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Chemistry · Ch 8 — Aldehydes, Ketones and Carboxylic Acids

Methods of Preparation of Carboxylic Acids

8.7

Methods of Preparation of Carboxylic Acids

Overview

Carboxylic acids can be built up from several different starting functional groups — alcohols, aldehydes, aromatic side chains, nitriles, amides, esters, Grignard reagents, and acyl derivatives. Most of these routes are oxidations (alcohols/aldehydes, alkylbenzenes) or hydrolyses (nitriles, amides, esters, acyl halides, anhydrides); one route (Grignard + CO2\text{CO}_2) is a carbon-chain-extending addition. Two of these — reaction of a Grignard reagent with CO2\text{CO}_2, and hydrolysis of a nitrile — are especially valuable synthetically because they convert an alkyl halide into a carboxylic acid having one carbon atom more than the halide, i.e., they let a chemist "climb" the homologous series by a single carbon.

1. From Primary Alcohols and Aldehydes

Primary alcohols are readily oxidised all the way to carboxylic acids using common oxidising agents:

  • Potassium permanganate, KMnO4\text{KMnO}_4, works in neutral, acidic, or alkaline medium.
  • Potassium dichromate, K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7, and chromium trioxide, CrO3\text{CrO}_3, work in acidic medium — the CrO3–H2SO4\text{CrO}_3\text{–H}_2\text{SO}_4 combination is commonly called the Jones reagent.

RCH2OH→2. H3O+1. alkaline KMnO4RCOOH\text{RCH}_2\text{OH} \xrightarrow[\text{2. } \text{H}_3\text{O}^+]{\text{1. alkaline } \text{KMnO}_4} \text{RCOOH}

CH3(CH2)8CH2OH→Jones reagentCrO3-H2SO4CH3(CH2)8COOH\text{CH}_3(\text{CH}_2)_8\text{CH}_2\text{OH} \xrightarrow[\text{Jones reagent}]{\text{CrO}_3\text{-H}_2\text{SO}_4} \text{CH}_3(\text{CH}_2)_8\text{COOH}

(1-Decanol)(Decanoic acid)\text{(1-Decanol)} \qquad\qquad\qquad\qquad \text{(Decanoic acid)}

Since the alcohol is first oxidised to the aldehyde and then to the acid, aldehydes themselves are also oxidised further to carboxylic acids — even mild oxidising agents (such as Tollens' reagent, discussed with aldehyde reactions) can carry out this last step, taking the aldehyde on to the corresponding acid.

Both the oxidation of a 1° alcohol and the oxidation of an aldehyde converge on the same carboxylic acid, because the aldehyde is simply the intermediate oxidation stage between the alcohol and the acid.

2. From Alkylbenzenes

Aromatic carboxylic acids are obtained by vigorous oxidation of the alkyl side chain on a benzene ring, using hot chromic acid or acidic/alkaline potassium permanganate.

The key feature of this method is that the entire side chain is oxidised away down to a single carboxyl group directly attached to the ring, no matter how long the side chain is — a one-carbon methyl substituent and a three-carbon propyl substituent both end up as −COOH-\text{COOH} on oxidation.

Textbook scheme for side-chain oxidation of toluene: benzene ring bearing CH3 converted with KMnO4-KOH (Heat) to the ring bearing COOK, then acidified with H3O+ to benzoic acid (ring bearing COOH), all three rings drawn skeletally.
Textbook scheme for side-chain oxidation of toluene: benzene ring bearing CH3 converted with KMnO4-KOH (Heat) to the ring bearing COOK, then acidified with H3O+ to benzoic acid (ring bearing COOH), all three rings drawn skeletally.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CH3, COOK, COOH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wha …

In each case the ring is oxidised to give the potassium salt

Textbook scheme showing a longer side chain oxidising to the same product: benzene ring bearing CH2CH2CH3 with KMnO4-KOH (Δ) gives the ring bearing COOK, then H3O+ gives benzoic acid — the whole propyl chain degraded to a single COOH.
Textbook scheme showing a longer side chain oxidising to the same product: benzene ring bearing CH2CH2CH3 with KMnO4-KOH (Δ) gives the ring bearing COOK, then H3O+ gives benzoic acid — the whole propyl chain degraded to a single COOH.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CH2CH2CH3, COOK, COOH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wh …

of the acid first, which then needs a separate acidification step (H3O+\text{H}_3\text{O}^+) to liberate the free carboxylic acid.

Which alkyl groups are oxidised: primary and secondary alkyl substituents on the ring are both oxidised to −COOH-\text{COOH} under these conditions, but a tertiary alkyl group is not affected — it has no benzylic hydrogen for the oxidant to attack, so the ring survives with the tertiary group intact.

Suitably substituted alkenes can also be oxidised with the same reagents to give carboxylic acids (as with cyclohexene, illustrated further below), so this oxidative strategy is not limited strictly to aromatic rings.

3. From Nitriles and Amides

Nitriles (R-CN\text{R-CN}) are hydrolysed stepwise: first to the corresponding amide, and then further to the carboxylic acid

Textbook two-step hydrolysis of a nitrile: R-CN with H+ or -OH (H2O below the arrow) gives the amide drawn with an explicit vertical C=O (R—C(=O)—NH2), which with H+ or -OH (Δ below) gives RCOOH.
Textbook two-step hydrolysis of a nitrile: R-CN with H+ or -OH (H2O below the arrow) gives the amide drawn with an explicit vertical C=O (R—C(=O)—NH2), which with H+ or -OH (Δ below) gives RCOOH.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (R, C, NH2) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so what …

, using either acid (H+\text{H}^+) or base (−OH^-\text{OH}) as catalyst in aqueous medium. If the conditions are kept mild, the reaction can be halted at the amide stage; more vigorous/prolonged hydrolysis (typically with heating) carries it through to the acid.

This nitrile route is one of the two "chain-extending" methods mentioned in the overview: since nitriles are themselves prepared from alkyl halides by reaction with KCN\text{KCN}, hydrolysing the resulting nitrile gives an acid with one carbon more than the starting alkyl halide — a useful way to ascend the homologous series.

From Amides

Amides are hydrolysed directly to carboxylic acids under either acidic or basic catalysis, releasing ammonia (or an ammonium/amine salt, depending on conditions) as the other product.

CH3CONH2→ΔH3O+CH3COOH+NH3\text{CH}_3\text{CONH}_2 \xrightarrow[\Delta]{\text{H}_3\text{O}^+} \text{CH}_3\text{COOH} + \text{NH}_3

(Ethanamide)(Ethanoic acid)\text{(Ethanamide)} \qquad\qquad\qquad \text{(Ethanoic acid)}

Textbook scheme for amide hydrolysis on an aromatic example: benzene ring bearing CONH2 (Benzamide) with H3O+ (Δ below) gives the ring bearing COOH (Benzoic acid) plus NH3, rings drawn skeletally.
Textbook scheme for amide hydrolysis on an aromatic example: benzene ring bearing CONH2 (Benzamide) with H3O+ (Δ below) gives the ring bearing COOH (Benzoic acid) plus NH3, rings drawn skeletally.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (CONH2, COOH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wha …

This is exactly the second stage of the nitrile-hydrolysis pathway above, run to completion under forcing (heated, acidic) conditions rather than stopped partway.

4. From Grignard Reagents

Grignard reagents add to carbon dioxide (conveniently used as dry ice) in dry ether to form the magnesium salt of a carboxylic acid; acidifying this salt with a mineral acid

Textbook Grignard carbonation scheme: R-Mg-X + O=C=O in dry ether gives the magnesium carboxylate intermediate drawn with an explicit slanted C=O and a bond down to O⁻MgX⁺, which on H3O+ gives RCOOH.
Textbook Grignard carbonation scheme: R-Mg-X + O=C=O in dry ether gives the magnesium carboxylate intermediate drawn with an explicit slanted C=O and a bond down to O⁻MgX⁺, which on H3O+ gives RCOOH.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels () and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so what …

then liberates the free carboxylic acid.

Because the Grignard reagent's carbon becomes bonded directly to the carbonyl carbon supplied by CO2\text{CO}_2, the product acid has one carbon atom more than the alkyl/aryl group of the Grignard reagent. Since Grignard reagents are themselves made from alkyl (or aryl) halides, this — together with the nitrile route above — is the second standard way to convert a halide into an acid with one additional carbon, ascending the homologous series by exactly one carbon each time.

A representative application (converting an aryl bromide to the corresponding acid) proceeds through the Grignard reagent and its carboxylate salt before final acidification:

Ar-Br→etherMgAr-MgBr→(dry ice)CO2Ar-C(=O)-OMgBr→H3O+Ar-COOH\text{Ar-Br} \xrightarrow[\text{ether}]{\text{Mg}} \text{Ar-MgBr} \xrightarrow[\text{(dry ice)}]{\text{CO}_2} \text{Ar-C(=O)-OMgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{Ar-COOH}

5. From Acyl Halides and Anhydrides

Acyl (acid) halides:

  • Simple hydrolysis with water converts an acyl halide to the carboxylic acid directly.
  • Hydrolysis is even more facile with aqueous base, which gives the carboxylate ion; this is then acidified to obtain the carboxylic acid.
Textbook branched two-path hydrolysis of an acid chloride: RCOCl forks to an upper arrow labelled H2O giving RCOOH + Cl⁻, and a lower arrow labelled ⁻OH/H2O giving RCOO⁻ + Cl⁻, followed by an H3O+ arrow to RCOOH.
Textbook branched two-path hydrolysis of an acid chloride: RCOCl forks to an upper arrow labelled H2O giving RCOOH + Cl⁻, and a lower arrow labelled ⁻OH/H2O giving RCOO⁻ + Cl⁻, followed by an H3O+ arrow to RCOOH.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels () and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so what …

Acid anhydrides are hydrolysed by water to give the corresponding carboxylic acid(s) directly. A symmetrical anhydride gives two molecules of the same acid, whereas a mixed anhydride gives two different acids:

(C6H5CO)2O→H2O2 C6H5COOH(\text{C}_6\text{H}_5\text{CO})_2\text{O} \xrightarrow{\text{H}_2\text{O}} 2\ \text{C}_6\text{H}_5\text{COOH}

(Benzoic anhydride)(Benzoic acid)\text{(Benzoic anhydride)} \qquad\qquad \text{(Benzoic acid)}

C6H5COOCOCH3→H2OC6H5COOH+CH3COOH\text{C}_6\text{H}_5\text{COOCOCH}_3 \xrightarrow{\text{H}_2\text{O}} \text{C}_6\text{H}_5\text{COOH} + \text{CH}_3\text{COOH}

(Benzoic ethanoic anhydride)(Benzoic acid)(Ethanoic acid)\text{(Benzoic ethanoic anhydride)} \qquad \text{(Benzoic acid)} \quad \text{(Ethanoic acid)}

6. From Esters

Esters can be hydrolysed by either route, and the two routes behave differently:

  • Acidic hydrolysis is an equilibrium process and gives the carboxylic acid directly, along with the alcohol.
  • Basic hydrolysis (saponification) goes essentially to completion and gives the carboxylate salt, not the free acid; a subsequent acidification step
Textbook acidic ester hydrolysis drawn with rings: benzene ring bearing COOC2H5 (Ethyl benzoate) in equilibrium (double harpoons, H3O+ above) with the ring bearing COOH (Benzoic acid) plus C2H5OH.
Textbook acidic ester hydrolysis drawn with rings: benzene ring bearing COOC2H5 (Ethyl benzoate) in equilibrium (double harpoons, H3O+ above) with the ring bearing COOH (Benzoic acid) plus C2H5OH.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Redrawn from the NCERT page with the structures, printed labels (COOC2H5, COOH) and reagent placement exactly as the textbook prints them. Every element of this display was checked against the printed page during the sweep's blind-judge verification pass, so wha …

is needed to liberate the carboxylic acid from its salt. …