MP Board · Class 12 · Chemistry · Aldehydes, Ketones and Carboxylic AcidsExplain the Aldol condensation reaction with a suitable chemical equation and mechanism steps for the reaction of acetaldehyde.

Aldol condensation is a characteristic reaction given by aldehydes and ketones containing at least one alpha-hydrogen atom. When such a compound, for instance, acetaldehyde ($CH_3CHO$), is treated with a dilute base (such as dilute $NaOH$), it undergoes self-condensation to form a $\beta$-hydroxyaldehyde, which is commonly known as an aldol. The term 'aldol' is derived from the combination of the functional groups present in the product: aldehyde ('ald') and alcohol ('ol'). \nChemical Equation: $2CH_3CHO \xrightarrow{dilute NaOH} CH_3-CH(OH)-CH_2-CHO$ (3-hydroxybutanal) \nReaction Mechanism Steps:\nStep 1: Formation of Enolate Ion\nThe dilute base provides hydroxide ions ($OH^-$), which abstract an acidic alpha-hydrogen atom from an acetaldehyde molecule, generating a resonance-stabilized enolate ion.\nEquation: $CH_3CHO + OH^- \rightleftharpoons [CH_2^-–CHO] + H_2O$ \nStep 2: Nucleophilic Addition\nThe nucleophilic enolate ion attacks the carbonyl carbon of another unreacted molecule of acetaldehyde in a nucleophilic addition step, forming an alkoxide intermediate ion.\nEquation: $CH_3CHO + CH_2^–CHO \rightarrow CH_3-CH(O^-)-CH_2-CHO$ \nStep 3: Protonation\nThe alkoxide intermediate abstracts a proton from water to form the final product, 3-hydroxybutanal (aldol), regenerating the hydroxide ion catalyst.\nEquation: $CH_3-CH(O^-)-CH_2-CHO + H_2O \rightarrow CH_3-CH(OH)-CH_2-CHO + OH^-$\nUpon heating, this aldol readily loses a water molecule to form an $\alpha,\beta$-unsaturated carbonyl compound (crotonaldehyde).