Aldehydes, Ketones & Carboxylic Acids - Distinguish between Aldehydes and Ketones

Aldehydes, Ketones & Carboxylic Acids

Distinguish between Aldehydes and Ketones

Aldehydes

Aldehydes have the functional group:

  • An aldehyde group consists of a carbon-oxygen double bond (>C=O) where the carbon atom is also bonded to a hydrogen atom (–H). Aldehydes have the following characteristics:
  • The general formula is RCHO.
  • Aldehydes are terminal carbonyl compounds as the carbonyl group is located at the end of the carbon chain.
  • They have a slightly higher boiling point compared to ketones due to the presence of a hydrogen atom that forms hydrogen bonding.
  • Some examples include formaldehyde (HCHO), acetaldehyde (CH3CHO), and benzaldehyde (C6H5CHO).

Ketones

Ketones have the functional group:

  • A ketone group consists of a carbon-oxygen double bond (>C=O) where the carbon atom is bonded to two other carbon atoms. Ketones have the following characteristics:
  • The general formula is RCOR'.
  • Ketones are non-terminal carbonyl compounds as the carbonyl group is located within the carbon chain.
  • They have a slightly lower boiling point compared to aldehydes as they do not have a hydrogen atom for hydrogen bonding.
  • Some examples include acetone (CH3COCH3), propanone (CH3COCH2CH3), and butanone (CH3COCH2CH2CH3).

Differences between Aldehydes and Ketones

  1. Aldehydes have a hydrogen atom bonded to the carbonyl carbon, while ketones have two carbon atoms bonded to the carbonyl carbon.
  1. Aldehydes are more reactive than ketones.
  1. Aldehydes undergo oxidation reactions more readily than ketones.
  1. Aldehydes have a higher boiling point than ketones due to the presence of a hydrogen atom for hydrogen bonding.
  1. Aldehydes are easily oxidized to carboxylic acids, while ketones are resistant to oxidation.
  1. Aldehydes are named by replacing the -e ending of the corresponding alkane with -al, while ketones are named by replacing the -e ending of the corresponding alkane with -one.

Oxidation of Aldehydes

Aldehydes can be oxidized to form carboxylic acids. Example: CH3CHO + [O] → CH3COOH In this reaction, the aldehyde loses a hydrogen atom from the carbonyl group, and the carbon atom gains an oxygen atom to form a carboxylic acid. Note: Ketones, on the other hand, do not undergo oxidation under normal conditions.

Chemical Tests for Aldehydes

  1. Fehling’s Test
  • Fehling’s reagent (blue solution of copper sulphate in sodium hydroxide) reacts with aldehydes to form a brick-red precipitate of copper(I) oxide. Example: Aldehyde + Fehling’s Reagent → Brick-red precipitate
  1. Tollens’ Test
  • Tollens’ reagent (ammoniacal silver nitrate) reacts with aldehydes to form a silver mirror on the inner surface of the test tube. Example: Aldehyde + Tollens’ Reagent → Silver mirror These tests help distinguish aldehydes from ketones as ketones do not yield positive results with these tests.

Physical Properties of Carboxylic Acids

Carboxylic acids have the following physical properties:

  • They are polar compounds due to the presence of the carbonyl group (–C=O) and the hydroxyl group (–OH).
  • They have higher boiling points compared to aldehydes and ketones due to strong hydrogen bonding between carboxylic acid molecules.
  • They are generally soluble in water.
  • They have a sharp, sour odor.
  • Some examples include acetic acid (CH3COOH), formic acid (HCOOH), and benzoic acid (C6H5COOH).

Reactions of Carboxylic Acids

  1. Acid-Base Reactions
  • Carboxylic acids react with bases to form salts and water. Example: CH3COOH + NaOH → CH3COONa + H2O
  1. Dehydration
  • Carboxylic acids can undergo dehydration reactions to form anhydrides. Example: CH3COOH → CH3CO2 + H2O
  1. Esterification
  • Carboxylic acids react with alcohols in the presence of an acid catalyst to form esters. Example: CH3COOH + C2H5OH → CH3COOC2H5 + H2O
  1. Reduction
  • Carboxylic acids can be reduced to primary alcohols using reducing agents. Example: CH3COOH + LiAlH4 → CH3CH2OH These are some of the important reactions of carboxylic acids.

Oxidation of Aldehydes

Aldehydes can be oxidized to form carboxylic acids. Example:

  • CH3CHO + [O] → CH3COOH In this reaction, the aldehyde loses a hydrogen atom from the carbonyl group, and the carbon atom gains an oxygen atom to form a carboxylic acid. Note: Ketones, on the other hand, do not undergo oxidation under normal conditions.

Chemical Tests for Aldehydes

  1. Fehling’s Test
  • Fehling’s reagent (blue solution of copper sulphate in sodium hydroxide) reacts with aldehydes to form a brick-red precipitate of copper(I) oxide.
  • Example: Aldehyde + Fehling’s Reagent → Brick-red precipitate
  1. Tollens’ Test
  • Tollens’ reagent (ammoniacal silver nitrate) reacts with aldehydes to form a silver mirror on the inner surface of the test tube.
  • Example: Aldehyde + Tollens’ Reagent → Silver mirror These tests help distinguish aldehydes from ketones as ketones do not yield positive results with these tests.

Physical Properties of Carboxylic Acids

Carboxylic acids have the following physical properties:

  • They are polar compounds due to the presence of the carbonyl group (–C=O) and the hydroxyl group (–OH).
  • They have higher boiling points compared to aldehydes and ketones due to strong hydrogen bonding between carboxylic acid molecules.
  • They are generally soluble in water.
  • They have a sharp, sour odor.
  • Some examples include acetic acid (CH3COOH), formic acid (HCOOH), and benzoic acid (C6H5COOH).

Reactions of Carboxylic Acids

  1. Acid-Base Reactions
  • Carboxylic acids react with bases to form salts and water.
  • Example: CH3COOH + NaOH → CH3COONa + H2O
  1. Dehydration
  • Carboxylic acids can undergo dehydration reactions to form anhydrides.
  • Example: CH3COOH → CH3CO2 + H2O
  1. Esterification
  • Carboxylic acids react with alcohols in the presence of an acid catalyst to form esters.
  • Example: CH3COOH + C2H5OH → CH3COOC2H5 + H2O
  1. Reduction
  • Carboxylic acids can be reduced to primary alcohols using reducing agents.
  • Example: CH3COOH + LiAlH4 → CH3CH2OH These are some of the important reactions of carboxylic acids.

Distinction Between Aldehydes and Ketones by Iodoform Test

The Iodoform test is used to distinguish between aldehydes and methyl ketones. Procedure:

  1. Add a small amount of iodine (I2) and sodium hydroxide (NaOH) to the unknown compound.
  1. Heat the mixture gently.
  1. If a yellow precipitate of iodoform (CHI3) forms, it indicates the presence of a methyl ketone.
  1. If no precipitate forms, it indicates the absence of a methyl ketone. Example:
  • CH3COR (Methyl Ketone) + 3I2 + 4NaOH → CHI3 + RCOONa + 3NaI + 3H2O This test is specific for methyl ketones and does not give a positive result for aldehydes.

Aldol Condensation Reaction

The Aldol condensation is a reaction involving the combination of an aldehyde or ketone with an enolate ion formed from the same or a different carbonyl compound. Steps of the Aldol condensation reaction:

  1. Formation of enolate ion: An aldehyde or ketone reacts with a base (e.g., NaOH or NaOEt) to form the enolate ion.
  1. Attack of the enolate ion: The enolate ion attacks another aldehyde or ketone molecule, forming a carbon-carbon bond.
  1. Deprotonation: A hydroxyl group is removed from the resulting compound to form the aldol product. Example:
  • CH3CHO + CH3COCH3 → CH3CH(OH)CH2COCH3 The aldol condensation reaction allows the formation of larger molecules by joining smaller ones together.

Decarboxylation of Carboxylic Acids

Decarboxylation is a reaction in which a carboxylic acid loses a carbon dioxide molecule to form an alkane.

Conditions for decarboxylation:

  • High temperature (around 600-800°C)
  • Presence of a catalyst (e.g., copper or zinc oxide) Example:
  • CH3COOH → CH4 + CO2 Decarboxylation is an important reaction in the synthesis of hydrocarbons.

Reimer-Tiemann Reaction

The Reimer-Tiemann reaction is a useful method for the synthesis of phenols from carbonyl compounds. Steps of the Reimer-Tiemann reaction:

  1. Reaction with chloroform and a strong base (e.g., NaOH).
  1. Formation of the dichlorocarbene intermediate.
  1. Rearrangement and hydrolysis to form the phenol product. Example:
  • CH3CHO + CHCl3 + NaOH → C6H5OH + NaCl + H2O The Reimer-Tiemann reaction is an important synthetic tool in organic chemistry.

Cannizzaro Reaction

The Cannizzaro reaction is a disproportionation reaction in which an aldehyde is simultaneously oxidized and reduced. Conditions for the Cannizzaro reaction:

  • Strong alkaline conditions (e.g., NaOH or KOH)
  • Lack of an α-hydrogen atom in the aldehyde molecule Example:
  • 2CH3CHO → CH3COOH + CH3CH2OH The Cannizzaro reaction is an important reaction for the preparation of alcohols and carboxylic acids.

Reactions of Carboxylic Acids (Continued)

  1. Decarboxylation
  • Carboxylic acids can undergo decarboxylation under high temperature and in the presence of a catalyst, such as copper or zinc oxide, to form an alkane. Example: CH3COOH → CH4 + CO2
  1. Reimer-Tiemann Reaction
  • The Reimer-Tiemann reaction is a method for the synthesis of phenols from carbonyl compounds. It involves the reaction of the carbonyl compound with chloroform and a strong base, followed by rearrangement and hydrolysis. Example: CH3CHO + CHCl3 + NaOH → C6H5OH + NaCl + H2O
  1. Cannizzaro Reaction
  • The Cannizzaro reaction is a disproportionation reaction in which an aldehyde is simultaneously oxidized and reduced in the presence of a strong alkali. Example: 2CH3CHO → CH3COOH + CH3CH2OH These are some other important reactions of carboxylic acids.

Fischer Esterification

  • Fischer esterification is a reaction in which a carboxylic acid reacts with an alcohol in the presence of an acid catalyst to form an ester. Steps of Fischer esterification:
  1. Protonation: The acid catalyst donates a proton to the carbonyl group of the carboxylic acid, making it more electrophilic.
  1. Nucleophilic attack: The alcohol acts as a nucleophile, attacking the carbonyl group and forming a tetrahedral intermediate.
  1. Proton transfer: A proton is transferred from the –OH group of the alcohol to the oxygen of the tetrahedral intermediate, forming water and regenerating the acid catalyst.
  1. Elimination: The tetrahedral intermediate collapses, eliminating water and forming the ester. Example: CH3COOH + C2H5OH → CH3COOC2H5 + H2O Fischer esterification is an important reaction in the synthesis of esters in the laboratory and industrial processes.

Hydrolysis of Esters

  • Hydrolysis of esters is a reaction in which an ester reacts with water to