SAChemistry

CBSE · 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.

Step-by-Step Solution

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).

💡 Study Guide: This question tests core syllabus concepts from Aldehydes, Ketones and Carboxylic Acids. For formulas, key summaries, and mock exam reference guides, read the full Aldehydes, Ketones and Carboxylic Acids Revision Notes.
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