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Aldehydes, Ketones and Carboxylic Acids - Preparation of Carboxylic Acids

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Carboxylic acids can be prepared from primary alcohols and aldehydes using strong oxidizing agents such as potassium permanganate (KMnO4KMnO_4) in acidic, alkaline, or neutral media, or Jones reagent (CrO3−H2SO4CrO_3 - H_2SO_4).

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Aromatic carboxylic acids are prepared by vigorous oxidation of alkylbenzenes with chromic acid or alkaline KMnO4KMnO_4. The entire side chain is oxidized to a carboxyl group regardless of the length of the chain, provided there is at least one benzylic hydrogen.

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Nitriles are hydrolyzed to amides and then to carboxylic acids in the presence of H+H^+ or OH−OH^- as a catalyst. The reaction is: R−CN→H2O/H+R−CONH2→H2O/H+,ΔR−COOHR-CN \xrightarrow{H_2O/H^+} R-CONH_2 \xrightarrow{H_2O/H^+, \Delta} R-COOH.

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Grignard reagents react with dry ice (solid CO2CO_2) in ethereal solution to form salts of carboxylic acids, which upon acidification with mineral acids yield corresponding carboxylic acids. This method is useful for increasing the carbon chain length by one unit.

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Acid chlorides (acyl halides) are hydrolyzed with water to give carboxylic acids, or more readily with aqueous base to give carboxylate ions, which on acidification provide carboxylic acids.

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Esters undergo nucleophilic substitution (hydrolysis) with dilute mineral acids to yield carboxylic acids and alcohols directly, or with aqueous alkalis (saponification) to yield carboxylates and alcohols.

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Acid anhydrides are hydrolyzed to carboxylic acids upon reaction with water. Symmetrical anhydrides yield two molecules of the same acid, while unsymmetrical ones yield two different acids.

📐Formulae

R−CH2OH→2.H3O+1.alkaline KMnO4R−COOHR-CH_2OH \xrightarrow[2. H_3O^+]{1. alkaline\ KMnO_4} R-COOH

Ar−CH3→2.H3O+1.KMnO4−KOH,ΔAr−COOHAr-CH_3 \xrightarrow[2. H_3O^+]{1. KMnO_4-KOH, \Delta} Ar-COOH

R−CN→H+/H2OR−CONH2→H+/H2O,ΔR−COOHR-CN \xrightarrow{H^+/H_2O} R-CONH_2 \xrightarrow{H^+/H_2O, \Delta} R-COOH

R−MgX+O=C=O→Dry etherR−COOMgX→H3O+R−COOHR-MgX + O=C=O \xrightarrow{Dry\ ether} R-COOMgX \xrightarrow{H_3O^+} R-COOH

R−COCl→H2OR−COOH+HClR-COCl \xrightarrow{H_2O} R-COOH + HCl

(RCO)2O→H2O2R−COOH(RCO)_2O \xrightarrow{H_2O} 2R-COOH

R−COOR′→H3O+R−COOH+R′OHR-COOR' \xrightarrow{H_3O^+} R-COOH + R'OH

💡Examples

Problem 1:

Write the chemical equation for the preparation of Benzoic acid from Ethylbenzene.

Solution:

C6H5CH2CH3→2.H3O+1.KMnO4/KOH,ΔC6H5COOHC_6H_5CH_2CH_3 \xrightarrow[2. H_3O^+]{1. KMnO_4 / KOH, \Delta} C_6H_5COOH

Explanation:

Alkyl groups on a benzene ring are oxidized to the carboxyl group (−COOH-COOH) by alkaline KMnO4KMnO_4 followed by acidification, provided there is at least one benzylic hydrogen. The length of the alkyl chain does not matter; it is always converted to a single −COOH-COOH group.

Problem 2:

How is Ethanoic acid prepared using a Grignard reagent?

Solution:

CH3MgBr+CO2→Dry etherCH3COOMgBr→H3O+CH3COOH+Mg(OH)BrCH_3MgBr + CO_2 \xrightarrow{Dry\ ether} CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)Br

Explanation:

Methylmagnesium bromide reacts with carbon dioxide (dry ice) to form an adduct (magnesium salt of acetic acid), which on subsequent acid hydrolysis yields Ethanoic acid.

Problem 3:

Convert Benzamide to Benzoic acid.

Solution:

C6H5CONH2+H2O→H+,ΔC6H5COOH+NH4+C_6H_5CONH_2 + H_2O \xrightarrow{H^+, \Delta} C_6H_5COOH + NH_4^+

Explanation:

Amides undergo hydrolysis in the presence of an acid catalyst and heat to produce the corresponding carboxylic acid and ammonium ion.