Properties of Carboxylic acids

Q: Why do carboxylic acids have higher boiling points than alcohols of similar mass?

Carboxylic acids form cyclic dimers held by two intermolecular hydrogen bonds, and these hydrogen bonds are stronger than those in alcohols. The dimer behaves like a molecule of double mass, so more energy is needed to boil the acid. Acetic acid boils at 118 °C, propan-1-ol at 97 °C.

Q: Why are carboxylic acids more acidic than alcohols and phenols?

The carboxylate ion has two identical resonance structures that share the negative charge equally between two oxygen atoms. An alkoxide ion has no resonance, and in the phenoxide ion the charge spreads onto less electronegative ring carbons. The better-stabilised carboxylate makes the acid stronger.

Q: How does a chlorine atom affect the acidity of acetic acid?

Chlorine withdraws electrons by the inductive effect, dispersing the negative charge of the carboxylate ion and stabilising it. The pKa falls from 4.76 for acetic acid to 2.86 for chloroacetic acid, and further with each extra chlorine. The effect weakens quickly as the chlorine moves away from -COOH.

Q: What is the ortho effect in benzoic acids?

Almost any substituent in the ortho position makes a benzoic acid more acidic than its meta and para isomers. Steric crowding twists the -COOH out of the ring plane, and groups such as -OH can hydrogen bond to the carboxylate. Salicylic acid (pKa 2.97) is therefore far stronger than benzoic acid.

Q: What is the Hell-Volhard-Zelinsky reaction?

It is the alpha-halogenation of a carboxylic acid with chlorine or bromine in the presence of red phosphorus. Phosphorus forms a little acyl halide, which enolises and reacts with the halogen at the alpha carbon. The acid must have at least one alpha hydrogen.

Q: Why do beta-keto acids lose carbon dioxide so easily?

The carboxyl hydrogen is hydrogen bonded to the beta carbonyl oxygen, allowing a six-membered cyclic transition state in which carbon dioxide leaves and an enol forms. The enol quickly tautomerises to a ketone. Acetoacetic acid gives acetone on gentle warming.

Q: Which reactions of carboxylic acids are most important for JEE?

For JEE Main and Advanced, focus on acidity order with inductive, ortho and hydrogen-bonding effects, the esterification mechanism, HVZ, Hunsdiecker, Schmidt and soda-lime decarboxylation, and effects of heat on beta-keto, hydroxy and dicarboxylic acids. Multi-step problems often combine two of these.

Q: What properties of carboxylic acids are asked in NEET?

NEET mostly tests NCERT points: hydrogen bonding and boiling points, acidity order of substituted acids, the NaHCO3 test, formation of acid chlorides, esters, amides and anhydrides, reduction with LiAlH4, decarboxylation, HVZ reaction and meta substitution in benzoic acid.

ChemistryCarboxylic Acids And Their DerivativesFor JEE aspirants

Properties of carboxylic acids come from the -COOH group. Strong intermolecular hydrogen bonding makes them form dimers and boil higher than alcohols, while resonance stabilisation of the carboxylate ion makes them the most acidic common organic compounds ( $p K_{a} \approx 4$ to 5). Chemically, carboxylic acids react at the O-H bond (salts, fizzing with $NaHC O_{3}$ ), at the C-OH bond (acid chlorides, esters, amides, anhydrides), at the carboxyl carbon (reduction, decarboxylation) and at the $α$ -carbon (HVZ reaction). These properties of carboxylic acids are heavily tested in JEE Main, JEE Advanced and NEET.

On this page1Physical properties2Reactive sites3Acidity4Salts5Derivatives6Reduction7Loss of

C O_{2}

8HVZ9Ring substitution10Effect of heat11Formic acid12Exam practice

Key Formulas - Quick Reference

$RCOOH ⇌ RCO O^{-} + H^{+}$ , $K_{a} = \frac{[ RCO O ^{-} ] [ H ^{+} ]}{[ RCOOH ]}$ , $p K_{a} = - lo g K_{a}$
★ Must learn Acidity: RCOOH > $H_{2} C O_{3}$ > $C_{6} H_{5} OH$ > $H_{2} O$ > ROH > HC≡CH > $N H_{3}$ > RH
★ Must learn Electron-withdrawing groups raise acidity: $C l_{3} CCOOH$ > $C l_{2} CHCOOH$ > $ClC H_{2} COOH$ > $C H_{3} COOH$
★ Must learn $RCOOH + NaHC O_{3} \to RCOONa + H_{2} O + C O_{2} ↑$ (test for -COOH)
$RCOOH + SOC l_{2} \to RCOCl + S O_{2} + HCl$
★ Must learn $RCOOH + R^{'} OH ⇌ H^{+} RCOO R^{'} + H_{2} O$
$RCOOH LiAl H_{4} RC H_{2} OH$ ( $NaB H_{4}$ does not reduce -COOH)
$RCOONa + NaOH CaO, Δ RH + N a_{2} C O_{3}$ (one carbon less)
$RCOOAg + B r_{2} CC l_{4}, Δ RBr + AgBr + C O_{2}$ (Hunsdiecker)
★ Must learn $RC H_{2} COOH X_{2} /red P RCH (X) COOH$ (HVZ, needs an $α$ -H)

1. Physical Properties of Carboxylic Acids

Solubility

Carboxylic acids are polar and form hydrogen bonds with water, so the lower members (up to about four carbons) are miscible with water. As the size of the alkyl group increases, the solubility of the acid decreases because the non-polar part grows and the overall polarity is reduced. Benzoic acid is almost insoluble in cold water.

Boiling points

Because of intermolecular hydrogen bonding, carboxylic acid molecules pair up into dimers, so the boiling point of a carboxylic acid is higher than expected. The effective molecular mass of the dimer is double the actual mass, and the hydrogen bonds in acids are stronger than in alcohols. Hence carboxylic acids have higher boiling points than alcohols of comparable molecular mass.

Figure 1: Cyclic hydrogen-bonded dimer of a carboxylic acid, the reason for high boiling points and a vapour molar mass of 120 for acetic acid.

Melting points

The melting points of aliphatic carboxylic acids do not show a regular pattern. The first ten members show an alternation effect: the melting point of an acid with an even number of carbon atoms is higher than those of the next lower and next higher homologues with an odd number of carbon atoms.

In even-carbon acids, the terminal methyl group and the carboxyl group lie on opposite sides of the zig-zag carbon chain, so the molecules fit better in the crystal lattice and intermolecular forces are stronger. In odd-carbon acids, both groups lie on the same side, the molecules fit poorly and the melting point is relatively lower.

Figure 2: Even-odd alternation in melting points. Even-carbon acids pack better in the crystal.

Aromatic carboxylic acids have higher melting and boiling points than aliphatic acids of comparable molecular mass, because the planar benzene ring lets the molecules pack closely in the crystal.

Key idea

Two intermolecular hydrogen bonds per dimer: that one picture explains the high boiling points and the vapour molar mass of 120 for acetic acid.

2. How Carboxylic Acids React

The characteristic chemical behaviour of carboxylic acids is determined by the carboxyl group, -COOH, which is made up of a carbonyl group (C=O) and a hydroxyl group (-OH). It is the -OH that actually undergoes nearly every reaction, by loss of $H^{+}$ or by replacement with another group, but it does so in a way that is possible only because of the effect of the C=O.

(a) Removal of $H^{+}$ : the O-H bond breaks when the acid reacts with a base.
(b) C-O bond cleavage: -OH is replaced by -Cl or $- N H_{2}$ using $PC l_{5}$ , $PC l_{3}$ , $SOC l_{2}$ or $N H_{3} /Δ$ .
(c) Nucleophilic attack at the carboxyl carbon: as in ester formation. Replacement of -OH by $- N H_{2}$ or -Cl is a substitution at this carbon.
(d) Halogenation at the $α$ -carbon: with P/ $B r_{2}$ (Hell-Volhard-Zelinsky reaction).
(e) Oxidation of the $α$ -methylene group: by $Se O_{2}$ , giving an $α$ -keto acid.

RC H_{2} COOH Se O_{2} R - CO - COOH + H_{2} O + Se

The main reactions are summarised in the map below and explained section by section.

Figure 3: Reaction map of carboxylic acids: reagents (orange) and products around

RCOOH

3. Acidity of Carboxylic Acids

The acidity of a carboxylic acid is due to the resonance stabilisation of its anion. Both the acid and its anion are stabilised by resonance, but the stabilisation is far greater for the anion, because the anion has two identical resonating structures while the acid's structures are not equivalent. Because of this resonance, both carbon-oxygen bonds in the carboxylate anion have identical bond lengths; in the carboxylic acid they are not identical.

Figure 4: Resonance in the acid is non-equivalent, while the carboxylate ion has two identical structures and equal C-O bonds (1.27 Å).

Effect of substituents on acidity

The acidity of a carboxylic acid depends strongly on the substituent attached to the -COOH group. Since acidity comes from resonance stabilisation of the anion, any substituent that stabilises the anion increases acidity, and any substituent that destabilises it decreases acidity. An electron-withdrawing group ( $- I$ effect) disperses the negative charge of the anion and makes it more stable, increasing acidity. An electron-releasing group ( $+ I$ effect) intensifies the negative charge, makes the anion less stable and decreases acidity. On this basis:

Number of halogens: more halogen atoms on the $α$ -carbon give a stronger acid: $C l_{3} CCOOH$ > $C l_{2} CHCOOH$ > $ClC H_{2} COOH$ > $C H_{3} COOH$ .
Distance of halogen: the inductive effect decreases with distance, so 2-chlorobutanoic acid > 3-chlorobutanoic acid > 4-chlorobutanoic acid.
Electronegativity of halogen: $FC H_{2} COOH$ > $ClC H_{2} COOH$ > $BrC H_{2} COOH$ > $IC H_{2} COOH$ .
Electron-releasing alkyl groups: acidity falls as the $+ I$ effect grows: $C H_{3} COOH$ > $(C H_{3})_{2} CHCOOH$ > $(C H_{3})_{3} CCOOH$ .
Hybridisation of the carbon joined to -COOH: an sp carbon is the most electronegative, then sp $^{2}$ , then sp $^{3}$ : $HC \equiv C - COOH$ ( $p K_{a}$ 1.84) > $C H_{2} = CH - COOH$ (4.25) > $C H_{3} - C H_{2} - COOH$ (4.88).

Figure 5:

p K_{a}

values showing the effect of number, electronegativity and distance of halogen substituents.

JEE Trick: rank acids in 10 seconds

Ask three questions in order: (1) Is there a $- I$ group, and how many? (2) How close is it to -COOH ( $α$ beats $β$ beats $γ$ )? (3) How electronegative is it (F > Cl > Br > I)? Alkyl groups work the opposite way and weaken the acid.

Solved Example 1

Which one of the following would be expected to be most highly ionised in water?
(A)

C H_{2} Cl - C H_{2} - C H_{2} - COOH

(B)

C H_{3} - CHCl - C H_{2} - COOH

(C)

C H_{3} - C H_{2} - CHCl - COOH

(D)

C H_{3} - C H_{2} - CC l_{2} - COOH

Solution:

The strongest acid ionises most. Option (D) has two chlorine atoms on the $α$ -carbon, giving the largest $- I$ effect at the shortest distance. Answer: (D)

Solved Example 2

Arrange in order of increasing acidity: (i) HCOOH,

ClC H_{2} COOH

C H_{3} COOH

(ii)

C H_{3} COOH

(C H_{3})_{2} CHCOOH

(C H_{3})_{3} CCOOH

(iii)

ClC H_{2} COOH

C l_{2} CHCOOH

C l_{3} CCOOH

(iv)

ClC H_{2} COOH

C H_{3} C H_{2} COOH

ClC H_{2} C H_{2} COOH

(C H_{3})_{2} CHCOOH

C H_{3} COOH

(v)

C H_{3} COOH

C l_{2} CHCOOH

C H_{3} C H_{2} COOH

C l_{3} CCOOH

ClC H_{2} COOH

Solution:

(i) $C H_{3} COOH$ < HCOOH < $ClC H_{2} COOH$ (the methyl group is $+ I$ ; Cl is $- I$ )

(ii) $(C H_{3})_{3} CCOOH$ < $(C H_{3})_{2} CHCOOH$ < $C H_{3} COOH$ (more alkyl groups, more $+ I$ )

(iii) $ClC H_{2} COOH$ < $C l_{2} CHCOOH$ < $C l_{3} CCOOH$ (more chlorines)

(iv) $(C H_{3})_{2} CHCOOH$ < $C H_{3} C H_{2} COOH$ < $C H_{3} COOH$ < $ClC H_{2} C H_{2} COOH$ < $ClC H_{2} COOH$

(v) $C H_{3} C H_{2} COOH$ < $C H_{3} COOH$ < $ClC H_{2} COOH$ < $C l_{2} CHCOOH$ < $C l_{3} CCOOH$

Solved Example 3

(a) Can the aromatic ring in benzoic acid stabilise the benzoate anion by

π

-electron delocalisation? (b) Discuss the electronic effect of the

p - N O_{2}

group in

p - N O_{2} C_{6} H_{4} CO O^{-}

Solution:

(a) Practically no. The only structure that delocalises electron density from $C_{6} H_{5} CO O^{-}$ into the ring puts a positive charge on an oxygen atom with only six electrons. Such an extremely high-energy structure contributes nothing, so the ring does not stabilise the anion by resonance.

(b) There is no direct resonance interaction between $- CO O^{-}$ and $- N O_{2}$ . However, resonance of the nitro group with the ring places some positive charge on the ring carbon that carries $- CO O^{-}$ . This creates a strong electron-withdrawing inductive effect, which stabilises the anion and makes p-nitrobenzoic acid a stronger acid than benzoic acid.

Solved Example 4

The

p K_{a}

of acetylsalicylic acid (aspirin) is 3.5. The pH of gastric juice in the human stomach is about 2 to 3, and the pH in the small intestine is about 8. Aspirin will be:
(a) unionised in the small intestine and in the stomach
(b) completely ionised in the stomach and almost unionised in the small intestine
(c) ionised in the stomach and almost unionised in the small intestine
(d) ionised in the small intestine and almost unionised in the stomach

Solution:

When pH is below $p K_{a}$ , the acid form dominates; when pH is above $p K_{a}$ , the anion dominates. In the stomach (pH 2 to 3, below 3.5) the high $H^{+}$ concentration suppresses ionisation, so aspirin is mostly unionised. In the intestine (pH 8, far above 3.5) it is almost fully ionised. Answer: (d)

Ortho effect in aromatic acids

An ortho-substituted benzoic acid is a stronger acid than its meta and para isomers, whether the substituent is electron-withdrawing or electron-releasing. This is called the ortho effect. It comes from the joint action of steric effects and intramolecular hydrogen bonding, which stabilise the carboxylate anion because the substituent is so close. Groups such as -OH, -Cl and $- N O_{2}$ give extra stabilisation through direct interaction, such as intramolecular hydrogen bonding in the salicylate ion.

Figure 6: Ortho effect: the salicylate ion is stabilised by an intramolecular hydrogen bond.

Maleic acid and fumaric acid

Hydrogen bonding that involves the acidic hydrogen weakens an acid, while hydrogen bonding in the conjugate base strengthens it. Both effects appear in the two isomeric butenedioic acids.

Figure 7: Maleate and fumarate monoanions: intramolecular hydrogen bonding explains

K_{1}

and

K_{2}

Solved Example 5

On the basis of hydrogen bonding, explain why the second ionisation constant

K_{2}

of fumaric acid is greater than that of maleic acid.

Solution:

Both dicarboxylic acids have two ionisable hydrogen atoms. Consider the second ionisation step. In the maleate monoanion, the remaining ionisable hydrogen takes part in an intramolecular hydrogen bond with the neighbouring $- CO O^{-}$ group. More energy is needed to remove this hydrogen because the hydrogen bond must be broken, so the maleate monoanion is the weaker acid. The fumarate monoanion (trans) cannot form this hydrogen bond, so its $K_{2}$ is larger.

Solved Example 6

Maleic acid (cis-butenedioic acid) is more acidic than fumaric acid (trans-butenedioic acid) in its first ionisation. Why?

Solution:

After the first ionisation of the cis acid, the anion is stabilised by an intramolecular hydrogen bond from the second -COOH group, which lies on the same side as $- CO O^{-}$ . The anion from the trans acid cannot be stabilised this way, because its -COOH group is on the opposite side. A more stable anion means easier ionisation, so $K_{1}$ (maleic) > $K_{1}$ (fumaric).

The ranking rules as a flowchart:

Figure 8: Three questions rank most acid pairs: ortho group, electron-withdrawing group, electron-releasing group; anything that stabilises

RCO O^{-}

strengthens the acid.

Key idea

Anything that spreads the negative charge of

RCO O^{-}

- I

group, an sp carbon, an intramolecular H-bond in the anion) makes the acid stronger.

4. Salt Formation

Carboxylic acids are weak acids, and their carboxylate anions are weak conjugate bases. Solutions of their salts are slightly alkaline because the carboxylate anion is hydrolysed. Compared with other species, the orders of acidity and of basicity of the corresponding conjugate bases are:

Acidity: RCOOH > HOH > ROH > HC≡CH > $N H_{3}$ > RH

Basicity: $RCO O^{-}$ < $H O^{-}$ < $R O^{-}$ < $HC \equiv C^{-}$ < $N H_{2}^{-}$ < $R^{-}$

Carboxylic acids react with active metals to liberate hydrogen, and they dissolve in both NaOH and $NaHC O_{3}$ solutions.

2C H_{3} COOH + 2Na ⟶ 2C H_{3} CO O^{-} N a^{+} + H_{2}

C H_{3} COOH + NaOH ⟶ C H_{3} CO O^{-} N a^{+} + H_{2} O

C H_{3} COOH + NaHC O_{3} ⟶ C H_{3} CO O^{-} N a^{+} + H_{2} O + C O_{2}

2C H_{3} COOH + Zn ⟶ (C H_{3} COO)_{2} Zn + H_{2}

C H_{3} (C H_{2})_{10} COOH + NaOH

⟶ C H_{3} (C H_{2})_{10} CO O^{-} N a^{+} + H_{2} O

Here zinc acetate and sodium laurate (from lauric acid) are formed. Benzoic acid similarly gives sodium benzoate with $NaHC O_{3}$ . Carboxylic acids also react with basic oxides and carbonates:

C_{6} H_{5} COOH + NaHC O_{3} ⟶ C_{6} H_{5} COONa + C O_{2} + H_{2} O

2RCOOH + CaO ⟶ (RCOO)_{2} Ca + H_{2} O

2RCOOH + CaC O_{3} ⟶ (RCOO)_{2} Ca + H_{2} O + C O_{2}

Figure 9: Acid strength ladder. Only acids stronger than

H_{2} C O_{3}

release

C O_{2}

from

NaHC O_{3}

Solved Example 7

Compound A,

C_{4} H_{8} O_{2}

, has these properties: (i) it reacts with sodium bicarbonate to liberate

C O_{2}

; (ii) on fusion with alkali it gives propane; (iii) with

Ca (OH)_{2}

it gives a product

C_{8} H_{14} O_{4} Ca

, which on heating decomposes to diisopropyl ketone. Identify A.

Solution:

(i) Liberation of $C O_{2}$ with $NaHC O_{3}$ shows a -COOH group, so A is $C_{3} H_{7} COOH$ .

(ii) $C_{3} H_{7} COOH NaOH + CaO C_{3} H_{8} + N a_{2} C O_{3}$ . Both butanoic acid and 2-methylpropanoic acid would give propane.

(iii) $C_{8} H_{14} O_{4} Ca$ is the calcium salt $(C_{3} H_{7} COO)_{2} Ca$ . On heating it gives diisopropyl ketone, $(C H_{3})_{2} CH - CO$ $- CH (C H_{3})_{2}$ , so the chain is branched. A is 2-methylpropanoic acid, $(C H_{3})_{2} CH - COOH$ .

[(C H_{3})_{2} CHCOO]_{2} Ca

Δ (C H_{3})_{2} CH - CO - CH (C H_{3})_{2} + CaC O_{3}

Quick Recall: tap to check

Which of phenol and benzoic acid gives effervescence with

NaHC O_{3}

Benzoic acid only; phenol (

p K_{a}

10) is weaker than carbonic acid (6.4).

Why is a solution of sodium acetate slightly alkaline?

Acetate is a weak base and is hydrolysed:

C H_{3} CO O^{-}

H_{2} O

⇌

C H_{3} COOH

O H^{-}

5. Conversion into Functional Derivatives

The -OH of a carboxylic acid can be replaced by Z = -Cl, -OR' or $- N H_{2}$ to give acid chlorides, esters and amides; removing water between two acid molecules gives an anhydride.

(a) Conversion into acid chlorides

Carboxylic acids react with thionyl chloride, phosphorus trichloride or phosphorus pentachloride to give acid chlorides.

RCOOH + SOC l_{2} ⟶ RCOCl + S O_{2} ↑ + HCl ↑

3RCOOH + PC l_{3} ⟶ 3RCOCl + H_{3} P O_{3}

C_{6} H_{5} COOH + PC l_{5} 100^{\circ} C C_{6} H_{5} COCl + POC l_{3} + HCl

NEET Trick: why SOCl2 is preferred

With $SOC l_{2}$ both by-products, $S O_{2}$ and HCl, are gases that escape, so the acid chloride is obtained pure without separating $POC l_{3}$ or $H_{3} P O_{3}$ .

Solved Example 8

Benzoyl chloride is prepared from benzoic acid by:
(A)

C l_{2}

h ν

(B)

S O_{2} C l_{2}

(C)

SOC l_{2}

(D)

C l_{2}

H_{2} O

Solution:

Replacing -OH by -Cl needs a reagent such as $SOC l_{2}$ , $PC l_{3}$ or $PC l_{5}$ . Chlorine alone cannot convert -COOH into -COCl. Answer: (C)

(b) Conversion into esters (esterification)

A carboxylic acid reacting with an alcohol in the presence of a dehydrating agent (conc. $H_{2} S O_{4}$ or dry HCl gas) gives an ester. The reaction is known as esterification.

R - COOH + R^{'} OH ⇌ H^{+} R - COO R^{'} + H_{2} O

This reaction is reversible, and the same catalyst ( $H^{+}$ ) that catalyses esterification necessarily catalyses the reverse reaction, hydrolysis. The equilibrium is particularly unfavourable when phenols (ArOH) are used instead of alcohols; yet if water is removed during the reaction, phenolic esters (RCOOAr) are obtained in high yield. A bulky group near the site of reaction, in either the alcohol or the acid, slows down esterification (and its reverse, hydrolysis).

Reactivity of alcohols: $C H_{3} OH$ > 1° > 2° > 3°

Reactivity of acids: HCOOH > $C H_{3} COOH$ > $RC H_{2} COOH$ > $R_{2} CHCOOH$ > $R_{3} CCOOH$

Mechanism. The steps for forming an ester from an acid and an alcohol are the reverse of the steps for acid-catalysed hydrolysis of an ester, so the reaction can go either way depending on the conditions. A carboxylic acid does not react with an alcohol unless a strong acid is used as a catalyst: protonation makes the carbonyl group more electrophilic and lets it react with the alcohol, which is a weak nucleophile.

Protonation of the carbonyl oxygen of the acid.
Nucleophilic attack of the alcohol on the carbonyl carbon, giving a tetrahedral intermediate.
Proton transfer to one of the -OH groups, turning it into $- O H_{2}^{+}$ , a good leaving group.
Loss of water, giving the protonated ester.
Loss of $H^{+}$ , giving the ester and regenerating the catalyst.

Figure 10: Mechanism of acid-catalysed (Fischer) esterification.

C_{6} H_{5} COOH + C H_{3} OH ⇌ H^{+} C_{6} H_{5} COOC H_{3} + H_{2} O

C H_{3} COOH + C_{6} H_{5} C H_{2} OH ⇌ H^{+} C H_{3} COOC H_{2} C_{6} H_{5} + H_{2} O

(C H_{3})_{3} CCOOH SOC l_{2} (C H_{3})_{3} CCOCl

C_{2} H_{5} OH (C H_{3})_{3} CCOO C_{2} H_{5}

These give methyl benzoate, benzyl acetate and ethyl trimethylacetate. For a crowded acid such as trimethylacetic acid, the acid chloride route avoids the slow direct esterification.

Solved Example 9

Assign a structure to A:

HOOC - (C H_{2})_{9} - COOH C H_{3} OH (excess) H_{2} S O_{4}, Δ A

Solution:

With excess methanol and acid catalyst, both -COOH groups are esterified. A is the diester $C H_{3} OOC - (C H_{2})_{9}$ $- COOC H_{3}$ (dimethyl undecanedioate).

Solved Example 10

What is the final product P when a carboxylic acid is treated with diazomethane?

R - COOH + C H_{2} N_{2} \to P

Solution:

The acid protonates the carbon of diazomethane, giving a carboxylate ion and the methyldiazonium ion $C H_{3} - N_{2}^{+}$ . The carboxylate then attacks the methyl carbon in an $S_{N} 2$ step, and nitrogen gas leaves. P is the methyl ester, $R - COOC H_{3}$ .

RCOOH + C H_{2} N_{2} ⟶ RCO O^{-} + C H_{3} N_{2}^{+}

S_{N} 2 RCOOC H_{3} + N_{2}

Fischer esterification

RCOOH + R'OH with $H^{+}$ , heat.

Reversible: use excess alcohol or remove water. Slow for crowded acids and 3° alcohols.

Diazomethane

RCOOH + $C H_{2} N_{2}$ in ether.

Gives only methyl esters, fast and irreversible because $N_{2}$ escapes.

JEE Advanced

Isotope labelling proves acyl-oxygen cleavage. When acetic acid is esterified with methanol labelled with oxygen-18, the label ends up in the ester, and the water formed contains ordinary oxygen:

$C H_{3} COOH$ + $C H_{3} -^{18} OH$ $⇌ H^{+}$ $C H_{3} CO -^{18} OC H_{3}$ + $H_{2} O$

So the acid loses its -OH and the alcohol loses only its H, exactly as the tetrahedral-intermediate mechanism predicts ( $A_{AC} 2$ ). Esters of tertiary alcohols can instead break the alkyl-oxygen bond through a carbocation.

(c) Conversion into amides

Amides are usually made through the acid chloride, which reacts readily with ammonia.

R - COOH SOC l_{2} R - COCl N H_{3} R - CON H_{2}

C_{6} H_{5} C H_{2} COOH SOC l_{2} C_{6} H_{5} C H_{2} COCl

N H_{3} C_{6} H_{5} C H_{2} CON H_{2}

Here phenylacetic acid gives phenylacetyl chloride and then phenylacetamide. An acid can also be heated with ammonia or an amine: the ammonium salt forms first and loses water on strong heating.

RCOOH + N H_{3} ⟶ RCO O^{-} N H_{4}^{+} Δ RCON H_{2} + H_{2} O

RCOOH + C H_{3} N H_{2} ⟶ RCO O^{-} C H_{3} N H_{3}^{+}

Δ, - H_{2} O RCONHC H_{3}

Solved Example 11

Discuss why the characteristic reaction of aldehydes and ketones is nucleophilic addition, while acyl compounds give nucleophilic substitution products.

Solution:

Figure 11: Nucleophilic addition (aldehydes, ketones) versus nucleophilic acyl substitution (acyl compounds).

The first step in both reactions is nucleophilic addition at the carbonyl carbon atom. The two reactions differ only after this attack. The tetrahedral intermediate from an aldehyde or ketone usually accepts a proton to form a stable addition product. By contrast, the intermediate from an acyl compound usually eliminates a leaving group: this regenerates the carbon-oxygen double bond and gives a substitution product. Acyl substitution is therefore a nucleophilic addition-elimination process, and acyl compounds react this way because they carry good leaving groups (-Cl, -OCOR, -OR, $- N H_{2}$ ) on the carbonyl carbon, while H and R in aldehydes and ketones are very poor leaving groups.

(d) Conversion into anhydrides

A carboxylic acid heated with a dehydrating agent such as $P_{2} O_{5}$ loses a water molecule between two acid molecules to form an anhydride. Dicarboxylic acids that can form five- or six-membered rings give cyclic anhydrides simply on heating.

2R - COOH P_{2} O_{5}, Δ (R - CO)_{2} O + H_{2} O

HOOC - C H_{2} C H_{2} - COOH Δ succinic anhydride + H_{2} O

Solved Example 12

When acetic acid reacts with ketene, the product formed is:
(a) ethyl acetate
(b) acetoacetic ester
(c) acetic anhydride
(d) no reaction

Solution:

The O-H of acetic acid adds across the C=C of ketene, and the acetyl group ends up bonded to the acetate oxygen.

C H_{3} COOH + C H_{2} = C = O ⟶ (C H_{3} CO)_{2} O

Answer: (c)

6. Reduction of Carboxylic Acids

Lithium aluminium hydride reduces carboxylic acids to primary alcohols.

R - COOH LiAl H_{4} R - C H_{2} OH (1^{\circ} alcohol)

4 (C H_{3})_{3} CCOOH + 3LiAl H_{4}

ether [(C H_{3})_{3} CC H_{2} O]_{4} AlLi + 2LiAl O_{2} + 4 H_{2}

Hydrolysis of the aluminium alkoxide then gives neopentyl alcohol, $(C H_{3})_{3} CC H_{2} OH$ (2,2-dimethylpropan-1-ol). In the same way, m-toluic acid is reduced to m-methylbenzyl alcohol.

JEE/NEET Trick: which reducing agent?

$LiAl H_{4}$ and diborane ( $B_{2} H_{6}$ ) reduce -COOH to $- C H_{2} OH$ . $NaB H_{4}$ does not reduce carboxylic acids, so it can reduce a ketone in the same molecule while leaving -COOH untouched.

Solved Example 13

Identify X, Y and Z:

C H_{2} = CH - Br Mg/ether X (i) C O_{2} (ii) H_{3} O^{+} Y LiAl H_{4} Z

. Z with HBr gives A, which on reductive ozonolysis gives formaldehyde as one of the products. Deduce the whole scheme.

Solution:

X is vinylmagnesium bromide, $C H_{2} = CH - MgBr$ . Carbonation and acidification give Y, acrylic acid, $C H_{2} = CH - COOH$ . $LiAl H_{4}$ reduces -COOH (not the C=C) to give Z, allyl alcohol, $C H_{2} = CH - C H_{2} OH$ .

With HBr, the -OH is protonated and lost as water, giving the resonance-stabilised allyl carbocation; bromide then attacks to give A, allyl bromide, $C H_{2} = CH - C H_{2} Br$ . Reductive ozonolysis ( $O_{3}$ , then Zn/ $H_{2} O$ ) splits the C=C:

C H_{2} = CH - C H_{2} Br

(i) O_{3} (ii) Zn/ H_{2} O HCHO + OHC - C H_{2} Br

Quick Recall: tap to check

Which reagents reduce -COOH to

- C H_{2} OH

LiAl H_{4}

or diborane (

B_{2} H_{6}

);

NaB H_{4}

does not.

What does

HOOC - C H_{2} - C H_{2}

- CO - C H_{3}

give with

NaB H_{4}

Only the ketone is reduced:

HOOC - C H_{2} - C H_{2}

- CH (OH) - C H_{3}

7. Reactions Involving Loss of the -COOH Group

(a) Decarboxylation with soda lime

Sodium salts of carboxylic acids heated with soda lime (NaOH and CaO) give alkanes with one carbon atom less than the parent acid.

R - COOH + NaOH ⟶ RCOONa + H_{2} O

RCOONa + NaOH CaO, Δ R - H + N a_{2} C O_{3}

For example, lactic acid, $C H_{3} - CH (OH) - COOH$ , gives ethanol, $C H_{3} C H_{2} OH$ , on heating with soda lime.

(b) Hunsdiecker reaction

The silver salt of a carboxylic acid heated with bromine in $CC l_{4}$ gives an alkyl bromide with one carbon less. The reaction proceeds by a free-radical chain mechanism.

R - COOH A g_{2} O R - CO O^{-} A g^{+}

B r_{2}, CC l_{4} R - Br + AgBr + C O_{2}

RCOOBr ⟶ RCO O^{∙} + B r^{∙} (initiation)

RCO O^{∙} ⟶ R^{∙} + C O_{2}

R^{∙} + RCOOBr ⟶ RBr + RCO O^{∙}

Solved Example 14

A salt (A),

C_{4} H_{5} O_{2} Ag

, on refluxing with bromine gives (B),

C_{3} H_{5} Br

. (B) on heating with alcoholic KOH gives (C),

C_{3} H_{4}

, which decolourises

B r_{2}

CC l_{4}

and cold dilute

KMn O_{4}

but does not react with ammoniacal

AgN O_{3}

C u_{2} C l_{2}

. (C) on ozonolysis gives (D),

C_{3} H_{4} O_{4}

, which on heating eliminates

C O_{2}

to give acetic acid. What are (A) to (D)?

Solution:

(D) loses $C O_{2}$ on heating to give acetic acid, so (D) is malonic acid, $HOOC - C H_{2} - COOH$ . Ozonolysis giving one malonic acid means (C) is a ring with one C=C: cyclopropene. It has no terminal alkyne hydrogen, which is why it does not react with ammoniacal $AgN O_{3}$ or $C u_{2} C l_{2}$ .

(B) is bromocyclopropane (dehydrobromination gives cyclopropene), and (A), which lost $C O_{2}$ in a Hunsdiecker reaction, is silver cyclopropanecarboxylate.

A = silver cyclopropanecarboxylate, B = bromocyclopropane, C = cyclopropene, D = malonic acid.

(c) Schmidt reaction

A carboxylic acid reacts with hydrazoic acid in the presence of conc. $H_{2} S O_{4}$ at about 90 °C to form a primary amine with one carbon less.

R - COOH + N_{3} H conc. H_{2} S O_{4} 90^{\circ} C R - N H_{2} + N_{2} + C O_{2}

The protonated acid is attacked by $H N_{3}$ ; after loss of water, nitrogen gas leaves as the R group shifts from carbon to nitrogen, giving an isocyanate ( $R - N = C = O$ ). Water converts the isocyanate into a carbamic acid, which loses $C O_{2}$ to give the amine. For example, benzoic acid gives aniline.

C_{6} H_{5} COOH + N_{3} H conc. H_{2} S O_{4} 90^{\circ} C C_{6} H_{5} N H_{2} + N_{2} + C O_{2}

(d) Reaction with organometallic compounds

A Grignard reagent only removes the acidic proton, giving the salt and an alkane. Excess methyllithium goes further: two moles add to give a dilithium salt, which on hydrolysis loses water to give a methyl ketone.

RCOOH + C H_{3} MgBr ⟶ RCO O^{-}^{+} MgBr + C H_{4}

RCOOH 2C H_{3} Li R - C (OLi)_{2} - C H_{3} H_{2} O R - CO - C H_{3}

(e) Easy decarboxylation of $β$ -keto acids

The loss of $C O_{2}$ from a carboxylic acid is favoured by the stability of carbon dioxide, but it is usually slow. Some groups make it rapid. Acids with a carbonyl group on the $β$ -carbon ( $β$ -keto acids) decarboxylate readily when heated to 100 to 150 °C.

R - CO - C H_{2} - COOH 100 - 150^{\circ} C R - CO - C H_{3} + C O_{2}

Two facts explain the ease. First, when the carboxylate ion decarboxylates, it forms a resonance-stabilised enolate anion, much more stable than the $R - C H_{2}^{-}$ anion formed from an ordinary carboxylate. Second, the free acid can decarboxylate through a six-membered cyclic transition state. The same happens for $Y - CO - C H_{2} - COOH$ when Y = OH (malonic acid) or H ( $β$ -aldehydo acid). A $β$ -halo acid decarboxylates in base by elimination instead.

Figure 12: Decarboxylation of a

β

-keto acid through a six-membered cyclic transition state.

C_{6} H_{5} - CH (Br) - C H_{2} - COOH + KOH heat C_{6} H_{5} - CH = C H_{2} + C O_{2} + KBr + H_{2} O

Solved Example 15

Cinnamic acid reacts with

B r_{2}

to give (A), which on reaction with hot

N a_{2} C O_{3}

solution is converted into (B). (B) on reaction with KOH gives an acetylenic compound (C). Give the reaction scheme.

Solution:

Bromine adds across the C=C of cinnamic acid, $C_{6} H_{5} - CH = CH - COOH$ , giving (A), 2,3-dibromo-3-phenylpropanoic acid. In hot $N a_{2} C O_{3}$ , the carboxylate loses $C O_{2}$ while bromide leaves from the $β$ -carbon (decarboxylative elimination), giving (B), $β$ -bromostyrene, $C_{6} H_{5} - CH = CH - Br$ . KOH then eliminates HBr to give (C), phenylacetylene.

C_{6} H_{5} CH = CHCOOH B r_{2} C_{6} H_{5} CH (Br) CH (Br) COOH

N a_{2} C O_{3} C_{6} H_{5} CH = CHBr

C_{6} H_{5} CH = CHBr KOH, Δ C_{6} H_{5} - C \equiv CH

Solved Example 16

Which of the following would be expected to decarboxylate when heated?
(a)

C H_{3} - CO - C H_{2} - COOH

(b)

C H_{3} - CO - C H_{2} - CO - C H_{3}

(c)

C H_{3} - CO - C H_{2}

- CO - CO - C H_{3}

(d)

C H_{3} - CO - C H_{2}

- C (OH) = C H_{2}

Solution:

Only (a) has a carboxyl group, and it is a $β$ -keto acid. It decarboxylates readily through the low-energy six-membered cyclic transition state, giving acetone. Answer: (a)

Solved Example 17

What is the end product when 2-acetylcyclohexanone is treated with (1)

I_{2}

+ NaOH,

Δ

and then (2)

H^{+}

Δ

?
(a) yellow ppt of

CH I_{3}

and 2-oxocyclohexane-1-carboxylic acid
(b) yellow ppt of

CH I_{3}

and 2-oxocyclohexane-1-carbaldehyde
(c) yellow ppt of

CH I_{3}

and cyclohexanone
(d) yellow ppt of

CH I_{3}

and hexanedioic acid

Solution:

The acetyl group (a methyl ketone) gives the iodoform reaction, forming $CH I_{3}$ and 2-oxocyclohexane-1-carboxylic acid after acidification. This intermediate is a $β$ -keto acid, so on heating it loses $C O_{2}$ to give cyclohexanone. Answer: (c)

(f) Kolbe electrolysis

Electrolysis of a concentrated aqueous solution of the sodium or potassium salt of a carboxylic acid also removes -COOH, but the alkyl radicals formed at the anode pair up. The hydrocarbon has twice the number of carbon atoms in the alkyl group of the acid.

2RCOONa + 2 H_{2} O electrolysis R - R + 2C O_{2} + H_{2} + 2NaOH

At the anode: $2RCO O^{-} \to 2RCO O^{∙} + 2 e^{-}$ , then $RCO O^{∙} \to R^{∙} + C O_{2}$ and $2 R^{∙} \to R - R$ . Hydrogen and NaOH form at the cathode. Sodium acetate gives ethane:

2C H_{3} COONa + 2 H_{2} O electrolysis C H_{3} - C H_{3} + 2C O_{2} + H_{2} + 2NaOH

Soda-lime decarboxylation

RCOONa + NaOH (CaO), heat.

Product R-H: one carbon fewer than the acid.

Kolbe electrolysis

Aqueous RCOONa, electrolysis.

Product R-R: twice the carbons of R (acetate gives ethane).

Key idea

Soda lime, Hunsdiecker and Schmidt each remove one carbon; Kolbe electrolysis joins two alkyl groups, and

β

-keto acids lose

C O_{2}

just on warming.

8. Substitution in the Alkyl Chain

Hell-Volhard-Zelinsky (HVZ) reaction

In the presence of phosphorus, aliphatic carboxylic acids react smoothly with chlorine or bromine to give a compound in which an $α$ -hydrogen has been replaced by halogen. The reaction does not stop at monosubstitution; with excess halogen, all $α$ -hydrogens are replaced.

C H_{3} COOH C l_{2} /P ClC H_{2} COOH

C l_{2} /P C l_{2} CHCOOH C l_{2} /P C l_{3} CCOOH

The function of phosphorus is to convert a little of the acid into the acid halide, and it is the acid halide, not the acid itself, that is halogenated. The acid halide enolises much more readily than the acid, and the enol attacks the halogen.

2P + 3 X_{2} ⟶ 2P X_{3}

3RC H_{2} COOH + P X_{3} ⟶ 3RC H_{2} COX + H_{3} P O_{3}

RC H_{2} COX + X_{2} ⟶ RCH (X) COX + HX

H_{2} O RCH (X) C O_{2} H

Figure 13: Hell-Volhard-Zelinsky mechanism and uses of the

α

-halo acid.

The halogen of these halogenated acids undergoes nucleophilic displacement and elimination just as it does in simple alkyl halides. Halogenation is therefore the first step in converting a carboxylic acid into many important substituted acids.

R - CH (Br) - COOH + N H_{3}

⟶ R - CH (N H_{2}) - COOH (α - amino acid)

R - CH (Cl) - COOH + NaOH

H^{+} R - CH (OH) - COOH (α - hydroxy acid)

(C H_{3})_{2} CHC H_{2} COOH B r_{2} /P (C H_{3})_{2} CHCH (Br) COOH

The first two equations give an $α$ -amino acid and an $α$ -hydroxy acid; the last converts isovaleric acid into $α$ -bromoisovaleric acid.

Solved Example 18

Which of the following will not undergo the HVZ reaction?
(A) 2,2-dimethylpropanoic acid
(B) propanoic acid
(C) acetic acid
(D) 2-methylpropanoic acid

Solution:

The HVZ reaction replaces an $α$ -hydrogen. 2,2-Dimethylpropanoic acid, $(C H_{3})_{3} C - COOH$ , has no hydrogen on its $α$ -carbon. Answer: (A)

Solved Example 19

Which of the following does not undergo the Hell-Volhard-Zelinsky reaction?
(a) HCOOH
(b)

CC l_{3} COOH

(c)

C_{6} H_{5} COOH

(d) all of these

Solution:

None of these acids has a hydrogen atom on an $α$ -carbon: formic acid has no $α$ -carbon, trichloroacetic acid's $α$ -carbon carries three Cl atoms, and in benzoic acid the -COOH is on a ring carbon with no H. Answer: (d)

Solved Example 20

Acetic acid reacts with chlorine in the presence of a catalyst (anhydrous

FeC l_{3}

) to give:
(A) acetyl chloride
(B) methyl chloride
(C) trichloroacetic acid
(D) chloral hydrate

Solution:

With a halogen carrier, chlorine substitutes the $α$ -hydrogens of acetic acid one after another, and with excess chlorine all three are replaced, giving $C l_{3} CCOOH$ . Answer: (C)

Solved Example 21

An organic acid (A),

C_{5} H_{10} O_{2}

, reacts with

B r_{2}

in the presence of red phosphorus to give (B). (B) contains an asymmetric carbon atom and yields (C) on dehydrobromination. (C) does not show geometrical isomerism and on decarboxylation gives an alkene (D), which on ozonolysis gives (E) and (F). (E) gives a positive Schiff's test but (F) does not. Give the structures of (A) to (F).

Solution:

(A) undergoes the HVZ reaction, so it is $R - C H_{2} - COOH$ with R = $C_{3} H_{7}$ . (B) is R-CHBr-COOH, with the asymmetric $α$ -carbon. Loss of HBr gives (C), which shows no geometrical isomerism, so the $β$ -carbon carries two identical groups: R must be isopropyl.

A = $(C H_{3})_{2} CH - C H_{2} - COOH$ (3-methylbutanoic acid); B = $(C H_{3})_{2} CH - CHBr - COOH$ ; C = $(C H_{3})_{2} C = CH - COOH$ (3-methylbut-2-enoic acid); D = $(C H_{3})_{2} C = C H_{2}$ (2-methylpropene); ozonolysis gives E = HCHO (an aldehyde, Schiff's test positive) and F = $(C H_{3})_{2} CO$ (acetone, a ketone, Schiff's test negative).

Solved Example 22

An organic acid A,

C_{3} H_{4} O_{3}

, is catalytically reduced in the presence of ammonia to give B,

C_{3} H_{7} N O_{2}

. B reacts with acetyl chloride, hydrochloric acid and alcohols. It also reacts with nitrous acid to give C,

C_{3} H_{6} O_{3}

, with evolution of nitrogen. What are A, B and C?

Solution:

A has one -COOH group; the remaining $C_{2} H_{3} O$ can only be $C H_{3} - CO$ -, so A is pyruvic acid, $C H_{3} - CO - COOH$ . With ammonia, the keto group forms an imine, which catalytic hydrogenation reduces to an amine (reductive amination), giving B, alanine, $C H_{3} - CH (N H_{2}) - COOH$ . As an amino acid, B reacts with acetyl chloride (at -NH2), HCl (forms a salt) and alcohols (at -COOH).

C H_{3} - CH (N H_{2}) - COOH + HN O_{2}

⟶ C H_{3} - CH (OH) - COOH + N_{2} + H_{2} O

C is lactic acid.

Oxidation at the $α$ - and $β$ -carbon

Mild oxidising agents such as $H_{2} O_{2}$ oxidise an acid at the $β$ -position (butanoic acid gives 3-hydroxybutanoic acid), while selenium dioxide oxidises the $α$ -carbon to give an $α$ -keto acid.

C H_{3} C H_{2} C H_{2} COOH + [O]

H_{2} O_{2} C H_{3} - CH (OH) - C H_{2} COOH

RC H_{2} COOH + [O] Se O_{2}, Δ R - CO - COOH + Se + H_{2} O

9. Ring Substitution in Aromatic Acids

The -COOH group deactivates the benzene ring and directs an incoming electrophile to the meta position, so nitration of benzoic acid gives m-nitrobenzoic acid.

C_{6} H_{5} COOH HN O_{3} / H_{2} S O_{4}, Δ m - O_{2} N - C_{6} H_{4} - COOH

Aromatic carboxylic acids do not undergo the Friedel-Crafts reaction. The -COOH group deactivates the ring, and the Lewis acid catalyst ( $AlC l_{3}$ ) binds to the carboxyl oxygen instead of generating the electrophile.

Solved Example 23

Hydrolysis of a compound A

(C_{7} H_{3} C l_{5})

gives an acid B,

C_{7} H_{4} C l_{2} O_{2}

. Decarboxylation of the acid gives a neutral substance C, whose nitration forms only one mononitro derivative D. Identify A, B, C and D.

Solution:

A $- CC l_{3}$ group on a ring hydrolyses to -COOH, so A is a dichlorobenzotrichloride and B is a dichlorobenzoic acid. Decarboxylation removes -COOH to give a dichlorobenzene C. Only p-dichlorobenzene gives a single mononitro product, because all four ring hydrogens are equivalent.

C = 1,4-dichlorobenzene; D = 1,4-dichloro-2-nitrobenzene; B = 2,5-dichlorobenzoic acid; A = 1,4-dichloro-2-(trichloromethyl)benzene.

10. Effect of Heat on Dicarboxylic, Hydroxy and Halo Acids

Dicarboxylic acid	On heating
Oxalic acid, HOOC-COOH	HCOOH + $C O_{2}$
Malonic acid, $HOOC - C H_{2} - COOH$	$C H_{3} COOH$ + $C O_{2}$ ( $β$ -carbonyl decarboxylation)
Succinic acid, $HOOC - (C H_{2})_{2} - COOH$	Succinic anhydride (5-membered ring) + $H_{2} O$
Glutaric acid, $HOOC - (C H_{2})_{3} - COOH$	Glutaric anhydride (6-membered ring) + $H_{2} O$

Hydroxy acids behave according to the distance between -OH and -COOH:

Figure 14: Effect of heat on

α

β

- and

γ

-hydroxy acids.

Halo acids with aqueous NaOH also depend on the position of the halogen: an $α$ -halo acid is substituted to the $α$ -hydroxy acid; a $β$ -halo acid undergoes elimination (E2) to the $α, β$ -unsaturated acid; a $γ$ -halo acid cyclises, as the carboxylate displaces bromide intramolecularly ( $S_{N} 2$ ), to give a $γ$ -lactone.

C H_{3} CH (Br) COOH O H^{-} C H_{3} CH (OH) CO O^{-}

C H_{3} CH (Br) C H_{2} COOH O H^{-}, E2 C H_{3} CH = CHCO O^{-}

BrC H_{2} C H_{2} C H_{2} COOH

O H^{-}, intramolecular S_{N} 2 γ - butyrolactone

Solved Example 24

What is the product of heating cyclohexane-1,1,2-tricarboxylic acid (a gem-dicarboxylic acid with a third -COOH on the adjacent carbon)?

Solution:

The gem-dicarboxylic carbon behaves like malonic acid and loses $C O_{2}$ , giving cyclohexane-1,2-dicarboxylic acid. Its two -COOH groups are on adjacent carbons, so further heating removes water to form the cyclic anhydride (a five-membered anhydride ring fused to cyclohexane).

Solved Example 25

Treatment of pentane-2,4-dione with KCN and

C H_{3} COOH

, followed by hydrolysis, gives two products (A) and (B), both dicarboxylic acids of formula

C_{7} H_{12} O_{6}

. (A) melts at 98 °C. On heating, (B) gives first a lactonic acid

(C_{7} H_{10} O_{5})

and finally a dilactone

(C_{7} H_{8} O_{4})

. (a) What structure must (B) have to form both a monolactone and a dilactone? (b) What is (A)?

Solution:

Each C=O adds HCN to form a cyanohydrin, and hydrolysis turns each -CN into -COOH, giving 2,4-dihydroxy-2,4-dimethylpentanedioic acid, $HOOC - C (OH) (C H_{3}) - C H_{2}$ $- C (OH) (C H_{3}) - COOH$ . With two identical stereocentres, it exists as a meso form and a racemic form.

(a) (B) is the racemic form. An -OH on one carbon and the -COOH on the other form a five-membered lactone. In this isomer the remaining -OH and -COOH end up cis on the lactone ring, so they can close a second lactone ring, giving the dilactone.

(b) (A) is the meso form. It also gives a monolactone, but its remaining -OH and -COOH are trans on the ring, so a second lactone cannot form.

Solved Example 26

What is the end product of

C H_{3} - CO - C H_{2}

- C H_{2} - C H_{2} - C O_{2} H (1) NaB H_{4} (2) H_{2} O, H^{+}

?
(a) 6-methyloxan-2-one (six-membered lactone with

C H_{3}

next to the ring O)
(b) 4-methyloxan-2-one
(c) oxan-2-one
(d) 5-hydroxyhexanal

Solution:

$NaB H_{4}$ reduces only the keto group, giving 5-hydroxyhexanoic acid, $C H_{3} - CH (OH) - C H_{2}$ $- C H_{2} - C H_{2} - COOH$ . This $δ$ -hydroxy acid cyclises in acid to a six-membered lactone with the methyl group on the carbon next to the ring oxygen. Answer: (a)

11. Abnormal Behaviour of Formic Acid

Formic acid, H-CO-OH, behaves differently from other carboxylic acids because its carboxyl carbon also carries a hydrogen, so it contains an aldehyde-like H-C=O unit. It therefore acts as a reducing agent, which other acids do not: it gives a silver mirror with Tollens' reagent, a red precipitate of $C u_{2} O$ with Fehling's solution, and a white precipitate of $H g_{2} C l_{2}$ (turning grey as Hg forms) with $HgC l_{2}$ .

HCOOH + A g_{2} O ⟶ C O_{2} + H_{2} O + 2Ag ↓

HCOOH + 2CuO ⟶ C O_{2} + H_{2} O + C u_{2} O ↓

HCOOH + 2HgC l_{2} ⟶ H g_{2} C l_{2} ↓ (white) + C O_{2} + 2HCl

H g_{2} C l_{2} + HCOOH ⟶ 2Hg ↓ (grey) + C O_{2} + 2HCl

NEET Trick: formic vs acetic acid

Both fizz with $NaHC O_{3}$ , but only formic acid gives a silver mirror with Tollens' reagent, a red precipitate with Fehling's solution, and decolourises acidified $KMn O_{4}$ . Use any of these to tell them apart.

Summary mind map

Figure 15: Mind map of the properties of carboxylic acids: react at O-H, at C-OH, at the carboxyl carbon and at the

α

-carbon.

Quick Recall: tap to check

Which acid does not undergo the HVZ reaction: propanoic or 2,2-dimethylpropanoic acid?

2,2-Dimethylpropanoic acid: it has no

α

-H.

What does a

γ

-hydroxy acid give on heating?

γ

-lactone (five-membered cyclic ester).

Which acid reduces Tollens' reagent?

Formic acid, because it contains an H-C=O unit.

12. Solved Examples: Exam Practice

Solved Example 27

Kolbe electrolysis of an aqueous solution of sodium propanoate gives mainly:
(A) ethane
(B) propane
(C) butane
(D) hexane

Solution:

Answer: (C). The alkyl group of propanoate is ethyl, $C_{2} H_{5} -$ . Two ethyl radicals join: $C_{2} H_{5} - C_{2} H_{5}$ , butane.

Solved Example 28

Which is the strongest acid?
(A)

C H_{3} COOH

(B)

HC \equiv C - COOH

(C)

C H_{2} = CH - COOH

(D)

C H_{3} C H_{2} COOH

Solution:

Answer: (B). The carboxyl group of propynoic acid is joined to an sp carbon, the most electronegative, which stabilises the anion ( $p K_{a}$ 1.84 against 4.25 for acrylic, 4.76 for acetic and 4.88 for propanoic acid).

Solved Example 29

Calculate the pH and the degree of ionisation of 0.10 M acetic acid (

K_{a} = 1.8 \times 1 0^{- 5}

Solution:

For a weak acid, $[H^{+}] \approx K_{a} C$ .

[H^{+}] = 1.8 \times 1 0^{- 5} \times 0.10 = 1.34 \times 1 0^{- 3} M

pH $= - lo g (1.34 \times 1 0^{- 3}) = 2.87$ . Degree of ionisation $α = \frac{1.34 \times 1 0 ^{- 3}}{0.10} = 0.0134$ , so only about 1.3% of the molecules are ionised.

Solved Example 30

How many moles of

C O_{2}

are released when 1 mol of citric acid,

HOOC - C H_{2} - C (OH) (COOH)

- C H_{2} - COOH

, reacts with excess

NaHC O_{3}

Solution:

Each -COOH releases one $C O_{2}$ ; the alcoholic -OH does not react. Citric acid has three -COOH groups, so 3 mol of $C O_{2}$ are released.

Practice Questions

Arrange in increasing acidity: $HCOOH$ , $C H_{3} COOH$ , $C_{6} H_{5} COOH$ .Answer: $C H_{3} COOH$ (4.76) < $C_{6} H_{5} COOH$ (4.19) < $HCOOH$ (3.75)
Complete: $C H_{3} COOH$ + $PC l_{5}$ →Answer: $C H_{3} COCl$ + $POC l_{3}$ + HCl
What is formed when sodium butanoate is heated with soda lime?Answer: propane
$C H_{3} C H_{2} COOH$ with $B r_{2}$ /red P, then water, gives?Answer: 2-bromopropanoic acid
Name the product of nitrating benzoic acid.Answer: 3-nitrobenzoic acid (m-nitrobenzoic acid)
What does malonic acid give on heating?Answer: acetic acid and $C O_{2}$
Which hydrocarbon forms at the anode in the Kolbe electrolysis of sodium acetate?Answer: ethane (with $C O_{2}$ ); hydrogen forms at the cathode

Common Mistakes to Avoid

Watch out

Saying carboxylic acid dimers are held by intramolecular hydrogen bonds. The dimer uses two intermolecular hydrogen bonds.
Calling carboxylate ions strong bases. Because RCOOH is a fairly strong organic acid, $RCO O^{-}$ is a weak conjugate base.
Expecting phenol to fizz with $NaHC O_{3}$ . Phenol ( $p K_{a}$ 10) is weaker than carbonic acid; only carboxylic acids (and stronger acids) release $C O_{2}$ .
Assuming p-hydroxybenzoic acid is stronger than benzoic acid. The +R effect of para -OH makes it weaker; only the ortho isomer is stronger.
Using $NaB H_{4}$ to reduce -COOH. It does not work; use $LiAl H_{4}$ or $B_{2} H_{6}$ .
Forgetting the carbon loss. Soda-lime decarboxylation, Hunsdiecker and Schmidt reactions all give products with one carbon fewer.
Applying HVZ to acids with no $α$ -hydrogen, such as HCOOH, $C_{6} H_{5} COOH$ , $(C H_{3})_{3} CCOOH$ and $CC l_{3} COOH$ .
Mixing up heating products: $α$ -hydroxy acids give lactides, $β$ -hydroxy acids give $α, β$ -unsaturated acids, $γ$ -hydroxy acids give $γ$ -lactones.
Ranking the Grignard reaction with RCOOH as a ketone synthesis. RMgX only deprotonates the acid; two equivalents of RLi are needed to reach a ketone.

Frequently Asked Questions

Why do carboxylic acids have higher boiling points than alcohols of similar mass?

Carboxylic acids form cyclic dimers held by two intermolecular hydrogen bonds, and these hydrogen bonds are stronger than those in alcohols. The dimer behaves like a molecule of double mass, so more energy is needed to boil the acid. Acetic acid boils at 118 °C, propan-1-ol at 97 °C.

Why are carboxylic acids more acidic than alcohols and phenols?

The carboxylate ion has two identical resonance structures that share the negative charge equally between two oxygen atoms. An alkoxide ion has no resonance, and in the phenoxide ion the charge spreads onto less electronegative ring carbons. The better-stabilised carboxylate makes the acid stronger.

How does a chlorine atom affect the acidity of acetic acid?

Chlorine withdraws electrons by the inductive effect, dispersing the negative charge of the carboxylate ion and stabilising it. The pKa falls from 4.76 for acetic acid to 2.86 for chloroacetic acid, and further with each extra chlorine. The effect weakens quickly as the chlorine moves away from -COOH.

What is the ortho effect in benzoic acids?

Almost any substituent in the ortho position makes a benzoic acid more acidic than its meta and para isomers. Steric crowding twists the -COOH out of the ring plane, and groups such as -OH can hydrogen bond to the carboxylate. Salicylic acid (pKa 2.97) is therefore far stronger than benzoic acid.

What is the Hell-Volhard-Zelinsky reaction?

It is the alpha-halogenation of a carboxylic acid with chlorine or bromine in the presence of red phosphorus. Phosphorus forms a little acyl halide, which enolises and reacts with the halogen at the alpha carbon. The acid must have at least one alpha hydrogen.

Why do beta-keto acids lose carbon dioxide so easily?

The carboxyl hydrogen is hydrogen bonded to the beta carbonyl oxygen, allowing a six-membered cyclic transition state in which carbon dioxide leaves and an enol forms. The enol quickly tautomerises to a ketone. Acetoacetic acid gives acetone on gentle warming.

Which reactions of carboxylic acids are most important for JEE?

For JEE Main and Advanced, focus on acidity order with inductive, ortho and hydrogen-bonding effects, the esterification mechanism, HVZ, Hunsdiecker, Schmidt and soda-lime decarboxylation, and effects of heat on beta-keto, hydroxy and dicarboxylic acids. Multi-step problems often combine two of these.

What properties of carboxylic acids are asked in NEET?

NEET mostly tests NCERT points: hydrogen bonding and boiling points, acidity order of substituted acids, the NaHCO3 test, formation of acid chlorides, esters, amides and anhydrides, reduction with LiAlH4, decarboxylation, HVZ reaction and meta substitution in benzoic acid.

Previous year questions on Properties of Carboxylic acids

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

Ready to master Carboxylic Acids And Their Derivatives?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. Physical Properties of Carboxylic Acids

Solubility

Boiling points

Melting points

2. How Carboxylic Acids React

3. Acidity of Carboxylic Acids

Effect of substituents on acidity

Ortho effect in aromatic acids

Maleic acid and fumaric acid

4. Salt Formation

5. Conversion into Functional Derivatives

(a) Conversion into acid chlorides

(b) Conversion into esters (esterification)

(c) Conversion into amides

(d) Conversion into anhydrides

6. Reduction of Carboxylic Acids

7. Reactions Involving Loss of the -COOH Group

(a) Decarboxylation with soda lime

(b) Hunsdiecker reaction

(c) Schmidt reaction

(d) Reaction with organometallic compounds

(e) Easy decarboxylation of β-keto acids

(f) Kolbe electrolysis

8. Substitution in the Alkyl Chain

Hell-Volhard-Zelinsky (HVZ) reaction

Oxidation at the α- and β-carbon

9. Ring Substitution in Aromatic Acids

10. Effect of Heat on Dicarboxylic, Hydroxy and Halo Acids

11. Abnormal Behaviour of Formic Acid

Summary mind map

12. Solved Examples: Exam Practice

Common Mistakes to Avoid

Frequently Asked Questions

Why do carboxylic acids have higher boiling points than alcohols of similar mass?

Why are carboxylic acids more acidic than alcohols and phenols?

How does a chlorine atom affect the acidity of acetic acid?

What is the ortho effect in benzoic acids?

What is the Hell-Volhard-Zelinsky reaction?

Why do beta-keto acids lose carbon dioxide so easily?

Which reactions of carboxylic acids are most important for JEE?

What properties of carboxylic acids are asked in NEET?

Previous year questions on Properties of Carboxylic acids

Ready to master Carboxylic Acids And Their Derivatives?

(e) Easy decarboxylation of $β$ -keto acids

Oxidation at the $α$ - and $β$ -carbon