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Q.(i) Write the reaction involved in Aldol condensation.

(1)
(ii) Predict the product of the given reaction.
(1)
CH3COONa --NaOH/CaO, delta--> ?
Himachal HpboseHPBOSE Plus Two Board 2026Subjective· 2mImportance★★★★★
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(i) Aldol condensation: two molecules of an aldehyde/ketone with an α-H combine under dilute base to give a β-hydroxy carbonyl compound (the 'aldol'). (ii) Sodium acetate on heating with soda lime (NaOH/CaO) undergoes decarboxylation to give methane.

(i) Aldol condensation

When an aldehyde (or ketone) possessing at least one α-hydrogen is treated with dilute alkali (e.g. dilute NaOH), two molecules combine: the base removes an α-H to form a resonance-stabilised carbanion (enolate), which then attacks the carbonyl carbon of a second aldehyde molecule, giving a β-hydroxy aldehyde ('aldol'):

2CH3CHO→dil. NaOHCH3−CH∣OH−CH2−CHO2CH_3CHO \xrightarrow{dil.\,NaOH} CH_3-\underset{OH}{\underset{|}{CH}}-CH_2-CHO

(3-hydroxybutanal, the 'aldol'). On further heating this can lose water (dehydrate) to give an α,β-unsaturated carbonyl compound — this combined two-step process (addition then dehydration) is called aldol condensation.

(ii) CH₃COONa --NaOH/CaO, Δ--> ?

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