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Q.Write a note on aldol condensation.

Odisha ChseOdisha CHSE +2 Science Board Exam 2025Subjective· 3mImportance★★★★★
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Base removes an acidic α-hydrogen to form a resonance-stabilised enolate, which attacks the carbonyl carbon of a second aldehyde/ketone molecule, forming a new C-C bond and a β-hydroxy carbonyl (aldol) product; heating dehydrates this to give a conjugated enone.

Step 1 (base-catalysed enolate formation): dilute base (e.g. dil. NaOH) removes a proton from the α-carbon (the carbon adjacent to the carbonyl group) of one molecule of aldehyde or ketone, since this hydrogen is unusually acidic (its conjugate base, the enolate ion, is stabilised by resonance with the carbonyl group):

CH3CHO + OH- → CH2=CH-O- (enolate) + H2O

Step 2 (nucleophilic addition): this nucleophilic enolate carbon attacks the electrophilic carbonyl carbon of a second molecule of aldehyde/ketone, forming a new carbon-carbon bond:

CH3CHO + CH2=CH-O- → CH3-CH(O-)-CH2-CHO

Step 3 (protonation): the resulting alkoxide is protonated (by water) to give the neutral product, a β-hydroxy aldehyde (or ketone), called an 'aldol' (ALDehyde + alcOHOL):

2CH3CHO --dil. NaOH--> CH3-CH(OH)-CH2-CHO (3-hydroxybutanal, an aldol)

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