Carboxylic Acid and its Derivatives

A carboxylic acid contains the carboxyl group, —COOH: a carbonyl group and a hydroxyl group on the same carbon. Neither group behaves as it would alone. The —OH loses its proton far more readily than an alcohol does, and the >C=O refuses the addition reactions of Unit 13. Both facts come from one cause — delocalisation in the carboxylate ion. This unit then follows the acid into its four derivatives: acid halides, anhydrides, esters and amides.

14.1 Introduction, nomenclature and isomerism

A monocarboxylic acid has one —COOH group; the saturated open-chain series is CₙH₂ₙO₂, general formula R—COOH. Acids with two carboxyl groups are dicarboxylic. When —COOH is attached directly to a benzene ring the acid is aromatic.

IUPAC rules. Replace the final -e of the parent alkane by -oic acid. The carboxyl carbon is always C-1, so it never takes a locant. Two carboxyl groups give -dioic acid (and the -e is kept). For the aromatic series the parent name is benzoic acid.

Formula	IUPAC name	Common name
HCOOH	methanoic acid	formic acid
CH₃COOH	ethanoic acid	acetic acid
CH₃CH₂COOH	propanoic acid	propionic acid
CH₃(CH₂)₂COOH	butanoic acid	butyric acid
(CH₃)₂CHCOOH	2-methylpropanoic acid	isobutyric acid
ClCH₂COOH	2-chloroethanoic acid	chloroacetic acid
HOOC—COOH	ethanedioic acid	oxalic acid
HOOC—CH₂—COOH	propanedioic acid	malonic acid
C₆H₅COOH	benzoic acid	benzoic acid
C₆H₅CH₂COOH	phenylethanoic acid	phenylacetic acid
o-HOC₆H₄COOH	2-hydroxybenzoic acid	salicylic acid

Isomerism. Carboxylic acids show:

1. Chain isomerism — butanoic acid and 2-methylpropanoic acid, both C₄H₈O₂.
2. Position isomerism of a substituent — 2-chlorobutanoic, 3-chlorobutanoic and 4-chlorobutanoic acid. In aromatic acids the ortho, meta and para isomers, e.g. the three methylbenzoic (toluic) acids.
3. Functional isomerism with esters and with hydroxy-carbonyl compounds. C₃H₆O₂ is propanoic acid, methyl ethanoate or ethyl methanoate.
4. Metamerism among the ester isomers — methyl propanoate, ethyl ethanoate and propyl methanoate are all C₄H₈O₂.

C₄H₈O₂ therefore has two acids and four esters (methyl propanoate, ethyl ethanoate, propyl methanoate, isopropyl methanoate).

14.2 Preparation of monocarboxylic acids

From aldehydes — oxidation

Any oxidising agent that will attack an aldehyde works; acidified dichromate is the standard laboratory reagent:

3CH₃CHO + K₂Cr₂O₇ + 4H₂SO₄ → 3CH₃COOH + K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O

Primary alcohols give the same acid, passing through the aldehyde: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O.

From nitriles — hydrolysis

A nitrile has the same number of carbons as the acid it gives. Hydrolysis is done with dilute mineral acid or with alkali followed by acidification:

CH₃CN + 2H₂O + HCl → CH₃COOH + NH₄Cl

CH₃CN + H₂O + NaOH → CH₃COONa + NH₃↑

Because the nitrile itself is made from a haloalkane (R—X + KCN → R—CN + KX), this is the standard way of adding one carbon to a chain.

From dicarboxylic acids — decarboxylation

Dicarboxylic acids lose CO₂ on gentle heating:

HOOC—COOH --423 K--> HCOOH + CO₂↑

HOOC—CH₂—COOH --413 K--> CH₃COOH + CO₂↑

From sodium alkoxide — carbonylation

Sodium alkoxide absorbs carbon monoxide under pressure to give the sodium salt of the next higher acid, which is then acidified:

CH₃ONa + CO --425 K, 3 atm--> CH₃COONa

CH₃COONa + HCl → CH₃COOH + NaCl

From trihaloalkanes — hydrolysis

Three halogens on the same carbon hydrolyse to an unstable orthoacid, which loses two molecules of water:

CHCl₃ + 4NaOH → HCOONa + 3NaCl + 2H₂O

CH₃CCl₃ + 4NaOH → CH₃COONa + 3NaCl + 2H₂O

14.3 Preparation of benzoic acid from an alkylbenzene

Hot alkaline potassium permanganate attacks the benzylic carbon and degrades the whole side chain to a single —COOH group, however long the side chain is — provided it carries at least one hydrogen on the carbon joined to the ring:

C₆H₅CH₃ + 2KMnO₄ → C₆H₅COOK + 2MnO₂↓ + KOH + H₂O

C₆H₅COOK + HCl → C₆H₅COOH↓ + KCl

Ethylbenzene, propylbenzene and isopropylbenzene all give benzoic acid too. tert-Butylbenzene, C₆H₅C(CH₃)₃, has no benzylic hydrogen and is not oxidised — a favourite examiner's trap.

14.4 Physical properties

The first three acids are pungent-smelling liquids miscible with water; butanoic acid smells of rancid butter; from about C₁₀ they are waxy, odourless solids. Pure ethanoic acid freezes at 16.6 °C to an ice-like solid, so it is called glacial acetic acid — in a Kathmandu winter a bottle on the shelf may set solid.

Carboxylic acids boil higher than alcohols of similar molar mass because they form two hydrogen bonds between each pair of molecules, giving a cyclic dimer:

Compound	M (g mol⁻¹)	Boiling point
Propane	44	−42 °C
Ethanal	44	21 °C
Propan-1-ol	60	97 °C
Ethanoic acid	60	118 °C

Solubility in water falls steeply as the chain grows: methanoic to butanoic acid are miscible, pentanoic is partly soluble, and the higher ones are insoluble but dissolve in ethanol and ether. Aromatic acids are only slightly soluble in cold water — benzoic acid crystallises out on cooling, which is how it is purified.

14.5 Chemical properties

Acidic reactions

A carboxylic acid is a weak acid: K_a of ethanoic acid is 1.75\times10^{-5}, so only about 1.3 % of the molecules are ionised in a 0.1 M solution. It is nevertheless a far stronger acid than an alcohol or a phenol, and it neutralises all four of the following:

CH₃COOH + NaOH → CH₃COONa + H₂O (with an alkali)

2CH₃COOH + CaO → (CH₃COO)₂Ca + H₂O (with a metal oxide)

2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂↑ (with a carbonate)

CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑ (with a bicarbonate)

2CH₃COOH + 2Na → 2CH₃COONa + H₂↑ (with an active metal)

Substitution of the —OH group

With PCl₃, PCl₅ or SOCl₂ the hydroxyl group is replaced by chlorine, giving an acid chloride:

3CH₃COOH + PCl₃ → 3CH₃COCl + H₃PO₃

CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl

CH₃COOH + SOCl₂ → CH₃COCl + SO₂↑ + HCl↑

Thionyl chloride is preferred because both by-products are gases and escape, leaving pure acid chloride.

With an alcohol (esterification), conc. H₂SO₄ as catalyst and dehydrating agent:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Reduction with LiAlH₄

Lithium aluminium hydride in dry ether reduces the carboxyl group all the way to a primary alcohol:

CH₃COOH + 4[H] --LiAlH₄, then H₂O--> CH₃CH₂OH + H₂O

Dehydration

Heating with P₂O₅ removes water from two molecules to give the anhydride:

2CH₃COOH --P₂O₅, Δ--> (CH₃CO)₂O + H₂O

Decarboxylation

The dry sodium salt heated with soda-lime loses CO₂ as carbonate:

CH₃COONa + NaOH --CaO, Δ--> CH₄↑ + Na₂CO₃

This shortens the chain by one carbon and is the standard laboratory preparation of methane.

14.6 Hell–Volhard–Zelinsky (HVZ) reaction

An aliphatic acid with at least one α-hydrogen is halogenated at the α carbon by chlorine or bromine in the presence of a small amount of red phosphorus:

CH₃COOH + Cl₂ --red P--> ClCH₂COOH + HCl

ClCH₂COOH + Cl₂ --red P--> Cl₂CHCOOH + HCl

Cl₂CHCOOH + Cl₂ --red P--> Cl₃CCOOH + HCl

The phosphorus is not a catalyst in the ordinary sense: it reacts with the halogen to form PCl₃, which converts a little of the acid into its acid chloride. The acid chloride enolises far more readily than the acid does, and it is the enol that the halogen attacks. The α-halo acid produced then exchanges with more of the starting acid.

The α-halo acids are valuable intermediates: with aqueous KOH they give α-hydroxy acids, with alcoholic KOH unsaturated acids, and with NH₃ they give α-amino acids — this is a standard preparation of glycine from ethanoic acid.

14.7 Electrophilic substitution of benzoic acid

The —COOH group is strongly electron-withdrawing (−I and −R). It therefore deactivates the ring towards electrophilic attack and directs the incoming group to the meta position, because the ortho and para positions carry a partial positive charge in the resonance structures.

Reaction	Reagent and conditions	Product
Bromination	Br₂ / FeBr₃, warm	3-bromobenzoic acid + HBr
Nitration	conc. HNO₃ + conc. H₂SO₄, 373 K	3-nitrobenzoic acid + H₂O
Sulphonation	fuming H₂SO₄ (oleum), Δ	3-sulphobenzoic acid + H₂O

C₆H₅COOH + Br₂ --FeBr₃--> m-BrC₆H₄COOH + HBr

C₆H₅COOH + HNO₃ --conc. H₂SO₄, 373 K--> m-O₂NC₆H₄COOH + H₂O

C₆H₅COOH + H₂SO₄ (fuming) → m-HO₃SC₆H₄COOH + H₂O

All three need harsher conditions than benzene itself — higher temperature, stronger catalyst — which is exactly what "deactivated ring" means. Benzoic acid does not undergo Friedel–Crafts alkylation or acylation at all, because the AlCl₃ complexes with the carboxyl group and the ring is too deactivated.

14.8 Effect of substituents on acid strength; methanoic acid

Why a carboxylic acid is acidic at all

Removing the proton gives the carboxylate ion, RCOO⁻, in which the negative charge is shared equally between the two oxygen atoms. Both carbon–oxygen bonds in ethanoate are the same length, 127 pm — between a C=O (122 pm) and a C—O (143 pm). This delocalisation lowers the energy of the anion a great deal, so the ionisation equilibrium lies much further to the right than it does for an alcohol, whose anion RO⁻ has the charge stuck on one atom.

The same argument, applied three times, gives the order that NEB asks for every year:

RCOOH > H₂CO₃ > C₆H₅OH > H₂O > ROH

Ethanoic acid (pK_a 4.76) beats phenol (pK_a 10.0) because in phenoxide the charge is spread over carbon atoms, which are less electronegative than oxygen, and only two of the contributing structures carry it. Ethanol (pK_a 16) has no delocalisation at all.

Effect of substituents

An electron-withdrawing group (−I) pulls charge away from the carboxylate ion, spreads it further and stabilises it, so the acid is stronger. An electron-releasing group (+I) pushes charge onto the ion, destabilises it, and makes the acid weaker. The effect dies away rapidly with distance — it is roughly one-third as strong at each successive carbon.

Read off the trends:

• Number of substituents: CH₃COOH 4.76 → ClCH₂COOH 2.87 → Cl₂CHCOOH 1.29 → Cl₃CCOOH 0.66. Trichloroethanoic acid is about 12 000 times stronger than ethanoic acid.
• Electronegativity of the halogen: F > Cl > Br > I (2.59 < 2.87 < 2.90 < 3.18), exactly the order of −I strength.
• Distance: 2-chlorobutanoic acid (pK_a 2.86) is far stronger than 4-chlorobutanoic acid (pK_a 4.52), which is barely stronger than butanoic acid itself (4.82).
• Alkyl groups weaken: HCOOH 3.75 < CH₃COOH 4.76 < CH₃CH₂COOH 4.87 < (CH₃)₃CCOOH 5.05.

For aromatic acids, benzoic acid (pK_a 4.20) is stronger than ethanoic acid because the sp² ring carbon is more electronegative than an sp³ alkyl carbon. Electron-withdrawing ring substituents (—NO₂, —Cl) strengthen it further, and electron-releasing ones (—CH₃, —OCH₃) weaken it.

Abnormal behaviour of methanoic acid

Methanoic acid, H—COOH, is the only carboxylic acid whose carboxyl group is attached to hydrogen and not to carbon. It therefore contains a complete aldehydic (—CHO) grouping, and behaves partly as an aldehyde:

1. It is the strongest of the simple aliphatic acids (pK_a 3.75) because there is no alkyl group to push electrons in.
2. It is a reducing agent. It gives a silver mirror with Tollens' reagent and a red precipitate with Fehling's solution, which no other carboxylic acid does:

HCOOH + 2[Ag(NH₃)₂]⁺ + 2OH⁻ → 2Ag↓ + CO₂↑ + 4NH₃ + 2H₂O

HCOOH + 2CuO → Cu₂O↓ + CO₂↑ + H₂O

3. It decolourises acidified KMnO₄ and reduces mercury(II) chloride to mercury(I) chloride.
4. It cannot form an anhydride. With concentrated sulphuric acid it simply dehydrates to carbon monoxide: HCOOH --conc. H₂SO₄, Δ--> CO↑ + H₂O. This is the laboratory preparation of CO.
5. It has no α-hydrogen, so it gives no HVZ reaction.
6. Heating its sodium salt gives sodium oxalate and hydrogen, not a ketone.

14.9 The four derivatives and their preparation from the acid

Replacing the —OH of —COOH by —Cl, —OCOR, —OR′ or —NH₂ gives the four derivatives. All of them can be hydrolysed back to the parent acid, which is why they are called derivatives of it.

Derivative	General formula	Example	Preparation from CH₃COOH
Acid chloride	RCOCl	ethanoyl chloride, CH₃COCl	CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl
Anhydride	(RCO)₂O	ethanoic anhydride, (CH₃CO)₂O	2CH₃COOH --P₂O₅, Δ--> (CH₃CO)₂O + H₂O
Ester	RCOOR′	ethyl ethanoate, CH₃COOC₂H₅	CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄)
Amide	RCONH₂	ethanamide, CH₃CONH₂	CH₃COOH + NH₃ → CH₃COONH₄ --Δ--> CH₃CONH₂ + H₂O

The anhydride can also be made more cleanly from the acid chloride and the sodium salt:

CH₃COCl + CH₃COONa → (CH₃CO)₂O + NaCl

14.10 Comparative physical and chemical properties

Physical properties

Property	CH₃COCl	(CH₃CO)₂O	CH₃COOC₂H₅	CH₃CONH₂
M (g mol⁻¹)	78.5	102	88	59
State at 25 °C	liquid	liquid	liquid	white solid
Boiling point	51 °C	140 °C	77 °C	221 °C (m.p. 82 °C)
Smell	sharp, irritating	sharp	pleasant, fruity	almost none
Action of water	fumes and hydrolyses at once	slow at 25 °C, fast on warming	very slow without a catalyst	needs boiling acid or alkali

Amides are much the highest boiling because the —NH₂ group forms intermolecular hydrogen bonds; the other three have no N—H or O—H at all and rely on dipole–dipole attraction.

The common mechanism: nucleophilic acyl substitution

All four derivatives react with nucleophiles by the same two-step path — addition then elimination. This is why the C=O of a derivative gives substitution while the C=O of an aldehyde gives addition: the aldehyde has no group that can leave as an anion.

Hydrolysis

Derivative	Reagent	Equation
Acid chloride	cold water, violent	CH₃COCl + H₂O → CH₃COOH + HCl
Anhydride	warm water	(CH₃CO)₂O + H₂O → 2CH₃COOH
Ester	dilute acid (reversible)	CH₃COOC₂H₅ + H₂O ⇌ CH₃COOH + C₂H₅OH
Ester	alkali (irreversible)	CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH
Amide	boiling dilute acid	CH₃CONH₂ + H₂O + HCl → CH₃COOH + NH₄Cl
Amide	boiling alkali	CH₃CONH₂ + NaOH → CH₃COONa + NH₃↑

Alkaline hydrolysis of an ester is called saponification — it is the reaction that makes soap from a triglyceride.

Ammonolysis and reaction with amines

CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl

(CH₃CO)₂O + 2NH₃ → CH₃CONH₂ + CH₃COONH₄

CH₃COOC₂H₅ + NH₃ → CH₃CONH₂ + C₂H₅OH

With a primary amine the product is an N-substituted amide:

CH₃COCl + C₂H₅NH₂ → CH₃CONHC₂H₅ + HCl

C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH (acetanilide)

Acylation of an amine or a phenol with an acid chloride in the presence of aqueous NaOH is the Schotten–Baumann reaction; the alkali mops up the HCl.

Alcoholysis

CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl

(CH₃CO)₂O + C₂H₅OH → CH₃COOC₂H₅ + CH₃COOH

Both give esters in one step and, unlike direct esterification, neither is an equilibrium — which is why acid chlorides are used when a high yield matters.

Reduction

Derivative	Reagent	Product
CH₃COCl	LiAlH₄ (4[H])	CH₃CH₂OH + HCl
CH₃COCl	H₂, Pd/BaSO₄ poisoned (Rosenmund)	CH₃CHO + HCl
(CH₃CO)₂O	LiAlH₄ (8[H])	2CH₃CH₂OH + H₂O
CH₃COOC₂H₅	LiAlH₄ (4[H])	CH₃CH₂OH + C₂H₅OH
CH₃CONH₂	LiAlH₄ (4[H])	CH₃CH₂NH₂ + H₂O

Note the odd one out: an amide is reduced to an amine, not to an alcohol, because nitrogen is a much poorer leaving group than oxygen.

14.11 Claisen condensation, Hofmann bromamide, amides and reactivity

Claisen condensation

An ester with an α-hydrogen, treated with its own sodium alkoxide, condenses with a second molecule of ester to give a β-keto ester:

2CH₃COOC₂H₅ --C₂H₅ONa, then H⁺--> CH₃COCH₂COOC₂H₅ + C₂H₅OH

The product, ethyl 3-oxobutanoate (ethyl acetoacetate), is a key synthetic intermediate. The mechanism parallels the aldol condensation of Unit 13 — the alkoxide removes an α-hydrogen to give an ester enolate, which attacks the carbonyl carbon of the second ester — but here the tetrahedral intermediate expels ethoxide, so the result is substitution, not addition. Esters with no α-hydrogen, such as ethyl benzoate, cannot undergo it.

Hofmann bromamide reaction

An amide treated with bromine in excess alkali loses its carbonyl carbon as carbonate and gives a primary amine with one carbon fewer:

CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Check the first: C 1 + 1 = 2 ✓, H 5 + 4 = 5 + 4 ✓, N 1 ✓, O 1 + 4 = 3 + 2 ✓, Na 4 = 2 + 2 ✓, Br 2 ✓. It is a descending reaction, the counterpart of the ascending nitrile route of Section 14.2.

Amphoteric nature of amides

An amide is almost neutral to litmus, yet it reacts with both acids and bases:

As a base — the nitrogen lone pair accepts a proton from a strong acid:

CH₃CONH₂ + HCl → CH₃CONH₂·HCl

As an acid — the N—H hydrogen can be removed by sodium or mercury(II) oxide:

2CH₃CONH₂ + 2Na → 2CH₃CONHNa + H₂↑

2CH₃CONH₂ + HgO → (CH₃CONH)₂Hg + H₂O

The reason for both is the same delocalisation: the nitrogen lone pair is pulled into the carbonyl group, which makes the nitrogen a much weaker base than an amine, and at the same time makes the N—H bond polar enough to be attacked by a strong base.

Relative reactivity of the derivatives

RCOCl > (RCO)₂O > RCOOR′ > RCOOH > RCONH₂

Two effects explain the order, and they reinforce each other:

• Leaving-group ability. Cl⁻ is the conjugate base of a strong acid and so a very weak base — it leaves easily. NH₂⁻ is the conjugate base of ammonia, a very strong base — it will not leave.
• Resonance donation into the carbonyl group. The lone pair on the attached atom feeds into the C=O and reduces the δ+ charge. Nitrogen (2p–2p overlap, low electronegativity) donates best, oxygen less, chlorine (3p–2p, poor overlap, high electronegativity) hardly at all.

This order also means a more reactive derivative can be converted into a less reactive one, but not the reverse: an acid chloride gives an anhydride, an ester or an amide directly, but an amide cannot be turned into an acid chloride in one step.

Chapter summary

• Carboxylic acids are R—COOH, CₙH₂ₙO₂; the carboxyl carbon is always C-1. They are functional isomers of esters (C₃H₆O₂ = propanoic acid, methyl ethanoate, ethyl methanoate).
• Preparations: oxidation of a 1° alcohol or aldehyde, hydrolysis of a nitrile (+1 C), decarboxylation of a diacid (−1 C), carbonylation of a sodium alkoxide, hydrolysis of a trihaloalkane; benzoic acid from any alkylbenzene with a benzylic H by hot alkaline KMnO₄.
• They form hydrogen-bonded dimers, so they boil above alcohols of the same molar mass (ethanoic acid 118 °C against propan-1-ol 97 °C).
• Acidity: RCOOH > H₂CO₃ > C₆H₅OH > H₂O > ROH, because RCOO⁻ is resonance stabilised with both C—O bonds 127 pm. Effervescence with NaHCO₃ distinguishes an acid from a phenol.
• −I substituents strengthen the acid (pK_a: CH₃COOH 4.76, ClCH₂COOH 2.87, Cl₂CHCOOH 1.29, Cl₃CCOOH 0.66); +I groups weaken it; the effect falls off sharply with distance.
• Methanoic acid is abnormal: strongest simple acid, reduces Tollens' and Fehling's, gives CO with conc. H₂SO₄, gives no anhydride and no HVZ reaction.
• HVZ (Cl₂/red P) halogenates the α-carbon; —COOH is meta-directing and deactivating on a benzene ring.
• Derivatives RCOCl, (RCO)₂O, RCOOR′, RCONH₂ all react by nucleophilic acyl substitution (addition then elimination) and all hydrolyse back to the acid. Reactivity RCOCl > (RCO)₂O > RCOOR′ > RCOOH > RCONH₂.
• Named reactions: Claisen (2 esters → β-keto ester), Hofmann bromamide (amide → 1° amine, one C fewer), Schotten–Baumann (acylation in aq. NaOH). Amides are amphoteric and are reduced by LiAlH₄ to amines.

Practice questions

Group A — Multiple choice (1 mark each)

1. Which reagent distinguishes ethanoic acid from phenol? [1]aNaOHbNaHCO₃cNa metaldneutral FeCl₃
2. The strongest acid among the following is [1]aCH₃COOHbClCH₂COOHcCl₂CHCOOHdCl₃CCOOH
3. Which carboxylic acid gives a silver mirror with Tollens' reagent? [1]aCH₃COOHbHCOOHcC₆H₅COOHdCH₃CH₂COOH
4. Reduction of ethanamide with LiAlH₄ gives [1]aethanolbethanalcethylaminedethanoic acid
5. The correct order of reactivity towards hydrolysis is [1]aRCONH₂ > RCOOR′ > (RCO)₂O > RCOClbRCOCl > (RCO)₂O > RCOOR′ > RCONH₂cRCOOR′ > RCOCl > RCONH₂ > (RCO)₂Od(RCO)₂O > RCOCl > RCONH₂ > RCOOR′
6. In the Hofmann bromamide reaction, the amine formed contains [1]athe same number of carbons as the amidebone carbon morecone carbon lessdtwo carbons less

Answers to Group A

1. (b) — only the acid is strong enough to liberate CO₂ from NaHCO₃; phenol is not.

2. (d) — three chlorine atoms give the largest total −I effect, pK_a 0.66.

3. (b) — methanoic acid contains a —CHO grouping and so acts as a reducing agent.

4. (c) — an amide is reduced to an amine, since N is a poor leaving group.

5. (b) — Cl⁻ is the weakest base and therefore the best leaving group.

6. (c) — the carbonyl carbon is lost as Na₂CO₃.

Group B — Short answer (5 marks each)

1. Explain, with resonance structures, why a carboxylic acid is more acidic than both phenol and ethanol. [5]
2. Arrange the following in increasing order of acid strength and justify: CH₃COOH, HCOOH, ClCH₂COOH, (CH₃)₃CCOOH, Cl₃CCOOH. [5]
3. What is the Hell–Volhard–Zelinsky reaction? Write the equations for the conversion of ethanoic acid into chloroethanoic acid and then into glycine, and state why methanoic acid does not undergo it. [5]
4. 3.00 g of ethanoic acid is treated with excess sodium carbonate solution. Calculate the volume of CO₂ evolved at STP and the mass of sodium ethanoate formed. [5]
5. Write the reactions of ethanoyl chloride with(a) water,(b) ethanol,(c) ammonia,(d) aniline and(e) H₂ over poisoned Pd, naming each product. [5]
6. Describe the abnormal behaviour of methanoic acid under four headings with equations. [5]

Answers to Group B

2. (CH₃)₃CCOOH (5.05) < CH₃COOH (4.76) < HCOOH (3.75) < ClCH₂COOH (2.87) < Cl₃CCOOH (0.66) in order of increasing strength. Alkyl groups are +I and destabilise the anion; H has no inductive effect; chlorine is −I and stabilises it, and three chlorines do so three times over.

3. CH₃COOH + Cl₂ --red P--> ClCH₂COOH + HCl; then ClCH₂COOH + 2NH₃ → H₂NCH₂COOH + NH₄Cl (glycine). Methanoic acid has no α-carbon, so there is no α-hydrogen to replace.

4. n(CH₃COOH) = 3.00/60 = 0.0500 mol. From 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂, the acid : CO₂ ratio is 2 : 1, so n(CO₂) = 0.0250 mol and V = 0.0250 × 22.4 = 0.560 dm³. The acid : salt ratio is 1 : 1, so n(CH₃COONa) = 0.0500 mol; M(CH₃COONa) = 82 g mol⁻¹, giving m = 0.0500 × 82 = 4.10 g.

5. (a) CH₃COCl + H₂O → CH₃COOH + HCl (ethanoic acid). (b) CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl (ethyl ethanoate). (c) CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl (ethanamide). (d) CH₃COCl + C₆H₅NH₂ → C₆H₅NHCOCH₃ + HCl (acetanilide). (e) CH₃COCl + H₂ --Pd/BaSO₄, S--> CH₃CHO + HCl (ethanal, Rosenmund reduction).

Group C — Long answer (8 marks each)

1. (a) Define a carboxylic acid and explain its acidic character using the resonance of the carboxylate ion. [3](b) Discuss the effect of electron-withdrawing and electron-releasing substituents on acid strength, using pK_a data. [3](c) Why is benzoic acid a stronger acid than ethanoic acid? [2]
2. (a) Name the four derivatives of ethanoic acid and give one equation for the preparation of each from ethanoic acid. [4](b) Compare their reactivity towards hydrolysis, explain the order, and write the equation for the alkaline hydrolysis of ethyl ethanoate and of ethanamide. [4]
3. An organic compound A of formula C₇H₈ is oxidised by hot alkaline KMnO₄ and then acidified to give B, C₇H₆O₂, which dissolves in NaHCO₃ with effervescence. B with SOCl₂ gives C; C with ammonia gives D, C₇H₇NO; D with Br₂ and excess NaOH gives E, C₆H₇N, which gives an orange dye when coupled after diazotisation. Identify A to E and write all the equations. [8]

Answers to Group C

1. (c) The carbon of the benzene ring is sp² hybridised and therefore more electronegative than the sp³ carbon of a methyl group. It withdraws electron density from the carboxyl group (−I) instead of releasing it, so benzoate is better stabilised than ethanoate: pK_a 4.20 against 4.76.

2. (b) RCOCl > (RCO)₂O > RCOOR′ > RCONH₂, because the weaker the base L⁻ the faster it leaves, and the less the atom L donates its lone pair into the C=O. CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH (saponification); CH₃CONH₂ + NaOH → CH₃COONa + NH₃.

3. C₇H₈ with a benzene ring and a side chain is toluene, C₆H₅CH₃ = A.

C₆H₅CH₃ + 2KMnO₄ → C₆H₅COOK + 2MnO₂ + KOH + H₂O; then C₆H₅COOK + HCl → C₆H₅COOH + KCl, so B = benzoic acid, C₇H₆O₂ ✓ (it effervesces with NaHCO₃, confirming —COOH).

C₆H₅COOH + SOCl₂ → C₆H₅COCl + SO₂ + HCl, so C = benzoyl chloride.

C₆H₅COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl, so D = benzamide, C₇H₇NO ✓.

C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O, so E = aniline, C₆H₇N ✓ — and aniline is diazotised and coupled to give azo dyes, as required.