Q.(a) Write a short note on: Aldol condensation.
(c)(i) Carry out the conversion: Ethanoic acid -> Propanone.
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Start your 14-day free trial to unlock the full solution →(a) Aldol condensation forms a new C-C bond then eliminates water. (b) NaHCO3 distinguishes an acid strong enough to release CO2 from one that isn't. (c) A classic decarboxylative route to acetone, and a crossed Cannizzaro between formaldehyde and benzaldehyde.
(a) Aldol condensation - short note.
An aldehyde or ketone possessing at least one alpha-hydrogen atom, when treated with a dilute base (e.g. dilute NaOH), undergoes self-addition: the base generates an enolate at the alpha-carbon of one molecule, which attacks the electrophilic carbonyl carbon of a second molecule of the same compound. This gives a beta-hydroxy aldehyde or ketone, called an 'aldol'. On further heating, this aldol readily loses a molecule of water (dehydration) to give an alpha,beta-unsaturated carbonyl compound. The overall two-stage process (addition then condensation/dehydration) is called aldol condensation.
Example: 2CH3CHO --dil. NaOH--> CH3-CH(OH)-CH2-CHO (an aldol) --heat, -H2O--> CH3-CH=CH-CHO (crotonaldehyde).
(b) Distinguishing benzoic acid and phenol.
Add aqueous sodium bicarbonate (NaHCO3) solution to each. Benzoic acid (a carboxylic acid, stronger than carbonic acid) reacts with NaHCO3 to give brisk effervescence of CO2 gas:
C6H5COOH + NaHCO3 -> C6H5COONa + H2O + CO2 (up arrow)
Phenol is a much weaker acid (weaker than carbonic acid) and does NOT react with NaHCO3 - no effervescence is observed. This reactivity difference distinguishes the two.
(c)(i) Ethanoic acid -> Propanone.
First neutralise acetic acid with calcium hydroxide (or calcium carbonate) to form calcium acetate:
2CH3COOH + Ca(OH)2 -> (CH3COO)2Ca + 2H2O
Then dry-distil (heat strongly, without air) calcium acetate; it decomposes to give acetone (propanone) and calcium carbonate:
(CH3COO)2Ca --heat (dry distillation)--> CH3-CO-CH3 (propanone) + CaCO3
(c)(ii) C6H5CHO + HCHO with 50% NaOH. …
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