks the \delta+ carbonyl carbon of ethanoyl chloride; the \ce{C=O} π electrons move onto oxygen, giving a tetrahedral intermediate.
\ce{CH3COCl + CH3OH -> CH3C(O^-)(Cl)(O^+HCH3)}
The \ce{C=O} bond re-forms and chloride, a good leaving group, is expelled.
\ce{-> CH3C(=O)O^+HCH3 + Cl-}
Chloride removes the proton from the positive oxygen, giving the ester and HCl.
\ce{CH3C(=O)O^+HCH3 + Cl- -> CH3COOCH3 + HCl}

Comparison with ethanol

With ethanol the product is ethyl ethanoate, \ce{CH3COOCH2CH3}. The acyl part (\ce{CH3CO-}) comes from the acyl chloride in both cases; only the alkyl part of the ester changes, from ethyl to methyl. Methanol is slightly less hindered, so it reacts at least as fast as ethanol.

\ce{CH3COCl + CH3OH -> CH3COOCH3 + HCl}

56 marks

Write the balanced chemical equations for the reactions of phenol with NaOH and sodium metal. Describe the products formed in each reaction.

Solution

Phenol with sodium hydroxide

\ce{C6H5OH + NaOH(aq) -> C6H5O^-Na+ + H2O}

Phenol is a weak acid (K_a=1.3\times10^{-10}), strong enough to be neutralised by the strong base \ce{NaOH}. It dissolves in the alkali, forming sodium phenoxide and water. Sodium phenoxide is an ionic salt, soluble in water, giving a colourless solution containing \ce{Na+} and \ce{C6H5O-} ions. The phenoxide ion is stabilised by delocalisation of its charge into the ring. Adding a strong acid such as \ce{HCl} turns it back into phenol, which separates out.

Phenol with sodium metal

\ce{2C6H5OH + 2Na -> 2C6H5O^-Na+ + H2 ^}

Sodium replaces the acidic hydrogen of the \ce{O-H} group. The products are sodium phenoxide (the same salt as with NaOH) and hydrogen gas, which is seen as effervescence and burns with a squeaky pop. The reaction is a redox reaction: sodium is oxidised to \ce{Na+} and hydrogen (+1) is reduced to \ce{H2} (0).

phenol

sodium phenoxide

What the reactions show

Both reactions show that phenol is acidic. Ethanol also reacts with sodium (\ce{2C2H5OH + 2Na -> 2C2H5ONa + H2}) but it does not react with \ce{NaOH}, because it is a weaker acid than water. This difference is a simple way to tell phenol from an alcohol.

62026 Annual I, Set D, Q.66 marks

Explain the reaction of phenol with bromine water.

Solution

Observation and equation

When bromine water is added to an aqueous solution of phenol at room temperature, the orange-brown colour of bromine disappears immediately and a white precipitate of 2,4,6-tribromophenol forms. Hydrogen bromide is the other product. No catalyst is needed.

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

phenol

2,4,6-tribromophenol

Explanation

This is an electrophilic substitution. A lone pair on the oxygen of the \ce{-OH} group is delocalised into the benzene ring, so the ring is very electron-rich, especially at positions 2, 4 and 6:

Resonance in phenol: the O lone pair is donated into the ring, making carbons 2, 4 and 6 electron-rich.

The electron-rich ring repels the electrons of a \ce{Br2} molecule and polarises it (\ce{Br^{\delta+}-Br^{\delta-}}); the \delta+ bromine is attacked directly by the ring, so no Lewis acid catalyst is needed.
The positive intermediate is stabilised by the oxygen lone pair, which lowers the activation energy.
All three activated positions (2, 4, 6) are substituted, so three bromine atoms enter.

Comparison with benzene

Benzene does not react with bromine water. It needs pure liquid bromine and a catalyst (\ce{FeBr3} or iron), and only one Br enters: \ce{C6H6 + Br2 ->[FeBr3] C6H5Br + HBr}.

76 marks

Describe the reaction of phenol with dilute nitric acid at room temperature. What are the products, and why do they form in the observed positions on the aromatic ring?

Solution

Reaction and products

Phenol reacts with dilute nitric acid at room temperature, without concentrated sulfuric acid. A mixture of 2-nitrophenol (ortho) and 4-nitrophenol (para) is formed, together with water. The reaction is electrophilic substitution of the ring.

\ce{2C6H5OH + 2HNO3 -> $\underset{\text{2-nitrophenol}}{\ce{C6H4(OH)NO2}}$ + $\underset{\text{4-nitrophenol}}{\ce{C6H4(OH)NO2}}$ + 2H2O}

2-nitrophenol

4-nitrophenol

Why positions 2 and 4

The oxygen of the \ce{-OH} group donates a lone pair into the ring. The resonance structures show that the extra electron density appears at carbons 2, 4 and 6 (ortho and para), not at 3 and 5 (meta):

Resonance in phenol: the O lone pair is donated into the ring, making carbons 2, 4 and 6 electron-rich.

The electrophile (\ce{NO2+}, formed in small amounts from dilute nitric acid) is attracted to these electron-rich carbons.
The intermediate formed by attack at C2 or C4 has a resonance form in which the positive charge sits on the carbon joined to O, where the O lone pair can stabilise it (\ce{C=O+H}). Attack at C3 cannot be stabilised this way.
So \ce{-OH} is an activating, 2,4-directing group. Positions 2 and 6 are equivalent, so they give the same product.

Why the conditions are milder than for benzene

Benzene needs concentrated \ce{HNO3} and concentrated \ce{H2SO4} at about 50\,^\circ\mathrm{C} to produce enough \ce{NO2+}. The activated ring of phenol reacts even with dilute acid. With concentrated nitric acid phenol gives 2,4,6-trinitrophenol (picric acid).

86 marks

Discuss the similarities and differences in the directing effects of the hydroxyl group in phenol and 1-naphthol.

Solution

Similarities

In both compounds the \ce{-OH} group donates a lone pair of its oxygen into the aromatic ring (resonance, +M effect), so it is an activating group: both react with bromine water and dilute nitric acid without a catalyst, at room temperature.
In both, the \ce{-OH} group is ortho, para-directing: the new group enters the positions next to (ortho) or opposite (para) the carbon that carries \ce{-OH}.
Both are acidic and form phenoxide or naphthoxide ions, which are even more strongly activating (used in coupling with diazonium salts).

Differences

	Phenol	1-Naphthol
Ring system	one benzene ring	two fused rings; OH on C1
Ortho positions available	2 and 6	2 only (the other neighbour is the ring-junction carbon, C8a)
Para position	4	4
Bromine water	2,4,6-tribromophenol (3 Br)	2-bromo- and 4-bromo-1-naphthol (book)
Nitration	2- and 4-nitrophenol	2- and 4-nitro-1-naphthol
Diazonium coupling	at 4; yellow-orange	at 4; orange-red

So in phenol three positions (2, 4, 6) are activated, and with bromine water all three are substituted at once. In 1-naphthol only two positions in the same ring, 2 and 4, are activated and available, and substitution stays in the ring that carries the \ce{-OH} group; the second ring is not substituted.

\ce{2C10H7OH + 2Br2(aq) -> C10H6Br(OH) + C10H6Br(OH) + 2HBr}

2-bromo-1-naphthol and 4-bromo-1-naphthol

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

2,4,6-tribromophenol

2-bromo-1-naphthol

4-bromo-1-naphthol

96 marks

Phenol is more acidic than ethanol. Analyze the structure of phenol to explain this difference in acidity.

Solution

Comparing the acids

Both compounds have an \ce{O-H} group, but K_a(phenol) =1.3\times10^{-10} while K_a(ethanol) is about 10^{-16}: phenol is about a million times stronger. Phenol dissolves in \ce{NaOH(aq)}; ethanol does not.

\ce{C6H5OH <=> C6H5O- + H+}\qquad\qquad \ce{C2H5OH <=> C2H5O- + H+}

Structure of phenol and its conjugate base

In phenol the \ce{-OH} group is attached to an sp^2 carbon of the benzene ring. A lone pair on oxygen overlaps with the ring π system, which pulls electron density from oxygen towards the ring. The \ce{O-H} bond becomes more polar and the proton is released more easily.
More importantly, in the phenoxide ion the negative charge is delocalised over the oxygen and carbons 2, 4 and 6. Delocalisation stabilises the ion, so it has less tendency to take back a proton.

Resonance structures of the phenoxide ion: the negative charge is spread over O and ring carbons 2, 4 and 6.

Structure of ethanol and its conjugate base

In ethanol the \ce{-OH} group is on an sp^3 carbon. There is no π system, so the oxygen lone pairs cannot be delocalised.
In the ethoxide ion the charge is localised on one oxygen atom, and the ethyl group releases electrons (+I effect): \ce{CH3-CH2}\!\rightarrow\!\ce{O^-}. This intensifies the negative charge and destabilises the ion, so ethoxide readily takes back a proton (it is a strong base).

Conclusion

Phenoxide (resonance-stabilised) is much more stable than ethoxide (destabilised by +I). The more stable the conjugate base, the stronger the acid, so phenol is more acidic than ethanol.

102026 Annual I, Set B, Q.2 (vii) OR2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (viii), old syllabus6 marks

Phenol reacts with bromine water to form a white precipitate without needing a catalyst. What does this suggest about the electron density of the aromatic ring in phenol?

Solution

What the reaction suggests

It shows that the aromatic ring in phenol has a much higher electron density than the ring in benzene. The ring is activated, so even a weak electrophilic reagent such as bromine water can substitute it.

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

Source of the extra electron density

The oxygen atom of the \ce{-OH} group has two lone pairs. One of them is in a p-type orbital that overlaps with the π electron cloud of the ring, so it is delocalised into the ring. The resonance structures show a negative charge on carbons 2, 4 and 6:

Resonance in phenol: the O lone pair is donated into the ring, making carbons 2, 4 and 6 electron-rich.

How this removes the need for a catalyst

In benzene the π electrons are not rich enough to break the \ce{Br-Br} bond, so \ce{FeBr3} must polarise \ce{Br2} to give a \ce{Br+} electrophile.
In phenol the very electron-rich ring polarises the approaching \ce{Br2} molecule itself (\ce{Br^{\delta+}-Br^{\delta-}}) and attacks the \delta+ end.
The intermediate is stabilised by the oxygen lone pair, so the reaction is fast at room temperature.

Why three bromine atoms

The electron density is concentrated at positions 2, 4 and 6, so all three are substituted at once, giving 2,4,6-tribromophenol as a white precipitate (it is insoluble in water). Benzene, in contrast, gives only bromobenzene, and only with liquid bromine and a catalyst.

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, Q.2 (vii) OR1+2 = 3 marks

Predict the product when phenol is treated with bromine water. Why does this reaction occur at room temperature, unlike benzene?

Solution

Product (1 mark)

Phenol decolourises bromine water at once and gives a white precipitate of 2,4,6-tribromophenol:

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

2,4,6-tribromophenol

Reason (2 marks)

A lone pair on the oxygen of the \ce{-OH} group is delocalised into the benzene ring, which greatly increases the electron density of the ring, especially at positions 2, 4 and 6. This activated ring polarises the \ce{Br2} molecule by itself and attacks it, so no catalyst or heating is needed. Benzene has no such electron-donating group; its ring is less electron-rich, so it needs liquid bromine with an \ce{FeBr3} catalyst to form the \ce{Br+} electrophile.

2026 Annual I, Set B, Q.63+3 = 6 marks

Justify the following order of acidic strength by drawing the resonance structures of their conjugate bases: Phenol > Water > Ethanol

Solution

Principle

The strength of an acid depends on the stability of its conjugate base. The more the negative charge of the conjugate base is spread out (delocalised), the more stable it is and the stronger the acid.

\begin{aligned}\ce{C6H5OH &<=> C6H5O- + H+}\\\ce{H2O &<=> OH- + H+}\\\ce{C2H5OH &<=> C2H5O- + H+}\end{aligned}

Phenol: resonance structures of phenoxide (3 marks)

In the phenoxide ion a lone pair of the negative oxygen is delocalised into the ring. Four resonance structures can be drawn; the charge is spread over O and carbons 2, 4 and 6:

Resonance structures of the phenoxide ion: the negative charge is spread over O and ring carbons 2, 4 and 6.

The phenoxide ion is therefore resonance-stabilised, and phenol is the strongest acid of the three (K_a=1.3\times10^{-10}).

Water and ethanol (3 marks)

Hydroxide ion, \ce{H-O^-}: only one structure can be drawn; the charge is localised on one oxygen atom and there is no resonance. \ce{OH-} is less stable than phenoxide, so water (K_a=1.8\times10^{-16}) is weaker than phenol.

Ethoxide ion, \ce{CH3-CH2-O^-}: again only one structure (no resonance). In addition the ethyl group is electron-releasing (+I effect) and pushes electron density onto the oxygen, intensifying the negative charge:

\ce{CH3-CH2}\!\rightarrow\!\ce{O^-}\qquad\text{vs}\qquad \ce{H-O^-}

So ethoxide is less stable than hydroxide, and ethanol (K_a\approx10^{-16}) is the weakest acid.

\text{stability of conjugate base: } \ce{C6H5O-} > \ce{OH-} > \ce{C2H5O-}\;\Rightarrow\;\text{acidity: phenol} > \text{water} > \text{ethanol}

2026 Annual I, Set D, Q.2 (ix) OR1+2 = 3 marks

Arrange water, phenol, and ethanol in order of increasing acidity and explain the reason for this order.

Solution

Order (1 mark)

\text{ethanol} < \text{water} < \text{phenol}

Reason (2 marks)

Phenol is the most acidic because its conjugate base, the phenoxide ion, is stabilised by resonance: the negative charge is delocalised from oxygen into the benzene ring (carbons 2, 4 and 6).
Water gives \ce{OH-}, in which the charge stays on one oxygen with no resonance, so it is less stable than phenoxide.
Ethanol gives \ce{C2H5O-}; the ethyl group releases electrons (+I effect) and intensifies the charge on oxygen, so ethoxide is the least stable and ethanol is the weakest acid.

The more stable the conjugate base, the stronger the acid.

2026 Annual I, Set D, Q.2 (x)1.5+1.5 = 3 marks

Write chemical equations to prepare phenol from aniline.

Solution

Step 1: diazotisation (1.5 marks)

Aniline (phenylamine) in dilute hydrochloric acid is treated with nitrous acid (made from \ce{NaNO2 + HCl}) at 0 to 5\,^\circ\mathrm{C} (below 10\,^\circ\mathrm{C}). Benzenediazonium chloride forms:

\ce{C6H5NH2 + NaNO2 + 2HCl ->[0\text{–}5\,^\circ C] C6H5N2+Cl- + NaCl + 2H2O}

Step 2: hydrolysis (1.5 marks)

The diazonium salt solution is warmed above 10\,^\circ\mathrm{C}. Nitrogen gas is given off and phenol is formed:

\ce{C6H5N2+Cl- + H2O ->[\text{warm}, > 10\,^\circ C] C6H5OH + N2 ^ + HCl}

aniline

benzenediazonium chloride

phenol

2026 Annual I, Set D, Q.52+2+2 = 6 marks

Justify the following order of acidic strength by drawing the resonance structures of their conjugate bases: Ethanoic acid > Phenol > Ethanol

Solution

An acid is stronger when its conjugate base is more stable. A negative charge is most stable when it is delocalised, and best of all when it is shared by electronegative atoms such as oxygen.

\begin{aligned}\ce{CH3COOH &<=> CH3COO- + H+}\\\ce{C6H5OH &<=> C6H5O- + H+}\\\ce{C2H5OH &<=> C2H5O- + H+}\end{aligned}

Ethanoic acid: ethanoate ion (2 marks)

The ethanoate ion has two resonance structures that are identical in energy. The negative charge is shared equally by two oxygen atoms, and both C–O bonds are the same length (between single and double). This is very effective stabilisation, so ethanoic acid is the strongest (K_a=1.8\times10^{-5}).

ethanoate (A)

ethanoate (B)

Phenol: phenoxide ion (2 marks)

The phenoxide ion is also resonance-stabilised: the charge is delocalised over O and ring carbons 2, 4 and 6.

Resonance structures of the phenoxide ion: the negative charge is spread over O and ring carbons 2, 4 and 6.

But in structures II to IV the charge sits on carbon, which is less electronegative than oxygen. These structures are less stable and contribute less, so most of the charge stays on one oxygen. Phenoxide is less stable than ethanoate, and phenol (K_a=1.3\times10^{-10}) is weaker than ethanoic acid.

Ethanol: ethoxide ion (2 marks)

The ethoxide ion has no resonance structures: the charge is localised on one oxygen. The ethyl group releases electrons (+I effect), which intensifies the charge and destabilises the ion:

\ce{CH3-CH2}\!\rightarrow\!\ce{O^-}

So ethanol (K_a\approx10^{-16}) is the weakest acid.

\ce{CH3COO-} > \ce{C6H5O-} > \ce{C2H5O-}\;\Rightarrow\;\text{ethanoic acid} > \text{phenol} > \text{ethanol}

2026 Annual I, Set D, Q.62+4 = 6 marks

When phenol is treated with bromine water, a white precipitate is formed. Write balanced equation. Compare this reaction with bromination of benzene.

Solution

Balanced equation (2 marks)

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

The white precipitate is 2,4,6-tribromophenol; the orange colour of bromine water disappears at once.

phenol

2,4,6-tribromophenol

Comparison with the bromination of benzene (4 marks)

\ce{C6H6 + Br2(l) ->[FeBr3][\text{warm}] C6H5Br + HBr}

	Phenol	Benzene
Bromine used	bromine water, \ce{Br2(aq)}	pure liquid bromine, \ce{Br2(l)}
Catalyst	not needed	\ce{FeBr3} (or Fe) needed
Conditions	room temperature, fast	warm, slow
Number of Br substituted	three (2, 4, 6)	one
Product	2,4,6-tribromophenol, white ppt	bromobenzene, a liquid

Reason. In phenol a lone pair on the oxygen of \ce{-OH} is delocalised into the ring. The ring becomes strongly activated, with extra electron density at positions 2, 4 and 6. It polarises \ce{Br2} by itself, so no catalyst is needed, and all three activated positions are substituted. Benzene has no activating group; its π electrons are less available, so \ce{FeBr3} must polarise \ce{Br2} to produce the electrophile, and substitution stops after one bromine.

Resonance in phenol: the O lone pair is donated into the ring, making carbons 2, 4 and 6 electron-rich.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 9, old syllabus1 mark

The oxidation of ethanol to ethanoic acid is an example of:

ASubstitution reaction
BRedox reactionAnswer
CAddition reaction
DElimination reaction

Why

Ethanol is oxidised (the carbon's oxidation number rises from -1 to +3) by acidified dichromate, which is reduced at the same time (\ce{Cr}: +6 \to +3). Oxidation and reduction together make a redox reaction.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 13, old syllabus1 mark

The functional group of alcohols is:

A\ce{-OH}Answer
B\ce{-COOH}
C\ce{-CHO}
D\ce{-O-}

Why

Alcohols contain the hydroxyl group \ce{-OH} on a saturated carbon. \ce{-COOH} is the carboxyl group of acids, \ce{-CHO} the aldehyde group and \ce{-O-} the ether link.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (vii) OR, old syllabus2+1 = 3 marks

Write the reaction of alcohol with sodium metal. What this indicates about their chemical nature?

Solution

Reaction (2 marks)

Alcohols react with sodium metal at room temperature. The hydrogen of the \ce{O-H} group is replaced by sodium, an alkoxide is formed and hydrogen gas is evolved (effervescence). For ethanol:

\ce{2C2H5OH + 2Na -> 2C2H5O^-Na+ + H2 ^}

\ce{2R-OH + 2Na -> 2R-O^-Na+ + H2 ^}

general equation; the product is a sodium alkoxide

What it shows (1 mark)

The reaction shows that alcohols are weakly acidic: the H of the \ce{O-H} group can be released as \ce{H+} and replaced by a reactive metal. They are weaker acids than water (the reaction is gentler than sodium with water), so they do not affect litmus and do not react with \ce{NaOH}.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (viii), old syllabus3 marks

Why phenol forms 2,4,6-tribromophenol with bromine water without the need for a catalyst?

Solution

\ce{C6H5OH + 3Br2(aq) -> C6H2Br3OH v + 3HBr}

The \ce{-OH} group of phenol is a strongly activating group. One lone pair on its oxygen atom overlaps with the π electrons of the ring and is delocalised into it. This raises the electron density of the ring, especially at the ortho (2, 6) and para (4) positions.

The electron-rich ring polarises an approaching \ce{Br2} molecule (\ce{Br^{\delta+}-Br^{\delta-}}) and attacks its \delta+ end directly, and the intermediate is stabilised by the O lone pair. So no Lewis acid catalyst is needed to make \ce{Br+}, unlike benzene.

Since all three positions 2, 4 and 6 are activated, three bromine atoms are substituted at once and insoluble 2,4,6-tribromophenol separates as a white precipitate.

Exam focus

Justify phenol > water > ethanol (or ethanoic acid > phenol > ethanol) with resonance structures of the conjugate bases (long)
Asked as a 6-mark Section C question in both 2026 sets.
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Ethanoic acid > phenol > ethanol with resonance structures (long)
2026 Set D Q.5: add ethanoate's two equivalent resonance forms.
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Phenol with bromine water: equation, white precipitate, comparison with benzene (long)
Asked in all three 2026 papers, as a short or long question.
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Why phenol needs no catalyst for bromination (electron density of the ring) (short)
Old and new papers ask for the activating effect of the OH lone pair.
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Phenol from aniline via the diazonium salt, with temperatures (short)
2026 Set D Q.2 (x); two equations, 1.5 marks each.
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Arrange water, phenol and ethanol in order of acidity with reasons (short)
Short-question form of the acidity topic (2026 Set D).
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Nitration of phenol with dilute HNO3 and the 2,4-directing effect (long)
Explains the positions of substitution; standard long question.
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Ester from ethanoyl chloride and an alcohol, with mechanism (long)
Section 10.1; the methanol version is in the exercise.
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Phenol and alcohols with Na and NaOH; what they show (short)
Old-syllabus 2026 question; tells phenol from ethanol.
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Reactions of 1-naphthol and 2-naphthol: positions of substitution (short)
Exercise short xi and long 3 and 8.
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