Question:

Which of the following compounds will undergo self aldol condensation in the presence of cold dilute alkali?

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Only aldehydes and ketones with \(\alpha\)-hydrogen undergo aldol condensation.
Updated On: May 22, 2025
  • CH2 = CH − CHO
  • CH ≡ C − CHO
  • C6H5CHO
  • CH3CH2CHO
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The Correct Option is D

Approach Solution - 1

Step 1: Understanding Aldol Condensation 
- Aldol condensation occurs in aldehydes/ketones that contain at least one \(\alpha\)-hydrogen. 
- The \(\alpha\)-hydrogen is removed by a base to form an enolate ion, which then attacks another carbonyl compound. 
Step 2: Analyzing the Given Compounds 
- \( CH_3CH_2CHO \) (Propionaldehyde) contains an \(\alpha\)-hydrogen, allowing aldol condensation. 
- \( CH_2 = CH - CHO \) (Acrolein) and \( CH \equiv C - CHO \) do not easily form enolates due to resonance stabilization. 
- \( C_6H_5CHO \) (Benzaldehyde) lacks \(\alpha\)-hydrogen, preventing aldol condensation. 
Final Answer: \( CH_3CH_2CHO \) undergoes aldol condensation. 
 

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Approach Solution -2

The compound that will undergo self aldol condensation in the presence of cold dilute alkali is CH₃CH₂CHO (Propionaldehyde).

Aldol condensation occurs when an aldehyde or ketone with at least one α-hydrogen reacts under basic conditions to form a β-hydroxy aldehyde or ketone, which can further dehydrate to form an α,β-unsaturated compound.

Propionaldehyde (CH₃CH₂CHO) has α-hydrogens adjacent to the carbonyl group, making it capable of forming an enolate ion under basic conditions. This enolate can attack another molecule of propionaldehyde leading to self aldol condensation.

On the other hand, compounds without α-hydrogens or with steric hindrance do not readily undergo this reaction. Therefore, CH₃CH₂CHO is the correct answer as it efficiently undergoes self aldol condensation when treated with cold dilute alkali.
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