Preparation of Carboxylic acids

ChemistryCarboxylic Acids And Their DerivativesFor NEET aspirants

Preparation of carboxylic acids uses three broad strategies: oxidation (of primary alcohols, aldehydes, alkylbenzenes, alkenes and alkynes), hydrolysis (of nitriles and acid derivatives such as esters, amides and acid chlorides), and new carbon-carbon bonds made with carbon dioxide (Grignard reagents) or carbon monoxide. The fastest way to choose a method is to count carbons: oxidation and hydrolysis of derivatives keep the carbon count, while the Grignard and nitrile routes add one carbon. The preparation of carboxylic acids is central to conversion questions in JEE Main, JEE Advanced and NEET.

On this page1Methods map2Oxidation3Haloform4Alkylbenzenes5Grignard6Nitriles7Acid derivatives8Carbon monoxide9Oxidative cleavage10Choosing a method11Exam practice

Key Formulas - Quick Reference

★ Must learn $RC H_{2} OH KMn O_{4} / H^{+} RCOOH$ (same carbon count)
$RCHO Tollens or Fehling RCOOH$ (only -CHO is oxidised)
★ Must learn $C_{6} H_{5} C H_{2} R hot KMn O_{4}, H^{+} C_{6} H_{5} COOH$ (needs a benzylic H)
★ Must learn $RMgX (i) C O_{2} (ii) H_{3} O^{+} RCOOH$ (+1 carbon)
★ Must learn $RX NaCN RCN H_{3} O^{+} RCOOH$ (+1 carbon)
$RCOZ + H_{2} O \to RCOOH + HZ$ (Z = Cl, OCOR, OR', $N H_{2}$ )
$RCH = CH R^{'} KMn O_{4} / H^{+}, Δ RCOOH + R^{'} COOH$ (chain splits)
$C H_{3} OH + CO catalyst, Δ, p C H_{3} COOH$ (+1 carbon)
★ Must learn $RCOC H_{3} I_{2} /NaOH RCOONa + CH I_{3}$ (loses 1 carbon)

Overview: Methods at a Glance

Before memorising individual reactions, see how the routes connect. Each route either keeps, adds or removes carbon atoms, and that single idea solves most "convert A into B" questions.

Figure 1: Roadmap of the preparation of carboxylic acids. The badge on each route shows how the carbon count changes.

1. Oxidation of Alcohols, Aldehydes and Ketones

Primary alcohols are oxidised by acidified potassium permanganate ( $KMn O_{4}$ / $H^{+}$ ) or acidified potassium dichromate ( $K_{2} C r_{2} O_{7}$ / $H^{+}$ ). The alcohol is first oxidised to an aldehyde, which is oxidised further to the carboxylic acid with the same number of carbon atoms.

R - C H_{2} OH + [O] KMn O_{4} / H^{+} K_{2} C r_{2} O_{7} / H^{+} R - CHO + H_{2} O

[O] R - COOH

Aldehydes are oxidised to carboxylic acids even by mild oxidising agents such as Tollens' reagent or Fehling's solution, again with no change in the number of carbon atoms.

R - CHO + [O] Tollens’ reagent R - CO O^{-}

H^{+} R - COOH

C_{6} H_{5} CHO + [O] Tollens’ reagent C_{6} H_{5} CO O^{-}

H^{+} C_{6} H_{5} COOH

Figure 2: Oxidation ladder: primary alcohol (

- 1

) to aldehyde (

+ 1

) to carboxylic acid (

+ 3

), with the reagents that control each step.

JEE Trick: selective oxidants

Use a mild reagent when the molecule has a group you want to keep. $Mn O_{2}$ oxidises allylic and benzylic alcohols only as far as the aldehyde, and Tollens' or Fehling's reagent oxidises only -CHO. A C=C double bond survives both, while hot $KMn O_{4}$ would cleave it.

Oxidation of ketones

Ketones resist mild oxidising agents. With strong oxidising agents and heat, a carbon-carbon bond next to the keto group breaks and both pieces are oxidised to acids, so the product acids have fewer carbon atoms than the ketone. In a keto acid, the keto carbon stays with the smaller group (Popoff's rule) and is converted into -COOH.

Figure 3: Careful oxidation of a keto acid breaks the C-C bond next to the keto group (Popoff's rule).

Solved Example 1

How will you prepare

C H_{3} C H_{2} COOH

from

C H_{3} - CH = C H_{2}

Solution:

Propanoic acid has the same 3 carbons as propene, so oxidation of a primary alcohol is the right route. We need propan-1-ol, the anti-Markovnikov alcohol, which hydroboration-oxidation gives. Acidified permanganate then oxidises it to the acid.

C H_{3} - CH = C H_{2} (i) B H_{3} /THF (ii) H_{2} O_{2} /O H^{-} C H_{3} - C H_{2} - C H_{2} - OH

Mn O_{4}^{-} / H^{+} C H_{3} - C H_{2} - COOH

Acid-catalysed hydration would give propan-2-ol instead, which oxidises only to acetone and cannot give propanoic acid.

Solved Example 2

Bring about the transformation: (cyclohex-1-en-1-yl)methanol

Mn O_{2}

(A)

[Ag (N H_{3})_{2}]^{+} O H^{-}, H^{+}

(B). Identify A and B.

Solution:

$Mn O_{2}$ selectively oxidises the allylic $- C H_{2} OH$ group to -CHO and leaves the ring C=C untouched. A is cyclohex-1-ene-1-carbaldehyde.

Tollens' reagent oxidises only the aldehyde group; acidification releases the free acid. B is cyclohex-1-ene-1-carboxylic acid, with the double bond still intact.

Haloform reaction of methyl ketones

A methyl ketone ( $RCOC H_{3}$ ) or an alcohol that oxidises to one gives the haloform reaction with a halogen and sodium hydroxide (sodium hypohalite). The three $α$ -hydrogens of the methyl group are replaced by halogen, and hydroxide then cleaves the $C - C X_{3}$ bond. The acid is formed as its salt with one carbon fewer than the ketone, and the lost carbon leaves as the haloform.

RCOC H_{3} + 3 I_{2} + 4NaOH RCOONa + CH I_{3} ↓ + 3NaI + 3 H_{2} O

Acidification releases the free acid. Iodoform ( $CH I_{3}$ ) is a yellow solid, so the same reaction is the iodoform test for the $C H_{3} CO -$ group. With sodium hypochlorite (bleach), acetophenone gives benzoic acid:

C_{6} H_{5} COC H_{3} + 3NaOCl C_{6} H_{5} COONa + CHC l_{3} + 2NaOH

Key idea

Oxidation keeps the carbon count (1° alcohol or aldehyde to acid); the haloform reaction removes exactly one carbon from a methyl ketone.

2. Oxidation of Alkylbenzenes

Although benzene and alkanes are quite unreactive towards the usual oxidising agents ( $KMn O_{4}$ , $K_{2} C r_{2} O_{7}$ and so on), the benzene ring makes an aliphatic side chain quite susceptible to oxidation. The side chain is oxidised down to the ring, and only a carboxyl group (-COOH) remains to show the position of the original side chain. Potassium permanganate is generally used, although potassium dichromate or dilute nitric acid can also be used. Oxidation of a side chain is more difficult than oxidation of an alkene and needs prolonged treatment with hot $KMn O_{4}$ .

C_{6} H_{5} - C H_{2} C H_{2} C H_{2} C H_{3}

hot KMn O_{4} C_{6} H_{5} - COOH + C O_{2}

Whatever the length of the side chain, the product is benzoic acid, provided the carbon attached to the ring (the benzylic carbon) carries at least one hydrogen. An alkyl group with no benzylic hydrogen, such as tert-butyl, is not oxidised to -COOH.

Figure 4: Side-chain oxidation needs at least one benzylic hydrogen; tert-butylbenzene is not oxidised.

This reaction is used for two purposes: (a) the synthesis of carboxylic acids and (b) the identification of alkylbenzenes, because the number and positions of -COOH groups in the product reveal the number and positions of the side chains.

NEET Trick

Count -COOH groups in the product to count side chains: o-xylene gives phthalic acid (benzene-1,2-dicarboxylic acid), and p-xylene gives terephthalic acid (benzene-1,4-dicarboxylic acid).

Quick Recall: tap to check

What does isopropylbenzene (cumene) give with hot

KMn O_{4}

Benzoic acid: the benzylic carbon carries one H.

Which side chain survives hot

KMn O_{4}

A tert-butyl group, because its benzylic carbon has no hydrogen.

What does p-xylene give?

Terephthalic acid (benzene-1,4-dicarboxylic acid).

3. Carbonation of Grignard Reagents

The Grignard synthesis of a carboxylic acid is carried out by bubbling gaseous $C O_{2}$ into the ether solution of the Grignard reagent, or by pouring the Grignard reagent onto crushed dry ice (solid $C O_{2}$ ). In the second method dry ice serves not only as the reagent but also as the cooling agent.

The Grignard reagent adds to the carbon-oxygen double bond of $C O_{2}$ just as it does with aldehydes and ketones. The product is the magnesium salt of the carboxylic acid, from which the free acid is liberated by treatment with a mineral acid.

R - MgX + C O_{2} ⟶ R - CO O^{-}^{+} MgX

H^{+} R - COOH + M g^{2 +} + X^{-}

Figure 5: Mechanism of Grignard carbonation. The carbanion-like carbon of

RMgX

attacks

C O_{2}

, adding one carbon.

The Grignard reagent can be prepared from primary, secondary, tertiary or aromatic halides. The method is limited only by the presence of other reactive groups in the molecule. The following synthesis shows its use:

(C H_{3})_{3} C - OH HCl (C H_{3})_{3} C - Cl

Mg (C H_{3})_{3} C - MgCl

C O_{2} H^{+} (C H_{3})_{3} C - COOH

The product, 2,2-dimethylpropanoic acid, is commonly called trimethylacetic acid.

Solved Example 3

Write A and B: (i) mesitylene

B r_{2} /Fe

Mg/ether

C O_{2} H^{+}

2,4,6-trimethylbenzoic acid. (ii) p-bromo-sec-butylbenzene

Mg/ether

C O_{2}

H^{+}

p-sec-butylbenzoic acid.

Solution:

(i) Mesitylene (1,3,5-trimethylbenzene) undergoes ring bromination. A is bromomesitylene (2-bromo-1,3,5-trimethylbenzene), and B is the Grignard reagent mesitylmagnesium bromide. Carbonation and acidification give mesitoic acid (2,4,6-trimethylbenzoic acid).

(ii) A is 4-sec-butylphenylmagnesium bromide. Adding $C O_{2}$ gives B, the bromomagnesium salt of 4-sec-butylbenzoic acid (Ar-COOMgBr), which acid releases as p-sec-butylbenzoic acid.

Why not oxidise instead? Hot $KMn O_{4}$ would also attack the sec-butyl side chain (it has a benzylic H), so the Grignard route is the only way to keep it.

Solved Example 4

Convert 2-bromopropane into 2-methylpropanoic acid.

Solution:

The product has one more carbon, attached where the bromine was, so use a Grignard reagent and carbon dioxide.

(C H_{3})_{2} CH - Br Mg/ether (C H_{3})_{2} CH - MgBr

C O_{2} (C H_{3})_{2} CH - C O_{2} MgBr

H_{3} O^{+} (C H_{3})_{2} CH - COOH

Solved Example 5

Compound (A),

C_{3} H_{7} Cl

, reacts with alcoholic KOH to form (B),

C_{3} H_{6}

, which discharges

B r_{2}

CC l_{4}

. Reaction of (A) with Mg in ether, followed by

C O_{2}

and dilute acid, gives (C),

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

, which liberates gas bubbles with aqueous

NaHC O_{3}

. What are (A) to (C)?

Solution:

(B) decolourises bromine, so it is an alkene: propene, $C H_{3} - CH = C H_{2}$ . (C) effervesces with $NaHC O_{3}$ , so it is a carboxylic acid, formed from (A) with one extra carbon through the Grignard route.

(A) is 2-chloropropane, $(C H_{3})_{2} CHCl$ , and (C) is 2-methylpropanoic acid, $(C H_{3})_{2} CH - COOH$ .

Strictly, 1-chloropropane also fits the data given: it too gives propene, and its Grignard route gives butanoic acid, which is also $C_{4} H_{8} O_{2}$ . An extra clue, such as the acid being branched, is needed to rule it out.

4. Hydrolysis of Nitriles

Aliphatic nitriles are prepared by treating alkyl halides with sodium cyanide in a solvent that dissolves both reactants. In dimethyl sulfoxide (DMSO), the reaction occurs rapidly and exothermically at room temperature. The resulting nitrile is then hydrolysed to the acid by boiling with aqueous acid or alkali.

R - Cl + NaCN DMSO R - C \equiv N

R - C \equiv N + 2 H_{2} O + H^{+} ⟶ R - COOH + N H_{4}^{+}

R - C \equiv N + H_{2} O + O H^{-} ⟶ R - CO O^{-} + N H_{3}

Figure 6: Nitrile route:

S_{N} 2

cyanation (+1 C) followed by acidic or basic hydrolysis, and which halides work.

Mechanism of acid hydrolysis of a nitrile

Protonation: the nitrogen of $R - C \equiv N$ takes up $H^{+}$ , making the carbon strongly electrophilic.
Attack by water: a lone pair on water attacks the nitrile carbon, and a pi bond shifts onto nitrogen.
Proton loss: the oxonium ion loses $H^{+}$ to give an imidic acid, $R - C (OH) = NH$ .
Tautomerism: the imidic acid rearranges to the more stable amide, $R - CON H_{2}$ .
Amide hydrolysis: hot aqueous acid hydrolyses the amide to $R - COOH$ and $N H_{4}^{+}$ .

Figure 7: Mechanism of acid-catalysed nitrile hydrolysis through the imidic acid and amide.

Solved Example 6

(i) n-butyl bromide

NaCN

\to

\to

C +

N H_{4}^{+}

. Write A, B and C. (ii) Benzyl chloride

NaCN

70% H_{2} S O_{4}, reflux

B +

N H_{4}^{+}

. Write A and B.

Solution:

(i) Cyanide displaces bromide: A is n-valeronitrile (pentanenitrile), $n - C_{4} H_{9} CN$ . Refluxing with aqueous alcoholic NaOH gives B, the pentanoate ion $n - C_{4} H_{9} CO O^{-}$ (with $N H_{3}$ ). Acidification gives C, n-valeric acid (pentanoic acid), $n - C_{4} H_{9} COOH$ .

n - C_{4} H_{9} Br NaCN n - C_{4} H_{9} CN

aq. alc. NaOH, reflux n - C_{4} H_{9} CO O^{-} + N H_{3}

H^{+} n - C_{4} H_{9} COOH + N H_{4}^{+}

(ii) A is phenylacetonitrile, $C_{6} H_{5} C H_{2} CN$ , and acid hydrolysis gives B, phenylacetic acid, $C_{6} H_{5} C H_{2} COOH$ .

C_{6} H_{5} C H_{2} Cl NaCN C_{6} H_{5} C H_{2} CN

70% H_{2} S O_{4}, reflux C_{6} H_{5} C H_{2} COOH + N H_{4}^{+}

Solved Example 7

R - C H_{2} - C H_{2} OH

can be converted into

R - C H_{2} C H_{2} COOH

. The correct sequence of reagents is:
(a)

PB r_{3}

, KCN,

H^{+}

(b)

PB r_{3}

, KCN,

H_{2}

H^{+}

(d) HCN,

PB r_{3}

H^{+}

Solution:

One carbon must be added at the carbon that carries -OH. First convert -OH into a good leaving group (-Br), then displace it with cyanide, then hydrolyse the nitrile. Option (b) would reduce the nitrile to an amine, and cyanide cannot displace -OH directly in (c) or (d).

RC H_{2} C H_{2} OH PB r_{3} RC H_{2} C H_{2} Br

KCN RC H_{2} C H_{2} CN H_{2} O/ H^{+} RC H_{2} C H_{2} COOH

Answer: (a)

Grignard route: RX, Mg,

C O_{2}

+1 carbon. Works for 1°, 2°, 3°, aryl and vinyl halides.

Fails if the molecule has -OH, -NH, -COOH or C=O.

Nitrile route: RX, NaCN,

H_{3} O^{+}

+1 carbon. Needs an $S_{N} 2$ halide: $C H_{3} X$ , 1°, benzyl.

Tolerates -OH and C=O elsewhere in the molecule.

Key idea

Both routes add one carbon. Pick the Grignard for 3°, aryl and vinyl halides; pick the nitrile when the molecule carries -OH or C=O.

5. Hydrolysis of Acid Derivatives

Acid chlorides, anhydrides, esters and amides are all hydrolysed to the parent carboxylic acid, with no change in the carbon skeleton. Esters and amides need dilute acid (or base followed by acidification) and heat.

R - COCl + H_{2} O ⟶ R - COOH + HCl

(R - CO)_{2} O + H_{2} O ⟶ 2 R - COOH

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

R - CON H_{2} + H_{2} O HCl R - COOH + N H_{4} Cl

Figure 8: Hydrolysis of acid derivatives

RCOZ

to carboxylic acids, from fastest (acid chloride) to slowest (amide).

The order acid chloride > anhydride > ester > amide follows the reactivity of acid derivatives towards nucleophiles, explained in the Carboxylic Acid Derivatives concept.

Quick Recall: tap to check

Which acid derivative is hydrolysed by water alone at room temperature?

The acid chloride, giving RCOOH and HCl.

What are the products of hydrolysing acetamide with dilute HCl?

Acetic acid and ammonium chloride.

6. Preparation Using Carbon Monoxide

(a) Use of alkoxides and methanol

Sodium alkoxides absorb carbon monoxide on heating to give the sodium salt of an acid containing one more carbon atom, and acidification releases the acid. With sodium hydroxide itself (R = H), sodium formate is formed. Industrially, methanol and carbon monoxide combine over a cobalt carbonyl catalyst to give acetic acid.

R O^{-} N a^{+} + CO Δ, pressure RCOONa H^{+} RCOOH

C H_{3} O^{-} N a^{+} + CO ⟶ C H_{3} COONa H^{+} C H_{3} COOH

C H_{3} OH + CO C o_{2} (CO)_{8} 210^{\circ} C, pressure C H_{3} COOH

(b) Carbonylation of alkenes

An alkene, carbon monoxide and steam combine in the presence of phosphoric acid at 300 to 400 °C to give a carboxylic acid with one extra carbon. With an unsymmetrical alkene, the -COOH group ends up on the more substituted carbon, as in Markovnikov addition.

C H_{2} = C H_{2} + CO + H_{2} O

H_{3} P O_{4} 300 - 400^{\circ} C C H_{3} C H_{2} COOH

C H_{3} CH = C H_{2} + CO + H_{2} O

H_{3} P O_{4} 300 - 400^{\circ} C (C H_{3})_{2} CH - COOH

Figure 9: Carbon monoxide routes, where CO becomes the carboxyl carbon.

JEE Advanced

Why carbonylation follows Markovnikov (Koch reaction). Phosphoric acid protonates the alkene to the more stable carbocation. Carbon monoxide, a neutral nucleophile through carbon, adds to it to give an acylium ion, which water captures:

$C H_{3} CH = C H_{2}$ $H^{+}$ $(C H_{3})_{2} C H^{+}$ $CO$ $(C H_{3})_{2} CH - C \equiv O^{+}$ $H_{2} O, - H^{+}$ $(C H_{3})_{2} CHCOOH$

So -COOH lands where the positive charge was. A carbocation can also rearrange first (hydride or methyl shift), which is why branched acids dominate.

7. Oxidative Cleavage of Alkenes, Alkynes and Cycloalkenes

Strong oxidising agents break carbon-carbon double and triple bonds completely. Each carbon of the multiple bond that carries a hydrogen or an alkyl group ends up as a -COOH carbon.

RC \equiv C R^{'} (i) O_{3} or KMn O_{4} (ii) H_{2} O RCOOH + R^{'} COOH

C H_{3} C \equiv CC H_{3} (i) O_{3} (ii) H_{2} O 2 C H_{3} COOH

RCH = CH R^{'} (i) conc. KMn O_{4} /O H^{-} (ii) H_{3} O^{+}, Δ RCOOH + R^{'} COOH

RCH = CH R^{'} (i) O_{3} (ii) H_{2} O_{2} RCOOH + R^{'} COOH

A cycloalkene does not split into two molecules. Its ring opens instead, giving a single dicarboxylic acid: cyclobutene gives succinic acid.

cyclobutene KMn O_{4} /Δ/ H^{+} HOOC - C H_{2} - C H_{2} - COOH

Figure 10: Oxidative cleavage of an alkene, an alkyne and a cycloalkene. The red line marks the bond that breaks.

JEE Trick: ozonolysis work-up decides the product

Ozonide + Zn/ $H_{2} O$ (reductive work-up) gives aldehydes and ketones. Ozonide + $H_{2} O_{2}$ (oxidative work-up) turns every aldehyde piece into a carboxylic acid. Ketone pieces stay ketones in both.

Key idea

In oxidative cleavage every carbon of the multiple bond that carries H or R ends up as -COOH; a terminal

= C H_{2}

becomes

C O_{2}

, and a ring gives one diacid.

8. Choosing the Right Method

Method	Starting material	Carbon change	Watch out for
Oxidation	1° alcohol, aldehyde	Same	2° alcohols give ketones, not acids
Side-chain oxidation	Alkylbenzene	Chain reduced to -COOH	Needs a benzylic H
Grignard + $C O_{2}$	Alkyl or aryl halide	+1	No -OH, -NH, -COOH or C=O allowed
Nitrile hydrolysis	Methyl, 1° or benzyl halide	+1	3°, aryl and vinyl halides fail
Hydrolysis of derivatives	Ester, amide, anhydride, acid chloride	Same	Amides need long heating
Oxidative cleavage	Alkene, alkyne, cycloalkene	Chain splits	Terminal $= C H_{2}$ becomes $C O_{2}$
Carbon monoxide	Alkoxide, methanol, alkene	+1	Needs heat, pressure or catalyst
Haloform	Methyl ketone	-1	Only for $C H_{3} CO$ - compounds

The same choice as a flowchart: count the carbons first.

Figure 11: Choose the method by counting carbons: same count (oxidise or hydrolyse), +1 (Grignard or nitrile), -1 (haloform), side chain on a ring (KMnO

_{4}

Solved Example 8

Compound A,

C_{5} H_{8} O_{3}

, when heated with soda lime gives B, which reacts with HCN to give C. C reacts with

PC l_{5}

to give D, which reacts with KCN to form E. E, on alkaline hydrolysis, gives a salt which is isolated and heated with soda lime to produce n-butane. A, on careful oxidation with

K_{2} C r_{2} O_{7}

, gives acetic acid and malonic acid. Give the structural formulae of A to E.

Solution:

On oxidation only two -COOH groups can be introduced, one on each carbon at the C-C bond that breaks. The products contain three -COOH groups in total, so one -COOH is already present in A. The remaining part, $C_{4} H_{7} O$ -, must be a keto-substituted alkyl group, $C_{3} H_{7} - CO$ -, so A is a keto acid.

A keto acid breaks at the keto group on careful oxidation, and the keto carbon becomes -COOH on the piece with fewer carbons. Acetic acid therefore comes from a $C H_{3} - CO$ - unit, so $C_{3} H_{7} CO$ - is $C H_{3} - CO - C H_{2} - C H_{2}$ - (Figure 3).

A = $C H_{3} - CO - C H_{2}$ $- C H_{2} - COOH$ (4-oxopentanoic acid)

B = $C H_{3} - CO - C H_{2} - C H_{3}$ (butanone), formed by decarboxylation with soda lime

C = $C H_{3} - C (OH) (CN)$ $- C H_{2} - C H_{3}$ , the cyanohydrin from addition of HCN to the ketone

D = $C H_{3} - C (Cl) (CN)$ $- C H_{2} - C H_{3}$ , where $PC l_{5}$ replaces -OH by -Cl

E = $C H_{3} - C (CN)_{2}$ $- C H_{2} - C H_{3}$ , where cyanide replaces -Cl

Alkaline hydrolysis of E gives $C H_{3} - C (COONa)_{2}$ $- C H_{2} - C H_{3}$ . Heating with soda lime removes both -COONa groups, giving $C H_{3} - C H_{2} - C H_{2} - C H_{3}$ (n-butane), which confirms the structure.

Solved Example 9

How will you effect the following conversions? (i) Toluene to m-chlorobenzoic acid (ii) But-2-yne to propanoic acid

Solution:

(i) Oxidise first. $- C H_{3}$ is ortho/para-directing, but -COOH is meta-directing, so chlorination must come after oxidation.

C_{6} H_{5} C H_{3} KMn O_{4} /O H^{-}, then H^{+} C_{6} H_{5} COOH

C l_{2} /Fe m - Cl C_{6} H_{4} COOH

(ii) Hydration of the alkyne gives a methyl ketone, and the iodoform (haloform) reaction then removes one carbon.

C H_{3} C \equiv CC H_{3} H_{2} O/ H^{+}, H g^{2 +} C H_{3} COC H_{2} C H_{3}

NaOH/ I_{2} C H_{3} C H_{2} COOH

Solved Example 10

An organic compound A

(C_{4} H_{8} O_{3})

, acidic in nature, is oxidised to give B, which on gentle heating produces C

(C_{3} H_{6} O)

and

C O_{2}

. A on heating yields D

(C_{4} H_{6} O_{2})

having an acid neutralisation equivalent of 86. What are A, B, C and D?

Solution:

D ( $C_{4} H_{6} O_{2}$ , molar mass 86) is a monobasic acid formed from A by loss of water, so heating causes dehydration and A must be a $β$ -hydroxy acid.

A = $C H_{3} - CH (OH) - C H_{2} - COOH$ (3-hydroxybutanoic acid) and D = $C H_{3} - CH = CH - COOH$ (but-2-enoic acid).

Oxidation of the secondary -OH gives B = $C H_{3} - CO - C H_{2} - COOH$ (3-oxobutanoic acid). Because the carbonyl group is $β$ to the carboxyl group, B decarboxylates on gentle heating to C = $C H_{3} - CO - C H_{3}$ (acetone).

Solved Example 11

Formic acid is obtained when:
(A) calcium acetate is heated with conc.

H_{2} S O_{4}

(B) glycerol is heated with oxalic acid
(C) acetaldehyde is oxidised with

K_{2} C r_{2} O_{7}

and

H_{2} S O_{4}

(D) calcium formate is heated with calcium acetate

Solution:

Heating glycerol with oxalic acid (about 100 to 110 °C) forms glyceryl monooxalate, which loses $C O_{2}$ to give glyceryl monoformate; this releases formic acid. This is the laboratory preparation of formic acid. Option (A) gives acetic acid, (C) gives acetic acid and (D) gives acetaldehyde. Answer: (B)

Summary mind map

Figure 12: Mind map of the preparation of carboxylic acids, with the carbon change of each route.

Quick Recall: tap to check

Which two routes add one carbon?

Grignard carbonation and nitrile hydrolysis (and the CO routes).

Which route removes one carbon?

The haloform reaction of a methyl ketone.

What does cyclobutene give on oxidative cleavage?

Succinic acid,

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

9. Solved Examples: Exam Practice

Solved Example 12

Which compound is not oxidised to benzoic acid by hot alkaline

KMn O_{4}

followed by acidification?
(A) toluene
(B) ethylbenzene
(C) tert-butylbenzene
(D) isopropylbenzene

Solution:

Answer: (C). Side-chain oxidation starts at a benzylic C-H. In tert-butylbenzene the benzylic carbon carries three methyl groups and no hydrogen, so the chain is not attacked.

Solved Example 13

Bromoethane can be converted into propanoic acid by:
(A) KCN, then

H_{3} O^{+}

(B) aqueous KOH, then

KMn O_{4}

H_{3} O^{+}

(D) alcoholic KOH, then

O_{3}

H_{2} O_{2}

Solution:

Answer: (A). Propanoic acid has one more carbon than bromoethane. Cyanide adds it ( $C H_{3} C H_{2} CN$ ), and hydrolysis gives $C H_{3} C H_{2} COOH$ . Route (B) gives acetic acid (same carbons), (C) gives propan-1-ol, and (D) gives ethene and then formic acid and $C O_{2}$ .

Solved Example 14

Complete: acetophenone

I_{2} /NaOH

A + B (yellow)

H^{+}

C. Identify A, B and C.

Solution:

The $C H_{3} CO -$ group gives the iodoform reaction. A is sodium benzoate, B is iodoform ( $CH I_{3}$ ) and C, after acidification, is benzoic acid. The acid has one carbon fewer than acetophenone (7 against 8).

C_{6} H_{5} COC H_{3} + 3 I_{2} + 4NaOH C_{6} H_{5} COONa + CH I_{3} ↓ + 3NaI + 3 H_{2} O

Solved Example 15

Convert propan-1-ol into butanoic acid.

Solution:

Butanoic acid has one more carbon, added where the -OH was. Turn -OH into a leaving group, add cyanide by $S_{N} 2$ , then hydrolyse.

C H_{3} C H_{2} C H_{2} OH PBr3 C H_{3} C H_{2} C H_{2} Br

C H_{3} C H_{2} C H_{2} Br NaCN, DMSO C H_{3} C H_{2} C H_{2} CN

C H_{3} C H_{2} C H_{2} CN H_{3} O^{+}, heat C H_{3} C H_{2} C H_{2} COOH

The Grignard route (Mg, then $C O_{2}$ , then $H_{3} O^{+}$ on the bromide) works equally well.

Practice Questions

What does p-xylene give with hot $KMn O_{4}$ , then acid?Answer: terephthalic acid (benzene-1,4-dicarboxylic acid)
$C H_{3} C H_{2} MgBr$ + $C O_{2}$ , then $H_{3} O^{+}$ gives?Answer: propanoic acid, $C H_{3} C H_{2} COOH$
Hex-3-ene is heated with $KMn O_{4}$ / $H^{+}$ . Name the product.Answer: propanoic acid (2 mol per mol of alkene)
Benzonitrile is boiled with dilute acid. Give the products.Answer: benzoic acid and $N H_{4}^{+}$
Cyclohexene with hot $KMn O_{4}$ / $H^{+}$ gives?Answer: hexanedioic acid (adipic acid)
Which of bromobenzene and benzyl bromide cannot be used in the nitrile route?Answer: bromobenzene (aryl halides do not undergo $S_{N} 2$ )
Butanone with $I_{2}$ /NaOH, then acid, gives which acid?Answer: propanoic acid, with iodoform

Common Mistakes to Avoid

Watch out

Forgetting the extra carbon. The Grignard and nitrile routes both give an acid with one more carbon than the alkyl halide.
Making a Grignard reagent from a molecule that contains -OH, -NH, -COOH or C=O. These groups destroy the reagent before $C O_{2}$ is added.
Using NaCN on a tertiary, aryl or vinyl halide. Tertiary halides eliminate and aryl or vinyl halides do not undergo $S_{N} 2$ ; use the Grignard route.
Expecting tert-butylbenzene to give benzoic acid. With no benzylic hydrogen, the side chain is not oxidised.
Thinking the side chain length is kept. Every oxidisable side chain, however long, ends up as a single -COOH on the ring.
Oxidising a secondary alcohol and expecting an acid. Secondary alcohols give ketones; only primary alcohols and aldehydes give acids with the same carbon count.
Writing aldehydes as the product of ozonolysis with $H_{2} O_{2}$ . Oxidative work-up gives carboxylic acids; Zn/ $H_{2} O$ gives aldehydes.
Putting -COOH on the end carbon in carbonylation of propene. It goes to the middle carbon, giving 2-methylpropanoic acid, not butanoic acid.
Chlorinating toluene before oxidising it when the meta product is required. Oxidise first so the -COOH group directs meta.

Frequently Asked Questions

What are the main methods for the preparation of carboxylic acids?

The main methods are oxidation of primary alcohols and aldehydes, side-chain oxidation of alkylbenzenes, carbonation of Grignard reagents, hydrolysis of nitriles, hydrolysis of acid derivatives such as esters and amides, oxidative cleavage of alkenes and alkynes, and reactions of carbon monoxide with alkoxides, methanol or alkenes.

Which methods increase the number of carbon atoms in a carboxylic acid?

Carbonation of a Grignard reagent, hydrolysis of a nitrile made from an alkyl halide, and carbon monoxide routes each add one carbon. Oxidation of alcohols or aldehydes and hydrolysis of esters, amides or acid chlorides keep the same count, while the haloform reaction removes one carbon.

Why is tert-butylbenzene not oxidised to benzoic acid by KMnO4?

Side-chain oxidation starts by removing a hydrogen from the benzylic carbon, the carbon joined to the ring. In tert-butylbenzene that carbon carries three methyl groups and no hydrogen, so the oxidation cannot begin and the side chain stays intact.

Why can't a Grignard reagent be used if the molecule has an -OH or -COOH group?

A Grignard reagent is a very strong base. It removes the acidic hydrogen of -OH, -NH or -COOH at once, turning R-MgX into the alkane R-H before it can react with carbon dioxide. It also adds to C=O groups, so such groups must be absent or protected.

What is the difference between acidic and basic hydrolysis of nitriles?

Acidic hydrolysis gives the free carboxylic acid and ammonium ion directly. Basic hydrolysis gives the carboxylate salt and ammonia gas, so the solution must be acidified afterwards to obtain the acid. Both pass through an amide intermediate.

Can ketones be oxidised to carboxylic acids?

Not with mild reagents such as Tollens' or Fehling's. Strong oxidising agents with heat break a carbon-carbon bond next to the keto group, giving acids with fewer carbons. Methyl ketones can also be converted to an acid with one carbon less by the haloform reaction.

Which preparation methods are most important for JEE Main and JEE Advanced?

JEE focuses on multi-step conversions, so the Grignard and nitrile routes (both +1 carbon), side-chain oxidation with the benzylic hydrogen rule, oxidative cleavage with KMnO4 or ozone, and the haloform reaction are the most tested. Keep a carbon-count check for every step.

Which preparation of carboxylic acids reactions are asked in NEET?

NEET follows NCERT: preparation from primary alcohols and aldehydes, alkylbenzenes, nitriles and amides, Grignard reagents, acyl halides and anhydrides, and esters. Questions usually ask for the product or reagent in a one- or two-step conversion.

Previous year questions on Preparation of Carboxylic acids

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

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