Carbonyl compounds

Class 12 Chemistry, unit 11, book pages 193–203

Practise 57 MCQs Take the unit quiz Review 48 cards

Carbonyl compounds contain the polar $C = O$ group, whose $δ +$ carbon attracts nucleophiles. This unit concentrates on carboxylic acids and acyl chlorides: making benzoic acid by side-chain oxidation, explaining acid strength through the stability of the conjugate base, replacing $- OH$ by $- Cl$ , and using the very reactive acyl chlorides to make acids, esters and amides by nucleophilic addition-elimination. It ends with why acyl chlorides hydrolyse far more easily than alkyl and aryl chlorides, plus the aldehyde and ketone reactions asked in the old-syllabus paper.

Learning outcomes (10)

State the reactions by which benzoic acid can be produced: reaction of an alkylbenzene with hot alkaline $KMnO X_{4}$ and then dilute acid, exemplified by methylbenzene.
Describe the reaction of carboxylic acids with $PCl X_{3}$ and heat, $PCl X_{5}$ , or $SOCl X_{2}$ to form acyl chlorides.
Recognise that some carboxylic acids can be further oxidised: (a) methanoic acid with Fehling's solution, Tollens' reagent, acidified $KMnO X_{4}$ or acidified $K X_{2} Cr X_{2} O X_{7}$ to carbon dioxide and water; (b) ethanedioic acid with warm acidified $KMnO X_{4}$ to carbon dioxide.
Explain the relative acidities of carboxylic acids, phenols and alcohols.
Explain the relative acidities of chlorine-substituted carboxylic acids.
Recall the reactions by which esters can be produced: reaction of alcohols with acyl chlorides, using the formation of ethyl ethanoate and phenyl benzoate as examples.
Recall the reactions (reagents and conditions) by which acyl chlorides can be produced: reaction of carboxylic acids with $PCl X_{3}$ and heat, $PCl X_{5}$ , or $SOCl X_{2}$ .
Describe the reactions of acyl chlorides at room temperature: (a) hydrolysis with water to give the carboxylic acid and HCl; (b) with an alcohol to give an ester and HCl; (c) with phenol to give an ester and HCl; (d) with ammonia to give an amide and HCl; (e) with a primary or secondary amine to give an amide and HCl.
Describe the addition-elimination mechanism of acyl chlorides in reactions.
Explain the relative ease of hydrolysis of acyl chlorides, alkyl chlorides and halogenoarenes (aryl chlorides).

IntroCarbonyl compounds and their classes

A carbonyl group is a carbon atom joined to an oxygen atom by a double bond, $C = O$ . Oxygen is more electronegative than carbon, so it pulls the shared electrons towards itself. The oxygen becomes slightly negative ( $δ -$ ) and the carbon slightly positive ( $δ +$ ). The carbonyl carbon is $s p^{2}$ hybridised, so the group is flat (trigonal planar) with bond angles of about $12 0^{\circ}$ .

> C X^{δ +} = O X^{δ -}

The $δ +$ carbon attracts nucleophiles (electron-pair donors). This one fact explains most reactions in this unit.

Carbonyl compounds are divided into classes by the atoms or groups attached to the carbonyl carbon. In the general formulas, $R$ and $R^{'}$ stand for alkyl or aryl groups.

Classes of carbonyl compounds. Acids, esters, amides, acyl chlorides and anhydrides are called carboxylic acid derivatives.
Class	General formula	What is attached to C=O	Examples (book)
Aldehydes	$RCHO$	at least one H atom, plus H, alkyl or aryl	formaldehyde $HCHO$ , acetaldehyde $CH X_{3} CHO$
Ketones	$RCO R^{'}$	two alkyl or aryl groups	acetone $CH X_{3} COCH X_{3}$ , butanone $CH X_{3} COC X_{2} H X_{5}$
Carboxylic acids	$RCOOH$	an $- OH$ group and an alkyl or aryl group	acetic acid $CH X_{3} COOH$ , benzoic acid $C X_{6} H X_{5} COOH$
Esters	$RCOO R^{'}$	an alkoxy group $- OR X^{'}$ and an alkyl or aryl group	ethyl acetate $CH X_{3} COOCH X_{2} CH X_{3}$ , methyl benzoate $C X_{6} H X_{5} COOCH X_{3}$
Amides	$RCON R^{'} R^{''}$	a nitrogen atom (with H atoms or alkyl groups)	acetamide $CH X_{3} CONH X_{2}$ , N,N-dimethylacetamide $CH X_{3} CON (CH X_{3}) X_{2}$
Acyl chlorides	$RCOCl$	a chlorine atom	ethanoyl chloride $CH X_{3} COCl$ , benzoyl chloride $C X_{6} H X_{5} COCl$
Acid anhydrides	$(RCO)_{2} O$	two acyl groups joined through one O atom	ethanoic anhydride $(CH X_{3} CO) X_{2} O$

methanal

ethanal

propanone

ethanoic acid

The four derivatives below all contain the acyl group $RCO X^{-}$ . They differ only in the group joined to it: $- Cl$ , $- OR X^{'}$ , $- NH X_{2}$ or $- OCOR$ .

ethanoyl chloride

ethyl ethanoate

ethanamide

ethanoic anhydride

The book uses common and IUPAC names side by side. Learn both.
Common name	IUPAC name	Formula
formaldehyde	methanal	$HCHO$
acetaldehyde	ethanal	$CH X_{3} CHO$
acetone	propanone	$CH X_{3} COCH X_{3}$
formic acid	methanoic acid	$HCOOH$
acetic acid	ethanoic acid	$CH X_{3} COOH$
oxalic acid	ethanedioic acid	$HOOC - COOH$
acetyl chloride	ethanoyl chloride	$CH X_{3} COCl$
acetamide	ethanamide	$CH X_{3} CONH X_{2}$
toluene	methylbenzene	$C X_{6} H X_{5} CH X_{3}$

This unit of the book concentrates on carboxylic acids and acyl chlorides: how benzoic acid is made, why carboxylic acids are acidic, how the $- OH$ group is replaced to make acyl chlorides, and how acyl chlorides react with water, alcohols, phenol, ammonia and amines. Aldehydes and ketones are only defined here. Their key reactions (nucleophilic addition and aldol condensation) were asked in the 2026 old-syllabus paper, so they are covered at the end of these notes.

11.1Acidity of carboxylic acidsOften examined

Carboxylic acids are the most acidic simple organic compounds, with $p K_{a}$ values around 5. They are still weak acids: in water only a small fraction of the molecules give up a proton. Mineral acids such as $HCl$ and $H X_{2} SO X_{4}$ are strong acids and ionise almost completely.

RCOOH + H X_{2} O RCOO X^{-} + H X_{3} O X^{+}

$RCOO X^{-}$ is the carboxylate ion, the conjugate base of the acid.

K_{a} = \frac{[ RCOO X ^{-} ] [ H X _{3} O X ^{+} ]}{[ RCOOH ]} p K_{a} = - lo g_{10} K_{a}

$K_{a}$ : acid dissociation constant (in $mol d m^{- 3}$ ); square brackets mean equilibrium concentrations. A larger $K_{a}$ means more ionisation. Because of the minus sign, the **lower the $p K_{a}$ , the stronger the acid**.

Table 11.1: $p K_{a}$ values of some organic acids (book values).
Acid	Structure	$p K_{a}$
Ethanoic acid	$CH X_{3} CO X_{2} H$	4.7
Propanoic acid	$CH X_{3} CH X_{2} CO X_{2} H$	4.9
Fluoroethanoic acid	$CH X_{2} FCO X_{2} H$	2.6
Chloroethanoic acid	$CH X_{2} ClCO X_{2} H$	2.9
Dichloroethanoic acid	$CHCl X_{2} CO X_{2} H$	1.3
Trichloroethanoic acid	$CCl X_{3} CO X_{2} H$	0.9
Nitroethanoic acid	$O X_{2} NCH X_{2} CO X_{2} H$	1.7

What decides the strength of a carboxylic acid

An acid is strong when its conjugate base (the anion left after losing $H X^{+}$ ) is stable. A stable anion has little tendency to take the proton back, so the equilibrium lies further to the right. Anything that spreads out (disperses) the negative charge of the carboxylate ion stabilises it and makes the acid stronger. Anything that concentrates the charge makes the acid weaker.

Chlorine-substituted ethanoic acids

CCl X_{3} COOH > CHCl X_{2} COOH > CH X_{2} ClCOOH > CH X_{3} COOH

Order of acid strength (strongest first).

Chlorine is more electronegative than carbon. A $C - Cl$ bond on the carbon next to the carboxyl group pulls electron density along the chain of $σ$ bonds towards chlorine. This electron-withdrawing pull through $σ$ bonds is called the negative inductive effect ( $- I$ effect).

In the acid, it pulls electrons away from the $O - H$ oxygen, so the $O - H$ bond becomes more polar and the proton is released more easily.
In the anion, it draws some of the negative charge away from the $COO X^{-}$ group. The charge is spread over more atoms, so the anion is more stable and less likely to bond with an $H X^{+}$ ion again.
Each extra chlorine atom adds to the pull. So acid strength rises from one Cl to three Cl atoms: $p K_{a}$ falls from 2.9 to 1.3 to 0.9.

ethanoic acid pKa 4.7

chloroethanoic acid pKa 2.9

dichloroethanoic acid pKa 1.3

trichloroethanoic acid pKa 0.9

Each Cl atom lowers the

p K_{a}

n_{Cl}

p K_{a}

CH X_{3} COOH (4.7)

CH X_{2} ClCOOH (2.9)

CHCl X_{2} COOH (1.3)

CCl X_{3} COOH (0.9)

n_{Cl}

is the number of Cl atoms on the carbon next to

COOH

. Lower

p K_{a}

means a stronger acid; the first Cl atom has the largest effect.

How much stronger is chloroethanoic acid?

Use Table 11.1 to find how many times stronger chloroethanoic acid (

p K_{a} = 2.9

) is than ethanoic acid (

p K_{a} = 4.7

Convert each $p K_{a}$ back to $K_{a}$ using $K_{a} = 1 0^{- p K_{a}}$ .
$K_{a} (CH X_{2} ClCOOH) K_{a} (CH X_{3} COOH) = 1 0^{- 2.9} = 1.26 \times 1 0^{- 3} = 1 0^{- 4.7} = 2.0 \times 1 0^{- 5}$
Divide. The ratio of $K_{a}$ values is $1 0^{Δ p K_{a}}$ .
$\frac{1.26 \times 1 0 ^{- 3}}{2.0 \times 1 0 ^{- 5}} = 1 0^{4.7 - 2.9} = 1 0^{1.8} \approx 63$

about 60 to 80 times stronger (book: about 100 times)

Electron-donating (alkyl) groups

Alkyl groups push electrons towards the carboxyl group (positive inductive effect, $+ I$ ). This strengthens the $O - H$ bond and concentrates more negative charge on the carboxylate ion, so the anion is less stable and accepts $H X^{+}$ more readily. Acid strength therefore decreases as alkyl groups are added:

p K_{a} = 3.75 HCOOH > 4.76 CH X_{3} COOH > 4.87 CH X_{3} CH X_{2} COOH

Methanoic (formic) acid has no alkyl group, so it is the strongest of the three.

Substituted benzoic acids (beyond the book)

The same idea works for aromatic acids. Benzoic acid has $p K_{a} = 4.20$ . A nitro group ( $- NO X_{2}$ ) at the 4-position (para) is strongly electron-withdrawing, both by the inductive effect and by resonance with the ring. It pulls electron density out of the ring and away from the $COO X^{-}$ group, spreading the negative charge of the 4-nitrobenzoate ion and stabilising it. So **4-nitrobenzoic acid ( $p K_{a} = 3.44$ ) is about 6 times stronger than benzoic acid**. An electron-donating methyl group has the opposite effect.

benzoic acid pKa 4.20

4-nitrobenzoic acid pKa 3.44

4-methylbenzoic acid pKa 4.37

Acid	Group at position 4	Effect of the group	$p K_{a}$
4-nitrobenzoic acid	$- NO X_{2}$	strongly electron-withdrawing	3.44
4-chlorobenzoic acid	$- Cl$	electron-withdrawing	3.98
benzoic acid	$- H$	reference	4.20
4-methylbenzoic acid	$- CH X_{3}$	electron-donating	4.37

11.2Preparation of benzoic acidOften examined

Benzoic acid is made by oxidising the side chain of an alkylbenzene. Methylbenzene (toluene) is heated under reflux with hot alkaline potassium manganate(VII) ( $KMnO X_{4}$ with a little $NaOH$ or $Na X_{2} CO X_{3}$ ). The methyl group is oxidised to a carboxylate group. The mixture is then acidified with dilute acid to release benzoic acid. The book writes the change with $[O]$ for oxygen supplied by the oxidising agent:

C X_{6} H X_{5} CH X_{3} + 3 [O] hot alkaline KMnO X_{4} then dilute acid C X_{6} H X_{5} COOH + H X_{2} O

methylbenzene (toluene)

benzoic acid

What happens in the flask

In alkaline solution the acid cannot exist as free $C X_{6} H X_{5} COOH$ : it is formed as the soluble potassium benzoate. The manganate(VII) ion is reduced to brown manganese(IV) oxide, $MnO X_{2}$ .

C X_{6} H X_{5} CH X_{3} + 2 KMnO X_{4} C X_{6} H X_{5} COOK + 2 MnO X_{2} ↓ + KOH + H X_{2} O

Oxidation in alkaline solution (heat under reflux). Purple colour fades and a brown precipitate forms.

C X_{6} H X_{5} COOK + HCl C X_{6} H X_{5} COOH ↓ + KCl

Acidification of the filtrate. Benzoic acid is only slightly soluble in cold water, so it separates as white crystals.

Preparing benzoic acid from methylbenzene (toluene).

How the side chain is oxidised

The attack happens at the benzylic carbon, the carbon joined directly to the ring. Its $C - H$ bonds are weaker than ordinary $C - H$ bonds because the intermediate formed when one H is removed is stabilised by the ring. The side chain is oxidised in stages, each stage adding oxygen or removing hydrogen:

C X_{6} H X_{5} CH X_{3} [O] C X_{6} H X_{5} CH X_{2} OH [O] C X_{6} H X_{5} CHO [O] C X_{6} H X_{5} COOH

methylbenzene, then phenylmethanol, then benzaldehyde, then benzoic acid (as benzoate in alkali).

The benzene ring itself is not attacked, because its delocalised $π$ electrons make it very stable. Any alkyl side chain with at least one hydrogen on the benzylic carbon is cut down to $- COOH$ , whatever its length. For example, ethylbenzene also gives benzoic acid (the extra carbon leaves as $CO X_{2}$ ). A side chain with no benzylic hydrogen, such as the tert-butyl group in $C X_{6} H X_{5} C (CH X_{3}) X_{3}$ , is not oxidised.

11.3, 11.4Reactivity of the carboxyl group

The carboxyl group, \ce{-COOH}, contains a carbonyl group \ce{C=O} and a hydroxyl group \ce{-OH} on the same carbon. So it shows some chemistry of both groups. In most reactions the carboxyl group is kept (it is not destroyed), and its reactivity comes mainly from the polar carbonyl group.

ethanoic acid the carboxyl group is COOH

The two groups also change each other. A lone pair on the \ce{-OH} oxygen is partly shared with the \ce{C=O} group, so the carbonyl carbon is less \delta+ than in an aldehyde or ketone. This is why carboxylic acids do not give the typical addition reactions of aldehydes and ketones, while the \ce{O-H} bond becomes easier to break, making the compound acidic.

Carboxylic acids undergo three types of reaction:

**Reactions of the H atom of \ce{-OH}** (acidic properties, salt formation): for example with \ce{Na}, \ce{NaOH} and \ce{NaHCO3}.
**Reactions in which \ce{-OH} is replaced** by another group: \ce{-Cl} (acyl chlorides), \ce{-OR} (esters) or \ce{-NH2} (amides).
Reactions of the carboxyl group as a whole: for example oxidation of methanoic and ethanedioic acids to \ce{CO2}.

11.4.1Reactions involving the H atom: acidic properties

Carboxylic acids are weaker than mineral acids, but they still release \ce{H+} ions in water. The \ce{O-H} bond breaks and the carboxylate ion is left behind:

\ce{R-COOH <=>[H2O] R-COO- + H+}

Ionisation (proton loss) of a carboxylic acid.

Because they are acids, they form salts with reactive metals, alkalis, carbonates and hydrogencarbonates:

\ce{2CH3COOH + 2Na -> 2CH3COONa + H2 ^}

Reactive metal: effervescence, hydrogen gas is released.

\ce{CH3COOH + NaOH -> CH3COONa + H2O}

Alkali: neutralisation gives sodium ethanoate and water.

\ce{CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2 ^}

Hydrogencarbonate: brisk effervescence of carbon dioxide (turns limewater milky).

\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2 ^}

Only carboxylic acids are strong enough to release CO2 from hydrogencarbonates.
Reagent	Carboxylic acid	Phenol	Ethanol
\ce{Na} metal	\ce{H2} gas	\ce{H2} gas	\ce{H2} gas (slowly)
\ce{NaOH(aq)}	reacts (salt + water)	reacts (sodium phenoxide)	no reaction
\ce{NaHCO3(aq)}	\ce{CO2} effervescence	no reaction	no reaction

11.4.2Replacing the OH group: preparation of acyl chloridesOften examined

In the second type of reaction, the \ce{-OH} group of the acid is replaced. The reagent first adds to the carboxyl group, and then \ce{-OH} is pushed out. Replacing \ce{-OH} by \ce{-Cl}, \ce{-OR} or \ce{-NH2} gives an acyl chloride, an ester or an amide. These are the carboxylic acid derivatives.

Acyl chlorides (acid chlorides) are made by treating the acid with a chlorinating agent: thionyl chloride, phosphorus(III) chloride or phosphorus(V) chloride. All three reactions need dry conditions, because the acyl chloride reacts with water.

1. With thionyl chloride, SOCl2

\ce{R-COOH + SOCl2 ->[\text{pyridine}] R-COCl + SO2 ^ + HCl ^}

\ce{CH3COOH + SOCl2 -> CH3COCl + SO2 ^ + HCl ^}

ethanoic acid gives ethanoyl chloride

The mixture is warmed. Pyridine, a weak base, is often added to remove the \ce{HCl} formed. Both by-products, \ce{SO2} and \ce{HCl}, are gases and escape, so the acyl chloride is left almost pure. This makes \ce{SOCl2} the most convenient reagent.

2. With phosphorus(III) chloride, PCl3

\ce{3CH3COOH + PCl3 -> 3CH3COCl + H3PO3}

phosphorous acid (\ce{H3PO3}) is formed as a by-product

One molecule of \ce{PCl3} converts three molecules of acid. The book says the reaction goes at room temperature; the SLO says 'with heat', and in practice the mixture is warmed. \ce{H3PO3} is not volatile, so the acyl chloride (b.p. of ethanoyl chloride 51^\circ\mathrm{C}) is separated by fractional distillation.

3. With phosphorus(V) chloride, PCl5

\ce{CH3COOH + PCl5 -> CH3COCl + POCl3 + HCl ^}

phosphorus oxychloride (\ce{POCl3}) and steamy fumes of \ce{HCl}

\ce{PCl5} is a solid that reacts with the acid even in the cold. Steamy white fumes of hydrogen chloride are given off; this is also a test for an \ce{-OH} group. The acyl chloride is separated from \ce{POCl3} (b.p. 107^\circ\mathrm{C}) by fractional distillation.

Comparing the three chlorinating agents.
Reagent	Conditions	By-products	Separation of RCOCl
\ce{SOCl2}	warm; pyridine may be added	\ce{SO2} and \ce{HCl} (both gases)	easy: gases escape
\ce{PCl3}	warm (book: room temperature)	\ce{H3PO3} (non-volatile)	fractional distillation
\ce{PCl5}	cold	\ce{POCl3} (liquid) and \ce{HCl} (gas)	fractional distillation

ethanoyl chloride

benzoyl chloride

11.4.2Esterification and its mechanismOften examined

When a carboxylic acid is heated with an alcohol and a few drops of concentrated sulphuric acid, an ester and water are formed. The \ce{-OH} of the acid is replaced by the \ce{-OR} group of the alcohol. The reaction is slow and reversible, so it reaches an equilibrium.

\ce{CH3COOH + CH3OH <=>[conc.\ H2SO4][\text{heat}] CH3COOCH3 + H2O}

ethanoic acid + methanol gives methyl ethanoate + water

\ce{CH3COOH + CH3CH2OH <=>[conc.\ H2SO4][\text{heat}] CH3COOCH2CH3 + H2O}

ethanoic acid + ethanol gives ethyl ethanoate + water

Concentrated \ce{H2SO4} is a catalyst (it supplies \ce{H+} to protonate the acid) and a dehydrating agent (it removes water and pushes the equilibrium to the right, Le Chatelier's principle).

Mechanism (acid-catalysed nucleophilic addition-elimination)

Methanol + ethanoic acid

Protonation. An \ce{H+} ion from the catalyst bonds to the carbonyl oxygen. The positive charge is shared with the carbonyl carbon, which becomes much more \delta+.
\ce{CH3-C(=O)-OH + H+ <=> CH3-C(OH)=OH+ }
Nucleophilic attack. A lone pair on the oxygen of methanol attacks the positive carbon. The \pi electrons of \ce{C=O} move onto oxygen, giving a tetrahedral intermediate.
\ce{CH3-C(OH)=OH+ + HOCH3 <=> CH3-C(OH)2-O^+H-CH3}
Proton transfer. A proton moves from the added \ce{CH3O^+H} group to one of the \ce{-OH} groups, turning it into \ce{-OH2+}, a good leaving group (water).
\ce{CH3-C(OH)2-O^+H-CH3 <=> CH3-C(OH)(OCH3)-OH2+}
Elimination of water. A lone pair on the remaining \ce{-OH} reforms the double bond and a water molecule leaves.
\ce{CH3-C(OH)(OCH3)-OH2+ <=> CH3-C(OCH3)=OH+ + H2O}
Deprotonation. The protonated ester loses \ce{H+}. The catalyst is regenerated, which is why only a little acid is needed.
\ce{CH3-C(OCH3)=OH+ <=> CH3-COOCH3 + H+}

protonated acid

tetrahedral intermediate

protonated ester

methyl ethanoate

11.4.3Oxidation of methanoic acid and ethanedioic acidOften examined

Most carboxylic acids resist oxidation. The carboxyl carbon is already highly oxidised (oxidation number +3), and oxidising it further would mean breaking a \ce{C-C} bond, which mild oxidising agents cannot do. Two acids are exceptions: methanoic acid and ethanedioic acid. Both are oxidised to carbon dioxide.

Methanoic acid, HCOOH

In methanoic acid the carboxyl carbon carries a hydrogen atom. So the molecule contains an aldehyde group (\ce{H-C=O}) as well as a carboxyl group, and it behaves as a reducing agent like an aldehyde. Its carbon (oxidation number +2) is oxidised to \ce{CO2} (+4) without any \ce{C-C} bond breaking, because there is none.

methanoic acid H-C=O part behaves like an aldehyde

\ce{HCOOH + [O] -> CO2 + H2O}

general change with any of the four oxidising agents below

(i) Tollens' reagent (ammoniacal silver nitrate, \ce{[Ag(NH3)2]+}). On warming, a silver mirror forms on the inside of the test tube. \ce{Ag+} is reduced to silver metal.

\ce{HCOOH + 2OH- + 2[Ag(NH3)2]+ -> CO2 + 2H2O + 2Ag v + 4NH3}

(ii) Fehling's solution (a blue complex of \ce{Cu^2+} in alkali). On warming, the blue colour fades and a red-brown precipitate of copper(I) oxide forms.

\ce{HCOOH + 4OH- + 2Cu^2+ -> CO2 + 3H2O + Cu2O v}

(iii) Acidified potassium manganate(VII). The purple colour is removed as \ce{MnO4-} is reduced to almost colourless \ce{Mn^2+}.

\ce{5HCOOH + 2MnO4- + 6H+ -> 5CO2 + 2Mn^2+ + 8H2O}

\ce{5HCOOH + 2KMnO4 + 3H2SO4 -> 5CO2 + 2MnSO4 + K2SO4 + 8H2O}

(iv) Acidified potassium dichromate(VI). The orange solution turns green as \ce{Cr2O7^2-} is reduced to \ce{Cr^3+}.

\ce{3HCOOH + Cr2O7^2- + 8H+ -> 3CO2 + 2Cr^3+ + 7H2O}

\ce{3HCOOH + K2Cr2O7 + 4H2SO4 -> 3CO2 + Cr2(SO4)3 + K2SO4 + 7H2O}

Ethanedioic acid, (COOH)2

Ethanedioic acid (oxalic acid) has two carboxyl groups joined directly. It is oxidised by warm acidified potassium manganate(VII) (about 60^\circ\mathrm{C}): the \ce{C-C} bond breaks and each carbon (+3) becomes \ce{CO2} (+4). The purple colour of \ce{KMnO4} changes to a very light pink (almost colourless) because of \ce{Mn^2+} ions.

ethanedioic acid

\ce{5(COOH)2 + 2KMnO4 + 3H2SO4 ->[\text{warm}] 10CO2 + 2MnSO4 + K2SO4 + 8H2O}

\ce{5H2C2O4 + 2MnO4- + 6H+ -> 10CO2 + 2Mn^2+ + 8H2O}

ionic equation

Acid	Oxidation number of C	Product	Why it can be oxidised
methanoic acid	+2	\ce{CO2 + H2O}	has an H on the carboxyl carbon (aldehyde-like)
ethanedioic acid	+3 (each C)	\ce{CO2}	the \ce{C-C} bond between two \ce{COOH} groups breaks easily
ethanoic acid	+3 (carboxyl C)	no reaction	no H on the carboxyl carbon; a strong \ce{C-C} bond would have to break

11.4.4Comparing carboxylic acids, phenols and alcoholsOften examined

Carboxylic acids, phenols and alcohols all contain an \ce{O-H} group, but their acid strengths are very different. The strength is compared using \mathrm{p}K_a: the smaller the \mathrm{p}K_a, the stronger the acid. The \mathrm{p}K_a depends on how stable the conjugate base is after the proton has left.

\begin{aligned}\ce{CH3COOH &<=> CH3COO- + H+} &\quad \mathrm{p}K_a&=4.76\\ \ce{C6H5OH &<=> C6H5O- + H+} &\quad \mathrm{p}K_a&=10\\ \ce{CH3CH2OH &<=> CH3CH2O- + H+} &\quad \mathrm{p}K_a&\approx16\end{aligned}

Lower \mathrm{p}K_a means stronger acid. Water (\mathrm{p}K_a 15.7) lies between phenol and ethanol.

Ethanoate ion: charge shared by two oxygen atoms

In the ethanoate ion the negative charge is delocalised by resonance over the whole \ce{COO-} group. The two resonance structures are identical, so the charge is shared equally between two electronegative oxygen atoms, and both \ce{C-O} bonds have the same length. This is very effective stabilisation.

\ce{CH3-C(=O)-O^- <-> CH3-C(-O^-)=O}

ethanoate ion resonance form 1

ethanoate ion resonance form 2

Phenoxide ion: charge spread into the ring

In the phenoxide ion, a lone pair on the oxygen overlaps with the \pi electrons of the ring. The negative charge spreads onto the carbon atoms at the 2, 4 and 6 positions. This stabilises the ion, but less well than in ethanoate, because the charge is shared with carbon atoms, which are less electronegative than oxygen. In most resonance structures the charge is still on the single oxygen. So phenol is a weaker acid than a carboxylic acid.

phenoxide ion

charge at C-2

charge at C-4

charge at C-6

Ethoxide ion: no delocalisation

In the ethoxide ion the negative charge is stuck on one oxygen atom: there is no \pi system to spread it into. The ethyl group also pushes electrons towards oxygen (+I effect), making the charge even more concentrated. The ethoxide ion is therefore very unstable and readily takes \ce{H+} back to reform ethanol. Ethanol is hardly acidic (it is even weaker than water).

\text{Acid strength:}\quad \ce{CH3COOH} > \ce{C6H5OH} > \ce{H2O} > \ce{CH3CH2OH}

	Ethanoic acid	Phenol	Ethanol
\mathrm{p}K_a	4.76	10	about 16
Conjugate base	\ce{CH3COO-}	\ce{C6H5O-}	\ce{CH3CH2O-}
Charge spread over	two O atoms (equally)	one O and three ring C atoms	one O atom only
Stability of anion	high	moderate	low
Reacts with \ce{NaHCO3}	yes, \ce{CO2} given off	no	no
Reacts with \ce{NaOH}	yes	yes	no

11.5Reactions of acyl chloridesOften examined

Acyl chlorides are the most reactive carboxylic acid derivatives. They react quickly at room temperature with any molecule that has a lone pair and an H atom it can lose: water, alcohols, phenols, ammonia and amines. The acyl group \ce{RCO-} replaces the H atom, and hydrogen chloride is always the second product (seen as steamy white fumes).

\ce{RCOCl + H-Nu -> RCO-Nu + HCl}

\ce{H-Nu}: \ce{H-OH}, \ce{H-OR}, \ce{H-OC6H5}, \ce{H-NH2}, \ce{H-NHR} or \ce{H-NR2}

1. With alcohols: esters

Acyl chlorides react vigorously with alcohols at room temperature to give esters. Unlike the acid-alcohol reaction, this is fast, needs no catalyst and does not reach an equilibrium.

\ce{CH3COCl + CH3CH2OH -> CH3COOCH2CH3 + HCl}

ethanoyl chloride + ethanol gives ethyl ethanoate

\ce{C6H5COCl + CH3CH2OH -> C6H5COOCH2CH3 + HCl}

benzoyl chloride + ethanol gives ethyl benzoate

ethyl ethanoate

ethyl benzoate

2. With phenol: phenyl esters

Phenol does not form esters easily with carboxylic acids, but it does react with acyl chlorides at room temperature. The reaction is slower than with alcohols, because the lone pair on the phenol oxygen is partly delocalised into the ring. With the less reactive benzoyl chloride, phenol is first dissolved in sodium hydroxide solution to form the phenoxide ion, a better nucleophile.

\ce{CH3COCl + C6H5OH -> CH3COOC6H5 + HCl}

phenyl ethanoate

\ce{C6H5COCl + C6H5OH -> C6H5COOC6H5 + HCl}

phenyl benzoate (SLO example)

\ce{C6H5COCl + C6H5O^-Na+ -> C6H5COOC6H5 + NaCl}

the practical method: shake benzoyl chloride with phenol in NaOH(aq)

phenyl ethanoate

phenyl benzoate

3. With water: hydrolysis

Acyl chlorides are hydrolysed by cold water at room temperature, giving the carboxylic acid and hydrochloric acid. Ethanoyl chloride fumes in moist air for this reason.

\ce{CH3COCl + H2O -> CH3COOH + HCl}

4. With ammonia: amides

Acyl chlorides react with ammonia at room temperature to give amides. The \ce{HCl} produced reacts at once with more ammonia to form ammonium chloride, so two moles of ammonia are used.

\ce{CH3COCl + NH3 -> CH3CONH2 + HCl}

ethanamide

\ce{HCl + NH3 -> NH4Cl}

\ce{CH3COCl + 2NH3 -> CH3CONH2 + NH4Cl}

overall

5. With primary and secondary amines: N-substituted amides

Primary and secondary amines react like ammonia at room temperature. The product is an N-substituted amide (the alkyl groups of the amine stay on the nitrogen). The \ce{HCl} reacts with excess amine to give an ammonium salt.

\ce{CH3COCl + CH3NH2 -> CH3CONHCH3 + HCl}

methylamine (methanamine) gives N-methylethanamide

\ce{HCl + CH3NH2 -> [CH3NH3]+Cl-}

methylammonium chloride

\ce{CH3COCl + (CH3)2NH -> CH3CON(CH3)2 + HCl}

dimethylamine (N-methylmethanamine) gives N,N-dimethylethanamide

\ce{HCl + (CH3)2NH -> [(CH3)2NH2]+Cl-}

dimethylammonium chloride

N-methylethanamide

N,N-dimethylethanamide

Summary of acyl chloride reactions.
Reagent (at room temperature)	Product	Example
water	carboxylic acid + HCl	\ce{CH3COOH}
alcohol	ester + HCl	\ce{CH3COOCH2CH3}
phenol	phenyl ester + HCl	\ce{CH3COOC6H5}
ammonia	amide + HCl (then NH4Cl)	\ce{CH3CONH2}
primary amine	N-substituted amide + HCl	\ce{CH3CONHCH3}
secondary amine	N,N-disubstituted amide + HCl	\ce{CH3CON(CH3)2}

11.5.1Mechanism: nucleophilic addition-eliminationOften examined

All the reactions in 11.5 follow one mechanism, called nucleophilic addition-elimination. The carbonyl carbon of an acyl chloride is attached to two electronegative atoms, \ce{O} and \ce{Cl}, which both pull electrons away from it. It is strongly \delta+. It is also flat (sp^2), so a nucleophile can easily approach it from above or below the plane.

\ce{CH3-\overset{\delta+}{C}(=O^{\delta-})-Cl^{\delta-}}

Nucleophilic addition-elimination: water adds to the \ce{C=O} carbon, then \ce{Cl-} and \ce{H+} are lost. Curly arrows show the movement of electron pairs.

Hydrolysis of an acyl chloride (book mechanism)

Nucleophilic addition. A lone pair on the oxygen of water attacks the \delta+ carbonyl carbon. At the same time the \pi electrons of \ce{C=O} move onto the oxygen. A tetrahedral intermediate forms, with \ce{O-} on one side and \ce{OH2+} on the other.
\ce{R-COCl + H2O -> R-C(O^-)(Cl)-OH2^+}
Proton transfer. A proton moves from the \ce{OH2+} group, so the intermediate becomes neutral or loses \ce{H+} to the solvent.
\ce{R-C(O^-)(Cl)-OH2^+ -> R-C(OH)(Cl)-OH}
Elimination. A lone pair on oxygen reforms the \ce{C=O} double bond and the \ce{C-Cl} bond breaks. The chlorine leaves as a chloride ion, \ce{Cl-}, a good leaving group.
\ce{R-C(OH)(Cl)-OH -> R-C(=OH^+)-OH + Cl^-}
Deprotonation. The chloride ion removes the extra proton, giving the carboxylic acid and \ce{HCl}.
\ce{R-C(=OH^+)-OH + Cl^- -> R-COOH + HCl}

ethanoyl chloride

tetrahedral intermediate

ethanoic acid

With an alcohol, phenol, ammonia or an amine the steps are the same; only the nucleophile changes. For example, with ammonia the lone pair on nitrogen attacks, the intermediate carries \ce{-NH3+}, and the product is an amide.

11.5.2Ease of hydrolysis: acyl, alkyl and aryl chloridesOften examined

Hydrolysis means breaking a bond by reaction with water. For compounds with a \ce{C-Cl} bond, hydrolysis replaces \ce{Cl} by \ce{OH}. The three types of chloride hydrolyse with very different ease:

\text{acyl chlorides} > \text{alkyl chlorides} > \text{aryl chlorides}

for example \ce{CH3COCl} > \ce{CH3CH2Cl} > \ce{C6H5Cl}

Why the three chlorides behave so differently.
	Acyl chloride, \ce{CH3COCl}	Alkyl chloride, \ce{CH3CH2Cl}	Aryl chloride, \ce{C6H5Cl}
Carbon bonded to Cl	sp^2 carbonyl carbon	sp^3 carbon	sp^2 ring carbon
Electronegative atoms on that carbon	two (O and Cl)	one (Cl)	one (Cl), but its lone pair is delocalised
Size of \delta+ on carbon	very large	moderate	small
Mechanism	addition-elimination	S_N1 or S_N2	none under ordinary conditions
Conditions for hydrolysis	cold water, room temperature, fast	heat under reflux with \ce{NaOH(aq)}	no reaction (only with \ce{NaOH} at about 350^\circ\mathrm{C} and high pressure)
\ce{AgNO3(aq)} test	white AgCl at once, in the cold	white AgCl slowly on warming	no precipitate

Acyl chlorides: hydrolysed most readily

The carbonyl carbon is joined to two electronegative atoms (O and Cl). Both pull electrons away, so this carbon is highly electron-deficient. It is also flat, so water (a weak nucleophile) can attack it easily. The \pi bond of \ce{C=O} can accept the electrons, forming a tetrahedral intermediate before the \ce{C-Cl} bond has to break. The reaction is fast even in the cold.

\ce{CH3-\overset{\delta+}{C}(=O)-\overset{\delta-}{Cl}}

Alkyl chlorides: slower

The carbon is attached to only one electronegative atom, so it is less electron-deficient. It is sp^3 and has no \pi bond to accept electrons, so the \ce{C-Cl} bond must break during the substitution itself. Water is too weak a nucleophile; the stronger nucleophile \ce{OH-} and heating are needed.

\ce{CH3CH2Cl + NaOH ->[\text{heat, reflux}] CH3CH2OH + NaCl}

\ce{CH3-\overset{\delta+}{C}H2-\overset{\delta-}{Cl}}

Aryl chlorides: not hydrolysed

In chlorobenzene a lone pair on chlorine overlaps with the \pi electrons of the ring and is delocalised over it. The \ce{C-Cl} bond gains partial double-bond character: it is shorter and stronger (about 400\,\mathrm{kJ\,mol^{-1}} compared with about 340\,\mathrm{kJ\,mol^{-1}} in chloroalkanes). The bond is hard to break even with a strong nucleophile, so hydrolysis does not occur under ordinary conditions.

chlorobenzene

charge at C-2

charge at C-4

charge at C-6

The ring carbon is sp^2; it holds the bonding electrons more tightly, which also strengthens the \ce{C-Cl} bond.
The cloud of \pi electrons above and below the ring repels an approaching nucleophile.
Back-side attack (S_N2) is impossible because the ring blocks it, and the phenyl cation needed for S_N1 is very unstable.

Aldehydes and ketones: nucleophilic addition (beyond the book)

The carbonyl group of aldehydes and ketones is polar: the carbon is \delta+ and the oxygen \delta-. The carbonyl carbon therefore behaves as an electrophile (electron-pair acceptor) and is attacked by nucleophiles. Aldehydes and ketones have no leaving group on the carbonyl carbon, so after the nucleophile adds, nothing is eliminated: the product keeps both new groups. This is nucleophilic addition, the typical reaction of aldehydes and ketones.

Mechanism: addition of HCN

Ethanal + HCN

Nucleophilic attack. The cyanide ion (from \ce{KCN} or \ce{NaCN}) uses its lone pair on carbon to attack the \delta+ carbonyl carbon. The \pi electrons of \ce{C=O} move onto oxygen, forming a tetrahedral alkoxide ion. This is the slow step.
\ce{CH3CHO + CN- -> CH3CH(O^-)CN}
Protonation. The alkoxide ion takes a proton from \ce{HCN} (or water), giving the cyanohydrin (hydroxynitrile) and regenerating \ce{CN-}, which is why \ce{KCN} acts as a catalyst.
\ce{CH3CH(O^-)CN + HCN -> CH3CH(OH)CN + CN-}

ethanal

alkoxide intermediate

2-hydroxypropanenitrile

Formaldehyde is more reactive than acetone

Reactivity towards nucleophilic addition falls in the order \ce{HCHO} > \ce{CH3CHO} > \ce{CH3COCH3}. Two reasons explain this:

Electronic (inductive) effect. Alkyl groups push electrons towards the carbonyl carbon (+I effect). In acetone two methyl groups do this, so the \delta+ charge on the carbonyl carbon is reduced and it attracts nucleophiles less. Formaldehyde has only H atoms, so its carbon keeps the largest \delta+ charge.
Steric effect. The two bulky methyl groups of acetone crowd the carbonyl carbon. They block the approach of the nucleophile and crowd the tetrahedral product, in which the bond angles shrink from 120^\circ to about 109.5^\circ. The small H atoms of formaldehyde cause almost no crowding.

	Formaldehyde, \ce{HCHO}	Acetone, \ce{CH3COCH3}
Groups on C=O	two H atoms	two methyl groups
\delta+ on carbonyl carbon	largest	reduced by two +I groups
Crowding (steric hindrance)	very little	large
Reactivity to nucleophiles	most reactive	least reactive
Hydrate formed in water	almost completely (over 99%)	hardly any (about 0.1%)

formaldehyde (methanal)

acetone (propanone)

Aldol condensation (beyond the book)

An aldehyde (or ketone) that has at least one hydrogen on the carbon next to \ce{C=O} (an **\alpha-hydrogen) reacts with itself in the presence of dilute alkali** (dilute \ce{NaOH}) in the cold. Two molecules join to form a **\beta-hydroxy aldehyde, called an aldol (aldehyde + alcohol**). On warming, the aldol loses water to give an unsaturated aldehyde. Losing water is why the overall process is called a condensation.

\ce{2CH3CHO ->[\text{dil. } NaOH][5-10^\circ C] CH3CH(OH)CH2CHO ->[\text{heat}][-H2O] CH3CH=CHCHO}

ethanal gives 3-hydroxybutanal (aldol), then but-2-enal (crotonaldehyde)

Mechanism for two molecules of acetaldehyde (ethanal)

Aldol condensation of ethanal

Formation of the carbanion. The hydroxide ion removes an \alpha-hydrogen from ethanal. The \alpha-hydrogen is acidic because the carbanion formed is stabilised by resonance with the \ce{C=O} group (enolate ion).
\ce{OH- + H-CH2-CHO <=> ^{-}CH2-CHO + H2O}
The negative charge is delocalised onto oxygen:
\ce{^{-}CH2-CH=O <-> CH2=CH-O^{-}}
Nucleophilic addition. The carbanion attacks the \delta+ carbonyl carbon of a second ethanal molecule. The \pi electrons of its \ce{C=O} move onto oxygen, giving an alkoxide ion. A new \ce{C-C} bond has formed.
\ce{CH3-CHO + ^{-}CH2-CHO -> CH3-CH(O^-)-CH2-CHO}
Protonation. The alkoxide ion takes a proton from water. The aldol is formed and \ce{OH-} is regenerated (it is a catalyst).
\ce{CH3-CH(O^-)-CH2-CHO + H2O -> CH3-CH(OH)-CH2-CHO + OH-}
Dehydration (condensation). On heating, \ce{OH-} removes another \alpha-hydrogen and the \ce{-OH} group leaves as water, forming a \ce{C=C} double bond conjugated with \ce{C=O}.
\ce{CH3-CH(OH)-CH2-CHO ->[\text{heat}] CH3-CH=CH-CHO + H2O}

ethanal

carbanion (enolate)

alkoxide ion

3-hydroxybutanal (aldol)

but-2-enal

Formula sheet

Acid dissociation constant

K_a=\dfrac{[\ce{RCOO-}][\ce{H3O+}]}{[\ce{RCOOH}]}

square brackets: equilibrium concentrations

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

pKa

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

lower \mathrm{p}K_a means a stronger acid

no unit

Comparing two acids

\dfrac{K_{a,1}}{K_{a,2}}=10^{\,\mathrm{p}K_{a,2}-\mathrm{p}K_{a,1}}

acid 1 is stronger when its \mathrm{p}K_a is lower

ratio, no unit

Benzoic acid from methylbenzene

\ce{C6H5CH3 + 3[O] ->[\text{hot {alk.} } KMnO4][\text{then } H+] C6H5COOH + H2O}

side-chain oxidation, then acidification

Acyl chloride with SOCl2

\ce{RCOOH + SOCl2 -> RCOCl + SO2 + HCl}

by-products are gases; pyridine may be added

Acyl chloride with PCl3

\ce{3RCOOH + PCl3 -> 3RCOCl + H3PO3}

warm

Acyl chloride with PCl5

\ce{RCOOH + PCl5 -> RCOCl + POCl3 + HCl}

cold; steamy fumes of HCl

Esterification

\ce{RCOOH + R'OH <=>[conc.\ H2SO4] RCOOR' + H2O}

reversible; acid catalyst

Methanoic acid with Tollens' reagent

\ce{HCOOH + 2OH- + 2[Ag(NH3)2]+ -> CO2 + 2H2O + 2Ag + 4NH3}

silver mirror

Methanoic acid with Fehling's solution

\ce{HCOOH + 4OH- + 2Cu^2+ -> CO2 + 3H2O + Cu2O}

red-brown precipitate

Methanoic acid with acidified KMnO4

\ce{5HCOOH + 2MnO4- + 6H+ -> 5CO2 + 2Mn^2+ + 8H2O}

purple to colourless

Ethanedioic acid with acidified KMnO4

\ce{5(COOH)2 + 2MnO4- + 6H+ -> 10CO2 + 2Mn^2+ + 8H2O}

warm; purple to colourless

Acyl chloride + water

\ce{RCOCl + H2O -> RCOOH + HCl}

room temperature

Acyl chloride + alcohol or phenol

\ce{RCOCl + R'OH -> RCOOR' + HCl}

\mathrm{R'} = alkyl or phenyl; room temperature

Acyl chloride + ammonia or amine

\ce{RCOCl + 2R'NH2 -> RCONHR' + R'NH3+Cl-}

\mathrm{R'} = H (ammonia) or alkyl (amine)

Aldol condensation

\ce{2CH3CHO ->[\text{dil. } NaOH] CH3CH(OH)CH2CHO ->[\text{heat}] CH3CH=CHCHO + H2O}

needs an \alpha-hydrogen

Key terms

Carbonyl group: A carbon atom double-bonded to an oxygen atom, \ce{C=O}; polar, with \delta+ carbon and \delta- oxygen.
Aldehyde: A carbonyl compound in which the carbonyl carbon carries at least one H atom; general formula \mathrm{RCHO}.
Ketone: A carbonyl compound in which the carbonyl carbon is joined to two alkyl or aryl groups; \mathrm{RCOR'}.
Carboxylic acid: A compound containing the carboxyl group \ce{-COOH}; general formula \mathrm{RCOOH}.
Carboxylate ion: The anion \ce{RCOO-} left when a carboxylic acid loses a proton; stabilised by resonance over two oxygen atoms.
Acyl group: The group \ce{RCO-} present in carboxylic acids and all their derivatives.
Acyl chloride: A carboxylic acid derivative in which \ce{-OH} is replaced by \ce{-Cl}; \mathrm{RCOCl}, for example ethanoyl chloride.
Ester: A derivative \mathrm{RCOOR'} formed from a carboxylic acid (or acyl chloride) and an alcohol or phenol.
Amide: A derivative \mathrm{RCONR'R''} in which the carbonyl carbon is joined to nitrogen.
Acid anhydride: A compound \mathrm{(RCO)_2O} with two acyl groups joined through an oxygen atom.
Acid dissociation constant, Ka: The equilibrium constant for the ionisation of a weak acid in water; a larger K_a means a stronger acid.
pKa: -\log_{10}K_a; the lower the \mathrm{p}K_a, the stronger the acid.
Conjugate base: The species left after an acid loses a proton, for example \ce{CH3COO-} from \ce{CH3COOH}.
Inductive effect: The shift of electron density along \sigma bonds caused by a more or less electronegative group: -I (withdrawing, such as Cl, \ce{NO2}) or +I (donating, such as alkyl groups).
Electron-withdrawing group: A group that pulls electron density towards itself (Cl, F, \ce{NO2}); it stabilises a carboxylate ion and increases acid strength.
Electron-donating group: A group that pushes electron density away from itself (alkyl groups); it decreases acid strength.
Resonance (delocalisation): Spreading of electrons, such as a negative charge, over several atoms; it lowers the energy of an ion or molecule.
Side-chain oxidation: Oxidation of an alkyl group on a benzene ring to \ce{-COOH} by hot alkaline \ce{KMnO4} followed by acid.
Benzylic carbon: The carbon atom of a side chain that is joined directly to a benzene ring.
Chlorinating agent: A reagent that replaces \ce{-OH} by \ce{-Cl}, such as \ce{SOCl2}, \ce{PCl3} or \ce{PCl5}.
Esterification: The reversible reaction of a carboxylic acid with an alcohol, catalysed by concentrated \ce{H2SO4}, giving an ester and water.
Tollens' reagent: Ammoniacal silver nitrate, \ce{[Ag(NH3)2]+}; mild oxidising agent that gives a silver mirror with reducing agents such as methanoic acid.
Fehling's solution: An alkaline solution of a copper(II) complex; reducing agents turn it from blue to a red-brown precipitate of \ce{Cu2O}.
Nucleophile: A species with a lone pair of electrons that attacks an electron-deficient (positive) centre, for example water, alcohols, ammonia, amines and the cyanide ion.
Electrophile: An electron-pair acceptor; the \delta+ carbonyl carbon behaves as an electrophile.
Nucleophilic addition-elimination: The mechanism of acyl chloride reactions: a nucleophile adds to the \ce{C=O} carbon to give a tetrahedral intermediate, then \ce{Cl-} (and \ce{H+}) are eliminated.
Tetrahedral intermediate: The short-lived species with four groups around the former carbonyl carbon, formed when a nucleophile adds to \ce{C=O}.
Hydrolysis: Breaking a compound by reaction with water; for a chloride, \ce{Cl} is replaced by \ce{OH}.
Nucleophilic addition: The typical reaction of aldehydes and ketones: a nucleophile adds to the carbonyl carbon and \ce{H+} adds to the oxygen.
Aldol condensation: The reaction of two aldehyde or ketone molecules having an \alpha-hydrogen, in dilute alkali, to give a \beta-hydroxy carbonyl compound (aldol), which loses water on heating.
Alpha-hydrogen: A hydrogen atom on the carbon next to a carbonyl group; it is slightly acidic.
Cyanohydrin (hydroxynitrile): The product of adding HCN to an aldehyde or ketone, containing \ce{-OH} and \ce{-CN} on the same carbon.

Multiple choice questionsp. 198–200

i1 mark

Which reagent would you use to convert toluene to benzoic acid?

A\ce{KMnO4} and dilute acidAnswer
BFehling's solution
C\ce{PCl5}
D\ce{SOCl2}

Why

Hot alkaline \ce{KMnO4} oxidises the methyl side chain to a benzoate salt, and dilute acid then releases benzoic acid. Fehling's solution is too weak an oxidant, and \ce{PCl5} and \ce{SOCl2} are chlorinating agents that convert \ce{-COOH} into \ce{-COCl}.

ii1 mark

What is the main product of the reaction between acetic acid and \ce{PCl5}?

AEthyl acetate
BAcetyl chlorideAnswer
CAcetic anhydride
DAcetone

Why

\ce{PCl5} replaces the \ce{-OH} group by \ce{-Cl}: \ce{CH3COOH + PCl5 -> CH3COCl + POCl3 + HCl}. The product is acetyl (ethanoyl) chloride.

iii1 mark

Which of the following is a strong oxidizing agent for the oxidation of methanoic acid?

A\ce{PCl5}
B\ce{SOCl2}
C\ce{KMnO4}Answer
D\ce{PCl3}

Why

Acidified \ce{KMnO4} oxidises methanoic acid to \ce{CO2} and water, and its purple colour disappears. The other three are chlorinating agents, not oxidising agents.

iv1 mark

In terms of acidity, which of the following is the most acidic?

AMethanol
BPhenol
CFormic acidAnswer
DBenzoic acid

Why

Formic (methanoic) acid has \mathrm{p}K_a=3.75, lower than benzoic acid (4.20), phenol (10) and methanol (about 15.5). Benzoic acid is the tempting distractor, but formic acid has no group that pushes electrons towards \ce{COOH}, so it is slightly stronger.

v1 mark

What happens when benzoyl chloride reacts with water?

ABenzoic acid and HCl are formed.Answer
BBenzene and water are formed.
CPhenol and water are formed.
DBenzoate and water are formed.

Why

Hydrolysis replaces \ce{-Cl} by \ce{-OH}: \ce{C6H5COCl + H2O -> C6H5COOH + HCl}. A benzoate salt would form only if an alkali were present.

vi1 mark

The mechanism by which acyl chlorides react with alcohols is:

AAddition
BElimination
CAddition-eliminationAnswer
DSubstitution

Why

The alcohol first adds to the \ce{C=O} carbon (tetrahedral intermediate), then \ce{Cl-} and \ce{H+} are eliminated. The overall change looks like a substitution, but the mechanism is nucleophilic addition-elimination.

vii1 mark

Which of the following acyl chlorides is expected to hydrolyze most readily?

AMethyl chloride
BAcetyl chlorideAnswer
CPhenyl chloride
DBenzyl chloride

Why

Acetyl chloride, \ce{CH3COCl}, is the only true acyl chloride in the list; its carbonyl carbon is bonded to both O and Cl and reacts with cold water. Methyl chloride and benzyl chloride (\ce{C6H5CH2Cl}) are alkyl-type chlorides and need warm alkali, and phenyl chloride (chlorobenzene) does not hydrolyse at all.

Note. The stem calls all four compounds acyl chlorides, but only acetyl chloride is one.

viii1 mark

When formic acid is oxidized, which of the following is formed?

AWater only
BCarbon dioxide only
CMethanol only
DWater and carbon dioxideAnswer

Why

\ce{HCOOH + [O] -> CO2 + H2O}. Both carbon dioxide and water are formed; methanol would be a reduction product.

ix1 mark

The relative acidity of chlorinated carboxylic acids is primarily due to:

AIncreased stability of the conjugate base.Answer
BElectron-donating effects of the chlorine atom.
CLowering of the carboxylate \mathrm{p}K_a.
DIncreased electron density around the carbonyl group.

Why

Chlorine withdraws electrons (-I effect) and spreads the negative charge of the carboxylate ion, so the conjugate base is more stable. A lower \mathrm{p}K_a (C) is the result of this, not its cause, and chlorine withdraws rather than donates electrons.

x1 mark

The addition-elimination reaction mechanism of acyl chlorides involves:

ANucleophilic attack on the carbonyl carbon.Answer
BAddition of HCl to the carbonyl group.
CElectrophilic substitution on the benzene ring.
DRadical addition to the carbonyl group.

Why

The first step is attack by a nucleophile (water, alcohol, ammonia or amine) on the \delta+ carbonyl carbon. HCl is a product, not a reagent, and no radicals or ring substitution are involved.

xi1 mark

Why are carboxylic acids more acidic than phenols and alcohols?

ACarboxylic acids have a lower molecular weight.
BCarboxylic acids are better at forming hydrogen bonds.
CThe carboxylate anion is resonance-stabilized.Answer
DCarboxylic acids have a higher boiling point.

Why

The negative charge of the carboxylate ion is shared equally by two oxygen atoms through resonance, so the anion is very stable. Molecular mass, hydrogen bonding and boiling point do not decide acid strength.

xii1 mark

How does chlorine substitution affect the acidity of carboxylic acids?

AIt increases acidity by donating electrons.
BIt decreases acidity by withdrawing electrons.
CIt increases acidity by withdrawing electrons.Answer

Why

Chlorine is electronegative and withdraws electrons by the inductive effect. This stabilises the carboxylate ion, so acidity increases: \mathrm{p}K_a of \ce{CH2ClCOOH} is 2.9 compared with 4.7 for \ce{CH3COOH}.

Note. The book prints only three options for this question.

xiii1 mark

Which of the following statements about the addition-elimination mechanism of acyl chlorides is correct?

AThe nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate.Answer
BThe nucleophile attacks the chlorine atom directly.
CThe reaction involves a free radical intermediate.
DThe acyl chloride first forms a carbocation intermediate.

Why

The nucleophile adds to the carbonyl carbon and the \ce{C=O} opens, giving a tetrahedral intermediate; \ce{Cl-} is lost afterwards. The chlorine atom is never attacked, and there is no radical or carbocation.

xiv1 mark

What is the major difference in the oxidation products of ethanedioic acid and methanoic acid when treated with warm acidified \ce{KMnO4}?

AEthanedioic acid produces carbon dioxide, while methanoic acid produces formaldehyde.
BEthanedioic acid produces carbon monoxide, while methanoic acid produces carbon dioxide and water.
CEthanedioic acid produces carbon dioxide, while methanoic acid produces carbon dioxide and water.Answer
DEthanedioic acid produces acetic acid, while methanoic acid produces carbon dioxide.

Why

\ce{5(COOH)2 + 2MnO4- + 6H+ -> 10CO2 + 2Mn^2+ + 8H2O} and \ce{5HCOOH + 2MnO4- + 6H+ -> 5CO2 + 2Mn^2+ + 8H2O}. Written as the acid alone, ethanedioic acid gives only \ce{CO2}, while the H atom of methanoic acid ends up as water.

Short answer questionsp. 200

i3 marks

Write the equation for the reaction of acetic acid with \ce{SOCl2}.

Solution

Acetic (ethanoic) acid reacts with thionyl chloride on warming to give acetyl (ethanoyl) chloride. The \ce{-OH} group of the acid is replaced by a \ce{-Cl} atom.

\ce{CH3COOH + SOCl2 ->[\text{warm, pyridine}] CH3COCl + SO2 ^ + HCl ^}

acetic acid

acetyl chloride

Sulphur dioxide and hydrogen chloride are both gases, so they escape and leave fairly pure acetyl chloride; this is why \ce{SOCl2} is the preferred reagent. Pyridine, a weak base, is often added to absorb the \ce{HCl}. The apparatus must be dry, because acetyl chloride reacts with water.

ii3 marks

Explain why methanoic acid can be oxidized while other carboxylic acids cannot.

Solution

Methanoic acid, \ce{HCOOH}, is the only carboxylic acid whose carboxyl carbon carries a hydrogen atom. Its molecule therefore contains an aldehyde group (\ce{H-C=O}) as well as a carboxyl group, so it acts as a reducing agent, just like an aldehyde.

methanoic acid

\ce{HCOOH + [O] -> CO2 + H2O}

The carbon (oxidation number +2) is oxidised to \ce{CO2} (+4) without breaking any \ce{C-C} bond. So methanoic acid reduces Tollens' reagent to silver, Fehling's solution to red \ce{Cu2O}, and decolourises acidified \ce{KMnO4}.

In other acids, such as ethanoic acid, the carboxyl carbon has no hydrogen and is already at +3; further oxidation would require breaking a strong \ce{C-C} bond, which these oxidising agents cannot do. Ethanedioic acid is the one exception, because its \ce{C-C} bond joins two carboxyl groups.

iii3 marks

What products are formed when ethanedioic acid is oxidized with warm acidified \ce{KMnO4}?

Solution

Warm (about 60^\circ\mathrm{C}) potassium manganate(VII), acidified with dilute sulphuric acid, oxidises ethanedioic acid to carbon dioxide and water. The \ce{C-C} bond breaks and each carbon (oxidation number +3) becomes \ce{CO2} (+4). Manganate(VII) is reduced to manganese(II), so the purple colour fades to a very light pink, almost colourless.

\ce{5(COOH)2 + 2KMnO4 + 3H2SO4 ->[\text{warm}] 10CO2 ^ + 2MnSO4 + K2SO4 + 8H2O}

\ce{5H2C2O4 + 2MnO4- + 6H+ -> 10CO2 + 2Mn^2+ + 8H2O}

ionic equation

So the products are \ce{CO2} (seen as effervescence), \ce{H2O}, manganese(II) sulphate and potassium sulphate.

Note. The book's equation leaves out the coefficient 5 in front of \ce{(COOH)2}; it is needed to balance carbon and oxygen.

iv3 marks

Describe the mechanism of acyl chloride hydrolysis with water.

Solution

Acyl chlorides are hydrolysed by nucleophilic addition-elimination. The carbonyl carbon is strongly \delta+ because it is bonded to two electronegative atoms, O and Cl, and water acts as the nucleophile.

Addition: a lone pair on the water oxygen attacks the carbonyl carbon; the \pi electrons of \ce{C=O} move onto oxygen, giving a tetrahedral intermediate.
\ce{CH3COCl + H2O -> CH3C(O^-)(Cl)OH2^+}
Elimination: the oxygen lone pair reforms \ce{C=O} and the chloride ion leaves.
\ce{CH3C(O^-)(Cl)OH2^+ -> CH3C(=O)OH2^+ + Cl^-}
Deprotonation: \ce{Cl-} removes \ce{H+} to give the acid and hydrogen chloride.
\ce{CH3C(=O)OH2^+ + Cl^- -> CH3COOH + HCl}

\ce{CH3COCl + H2O -> CH3COOH + HCl}

overall; fast at room temperature

v3 marks

What is the difference in reactivity between acyl chlorides and alkyl chlorides?

Solution

Acyl chlorides are much more reactive. Ethanoyl chloride reacts vigorously with cold water at room temperature and gives steamy fumes of \ce{HCl}, while chloroethane must be heated under reflux with aqueous NaOH (a stronger nucleophile) before it is hydrolysed.

\begin{aligned}\ce{CH3COCl + H2O &-> CH3COOH + HCl}\\\ce{CH3CH2Cl + NaOH &->[\text{heat}] CH3CH2OH + NaCl}\end{aligned}

In an acyl chloride the carbonyl carbon is bonded to two electronegative atoms (O and Cl), so it is far more \delta+ than the carbon of an alkyl chloride, which has only one.
The acyl carbon is sp^2 and flat, with a \ce{C=O} \pi bond that can accept the nucleophile's electrons. A tetrahedral intermediate forms before the \ce{C-Cl} bond breaks (addition-elimination). The sp^3 carbon of an alkyl chloride has no \pi bond, so the \ce{C-Cl} bond must break during the substitution (S_N1 or S_N2), which needs more energy.

vi2026 Annual I, Set B, Q.2 (viii)3 marks

Explain the influence of chlorine substitution on the acidity of carboxylic acids.

Solution

Chlorine substitution increases the acidity of a carboxylic acid. Chlorine is more electronegative than carbon, so it withdraws electron density along the \sigma bonds (**-I inductive effect**).

This pulls negative charge away from the \ce{COO-} group of the carboxylate ion and spreads it out, so the conjugate base is more stable and has less tendency to take back \ce{H+}. The pull also weakens the \ce{O-H} bond of the acid. The more chlorine atoms, the stronger the effect:

\underset{0.9}{\ce{CCl3COOH}} > \underset{1.3}{\ce{CHCl2COOH}} > \underset{2.9}{\ce{CH2ClCOOH}} > \underset{4.7}{\ce{CH3COOH}}\qquad(\mathrm{p}K_a)

The effect is larger for a more electronegative halogen (\ce{CH2FCOOH}, \mathrm{p}K_a 2.6) and smaller when the chlorine is farther from the \ce{COOH} group.

vii3 marks

What product results from the reaction of acyl chlorides with phenol?

Solution

An acyl chloride reacts with phenol at room temperature to give a phenyl ester and hydrogen chloride. The acyl group \ce{RCO-} replaces the hydrogen of the phenol's \ce{-OH} group.

\ce{CH3COCl + C6H5OH -> CH3COOC6H5 + HCl}

phenyl ethanoate

\ce{C6H5COCl + C6H5OH -> C6H5COOC6H5 + HCl}

phenyl benzoate

phenyl ethanoate

phenyl benzoate

The reaction is slower than with an alcohol because the lone pair on the phenol oxygen is partly delocalised into the ring. It goes faster if the phenol is first converted to sodium phenoxide with \ce{NaOH(aq)}. Phenyl esters cannot be made easily from a carboxylic acid and phenol, so acyl chlorides are used.

viii3 marks

Methylbenzene can be oxidized to benzoic acid using hot, alkaline \ce{KMnO4} and then dilute acid.

Solution

a. Describe the role of potassium permanganate in the oxidation of methylbenzene.

\ce{KMnO4} is the oxidising agent. In hot alkaline solution it supplies oxygen ([\ce{O}]) that attacks the benzylic carbon and oxidises the whole methyl side chain, step by step, to a carboxylate group (\ce{-COO-}). The benzene ring itself is not attacked. The manganate(VII) ion is reduced to brown manganese(IV) oxide, so the purple colour fades and a brown precipitate forms.

\ce{C6H5CH3 + 2KMnO4 -> C6H5COOK + 2MnO2 v + KOH + H2O}

b. Explain why the reaction mixture needs to be acidified after oxidation.

In alkaline solution the product is not free benzoic acid but its salt, potassium benzoate, which is soluble in water. Adding dilute acid (a stronger acid than benzoic acid) supplies \ce{H+} ions that convert the benzoate ion into benzoic acid. Benzoic acid is only slightly soluble in cold water, so it separates as white crystals that can be filtered off.

\begin{aligned}\ce{C6H5COO- + H+ &-> C6H5COOH v}\\\ce{C6H5COOK + HCl &-> C6H5COOH + KCl}\end{aligned}

ix3 marks

Carboxylic acids can be converted to acyl chlorides using \ce{PCl3}, \ce{PCl5} or \ce{SOCl2}.

Solution

a. Write the balanced chemical equation for the reaction of acetic acid with \ce{PCl5}.

\ce{CH3COOH + PCl5 -> CH3COCl + POCl3 + HCl ^}

The reaction happens in the cold and gives steamy fumes of hydrogen chloride. Acetyl chloride (b.p. 51^\circ\mathrm{C}) is separated from phosphorus oxychloride, \ce{POCl3} (b.p. 107^\circ\mathrm{C}), by fractional distillation.

b. Compare the by-products formed when using \ce{PCl3} and \ce{SOCl2} for this conversion.

\begin{aligned}\ce{3CH3COOH + PCl3 &-> 3CH3COCl + H3PO3}\\\ce{CH3COOH + SOCl2 &-> CH3COCl + SO2 ^ + HCl ^}\end{aligned}

With \ce{PCl3} the by-product is phosphorous acid, \ce{H3PO3}, which is not volatile; it stays in the flask and the acyl chloride must be separated by fractional distillation. With \ce{SOCl2} both by-products, \ce{SO2} and \ce{HCl}, are gases that escape, leaving the acyl chloride almost pure. So \ce{SOCl2} gives a cleaner product with less work.

x3 marks

Methanoic acid can be oxidized by Fehling's solution, Tollen's reagent, acidified \ce{KMnO4}, or acidified \ce{K2Cr2O7}.

Solution

a. Outline the changes observed when methanoic acid is treated with Tollen's reagent.

Tollens' reagent is a colourless solution of the diamminesilver(I) ion, \ce{[Ag(NH3)2]+}. When methanoic acid is added and the tube is warmed in a water bath, a shiny silver mirror forms on the inside of the tube (or a grey-black precipitate of silver). Methanoic acid is oxidised to carbon dioxide and water, while \ce{Ag+} ions are reduced to silver metal.

\ce{HCOOH + 2OH- + 2[Ag(NH3)2]+ -> CO2 + 2H2O + 2Ag v + 4NH3}

b. Write the balanced chemical equation for the oxidation of methanoic acid with acidified \ce{KMnO4}.

\ce{5HCOOH + 2KMnO4 + 3H2SO4 -> 5CO2 ^ + 2MnSO4 + K2SO4 + 8H2O}

\ce{5HCOOH + 2MnO4- + 6H+ -> 5CO2 + 2Mn^2+ + 8H2O}

ionic equation

The purple colour disappears and bubbles of \ce{CO2} appear. Check: Mn falls from +7 to +2 (2 atoms gain 10 electrons) and C rises from +2 to +4 (5 atoms lose 10 electrons), so the equation is balanced.

xi3 marks

Carboxylic acids are generally more acidic than phenols and alcohols.

Solution

a. Explain why carboxylic acids are more acidic than alcohols.

Acid strength depends on how stable the anion is after the acid loses \ce{H+}. The carboxylate ion (for example ethanoate) is stabilised by resonance: its negative charge is shared equally between two electronegative oxygen atoms.

\ce{CH3-C(=O)-O^- <-> CH3-C(-O^-)=O}

The alkoxide ion (\ce{CH3CH2O-}) has no resonance; its charge stays on one oxygen, and the +I effect of the ethyl group concentrates it further. So ethoxide readily takes \ce{H+} back. Hence ethanoic acid (\mathrm{p}K_a 4.76) is about 10^{11} times more acidic than ethanol (\mathrm{p}K_a about 16).

b. Discuss the effect of electron-withdrawing groups on the acidity of carboxylic acids.

Electron-withdrawing groups such as \ce{-Cl}, \ce{-F} and \ce{-NO2} increase acidity. By their -I effect they pull electron density away from the \ce{COO-} group, spread its negative charge and stabilise the conjugate base; they also weaken the \ce{O-H} bond.

\mathrm{p}K_a:\quad \ce{CH3COOH}\ 4.7,\quad \ce{CH2ClCOOH}\ 2.9,\quad \ce{CHCl2COOH}\ 1.3,\quad \ce{CCl3COOH}\ 0.9

The effect grows with the number of such groups and with their electronegativity (\ce{CH2FCOOH}, 2.6, is stronger than \ce{CH2ClCOOH}, 2.9) and becomes weaker as the group moves farther from \ce{COOH}. Electron-donating alkyl groups have the opposite effect.

Long answer questionsp. 201

Describe the process of converting toluene to benzoic acid using hot alkaline \ce{KMnO4} and then dilute acid. Include the mechanism and discuss the reasons for the choice of reagent.

Solution

Overall reaction

Toluene (methylbenzene) is converted into benzoic acid by side-chain oxidation. The methyl group is oxidised to a carboxyl group by hot alkaline potassium manganate(VII); dilute acid is then added to free the acid from its salt.

\ce{C6H5CH3 + 3[O] ->[\text{hot alkaline } KMnO4][\text{then dilute } HCl] C6H5COOH + H2O}

toluene

benzoic acid

Procedure

Toluene is heated under reflux for about an hour with \ce{KMnO4} solution made alkaline with a little \ce{Na2CO3} or \ce{NaOH}. Reflux stops the volatile toluene (b.p. 111^\circ\mathrm{C}) from escaping while the slow reaction proceeds.
The purple colour of \ce{KMnO4} fades and a brown precipitate of manganese(IV) oxide, \ce{MnO2}, forms.
The hot mixture is filtered to remove \ce{MnO2}. The filtrate contains potassium benzoate.
The filtrate is cooled and dilute \ce{HCl} (or \ce{H2SO4}) is added until it is acidic. White crystals of benzoic acid separate; they are filtered, washed with cold water and recrystallised from hot water.

Preparing benzoic acid from methylbenzene (toluene).

Equations

\ce{C6H5CH3 + 2KMnO4 -> C6H5COOK + 2MnO2 v + KOH + H2O}

oxidation in alkaline solution

\ce{C6H5COOK + HCl -> C6H5COOH v + KCl}

acidification

Mechanism of the side-chain oxidation

The oxidation begins at the benzylic carbon (the carbon attached to the ring). Its \ce{C-H} bonds are weaker than normal, because the species formed when one H is removed is stabilised by delocalisation into the ring. The side chain is then oxidised in stages, each stage adding oxygen or removing hydrogen:

\ce{C6H5CH3 ->[{[O]}] C6H5CH2OH ->[{[O]}] C6H5CHO ->[{[O]}] C6H5COO-}

At the same time the manganate(VII) ion gains electrons and is reduced to \ce{MnO2}:

\ce{MnO4- + 2H2O + 3e- -> MnO2 + 4OH-}

The benzene ring is not oxidised because its delocalised \pi electrons give it extra stability. Any side chain with at least one benzylic hydrogen is cut down to \ce{-COOH}; for example ethylbenzene also gives benzoic acid.

Reasons for the choice of reagent

\ce{KMnO4} is a powerful oxidising agent that can oxidise the whole side chain in one operation. Milder oxidants (Fehling's solution, Tollens' reagent) cannot oxidise an alkyl group.
It attacks the benzylic position selectively and leaves the aromatic ring intact.
In alkaline solution the product stays dissolved as potassium benzoate, while the reduced manganese forms insoluble \ce{MnO2} that is easily filtered off. This gives a clean separation.
Dilute acid is added last because benzoic acid is a weak acid: a stronger acid converts benzoate ions into benzoic acid, which is only slightly soluble in cold water and so precipitates.

The product is identified by its melting point (122^\circ\mathrm{C}) and by effervescence with \ce{NaHCO3} solution.

Explain the relative acidities of carboxylic acids, phenols, and alcohols. Provide examples and discuss how substituents can affect acidity.

Solution

Order of acid strength

All three classes contain an \ce{O-H} group that can lose \ce{H+}, but to very different extents. Using \mathrm{p}K_a values (smaller \mathrm{p}K_a, stronger acid):

Compound	Example	\mathrm{p}K_a
carboxylic acid	ethanoic acid, \ce{CH3COOH}	4.76
phenol	phenol, \ce{C6H5OH}	10
alcohol	ethanol, \ce{CH3CH2OH}	about 16

\ce{CH3COOH} > \ce{C6H5OH} > \ce{CH3CH2OH}

Reason: stability of the conjugate base

The more stable the anion left after the proton is lost, the stronger the acid, because a stable anion has little tendency to take the proton back.

Carboxylic acids

In the carboxylate ion the negative charge is delocalised by resonance over two equivalent oxygen atoms. Oxygen is highly electronegative and holds negative charge well, so the ion is very stable.

\ce{CH3-C(=O)-O^- <-> CH3-C(-O^-)=O}

Phenols

In the phenoxide ion the negative charge is spread into the benzene ring, onto the carbon atoms at positions 2, 4 and 6. This stabilises the ion, but less than in a carboxylate, because the charge is shared with carbon, which is less electronegative than oxygen.

phenoxide

C-2

C-4

C-6

Alcohols

In the alkoxide ion (\ce{CH3CH2O-}) there is no resonance. The charge stays on one oxygen, and the alkyl group pushes more electron density onto it (+I effect). The ion is unstable, so ethanol hardly ionises.

Evidence from reactions

Reagent	Ethanoic acid	Phenol	Ethanol
\ce{Na}	\ce{H2}	\ce{H2}	\ce{H2} (slow)
\ce{NaOH(aq)}	reacts	reacts	no reaction
\ce{NaHCO3(aq)}	\ce{CO2} given off	no reaction	no reaction

Effect of substituents

Electron-withdrawing groups (\ce{-Cl}, \ce{-F}, \ce{-NO2}) pull electron density away from the \ce{O-} group, disperse the negative charge and stabilise the anion, so they increase acidity. Electron-donating groups (alkyl groups) push electron density in and decrease acidity.

Carboxylic acids: \ce{CCl3COOH} (0.9) > \ce{CHCl2COOH} (1.3) > \ce{CH2ClCOOH} (2.9) > \ce{CH3COOH} (4.7) > \ce{CH3CH2COOH} (4.9); also 4-nitrobenzoic acid (3.44) > benzoic acid (4.20).
Phenols: 4-nitrophenol (\mathrm{p}K_a 7.2) is much stronger than phenol (10), and 2,4,6-trinitrophenol (picric acid, 0.4) is as strong as many mineral acids, because the nitro groups help delocalise the negative charge. 4-methylphenol (10.3) is slightly weaker than phenol.
Alcohols: 2,2,2-trifluoroethanol, \ce{CF3CH2OH} (\mathrm{p}K_a 12.4), is much more acidic than ethanol (about 16).

The effect is larger when there are more withdrawing groups, when they are more electronegative, and when they are closer to the acidic group.

iii2026 Annual I, Set B, Q.6 OR

Compare the reactivity and hydrolysis of acyl chlorides with alkyl and aryl chlorides. Explain the factors that affect their reactivity.

Solution

Order of reactivity

\text{acyl chloride }(\ce{CH3COCl}) > \text{alkyl chloride }(\ce{CH3CH2Cl}) > \text{aryl chloride }(\ce{C6H5Cl})

Hydrolysis and conditions

\ce{CH3COCl + H2O -> CH3COOH + HCl}

acyl chloride: cold water, fast, steamy fumes of HCl

\ce{CH3CH2Cl + NaOH ->[\text{heat, reflux}] CH3CH2OH + NaCl}

alkyl chloride: needs the stronger nucleophile OH- and heat

\ce{C6H5Cl + NaOH ->[350^\circ C][200\ atm] C6H5OH + NaCl}

aryl chloride: no reaction under ordinary conditions

Factors that affect reactivity

**1. Size of the \delta+ charge on carbon.** In an acyl chloride the carbonyl carbon is bonded to two electronegative atoms (O and Cl), so it is strongly electron-deficient and attracts even a weak nucleophile like water. The carbon of an alkyl chloride is bonded to only one electronegative atom and is less \delta+.

2. Mechanism available (hybridisation). The acyl carbon is sp^2 and planar, and its \ce{C=O} \pi bond can take the nucleophile's electrons. The nucleophile adds first and the \ce{C-Cl} bond breaks later (addition-elimination), a low-energy path. The alkyl carbon is sp^3 with no \pi bond, so the \ce{C-Cl} bond must break in the substitution itself (S_N1 or S_N2), which needs heat and \ce{OH-}.

3. Strength of the C-Cl bond (resonance). In chlorobenzene a lone pair of chlorine is delocalised into the ring. The \ce{C-Cl} bond gets partial double-bond character; it is shorter (about 169\,\mathrm{pm} against 177\,\mathrm{pm}) and stronger, so it is very hard to break.

chlorobenzene

charge at C-2

charge at C-4

charge at C-6

4. Repulsion and blocking. The \pi electron cloud of the ring repels approaching nucleophiles, the ring blocks back-side (S_N2) attack, and the phenyl cation needed for S_N1 is very unstable.

5. Strength of the nucleophile. Water is enough for acyl chlorides; alkyl chlorides need \ce{OH-}; aryl chlorides resist even \ce{OH-} unless the temperature and pressure are very high.

Experimental comparison

With aqueous silver nitrate, ethanoyl chloride gives a white precipitate of \ce{AgCl} immediately in the cold, chloroethane gives it slowly on warming, and chlorobenzene gives none. This confirms the order acyl > alkyl > aryl.

Esters can be formed from alcohols and acyl chlorides, exemplified by the formation of phenyl benzoate. Write the balanced chemical equation for the formation of phenyl benzoate from benzoyl chloride and phenol.

Solution

Balanced equation

\ce{C6H5COCl + C6H5OH -> C6H5COOC6H5 + HCl}

benzoyl chloride

phenol

phenyl benzoate

The acyl group \ce{C6H5CO-} of benzoyl chloride replaces the hydrogen of the phenol \ce{-OH} group. The equation is balanced: \ce{C13}, \ce{H11}, \ce{O2} and \ce{Cl} on each side.

Conditions and practical method

The reaction occurs at room temperature. Phenol is a weaker nucleophile than an alcohol (its oxygen lone pair is partly delocalised into the ring), and benzoyl chloride is less reactive than ethanoyl chloride. So in practice phenol is first dissolved in sodium hydroxide solution to form the phenoxide ion, a much better nucleophile. Benzoyl chloride is added and the mixture is shaken; solid phenyl benzoate separates (m.p. about 70^\circ\mathrm{C}). The alkali also neutralises the \ce{HCl} formed.

\ce{C6H5COCl + C6H5OH + NaOH -> C6H5COOC6H5 + NaCl + H2O}

Mechanism (addition-elimination)

The oxygen lone pair of phenol (or phenoxide) attacks the \delta+ carbonyl carbon of benzoyl chloride, and the \ce{C=O} \pi electrons move onto oxygen, forming a tetrahedral intermediate.
The \ce{C=O} bond reforms and \ce{Cl-} is expelled.
The extra proton on the phenol oxygen is lost (to \ce{Cl-} or \ce{OH-}), giving phenyl benzoate.

Why an acyl chloride is used

Phenyl esters are hard to make from a carboxylic acid and phenol, because the reaction is slow and reversible. The acyl chloride route is fast, works at room temperature and goes to completion, since \ce{HCl} escapes or is neutralised. The same method gives ethyl ethanoate from an alcohol:

\ce{CH3COCl + CH3CH2OH -> CH3COOCH2CH3 + HCl}

Acyl chlorides hydrolyze rapidly in water, while aryl chlorides are resistant. Analyze the electronic structures of these compounds to explain this contrast in reactivity.

Solution

The observation

\ce{CH3COCl + H2O -> CH3COOH + HCl}\qquad \ce{C6H5Cl + H2O ->} \text{no reaction}

Ethanoyl chloride reacts with cold water at once, giving steamy fumes of \ce{HCl}; chlorobenzene can be shaken with water, or even boiled with \ce{NaOH(aq)}, without change.

Electronic structure of an acyl chloride

The carbonyl carbon is sp^2 hybridised and planar. It forms \sigma bonds to R, O and Cl and a \pi bond to O.
Both oxygen and chlorine are more electronegative than carbon and pull electron density away from it (inductive effect), and the \ce{C=O} \pi bond is strongly polarised towards oxygen. So the carbon carries a large \delta+ charge.
The lone pairs of chlorine are in large 3p orbitals that overlap poorly with the 2p orbitals of the carbonyl group, so chlorine gives back very little electron density by resonance. The \delta+ charge is not reduced.
The \pi bond can accept the electron pair of a nucleophile, and \ce{Cl-} is a good leaving group.

Nucleophilic addition-elimination: water adds to the \ce{C=O} carbon, then \ce{Cl-} and \ce{H+} are lost. Curly arrows show the movement of electron pairs.

So water adds easily to form a tetrahedral intermediate, and \ce{Cl-} is then eliminated. The whole process has a low activation energy and is fast at room temperature.

Electronic structure of an aryl chloride

The carbon bonded to chlorine is an sp^2 ring carbon. A lone pair on chlorine sits in a p orbital parallel to the ring's p orbitals and is delocalised into the \pi system.
Resonance gives the \ce{C-Cl} bond partial double-bond character. It is shorter (about 169\,\mathrm{pm} compared with 177\,\mathrm{pm} in chloroalkanes) and stronger, and the carbon is hardly \delta+.
The ring's \pi electron cloud repels an approaching nucleophile, back-side attack is blocked by the ring, and the phenyl cation needed for an S_N1 route is very unstable.
There is no \pi bond on the carbon that could accept the nucleophile first, so any substitution would require breaking the strong \ce{C-Cl} bond directly.

chlorobenzene

charge at C-2

charge at C-4

charge at C-6

Conclusion

Feature	Acyl chloride	Aryl chloride
\delta+ on the carbon	large (O and Cl both withdraw)	small (Cl lone pair delocalised)
C-Cl bond	normal single bond	partial double bond, stronger
Route for the nucleophile	adds to C=O first (addition-elimination)	no easy route
Hydrolysis	fast in cold water	does not occur normally

The electron-deficient, planar carbonyl carbon makes acyl chlorides very reactive, while delocalisation of the chlorine lone pair into the ring makes aryl chlorides resistant to hydrolysis.

Given three compounds: ethanoyl chloride, chloromethane, and chlorobenzene, predict which one hydrolyzes the fastest and explain your reasoning with reference to bond polarity and resonance.

Solution

Prediction

Ethanoyl chloride hydrolyses fastest. The order of ease of hydrolysis is:

\ce{CH3COCl} > \ce{CH3Cl} > \ce{C6H5Cl}

Bond polarity

\ce{CH3-\overset{\delta+}{C}(=O^{\delta-})-Cl^{\delta-}}\qquad \ce{\overset{\delta+}{C}H3-Cl^{\delta-}}\qquad \ce{C6H5-Cl}\ (\text{weakly polar})

In ethanoyl chloride the carbonyl carbon is bonded to two electronegative atoms, O and Cl. Both the \ce{C=O} and \ce{C-Cl} bonds are polarised away from this carbon, so it is very strongly \delta+ and is attacked even by water, a weak nucleophile.

In chloromethane the carbon is bonded to only one electronegative atom, so it is less \delta+. The \ce{C-Cl} bond is polar, but it must break during the substitution. Water reacts only very slowly; hydroxide ions and heating are needed (S_N2).

\ce{CH3Cl + OH- ->[\text{heat}] CH3OH + Cl-}

Resonance

In chlorobenzene a lone pair on chlorine is delocalised into the benzene ring:

chlorobenzene

charge at C-2

charge at C-4

charge at C-6

This gives the \ce{C-Cl} bond partial double-bond character, making it shorter and stronger, and it reduces the polarity of the bond. The ring's \pi cloud also repels nucleophiles. So chlorobenzene is not hydrolysed under normal conditions.

In ethanoyl chloride, resonance donation from chlorine into \ce{C=O} is very weak, because the large 3p orbital of chlorine overlaps poorly with the 2p orbitals of carbon. The carbonyl carbon therefore stays strongly \delta+. The \ce{C=O} \pi bond can accept the nucleophile's electrons, so a tetrahedral intermediate forms before \ce{Cl-} leaves (addition-elimination). This low-energy path explains the very fast reaction:

\ce{CH3COCl + H2O -> CH3COOH + HCl}

room temperature, vigorous

Conclusion

Greatest bond polarity plus an easy addition-elimination path makes ethanoyl chloride the fastest. Chloromethane is slower because its carbon is less \delta+ and must undergo direct substitution. Chlorobenzene is slowest (practically unreactive) because resonance strengthens its \ce{C-Cl} bond.

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 (viii)1+2 = 3 marks

Compare the acidities of \ce{CH3COOH} and \ce{CH2ClCOOH} and explain the difference.

Solution

Comparison (1 mark)

Chloroacetic acid, \ce{CH2ClCOOH} (\mathrm{p}K_a=2.9), is a stronger acid than acetic acid, \ce{CH3COOH} (\mathrm{p}K_a=4.7). It is roughly 100 times stronger (the K_a ratio is 10^{1.8}, about 60 to 80).

Explanation (2 marks)

\begin{aligned}\ce{CH3COOH &<=> CH3COO- + H+}\\\ce{ClCH2COOH &<=> ClCH2COO- + H+}\end{aligned}

Chlorine is more electronegative than carbon, so the \ce{C-Cl} bond pulls electron density towards chlorine along the chain (-I inductive effect). In the chloroacetate ion this pulls part of the negative charge away from the \ce{COO-} group, spreading it over more atoms. The anion is therefore more stable and has less tendency to recombine with \ce{H+}, so the acid ionises more. The pull also makes the \ce{O-H} bond more polar and easier to break.

Acetic acid has a methyl group instead, which pushes electrons towards \ce{COO-} (+I effect). This concentrates the negative charge, makes the acetate ion less stable, and so acetic acid is the weaker acid.

acetate ion charge concentrated

chloroacetate ion charge spread by Cl

2026 Annual I, Set B, Q.6 OR2+2+2 = 6 marks

Justify the given order of ease of hydrolysis: Acyl chloride > Alkyl chloride > Aryl chloride.

Solution

The ease of hydrolysis depends on how electron-deficient the carbon bonded to chlorine is, on the mechanism available, and on the strength of the \ce{C-Cl} bond.

Acyl chloride: hydrolysed most easily (2 marks)

In \ce{CH3COCl} the carbonyl carbon is attached to two electronegative atoms, O and Cl, which both pull electrons away. The carbon is highly \delta+. It is also planar (sp^2) with a \ce{C=O} \pi bond that can take the nucleophile's electron pair. Even water attacks it at room temperature by nucleophilic addition-elimination, giving steamy fumes of \ce{HCl}.

\ce{CH3-\overset{\delta+}{C}(=O)-\overset{\delta-}{Cl} + H2O -> CH3COOH + HCl}

Alkyl chloride: slower (2 marks)

In \ce{CH3CH2Cl} the carbon is attached to only one electronegative atom, so it is less electron-deficient. It is sp^3 with no \pi bond, so the \ce{C-Cl} bond must break during the substitution. Water is too weak a nucleophile; the reaction needs \ce{OH-} and heating under reflux.

\ce{CH3CH2Cl + NaOH ->[\text{heat}] CH3CH2OH + NaCl}

Aryl chloride: not hydrolysed (2 marks)

In chlorobenzene a lone pair of chlorine is delocalised into the benzene ring by resonance. The \ce{C-Cl} bond gains partial double-bond character, so it is shorter and stronger and hard to break even by a strong nucleophile. The ring's \pi cloud also repels nucleophiles. So hydrolysis does not occur under ordinary conditions.

chlorobenzene

charge at C-2

charge at C-4

charge at C-6

Hence: acyl chloride > alkyl chloride > aryl chloride.

2026 Annual I, Set D, MCQ 111 mark

Which of the following is more acidic?

A\ce{CH3CH2COOH}
B\ce{Cl2CHCOOH}Answer
C\ce{ClCH2COOH}
D\ce{CH3COOH}

Why

Dichloroacetic acid has two chlorine atoms on the carbon next to \ce{COOH}. Their combined -I effect stabilises the carboxylate ion the most, so it is the strongest acid here: \mathrm{p}K_a values are \ce{Cl2CHCOOH} 1.3, \ce{ClCH2COOH} 2.9, \ce{CH3COOH} 4.7, \ce{CH3CH2COOH} 4.9.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 14, old syllabus1 mark

In nucleophilic addition reactions, the carbonyl group behaves as:

ANucleophile
BRadical
CElectrophileAnswer
DLeaving group

Why

The carbonyl carbon is \delta+ and accepts the electron pair of the attacking nucleophile, so the carbonyl group acts as an electrophile. It is not a nucleophile, a radical or a leaving group.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 15, old syllabus1 mark

Carboxylic acids react with metals like Na to produce:

AAlcohols
BEsters
CAldehydes
DHydrogen gasAnswer

Why

Like other acids, carboxylic acids react with reactive metals to give a salt and hydrogen gas: \ce{2CH3COOH + 2Na -> 2CH3COONa + H2}.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (viii) OR, old syllabus1+2 = 3 marks

Compare the reactivity of formaldehyde and acetone towards nucleophilic addition reactions. Give the reason for the difference.

Solution

Comparison (1 mark)

**Formaldehyde (\ce{HCHO}) is more reactive** towards nucleophilic addition than acetone (\ce{CH3COCH3}).

formaldehyde

acetone

Reasons (2 marks)

Inductive effect: the two methyl groups of acetone push electrons towards the carbonyl carbon (+I effect) and reduce its \delta+ charge, so it attracts nucleophiles less. Formaldehyde has only hydrogen atoms, so its carbonyl carbon keeps a larger \delta+ charge.

Steric effect: the bulky methyl groups crowd the carbonyl carbon of acetone and hinder the approach of the nucleophile; they also crowd the tetrahedral product. The small H atoms of formaldehyde give almost no hindrance. For example, formaldehyde adds water almost completely in solution, while acetone hardly does.

Note. The paper prints 'neucleophilic'; corrected to 'nucleophilic'.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xi), old syllabus1x3 = 3 marks

Write mechanism of the reaction when methanol reacts with acetic acid in the presence of an acid catalyst.

Solution

\ce{CH3COOH + CH3OH <=>[H+] CH3COOCH3 + H2O}

esterification: methyl acetate (methyl ethanoate) is formed

Step 1: protonation (1 mark). \ce{H+} from the acid catalyst (conc. \ce{H2SO4}) bonds to the carbonyl oxygen, making the carbonyl carbon strongly positive.
\ce{CH3C(=O)OH + H+ <=> CH3C(OH)=OH+}
Step 2: nucleophilic attack and proton transfer (1 mark). A lone pair on the methanol oxygen attacks the carbonyl carbon, forming a tetrahedral intermediate. A proton then moves to one \ce{-OH} group, turning it into \ce{-OH2+}.
\ce{CH3C(OH)=OH+ + CH3OH <=> CH3C(OH)(OCH3)OH2+}
**Step 3: loss of water and \ce{H+} (1 mark).** The \ce{C=O} bond reforms as water leaves, and the protonated ester loses \ce{H+}, regenerating the catalyst.
\ce{CH3C(OH)(OCH3)OH2+ -> CH3C(OCH3)=OH+ + H2O -> CH3COOCH3 + H+}

The \ce{-OH} removed as water comes from the acid and the \ce{H} from methanol. The reaction is reversible; concentrated \ce{H2SO4} also absorbs water and increases the yield of ester.

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

A student observes that Benzoic acid is less acidic than 4-NitroBenzoic acid. Explain this observation.

Solution

4-Nitrobenzoic acid (\mathrm{p}K_a=3.44) is about 6 times stronger than benzoic acid (\mathrm{p}K_a=4.20). Acid strength depends on the stability of the anion (conjugate base) formed after losing \ce{H+}.

\ce{O2N-C6H4-COOH <=> O2N-C6H4-COO- + H+}

benzoic acid

4-nitrobenzoate ion

The nitro group is strongly electron-withdrawing. It pulls electrons through the \sigma bonds (inductive effect, -I) and also out of the \pi system of the ring by resonance (-R effect), because its N atom carries a positive charge. This draws electron density away from the \ce{COO-} group and spreads its negative charge, so the 4-nitrobenzoate ion is more stable than the benzoate ion. A more stable conjugate base means a stronger acid. Benzoic acid has only H at position 4 and gets no such stabilisation.

2026 Annual I, Old syllabus (Curriculum 2006), Q.6 OR, old syllabus6 marks

Write mechanism of the reaction for the aldol condensation between two molecules of acetaldehyde.

Solution

Reaction and conditions

Two molecules of acetaldehyde (ethanal) join in the presence of dilute NaOH in the cold to give 3-hydroxybutanal (aldol). On heating, the aldol loses water to give but-2-enal (crotonaldehyde). Acetaldehyde can do this because it has \alpha-hydrogens (on the \ce{CH3} next to \ce{C=O}).

\ce{2CH3CHO ->[\text{dil. } NaOH] CH3CH(OH)CH2CHO ->[\text{heat}] CH3CH=CHCHO + H2O}

Step 1: formation of the carbanion

The hydroxide ion removes an acidic \alpha-hydrogen from one acetaldehyde molecule. The carbanion (enolate ion) formed is stabilised by resonance with the carbonyl group.

\ce{HO- + H-CH2-CHO <=> ^{-}CH2-CHO + H2O}

\ce{^{-}CH2-CH=O <-> CH2=CH-O^{-}}

Step 2: nucleophilic attack

The carbanion acts as a nucleophile and attacks the \delta+ carbonyl carbon of a second acetaldehyde molecule. The \ce{C=O} \pi electrons move onto oxygen, giving an alkoxide ion and a new \ce{C-C} bond.

\ce{CH3-\overset{\delta+}{C}H=O + ^{-}CH2-CHO -> CH3-CH(O^-)-CH2-CHO}

Step 3: protonation (aldol formed)

The alkoxide ion takes a proton from water, giving 3-hydroxybutanal and regenerating the hydroxide ion (catalyst).

\ce{CH3-CH(O^-)-CH2-CHO + H2O -> CH3-CH(OH)-CH2-CHO + OH-}

Step 4: dehydration (condensation)

On heating, \ce{OH-} removes an \alpha-hydrogen from the aldol and the \ce{-OH} group leaves as water, forming a \ce{C=C} bond conjugated with \ce{C=O}.

\ce{CH3-CH(OH)-CH2-CHO ->[\text{heat}] CH3-CH=CH-CHO + H2O}

acetaldehyde

carbanion

alkoxide ion

3-hydroxybutanal

but-2-enal

The overall process is a condensation because two molecules combine and a small molecule (water) is lost.

Exam focus

Compare and explain the acidities of \ce{CH3COOH}, \ce{CH2ClCOOH}, \ce{CHCl2COOH} and \ce{CCl3COOH}. (short)
Asked in 2026 (Set B Q.2 (viii), Set D MCQ 11); the inductive effect and anion stability earn the marks.
Go to answer
Justify the order of ease of hydrolysis: acyl chloride > alkyl chloride > aryl chloride. (long)
A 6-mark 2026 question (Set B Q.6 OR) and three book long questions ask the same idea.
Go to answer
Write the nucleophilic addition-elimination mechanism for the hydrolysis of an acyl chloride. (short)
SLO mechanism; curly arrows and the tetrahedral intermediate are marked point by point.
Go to answer
Reactions of acyl chlorides with water, alcohols, phenol, ammonia and amines (equations and names). (short)
Every SLO reaction is a likely equation question; phenyl benzoate and ethyl ethanoate are the named examples.
Go to answer
Preparation of acyl chlorides with \ce{SOCl2}, \ce{PCl3} and \ce{PCl5}, and why \ce{SOCl2} is preferred. (short)
Balanced equations and by-products are asked directly in the book's short questions.
Go to answer
Oxidation of methanoic acid (Tollens', Fehling's, KMnO4) and ethanedioic acid (warm acidified KMnO4) with balanced equations. (short)
Balancing the KMnO4 equations is a common trap; the book itself misprints the ethanedioic acid equation.
Go to answer
Convert toluene to benzoic acid with hot alkaline KMnO4 and dilute acid; explain the role of each reagent. (long)
First SLO of the unit and a book long question.
Go to answer
Explain the relative acidities of carboxylic acids, phenols and alcohols with resonance structures. (long)
Asked in both new 2026 sets (Set B Q.6, Set D Q.5) with resonance structures of the conjugate bases.
Go to answer
Why is 4-nitrobenzoic acid stronger than benzoic acid? (short)
Old-syllabus 2026 Q.2 (xiii); tests the same substituent idea on an aromatic acid.
Go to answer
Mechanism of acid-catalysed esterification (methanol + acetic acid). (short)
Old-syllabus 2026 Q.2 (xi), 3 marks for three mechanism stages.
Go to answer
Mechanism of the aldol condensation of acetaldehyde. (long)
Old-syllabus 2026 Q.6 OR, 6 marks.
Go to answer
Why is formaldehyde more reactive than acetone towards nucleophilic addition? (short)
Old-syllabus 2026 Q.2 (viii) OR; inductive and steric reasons.
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