Chapter 8 · Aldehydes, Ketones and Carboxylic Acids

8.3 Carboxylic Acids: Preparation, Acidity and Reactions

Join an OH to a carbonyl carbon and you get the carboxyl group, –COOH. How carboxylic acids are named and made, why they form hydrogen-bonded dimers, why they are far stronger acids than alcohols and phenols, how substituents tune that strength, and their reactions.

Vinegar is a dilute solution of ethanoic (acetic) acid; the sting of an ant or a nettle is methanoic (formic) acid; and the rancid smell of old butter is butanoic acid. Citric acid gives lemons their sourness, sodium benzoate keeps jams and soft drinks from spoiling, and every bar of soap is the sodium salt of a long-chain carboxylic acid such as octadecanoic (stearic) acid. On an industrial scale, ethanoic acid and hexanedioic acid are made by the million tonnes, the first for solvents and plastics, the second as a raw material for nylon.

The carboxyl group, $\mathrm{{-}COOH}$, combines a carbonyl group and a hydroxyl group on the same carbon, but it behaves like neither of them. Its $\mathrm{C{=}O}$ shows almost none of the nucleophilic addition reactions of Section 8.2, and its $\mathrm{O{-}H}$ is about $10^{5}$ times more acidic than that of phenol and $10^{11}$ times more acidic than that of ethanol. This section explains why, using resonance and inductive effects, and then works through the preparation, physical properties and reactions of carboxylic acids.

Carboxylic Acids: Names, Structure and Preparation

📘 Definition 8.10 — Carboxylic acids

A carboxylic acid contains the carboxyl group, $\mathrm{{-}COOH}$, a carbonyl group bonded to a hydroxyl group. General formula $\mathrm{R{-}COOH}$, where R is H, an alkyl or an aryl group. Acids with one, two or three carboxyl groups are mono-, di- and tricarboxylic acids. Aliphatic acids with long unbranched chains (about $\mathrm{C_{12}}$ to $\mathrm{C_{18}}$), which occur as esters in fats and oils, are called fatty acids.

Replacing the $\mathrm{OH}$ of the carboxyl group by another group gives the acid derivatives: acyl chlorides $\mathrm{RCOCl}$, anhydrides $\mathrm{(RCO)_{2}O}$, esters $\mathrm{RCOOR'}$ and amides $\mathrm{RCONH_{2}}$. Saturated open-chain monocarboxylic acids have the general formula $\mathrm{C_{n}H_{2n}O_{2}}$, the same as their esters.

📐 Rule 8.14 — Naming carboxylic acids; structure of the carboxyl group
  1. IUPAC names. Replace the final "-e" of the alkane by -oic acid; the $\mathrm{COOH}$ carbon is C-1: ethanoic acid, 2-methylpropanoic acid. For two $\mathrm{COOH}$ groups the "e" is kept: hexanedioic acid. A $\mathrm{COOH}$ group joined to a ring is named with the suffix -carboxylic acid: cyclohexanecarboxylic acid, benzenecarboxylic acid (benzoic acid).
  2. Common names come from natural sources: formic acid from ants (Latin formica), acetic acid from vinegar (acetum), butyric acid from butter (butyrum). Positions are marked α, β, γ… from the carbon next to $\mathrm{COOH}$, as for aldehydes.
  3. Structure. The carboxyl carbon is $sp^{2}$, and its three σ bonds lie in a plane at about $120^{\circ}$. The carbonyl group is in resonance with a lone pair of the $\mathrm{OH}$ oxygen, $\mathrm{R{-}C({=}O){-}OH\longleftrightarrow R{-}C(O^{-}){=}\overset{+}{O}H}$, so the carboxyl carbon is less electrophilic than a carbonyl carbon in an aldehyde or ketone. That is why carboxylic acids do not give the typical addition reactions of $\mathrm{C{=}O}$ (no 2,4-DNP precipitate, no bisulphite or cyanohydrin formation).
StructureCommon nameIUPAC name
$\mathrm{HCOOH}$formic acidmethanoic acid
$\mathrm{CH_{3}COOH}$acetic acidethanoic acid
$\mathrm{CH_{3}CH_{2}COOH}$propionic acidpropanoic acid
$\mathrm{CH_{3}CH_{2}CH_{2}COOH}$butyric acidbutanoic acid
$\mathrm{(CH_{3})_{2}CHCOOH}$isobutyric acid2-methylpropanoic acid
$\mathrm{HOOC{-}COOH}$oxalic acidethanedioic acid
$\mathrm{HOOC{-}CH_{2}{-}COOH}$malonic acidpropanedioic acid
$\mathrm{HOOC(CH_{2})_{2}COOH}$succinic acidbutanedioic acid
$\mathrm{HOOC(CH_{2})_{4}COOH}$adipic acidhexanedioic acid
$\mathrm{C_{6}H_{5}COOH}$benzoic acidbenzenecarboxylic acid (benzoic acid)
$\mathrm{C_{6}H_{5}CH_{2}COOH}$phenylacetic acid2-phenylethanoic acid
$\mathrm{1,2{-}C_{6}H_{4}(COOH)_{2}}$phthalic acidbenzene-1,2-dicarboxylic acid
📐 Rule 8.15 — Preparation of carboxylic acids

1. Oxidation of primary alcohols and aldehydes with $\mathrm{KMnO_{4}}$ (neutral, acidic or alkaline), $\mathrm{K_{2}Cr_{2}O_{7}}$ or $\mathrm{CrO_{3}}$ in acid:

$$\mathrm{RCH_{2}OH\xrightarrow[\text{(ii) }H_{3}O^{+}]{\text{(i) alkaline }KMnO_{4}}RCOOH},\qquad \mathrm{RCHO\xrightarrow{[O]}RCOOH}$$

2. Oxidation of alkylbenzenes. Hot alkaline $\mathrm{KMnO_{4}}$ or chromic acid oxidises a side chain attached to a benzene ring all the way to $\mathrm{{-}COOH}$, whatever its length, as long as the carbon attached to the ring carries a hydrogen. Methylbenzene, ethylbenzene and propylbenzene all give benzoic acid:

$$\mathrm{C_{6}H_{5}CH_{2}CH_{2}CH_{3}\xrightarrow[\text{(ii) }H_{3}O^{+}]{\text{(i) }KMnO_{4}{-}KOH,\ \Delta}C_{6}H_{5}COOH}$$

3. Hydrolysis of nitriles and amides. Heating with aqueous acid or alkali converts a nitrile, through the amide, into the acid (with alkali, the salt is formed and acidified at the end); under mild conditions the reaction can be stopped at the amide:

$$\mathrm{R{-}CN\xrightarrow{H^{+}\text{ or }OH^{-},\ H_{2}O}R{-}CONH_{2}\xrightarrow{H^{+}\text{ or }OH^{-},\ H_{2}O}R{-}COOH}$$

4. Grignard reagents and carbon dioxide. A Grignard reagent adds to $\mathrm{CO_{2}}$ (dry ice), giving the magnesium salt of the acid, which is hydrolysed by dilute acid. The acid has one carbon more than the alkyl halide:

$$\mathrm{R{-}MgX+O{=}C{=}O\xrightarrow{\text{dry ether}}R{-}COOMgX\xrightarrow{H_{3}O^{+}}R{-}COOH}$$

5. Hydrolysis of acyl chlorides and anhydrides by water (or by aqueous alkali followed by acid): $\mathrm{RCOCl+H_{2}O\rightarrow RCOOH+HCl}$, $\ \mathrm{(RCO)_{2}O+H_{2}O\rightarrow2RCOOH}$.

6. Hydrolysis of esters. Acid hydrolysis gives the acid directly; alkaline hydrolysis (saponification) gives the carboxylate salt, which is then acidified: $\mathrm{C_{6}H_{5}COOC_{2}H_{5}\xrightarrow{NaOH,\ H_{2}O}C_{6}H_{5}COONa+C_{2}H_{5}OH}$, $\ \mathrm{C_{6}H_{5}COONa\xrightarrow{H_{3}O^{+}}C_{6}H_{5}COOH}$.

✏️ Example 1 — Naming carboxylic acids
Give IUPAC names: (a) $\mathrm{CH_{3}CH(Cl)COOH}$; (b) $\mathrm{(CH_{3})_{3}CCOOH}$; (c) $\mathrm{CH_{3}CH{=}CHCOOH}$; (d) $\mathrm{HOOCCH_{2}CH(CH_{3})COOH}$; (e) $\mathrm{4{-}O_{2}NC_{6}H_{4}COOH}$; (f) $\mathrm{C_{6}H_{11}COOH}$ (a cyclohexane ring); (g) $\mathrm{CH_{3}COCH_{2}COOH}$.
(a), (b) (a) $\mathrm{COOH}$ is C-1, Cl on C-2: 2-chloropropanoic acid (α-chloropropionic acid). (b) Three-carbon chain with two methyls on C-2: 2,2-dimethylpropanoic acid.
(c) Four carbons, double bond between C-2 and C-3: but-2-enoic acid (crotonic acid).
(d) Both $\mathrm{COOH}$ carbons belong to the main chain: $\mathrm{HOOC{-}CH_{2}{-}CH(CH_{3}){-}COOH}$ is a four-carbon diacid with a methyl on C-2 (numbering from either end gives 2 or 3; choose 2): 2-methylbutanedioic acid.
(e), (f) (e) 4-nitrobenzoic acid (4-nitrobenzenecarboxylic acid); (f) cyclohexanecarboxylic acid.
(g) $\mathrm{COOH}$ outranks the ketone, which becomes the prefix oxo: $\mathrm{C{-}1}$ is $\mathrm{COOH}$, the $\mathrm{C{=}O}$ is on C-3: 3-oxobutanoic acid (acetoacetic acid).
✏️ Example 2 — Choosing a preparation
How would you prepare (a) benzoic acid from bromobenzene; (b) butanoic acid from 1-bromopropane, by two routes; (c) benzoic acid from propylbenzene; (d) benzoic acid from ethyl benzoate; (e) ethanoic acid from ethanoyl chloride? (f) Which of (b)'s routes would fail for preparing 2,2-dimethylpropanoic acid from 2-bromo-2-methylpropane, and why?
(a) $$\mathrm{C_{6}H_{5}Br\xrightarrow{Mg,\ dry\ ether}C_{6}H_{5}MgBr\xrightarrow{CO_{2}}C_{6}H_{5}COOMgBr\xrightarrow{H_{3}O^{+}}C_{6}H_{5}COOH}$$ (Aryl halides do not undergo $\mathrm{CN^{-}}$ substitution easily, so the nitrile route is not used.)
(b) Both routes add one carbon (3 → 4). Grignard: $$\mathrm{CH_{3}CH_{2}CH_{2}Br\xrightarrow{Mg}CH_{3}CH_{2}CH_{2}MgBr\xrightarrow{CO_{2};\ H_{3}O^{+}}CH_{3}CH_{2}CH_{2}COOH}$$ Nitrile: $$\mathrm{CH_{3}CH_{2}CH_{2}Br\xrightarrow{KCN}CH_{3}CH_{2}CH_{2}CN\xrightarrow{H_{3}O^{+},\ \Delta}CH_{3}CH_{2}CH_{2}COOH}$$
(c), (d), (e) (c) Hot alkaline $\mathrm{KMnO_{4}}$, then acid: the whole propyl chain becomes $\mathrm{COOH}$ (and two carbons are lost as $\mathrm{CO_{2}}$ and other fragments). (d) Boil with aqueous $\mathrm{NaOH}$ (saponification), then acidify. (e) Water: $\mathrm{CH_{3}COCl+H_{2}O\rightarrow CH_{3}COOH+HCl}$.
(f) The nitrile route: a tertiary halide reacts with $\mathrm{CN^{-}}$ mainly by elimination (giving 2-methylpropene), not substitution (Chapter 6). The Grignard route works: $\mathrm{(CH_{3})_{3}CMgBr+CO_{2}}$, then $\mathrm{H_{3}O^{+}}$, gives $\mathrm{(CH_{3})_{3}CCOOH}$.

Physical Properties and Acidity

Hydrogen-bonded dimers. The $\mathrm{O{-}H}$ of one carboxyl group hydrogen-bonds strongly to the $\mathrm{C{=}O}$ of another, and two molecules join through two hydrogen bonds into a ring-shaped dimer. The dimers persist in the vapour and in non-polar (aprotic) solvents, so measured molar masses of ethanoic acid in benzene or in the gas phase come out close to $120\ \mathrm{g\,mol^{-1}}$ instead of $60$ (an abnormal molar mass, Section 1.3). As a result, carboxylic acids boil even higher than alcohols of similar molar mass: ethanoic acid ($M=60$) boils at $391\ \mathrm{K}$, propan-1-ol ($M=60$) at $370\ \mathrm{K}$.

Solubility and smell. The acids with up to four carbon atoms mix with water in all proportions, because they hydrogen-bond to water. Solubility falls as the hydrocarbon part grows; benzoic acid is almost insoluble in cold water but dissolves in hot water. The lower acids have sharp, unpleasant smells (vinegar, rancid butter, goats); the higher fatty acids are waxy, almost odourless solids.

📐 Rule 8.16 — Acidity of carboxylic acids

Carboxylic acids are weak acids in water, but much stronger than phenols and alcohols:

$$\mathrm{RCOOH+H_{2}O\rightleftharpoons RCOO^{-}+H_{3}O^{+}},\qquad K_{a}=\frac{[\mathrm{RCOO^{-}}][\mathrm{H_{3}O^{+}}]}{[\mathrm{RCOOH}]},\qquad \mathrm{p}K_{a}=-\log K_{a}$$
  • The smaller the $\mathrm{p}K_{a}$, the stronger the acid. Most carboxylic acids have $\mathrm{p}K_{a}$ between about $2$ and $5$ (ethanoic acid $4.76$), compared with about $10$ for phenol and about $16$ for ethanol; strong mineral acids such as $\mathrm{HCl}$ have $\mathrm{p}K_{a}\lt1$.
  • Why. In the carboxylate ion the negative charge is shared equally by two electronegative oxygen atoms: its two resonance structures are equivalent, and both $\mathrm{C{-}O}$ bonds have the same length (about $127\ \mathrm{pm}$, between $\mathrm{C{=}O}$ and $\mathrm{C{-}O}$). In the phenoxide ion the charge is spread over one oxygen and three less electronegative ring carbons, and the structures are not equivalent, so it is less stabilised. An alkoxide ion has no resonance at all.
  • Carboxylic acids are strong enough to liberate $\mathrm{CO_{2}}$ from sodium hydrogencarbonate; phenols are not. This is the standard test that tells a carboxylic acid from a phenol.
📐 Rule 8.17 — Effect of substituents on acid strength
  • Electron-withdrawing groups (–I or –R: $\mathrm{{-}F}$, $\mathrm{{-}Cl}$, $\mathrm{{-}Br}$, $\mathrm{{-}NO_{2}}$, $\mathrm{{-}CN}$, $\mathrm{{-}CF_{3}}$, $\mathrm{C_{6}H_{5}}$) pull electron density away from the carboxylate group, spread its negative charge further and increase the acid strength. Electron-releasing groups (+I: alkyl groups) concentrate the charge and decrease it: methanoic acid is stronger than ethanoic acid, which is stronger than propanoic acid.
  • The effect adds up with the number of groups ($\mathrm{Cl_{3}CCOOH\gt Cl_{2}CHCOOH\gt ClCH_{2}COOH}$), is larger for more electronegative halogens ($\mathrm{F\gt Cl\gt Br\gt I}$), and falls off quickly with distance from $\mathrm{COOH}$ (2-chlorobutanoic $\gt$ 3-chlorobutanoic $\gt$ 4-chlorobutanoic acid).
  • A phenyl or vinyl group attached directly to $\mathrm{COOH}$ increases acidity (its $sp^{2}$ carbon is electron-withdrawing): benzoic acid is stronger than ethanoic acid. In substituted benzoic acids, electron-withdrawing groups at the para position increase acidity ($\mathrm{4{-}NO_{2}}$) and electron-releasing groups decrease it ($\mathrm{4{-}CH_{3}}$, $\mathrm{4{-}OCH_{3}}$).
Acid$\mathrm{p}K_{a}$Acid$\mathrm{p}K_{a}$
$\mathrm{CF_{3}COOH}$0.23$\mathrm{4{-}O_{2}NC_{6}H_{4}COOH}$3.41
$\mathrm{CCl_{3}COOH}$0.64$\mathrm{HCOOH}$3.75
$\mathrm{CHCl_{2}COOH}$1.26$\mathrm{C_{6}H_{5}COOH}$4.19
$\mathrm{O_{2}NCH_{2}COOH}$1.68$\mathrm{4{-}CH_{3}OC_{6}H_{4}COOH}$4.47
$\mathrm{NCCH_{2}COOH}$2.47$\mathrm{CH_{3}COOH}$4.76
$\mathrm{FCH_{2}COOH}$2.59$\mathrm{CH_{3}CH_{2}COOH}$4.88
$\mathrm{ClCH_{2}COOH}$2.87phenol, $\mathrm{C_{6}H_{5}OH}$10.0
$\mathrm{BrCH_{2}COOH}$2.90ethanol, $\mathrm{C_{2}H_{5}OH}$15.9

Approximate $\mathrm{p}K_{a}$ values in water at $298\ \mathrm{K}$, rounded; published values differ slightly between sources.

✏️ Example 3 — Ranking acid strength
Arrange in increasing order of acid strength, giving reasons: (a) $\mathrm{ClCH_{2}COOH}$, $\mathrm{CH_{3}COOH}$, $\mathrm{Cl_{2}CHCOOH}$, $\mathrm{Cl_{3}CCOOH}$; (b) butanoic acid, 2-chlorobutanoic acid, 3-chlorobutanoic acid, 4-chlorobutanoic acid; (c) benzoic acid, 4-nitrobenzoic acid, 4-methoxybenzoic acid; (d) ethanol, phenol, ethanoic acid, water.
(a) Each Cl adds a –I effect: $\mathrm{CH_{3}COOH}$ < $\mathrm{ClCH_{2}COOH}$ < $\mathrm{Cl_{2}CHCOOH}$ < $\mathrm{Cl_{3}CCOOH}$ ($\mathrm{p}K_{a}$ $4.76$, $2.87$, $1.26$, $0.64$).
(b) The –I effect fades with distance: butanoic $\lt$ 4-chloro $\lt$ 3-chloro $\lt$ 2-chlorobutanoic acid (approximate $\mathrm{p}K_{a}$ $4.8$, $4.5$, $4.1$, $2.9$). A chlorine on C-2 has by far the largest effect.
(c) $\mathrm{{-}OCH_{3}}$ at the para position releases electrons by resonance (+R), $\mathrm{{-}NO_{2}}$ withdraws them (–R, –I): 4-methoxybenzoic $\lt$ benzoic $\lt$ 4-nitrobenzoic acid ($4.47$, $4.19$, $3.41$).
(d) Ethanol ($\approx16$) $\lt$ water ($\approx15.7$) $\lt$ phenol ($10.0$) $\lt$ ethanoic acid ($4.76$): no resonance in ethoxide; resonance onto ring carbons in phenoxide; two equivalent structures with the charge on two oxygens in ethanoate.
✏️ Example 4 — Putting numbers on acidity
(a) Find $K_{a}$ for ethanoic acid ($\mathrm{p}K_{a}=4.76$) and chloroethanoic acid ($\mathrm{p}K_{a}=2.87$). How many times stronger is the second? (b) Find the pH and the percentage ionisation of $0.10\ \mathrm{mol\,L^{-1}}$ solutions of each.
(a) $K_{a}=10^{-\mathrm{p}K_{a}}$: ethanoic $10^{-4.76}=1.74\times10^{-5}$; chloroethanoic $10^{-2.87}=1.35\times10^{-3}$. Ratio $=10^{4.76-2.87}=10^{1.89}=78$: about $78$ times stronger.
(b) Ethanoic With $x=[\mathrm{H_{3}O^{+}}]$, $\dfrac{x^{2}}{0.10-x}=1.74\times10^{-5}$. Since $x\ll0.10$: $x=\sqrt{1.74\times10^{-6}}=1.32\times10^{-3}\ \mathrm{mol\,L^{-1}}$, $\mathrm{pH}=2.88$, ionisation $=1.3\%$.
(b) Chloroethanoic Here $x$ is not small compared with $0.10$, so solve the quadratic $x^{2}+K_{a}x-0.10K_{a}=0$: $$x=\frac{-K_{a}+\sqrt{K_{a}^{2}+0.40K_{a}}}{2}=1.10\times10^{-2}\ \mathrm{mol\,L^{-1}},$$ so $\mathrm{pH}=1.96$ and ionisation $=11\%$.
Check The approximation for chloroethanoic acid would give $x=\sqrt{1.35\times10^{-4}}=1.16\times10^{-2}$, only about $6\%$ off, but it is not justified when more than about $5\%$ is ionised. The pH difference ($0.92$) is much less than the $\mathrm{p}K_{a}$ difference ($1.89$), because $[\mathrm{H_{3}O^{+}}]\propto\sqrt{K_{a}}$ for a weak acid.
Interactive 1 — The Acidity Ladder

Every bar is an acid, placed by its approximate $\mathrm{p}K_{a}$ in water; the further left, the stronger the acid. Choose a compound (or press Play to climb the ladder). The readouts give $K_{a}$, its strength compared with ethanoic acid, and the pH and percentage ionisation of a $0.10\ \mathrm{mol\,L^{-1}}$ solution, found by solving $x^{2}/(0.10-x)=K_{a}$ exactly. The beaker shows the fraction of molecules that have given up their proton. Notice how the gap between phenol and the carboxylic acids dwarfs every substituent effect.

$\mathrm{p}K_{a}$ (approx.)—
$K_{a}$—
vs ethanoic acid—
pH at $0.10\ \mathrm{mol\,L^{-1}}$—
ionised—

Reactions of the O–H and C–OH Bonds

📐 Rule 8.18 — Reactions involving the O–H bond: metals, alkalis, carbonates

Carboxylic acids form salts with active metals (hydrogen is evolved), with alkalis, and with carbonates and hydrogencarbonates (carbon dioxide is evolved, with brisk effervescence):

$$\mathrm{2RCOOH+2Na\rightarrow2RCOONa+H_{2}}$$ $$\mathrm{RCOOH+NaOH\rightarrow RCOONa+H_{2}O}$$ $$\mathrm{RCOOH+NaHCO_{3}\rightarrow RCOONa+H_{2}O+CO_{2}\uparrow}$$

Phenols react with sodium and with $\mathrm{NaOH}$ but are too weak to decompose $\mathrm{NaHCO_{3}}$ (carbonic acid, $\mathrm{p}K_{a}\approx6.4$, is a stronger acid than phenol, $\mathrm{p}K_{a}\approx10$). Alcohols react only with the metal. The $\mathrm{NaHCO_{3}}$ test therefore picks out carboxylic acids.

📐 Rule 8.19 — Reactions involving the C–OH bond

1. Anhydride formation. Heating with a dehydrating agent ($\mathrm{P_{2}O_{5}}$ or concentrated $\mathrm{H_{2}SO_{4}}$) joins two molecules with loss of water:

$$\mathrm{2CH_{3}COOH\xrightarrow[\Delta]{P_{2}O_{5}}(CH_{3}CO)_{2}O+H_{2}O}$$

2. Esterification. With an alcohol and a little concentrated sulphuric acid or dry $\mathrm{HCl}$, an acid forms an ester. The reaction is reversible:

$$\mathrm{RCOOH+R'OH\overset{H^{+}}{\rightleftharpoons}RCOOR'+H_{2}O}$$

Mechanism: (i) $\mathrm{H^{+}}$ protonates the carbonyl oxygen, making the carbon more electrophilic; (ii) the alcohol oxygen adds to the carbonyl carbon, giving a tetrahedral intermediate; (iii) a proton moves to one of the $\mathrm{OH}$ groups, turning it into $\mathrm{{-}OH_{2}^{+}}$, a good leaving group; (iv) water leaves and $\mathrm{H^{+}}$ is lost, giving the ester and regenerating the catalyst. So the $\mathrm{C{-}OH}$ bond of the acid breaks, and the ester oxygen comes from the alcohol (shown by labelling the alcohol with $^{18}\mathrm{O}$). The yield is raised by using excess alcohol or by removing water as it forms (Le Chatelier's principle).

3. With $\mathrm{PCl_{5}}$, $\mathrm{PCl_{3}}$ and $\mathrm{SOCl_{2}}$ the $\mathrm{OH}$ is replaced by $\mathrm{Cl}$, giving acyl chlorides:

$$\mathrm{RCOOH+PCl_{5}\rightarrow RCOCl+POCl_{3}+HCl}$$ $$\mathrm{3RCOOH+PCl_{3}\rightarrow3RCOCl+H_{3}PO_{3}}$$ $$\mathrm{RCOOH+SOCl_{2}\rightarrow RCOCl+SO_{2}+HCl}$$

Thionyl chloride is preferred because both by-products are gases and escape, leaving the acyl chloride easy to purify.

4. With ammonia an acid first forms the ammonium salt, which loses water on strong heating to give an amide:

$$\mathrm{CH_{3}COOH+NH_{3}\rightleftharpoons CH_{3}COO^{-}NH_{4}^{+}\xrightarrow[-H_{2}O]{\Delta}CH_{3}CONH_{2}}$$

Benzoic acid gives benzamide in the same way; benzene-1,2-dicarboxylic (phthalic) acid gives phthalimide on strong heating, which is used in Chapter 9.

✏️ Example 5 — Reactions of the carboxyl group in numbers
(a) What volume of $\mathrm{CO_{2}}$ (at STP, $22.7\ \mathrm{L\,mol^{-1}}$) is released when $0.60\ \mathrm{g}$ of ethanoic acid reacts with excess $\mathrm{NaHCO_{3}}$? (b) Write the reactions of ethanoic acid with $\mathrm{PCl_{5}}$ and of benzoic acid with $\mathrm{SOCl_{2}}$. (c) For $\mathrm{CH_{3}COOH+C_{2}H_{5}OH\rightleftharpoons CH_{3}COOC_{2}H_{5}+H_{2}O}$, take $K_{c}\approx4$ (a typical value for this esterification). What fraction of the acid is converted to ester at equilibrium if 1 mol of acid is mixed with (i) 1 mol and (ii) 3 mol of ethanol? (d) Name the ester.
(a) 1 mol acid gives 1 mol $\mathrm{CO_{2}}$. $n=\dfrac{0.60}{60}=0.0100\ \mathrm{mol}$; $V=0.0100\times22.7=0.227\ \mathrm{L}=227\ \mathrm{mL}$.
(b) $$\mathrm{CH_{3}COOH+PCl_{5}\rightarrow CH_{3}COCl+POCl_{3}+HCl}$$ $$\mathrm{C_{6}H_{5}COOH+SOCl_{2}\rightarrow C_{6}H_{5}COCl+SO_{2}+HCl}$$
(c)(i) If $x$ mol of ester forms (the volume cancels): $\dfrac{x^{2}}{(1-x)^{2}}=4$, so $\dfrac{x}{1-x}=2$ and $x=\dfrac{2}{3}=0.67$: two-thirds of the acid.
(c)(ii) $\dfrac{x^{2}}{(1-x)(3-x)}=4$ gives $3x^{2}-16x+12=0$, so $x=\dfrac{16-\sqrt{112}}{6}=0.90$ (the other root, $4.4$, is impossible): $90\%$ of the acid. Excess alcohol pushes the equilibrium to the right.
(d) $\mathrm{CH_{3}COOC_{2}H_{5}}$, ethyl ethanoate (ethyl acetate): the alcohol part is named first as an alkyl group, then the acid part with "-oate".
Interactive 2 — The Unknown Compound: a Test-Tube Lab

A bottle has lost its label. It contains one of the nine compounds in the answer list. Run the five tests in any order (each costs one sample) and watch the tubes; the lab keeps track of which compounds are still consistent with what you have seen. When you are sure, choose your answer and press Check. Try to identify it with as few tests as possible. Press Play to watch a demonstration on a random unknown.

tests used—
still possible—
last result—

Reactions of the –COOH Group and of the Ring; Uses

📐 Rule 8.20 — Reduction and decarboxylation

1. Reduction. Lithium aluminium hydride (or diborane, $\mathrm{B_{2}H_{6}}$) reduces carboxylic acids to primary alcohols; $\mathrm{NaBH_{4}}$ does not reduce the carboxyl group:

$$\mathrm{RCOOH\xrightarrow[\text{(ii) }H_{3}O^{+}]{\text{(i) }LiAlH_{4}/\text{ether}}RCH_{2}OH}$$

2. Decarboxylation with soda lime. Heating the sodium salt with sodalime ($\mathrm{NaOH}$ and $\mathrm{CaO}$, $3:1$) removes the carboxyl group as carbonate, giving an alkane with one carbon fewer:

$$\mathrm{RCOONa\xrightarrow[\Delta]{NaOH\ \&\ CaO}R{-}H+Na_{2}CO_{3}}$$

3. Kolbe electrolysis. Electrolysis of a concentrated aqueous solution of the sodium or potassium salt gives an alkane with twice the number of carbons of R at the anode: $\mathrm{2CH_{3}COONa+2H_{2}O\rightarrow CH_{3}CH_{3}+2CO_{2}+H_{2}+2NaOH}$ (ethane and $\mathrm{CO_{2}}$ at the anode, $\mathrm{H_{2}}$ at the cathode).

📘 Definition 8.11 — Hell–Volhard–Zelinsky reaction

Carboxylic acids that have an α-hydrogen are halogenated at the α-position by chlorine or bromine in the presence of a small amount of red phosphorus; hydrolysis of the product gives the α-halocarboxylic acid:

$$\mathrm{R{-}CH_{2}{-}COOH\xrightarrow[\text{(ii) }H_{2}O]{\text{(i) }X_{2}/\text{red P}}R{-}CH(X){-}COOH}\qquad(\mathrm{X=Cl,\ Br})$$

For example ethanoic acid gives 2-chloroethanoic acid, and further chlorination gives the di- and trichloro acids. The α-halo acids are useful because the halogen can be replaced by nucleophiles (Chapter 6), giving, for example, α-hydroxy and α-amino acids.

📐 Rule 8.21 — Ring substitution in aromatic acids
The $\mathrm{{-}COOH}$ group withdraws electrons from the ring, so it is deactivating and meta-directing. Benzoic acid is nitrated and brominated at the 3-position: $$\mathrm{C_{6}H_{5}COOH\xrightarrow{conc.\ HNO_{3}/conc.\ H_{2}SO_{4}}3{-}O_{2}NC_{6}H_{4}COOH},\qquad \mathrm{C_{6}H_{5}COOH\xrightarrow{Br_{2}/FeBr_{3}}3{-}BrC_{6}H_{4}COOH}$$ Aromatic carboxylic acids do not undergo Friedel–Crafts reactions: the ring is deactivated, and the catalyst $\mathrm{AlCl_{3}}$ (a Lewis acid) binds to the carboxyl group.

Uses. Methanoic acid is used in the rubber, textile, dyeing, leather and electroplating industries. Ethanoic acid is a solvent and, as vinegar, a food additive and preservative. Hexanedioic acid is a raw material for nylon-6,6. Esters of benzoic acid are used in perfumes; sodium benzoate is a common food preservative. Higher fatty acids are used to make soaps and detergents.

✏️ Example 6 — Conversions
Show how to carry out each conversion, giving reagents: (a) ethanoic acid → 2-chloroethanoic acid; (b) sodium ethanoate → methane; (c) sodium propanoate → butane; (d) benzoic acid → 3-nitrobenzoic acid; (e) benzoic acid → phenylmethanol (benzyl alcohol); (f) ethanoic acid → ethanamide; (g) propanoic acid → propene in two steps.
(a) HVZ: $\mathrm{Cl_{2}}$ with a little red phosphorus, then $\mathrm{H_{2}O}$: $\mathrm{CH_{3}COOH\rightarrow ClCH_{2}COOH}$.
(b), (c) (b) Heat with sodalime: $\mathrm{CH_{3}COONa+NaOH\rightarrow CH_{4}+Na_{2}CO_{3}}$. (c) Kolbe electrolysis of aqueous sodium propanoate: two ethyl groups join, $\mathrm{2C_{2}H_{5}COO^{-}\rightarrow C_{4}H_{10}+2CO_{2}+2e^{-}}$ at the anode.
(d), (e) (d) Concentrated $\mathrm{HNO_{3}}$ and $\mathrm{H_{2}SO_{4}}$ (meta substitution). (e) (i) $\mathrm{LiAlH_{4}}$ in dry ether, (ii) $\mathrm{H_{3}O^{+}}$: $\mathrm{C_{6}H_{5}COOH\rightarrow C_{6}H_{5}CH_{2}OH}$.
(f) $\mathrm{NH_{3}}$, then strong heating of the ammonium salt: $\mathrm{CH_{3}COONH_{4}\rightarrow CH_{3}CONH_{2}+H_{2}O}$. (Alternatively $\mathrm{SOCl_{2}}$ to the acyl chloride, then $\mathrm{NH_{3}}$.)
(g) $$\mathrm{CH_{3}CH_{2}COOH\xrightarrow{LiAlH_{4};\ H_{3}O^{+}}CH_{3}CH_{2}CH_{2}OH\xrightarrow{conc.\ H_{2}SO_{4},\ 443\ K}CH_{3}CH{=}CH_{2}}$$ Reduction to propan-1-ol, then acid-catalysed dehydration (Chapter 7).
⚠️ Common mistakes with carboxylic acids

1. Counting carbons. Grignard + $\mathrm{CO_{2}}$ and $\mathrm{KCN}$ then hydrolysis add one carbon; sodalime decarboxylation removes one; Kolbe electrolysis doubles the alkyl group.

2. $\mathrm{NaHCO_{3}}$ versus $\mathrm{NaOH}$. Both phenols and carboxylic acids dissolve in $\mathrm{NaOH}$; only carboxylic acids give $\mathrm{CO_{2}}$ with $\mathrm{NaHCO_{3}}$.

3. Reducing agents. $\mathrm{NaBH_{4}}$ reduces aldehydes and ketones but not acids; $\mathrm{LiAlH_{4}}$ reduces acids to primary alcohols, not to aldehydes.

4. HVZ needs an α-hydrogen. 2,2-Dimethylpropanoic acid and benzoic acid do not undergo it.

5. Distance matters. A $\mathrm{Cl}$ on C-2 increases acid strength a lot, on C-4 only a little: count the bonds between the substituent and $\mathrm{COOH}$.

6. Carboxylic acids are not carbonyl compounds in the test-tube sense. They give no precipitate with 2,4-DNP, and (except methanoic acid, which also contains an aldehyde-like $\mathrm{H{-}C{=}O}$) they do not reduce Tollens' or Fehling's reagents.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — the ortho effect in benzoic acids

In JEE Main, NEET and JEE Advanced 2026 (acidic strength of carboxylic acids and factors affecting it); not required by the CBSE syllabus.

Almost any group in the ortho position makes a benzoic acid stronger, even an electron-releasing one: 2-methylbenzoic acid ($\mathrm{p}K_{a}\approx3.9$) is stronger than benzoic acid ($4.19$), while 4-methylbenzoic acid ($\approx4.4$) is weaker. The usual explanation is steric: an ortho group pushes the $\mathrm{COOH}$ out of the plane of the ring, which stops the ring from donating electrons into the carboxyl group by resonance (that donation slightly destabilises the anion), so the acid becomes stronger. For 2-hydroxybenzoic (salicylic) acid ($\mathrm{p}K_{a}\approx3.0$), intramolecular hydrogen bonding from the $\mathrm{OH}$ to the carboxylate oxygen also stabilises the anion. A typical JEE order is therefore 2-nitrobenzoic $\gt$ 4-nitrobenzoic $\gt$ 3-nitrobenzoic $\gt$ benzoic acid (approximately $2.2$, $3.4$, $3.5$, $4.2$).

Exercises

I
Names and preparation.
  • 1 Give IUPAC names: (a) $\mathrm{CH_{3}CH_{2}CH(CH_{3})COOH}$; (b) $\mathrm{(CH_{3})_{2}CHCH_{2}COOH}$; (c) $\mathrm{HOOCCH{=}CHCOOH}$; (d) $\mathrm{C_{6}H_{5}CH_{2}CH_{2}COOH}$.
  • 2 Write structures: (a) 3-bromobutanoic acid; (b) 2-hydroxypropanoic acid (lactic acid); (c) 4-methylbenzoic acid; (d) pentanedioic acid.
  • 3 Draw and name the carboxylic acids and the esters with molecular formula $\mathrm{C_{4}H_{8}O_{2}}$. How many of each are there?
  • 4 Give the product of vigorous oxidation (hot alkaline $\mathrm{KMnO_{4}}$, then acid) of (a) butan-1-ol; (b) 1-ethyl-4-methylbenzene; (c) (1,1-dimethylethyl)benzene (tert-butylbenzene).
  • 5 Give the acid formed from the Grignard reagent of (a) 2-bromopropane and (b) bromocyclohexane, with $\mathrm{CO_{2}}$ and then $\mathrm{H_{3}O^{+}}$.
  • 6 Benzonitrile is boiled with aqueous acid. Give the intermediate and the final product.
  • 7 Which route, Grignard + $\mathrm{CO_{2}}$ or $\mathrm{KCN}$ then hydrolysis, would you use to make benzoic acid from bromobenzene? Explain.
  • 8 Ethyl propanoate is boiled with aqueous $\mathrm{NaOH}$, and the solution is then acidified. Give the products of each stage.
1 (a) 2-methylbutanoic acid; (b) 3-methylbutanoic acid; (c) but-2-enedioic acid; (d) 3-phenylpropanoic acid.
2 (a) $\mathrm{CH_{3}CH(Br)CH_{2}COOH}$; (b) $\mathrm{CH_{3}CH(OH)COOH}$; (c) $\mathrm{4{-}CH_{3}C_{6}H_{4}COOH}$; (d) $\mathrm{HOOC(CH_{2})_{3}COOH}$.
3 Acids (2): $\mathrm{CH_{3}CH_{2}CH_{2}COOH}$ butanoic acid, $\mathrm{(CH_{3})_{2}CHCOOH}$ 2-methylpropanoic acid. Esters (4): $\mathrm{HCOOCH_{2}CH_{2}CH_{3}}$ propyl methanoate, $\mathrm{HCOOCH(CH_{3})_{2}}$ 1-methylethyl (isopropyl) methanoate, $\mathrm{CH_{3}COOC_{2}H_{5}}$ ethyl ethanoate, $\mathrm{CH_{3}CH_{2}COOCH_{3}}$ methyl propanoate.
4 (a) $\mathrm{CH_{3}CH_{2}CH_{2}COOH}$, butanoic acid; (b) both side chains are oxidised: benzene-1,4-dicarboxylic acid (terephthalic acid); (c) no side-chain oxidation: the carbon attached to the ring has no hydrogen.
5 (a) $\mathrm{(CH_{3})_{2}CHCOOH}$, 2-methylpropanoic acid; (b) $\mathrm{C_{6}H_{11}COOH}$, cyclohexanecarboxylic acid. One carbon is added each time.
6 $\mathrm{C_{6}H_{5}CN\rightarrow C_{6}H_{5}CONH_{2}}$ (benzamide) $\rightarrow\mathrm{C_{6}H_{5}COOH}$ (benzoic acid) + $\mathrm{NH_{4}^{+}}$.
7 Grignard + $\mathrm{CO_{2}}$. The C–Br bond of an aryl halide has partial double-bond character and does not undergo nucleophilic substitution by $\mathrm{CN^{-}}$ under ordinary conditions (Chapter 6), but it does form a Grignard reagent.
8 Saponification: $\mathrm{CH_{3}CH_{2}COOC_{2}H_{5}+NaOH\rightarrow CH_{3}CH_{2}COONa+C_{2}H_{5}OH}$; acidification: $\mathrm{CH_{3}CH_{2}COONa+H^{+}\rightarrow CH_{3}CH_{2}COOH+Na^{+}}$ (propanoic acid).
II
Physical properties and acid strength (use the $\mathrm{p}K_{a}$ table).
  • 1 Why do carboxylic acids boil at higher temperatures than alcohols of similar molar mass?
  • 2 The molar mass of ethanoic acid found from the freezing point of its solution in benzene is about $120\ \mathrm{g\,mol^{-1}}$. Explain, and give the van 't Hoff factor.
  • 3 Arrange in increasing acid strength: $\mathrm{CH_{3}CH_{2}COOH}$, $\mathrm{CH_{3}COOH}$, $\mathrm{HCOOH}$, $\mathrm{ClCH_{2}COOH}$.
  • 4 Arrange in increasing acid strength: $\mathrm{FCH_{2}COOH}$, $\mathrm{ClCH_{2}COOH}$, $\mathrm{BrCH_{2}COOH}$, $\mathrm{ICH_{2}COOH}$.
  • 5 Arrange in increasing acid strength: ethanoic acid, benzoic acid, 4-methylbenzoic acid, 4-nitrobenzoic acid.
  • 6 Benzoic acid has $\mathrm{p}K_{a}=4.19$. Find $K_{a}$, and the pH of a $0.010\ \mathrm{mol\,L^{-1}}$ solution (solve the quadratic).
  • 7 The two $\mathrm{C{-}O}$ bonds in sodium methanoate have the same length, but in methanoic acid they differ. Explain.
  • 8 Explain why $\mathrm{CCl_{3}COOH}$ is a stronger acid than $\mathrm{CH_{3}COOH}$, whereas $\mathrm{(CH_{3})_{3}CCOOH}$ ($\mathrm{p}K_{a}\approx5.0$) is weaker.
1 Each molecule forms two strong hydrogen bonds with a neighbour (the cyclic dimer), and the dimers persist even in the vapour; more energy is needed to separate the molecules than for alcohols, which form only single, weaker hydrogen-bond chains.
2 In benzene (non-polar, aprotic) the acid exists as hydrogen-bonded dimers, so the number of solute particles is halved: $i=\dfrac{60}{120}=0.5$ (association, Section 1.3).
3 $\mathrm{CH_{3}CH_{2}COOH\lt CH_{3}COOH\lt HCOOH\lt ClCH_{2}COOH}$ ($4.88$, $4.76$, $3.75$, $2.87$): +I of alkyl groups decreases, –I of Cl increases acidity.
4 $\mathrm{ICH_{2}COOH\lt BrCH_{2}COOH\lt ClCH_{2}COOH\lt FCH_{2}COOH}$: the –I effect follows electronegativity, $\mathrm{F\gt Cl\gt Br\gt I}$.
5 Ethanoic ($4.76$) $\lt$ 4-methylbenzoic ($4.37$) $\lt$ benzoic ($4.19$) $\lt$ 4-nitrobenzoic acid ($3.41$).
6 $K_{a}=10^{-4.19}=6.46\times10^{-5}$. $x^{2}+K_{a}x-0.010K_{a}=0$: $x=\dfrac{-6.46\times10^{-5}+\sqrt{(6.46\times10^{-5})^{2}+2.58\times10^{-6}}}{2}=7.72\times10^{-4}\ \mathrm{mol\,L^{-1}}$, $\mathrm{pH}=3.11$ (about $7.7\%$ ionised).
7 In the methanoate ion, the two resonance structures are equivalent, so each $\mathrm{C{-}O}$ bond is the same, halfway between single and double. In the acid, one oxygen is $\mathrm{C{=}O}$ and the other $\mathrm{C{-}OH}$; resonance there involves charge separation and contributes little.
8 Three Cl atoms (–I) withdraw electron density, spreading the negative charge of the carboxylate and stabilising it. The tert-butyl group releases electrons (+I) more than methyl does, concentrating the charge and destabilising the anion.
III
Reactions of the O–H and C–OH bonds. Use $\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$.
  • 1 Write balanced equations for ethanoic acid with sodium, sodium hydroxide and sodium carbonate.
  • 2 Give one chemical test to distinguish (a) phenol and benzoic acid; (b) ethanol and ethanoic acid.
  • 3 Write balanced equations for propanoic acid with $\mathrm{PCl_{5}}$, $\mathrm{PCl_{3}}$ and $\mathrm{SOCl_{2}}$. Which is the best reagent, and why?
  • 4 Write the esterification of benzoic acid with methanol and name the ester. What does the sulphuric acid do?
  • 5 Give the product of heating ethanoic acid with $\mathrm{P_{2}O_{5}}$, with its molecular formula and molar mass.
  • 6 What is formed when ammonium benzoate is heated strongly?
  • 7 $6.0\ \mathrm{g}$ of ethanoic acid is heated with excess ethanol and a little $\mathrm{H_{2}SO_{4}}$; $67\%$ of the acid is converted. What mass of ethyl ethanoate is formed?
  • 8 Ethanoic acid is esterified with ethanol labelled with oxygen-18, $\mathrm{C_{2}H_{5}{-}{}^{18}OH}$. Where does the $^{18}\mathrm{O}$ end up, and what does this show about the mechanism?
1 $\mathrm{2CH_{3}COOH+2Na\rightarrow2CH_{3}COONa+H_{2}}$; $\ \mathrm{CH_{3}COOH+NaOH\rightarrow CH_{3}COONa+H_{2}O}$; $\ \mathrm{2CH_{3}COOH+Na_{2}CO_{3}\rightarrow2CH_{3}COONa+H_{2}O+CO_{2}}$.
2 (a) $\mathrm{NaHCO_{3}}$ solution: benzoic acid gives brisk effervescence of $\mathrm{CO_{2}}$, phenol none. (Neutral $\mathrm{FeCl_{3}}$ gives a violet colour with phenol, Chapter 7.) (b) $\mathrm{NaHCO_{3}}$: effervescence with ethanoic acid only. (Iodoform: positive for ethanol only.)
3 $\mathrm{C_{2}H_{5}COOH+PCl_{5}\rightarrow C_{2}H_{5}COCl+POCl_{3}+HCl}$; $\ \mathrm{3C_{2}H_{5}COOH+PCl_{3}\rightarrow3C_{2}H_{5}COCl+H_{3}PO_{3}}$; $\ \mathrm{C_{2}H_{5}COOH+SOCl_{2}\rightarrow C_{2}H_{5}COCl+SO_{2}+HCl}$. $\mathrm{SOCl_{2}}$ is best: the by-products are gases, so propanoyl chloride is easily obtained pure.
4 $\mathrm{C_{6}H_{5}COOH+CH_{3}OH\overset{H^{+}}{\rightleftharpoons}C_{6}H_{5}COOCH_{3}+H_{2}O}$, methyl benzoate. $\mathrm{H_{2}SO_{4}}$ protonates the carbonyl oxygen (catalyst) and also absorbs the water formed, shifting the equilibrium towards the ester.
5 Ethanoic anhydride, $\mathrm{(CH_{3}CO)_{2}O}=\mathrm{C_{4}H_{6}O_{3}}$, $M=48+6+48=102\ \mathrm{g\,mol^{-1}}$.
6 Benzamide: $\mathrm{C_{6}H_{5}COONH_{4}\xrightarrow{\Delta}C_{6}H_{5}CONH_{2}+H_{2}O}$.
7 $n=\dfrac{6.0}{60}=0.10\ \mathrm{mol}$; ester $=0.10\times0.67=0.067\ \mathrm{mol}$; $M(\mathrm{C_{4}H_{8}O_{2}})=88\ \mathrm{g\,mol^{-1}}$; mass $=5.9\ \mathrm{g}$.
8 In the ester, $\mathrm{CH_{3}CO{-}{}^{18}O{-}C_{2}H_{5}}$; the water formed is unlabelled. So the alcohol keeps its oxygen and attacks the carbonyl carbon, while the acid loses its $\mathrm{OH}$: the $\mathrm{C{-}OH}$ bond of the acid breaks, not the $\mathrm{O{-}H}$ bond.
IV
Reactions of the –COOH group and of the ring.
  • 1 Give the product of reducing hexanedioic acid with excess $\mathrm{LiAlH_{4}}$ (then $\mathrm{H_{3}O^{+}}$).
  • 2 Give the hydrocarbon formed when each salt is heated with sodalime: (a) sodium benzoate; (b) sodium 2-methylpropanoate.
  • 3 Give the products of Kolbe electrolysis of aqueous (a) sodium ethanoate, (b) potassium propanoate, at each electrode.
  • 4 Propanoic acid is treated with $\mathrm{Br_{2}}$ and red phosphorus, and the product with aqueous $\mathrm{KOH}$ followed by acid. Give both products.
  • 5 Which of these do not undergo the HVZ reaction: ethanoic acid, 2,2-dimethylpropanoic acid, benzoic acid, 2-methylpropanoic acid?
  • 6 Give the product of bromination of benzoic acid ($\mathrm{Br_{2}/FeBr_{3}}$). Why does benzoic acid not undergo Friedel–Crafts alkylation?
  • 7 Convert methylbenzene into benzaldehyde through benzoic acid, in three steps.
  • 8 Food science. (Illustrative level.) A fruit drink contains $0.10\%$ by mass of sodium benzoate. (a) How many moles are there in $500\ \mathrm{g}$ of drink? (b) Only un-ionised benzoic acid ($\mathrm{p}K_{a}=4.19$) kills microbes well. What fraction of the benzoate is present as benzoic acid at pH $3.0$ and at pH $6.0$? Why are benzoate preservatives used in acidic drinks?
1 $\mathrm{HOOC(CH_{2})_{4}COOH\rightarrow HOCH_{2}(CH_{2})_{4}CH_{2}OH}$, hexane-1,6-diol.
2 (a) $\mathrm{C_{6}H_{5}COONa+NaOH\rightarrow C_{6}H_{6}+Na_{2}CO_{3}}$: benzene; (b) $\mathrm{(CH_{3})_{2}CHCOONa}$ → $\mathrm{CH_{3}CH_{2}CH_{3}}$, propane.
3 (a) Anode: ethane and $\mathrm{CO_{2}}$; cathode: $\mathrm{H_{2}}$ (with $\mathrm{NaOH}$ in solution). (b) Anode: butane, $\mathrm{C_{2}H_{5}{-}C_{2}H_{5}}$, and $\mathrm{CO_{2}}$; cathode: $\mathrm{H_{2}}$ (with $\mathrm{KOH}$).
4 $\mathrm{CH_{3}CH_{2}COOH\rightarrow CH_{3}CH(Br)COOH}$ (2-bromopropanoic acid), then $\mathrm{CH_{3}CH(OH)COOH}$ (2-hydroxypropanoic acid, lactic acid) by substitution of Br by OH.
5 2,2-Dimethylpropanoic acid and benzoic acid: neither has an α-hydrogen.
6 3-Bromobenzoic acid. The $\mathrm{COOH}$ group deactivates the ring, and $\mathrm{AlCl_{3}}$ forms a complex with the carboxyl group, so no Friedel–Crafts reaction occurs.
7 $\mathrm{C_{6}H_{5}CH_{3}\xrightarrow{KMnO_{4}/KOH,\ \Delta;\ H_{3}O^{+}}C_{6}H_{5}COOH\xrightarrow{SOCl_{2}}C_{6}H_{5}COCl\xrightarrow{H_{2},\ Pd{-}BaSO_{4}}C_{6}H_{5}CHO}$ (Rosenmund, Section 8.1).
8 (a) $0.10\%$ of $500\ \mathrm{g}$ $=0.50\ \mathrm{g}$; $M(\mathrm{C_{6}H_{5}COONa})=144\ \mathrm{g\,mol^{-1}}$; $n=3.47\times10^{-3}\ \mathrm{mol}$. (b) Fraction un-ionised $=\dfrac{1}{1+10^{\mathrm{pH}-\mathrm{p}K_{a}}}$: at pH 3.0, $\dfrac{1}{1+10^{-1.19}}=0.94$; at pH 6.0, $\dfrac{1}{1+10^{1.81}}=0.015$. In acidic drinks most of it is present as the active benzoic acid.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks (a) Why is ethanoic acid a stronger acid than phenol? (b) Although carboxylic acids contain a $\mathrm{C{=}O}$ group, they do not give the characteristic reactions of aldehydes and ketones. Explain. (c) Give a test to distinguish benzoic acid from phenol.

(a) In ethanoate the negative charge is shared equally by two electronegative oxygen atoms (two equivalent resonance structures); in phenoxide it is spread onto less electronegative ring carbons, and the structures are not equivalent. Ethanoate is more stable, so ethanoic acid ionises more. (1 mark)
(b) The lone pair on the $\mathrm{OH}$ oxygen is delocalised onto the carbonyl group (resonance), which reduces the electrophilic character of the carboxyl carbon, so nucleophilic addition does not occur. (1 mark)
(c) Add $\mathrm{NaHCO_{3}}$ solution: benzoic acid gives brisk effervescence of $\mathrm{CO_{2}}$; phenol does not. (1 mark)

E2. JEE Main The correct order of acid strength is (A) $\mathrm{CH_{3}COOH\gt HCOOH\gt C_{6}H_{5}COOH\gt CH_{3}CH_{2}COOH}$  (B) $\mathrm{HCOOH\gt C_{6}H_{5}COOH\gt CH_{3}COOH\gt CH_{3}CH_{2}COOH}$  (C) $\mathrm{C_{6}H_{5}COOH\gt HCOOH\gt CH_{3}CH_{2}COOH\gt CH_{3}COOH}$  (D) $\mathrm{HCOOH\gt CH_{3}COOH\gt CH_{3}CH_{2}COOH\gt C_{6}H_{5}COOH}$

Idea H (no +I) $\gt$ phenyl ($sp^{2}$, electron-withdrawing) $\gt$ methyl $\gt$ ethyl (+I increases): $\mathrm{p}K_{a}$ $3.75\lt4.19\lt4.76\lt4.88$.
Answer (B).

E3. NEET Sodium propanoate is heated with sodalime. The gas evolved is (A) propane  (B) ethane  (C) butane  (D) methane

Idea Decarboxylation replaces $\mathrm{COONa}$ by H: $\mathrm{C_{2}H_{5}COONa+NaOH\rightarrow C_{2}H_{6}+Na_{2}CO_{3}}$, one carbon fewer. Butane (C) would come from Kolbe electrolysis.
Answer (B) ethane.

E4. JEE Advanced Integer type. How many of the following liberate $\mathrm{CO_{2}}$ from aqueous $\mathrm{NaHCO_{3}}$? ethanoic acid; phenol; 2,4,6-trinitrophenol ($\mathrm{p}K_{a}\approx0.4$); benzoic acid; ethanol; methanoic acid; 4-nitrophenol ($\mathrm{p}K_{a}\approx7.1$); 2-hydroxybenzoic acid. (Carbonic acid: $\mathrm{p}K_{a1}\approx6.4$.)

Rule An acid displaces $\mathrm{CO_{2}}$ from $\mathrm{HCO_{3}^{-}}$ if it is stronger than carbonic acid, i.e. its $\mathrm{p}K_{a}$ is clearly below about $6.4$.
Sort Yes: ethanoic ($4.76$), 2,4,6-trinitrophenol ($0.4$), benzoic ($4.19$), methanoic ($3.75$), 2-hydroxybenzoic ($\approx3.0$). No: phenol ($10$), ethanol ($\approx16$), 4-nitrophenol ($7.1$).
Answer $\mathbf{5}$.

E5. JEE Main Numerical value. $7.85\ \mathrm{g}$ of bromobenzene is converted into its Grignard reagent, which is poured onto dry ice and the product acidified. If the overall yield is $80\%$, the mass of benzoic acid obtained (nearest integer, in g) is ______. ($\mathrm{C}=12$, $\mathrm{H}=1$, $\mathrm{O}=16$, $\mathrm{Br}=80$.)

Moles $M(\mathrm{C_{6}H_{5}Br})=157\ \mathrm{g\,mol^{-1}}$: $n=\dfrac{7.85}{157}=0.0500\ \mathrm{mol}$. $\mathrm{C_{6}H_{5}Br\rightarrow C_{6}H_{5}MgBr\rightarrow C_{6}H_{5}COOH}$ is 1 : 1, so $0.0400\ \mathrm{mol}$ at $80\%$.
Answer $0.0400\times122=4.88\ \mathrm{g}$, nearest integer $\mathbf{5}$.

E6. Board · Case study Read the passage and answer. Ethanoic acid is the acid in vinegar. Replacing hydrogen atoms of its methyl group by halogens gives much stronger acids: chloroethanoic acid is about 78 times stronger, and trichloroethanoic acid is nearly as strong as a mineral acid. Moving the halogen further away along a chain has a much smaller effect. (a) Explain why chloroethanoic acid is stronger than ethanoic acid, and trichloroethanoic acid stronger still. (2 marks) (b) Arrange $\mathrm{FCH_{2}COOH}$, $\mathrm{ClCH_{2}COOH}$ and $\mathrm{BrCH_{2}COOH}$ in increasing acid strength. (1 mark) (c) Which is the stronger acid, 2-chlorobutanoic acid or 4-chlorobutanoic acid? Why? (1 mark)

(a) Chlorine is electron-withdrawing (–I effect): it pulls electron density from the carboxylate group, disperses the negative charge and stabilises the anion, so ionisation is favoured. Three chlorines have a combined, larger –I effect, so $\mathrm{Cl_{3}CCOO^{-}}$ is even more stable.
(b) $\mathrm{BrCH_{2}COOH\lt ClCH_{2}COOH\lt FCH_{2}COOH}$ (–I effect increases with electronegativity).
(c) 2-Chlorobutanoic acid: the inductive effect weakens rapidly with distance, and in the 4-chloro acid the chlorine is three carbons away from $\mathrm{COOH}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Does a bottle of supermarket vinegar contain as much ethanoic acid as its label claims?

Why it matters. Food-standards agencies in many countries (for example in the USA and the European Union) expect vinegar sold as such to contain at least about $4\ \mathrm{g}$ of ethanoic acid per $100\ \mathrm{mL}$; pickling recipes rely on that strength to stop the growth of bacteria. Quality-control laboratories check it by titration with standard alkali, the same acid–base chemistry as Rule 8.18.

What to do.

  1. (Illustrative data, constructed for this exercise.) $10.0\ \mathrm{mL}$ of vinegar was diluted to $100.0\ \mathrm{mL}$ in a volumetric flask. $25.0\ \mathrm{mL}$ portions of the diluted solution needed a mean titre of $20.5\ \mathrm{mL}$ of $0.100\ \mathrm{mol\,L^{-1}}$ $\mathrm{NaOH}$ (phenolphthalein indicator). Write the equation and find the concentration of ethanoic acid in the original vinegar in $\mathrm{mol\,L^{-1}}$ and in $\mathrm{g}$ per $100\ \mathrm{mL}$.
  2. Does the vinegar meet a $4\ \mathrm{g}$ per $100\ \mathrm{mL}$ standard?
  3. Estimate the pH of the undiluted vinegar ($\mathrm{p}K_{a}=4.76$; other acids in vinegar may be ignored).
  4. Explain why phenolphthalein (colour change about pH 8.3–10) is a better indicator than methyl orange (about pH 3.1–4.4) for this titration.

Tools: A calculator; Interactive 1 (try ethanoic acid); the $K_{a}$ expression of Rule 8.16.

Hand in: Your calculation, a one-line verdict on the label, and a short paragraph on the main sources of error in the titration.

Going further. Repeat the analysis on two real vinegars (malt, white, cider, or coconut toddy vinegar) if your school laboratory allows, and compare them with their labels.

Instructor note (the titration data are illustrative, constructed for this exercise): $\mathrm{CH_{3}COOH+NaOH\rightarrow CH_{3}COONa+H_{2}O}$. Moles in 25.0 mL $=2.05\times10^{-3}$; in 100 mL of diluted solution (= 10.0 mL vinegar) $8.20\times10^{-3}$; concentration $0.820\ \mathrm{mol\,L^{-1}}$, i.e. $0.820\times60/10=4.92\ \mathrm{g}$ per 100 mL: meets the standard. pH: $[\mathrm{H_{3}O^{+}}]\approx\sqrt{1.74\times10^{-5}\times0.820}=3.78\times10^{-3}$, pH $\approx2.42$. At the end point the solution contains ethanoate, a weak base, so the equivalence pH is above 7 (about 8.7 here): phenolphthalein changes colour there, while methyl orange would change far too early.

🧪 Self-check quiz

0/5 correct

Q1. The IUPAC name of $\mathrm{(CH_{3})_{2}CHCOOH}$ is

Q2. Which is the strongest acid?

Q3. Which reagent distinguishes benzoic acid from phenol?

Q4. Ethanoic acid is converted into ethanol by

Q5. The Hell–Volhard–Zelinsky reaction of propanoic acid with $\mathrm{Cl_{2}}$ and red phosphorus gives

← 8.2 Reactions of Aldehydes and KetonesChapter 9 — Amines →