Question:
Aldehydes containing at least one \(\alpha\)-hydrogen atom undergo a reaction in presence of dilute aq. NaOH to form \(\beta\)-hydroxy aldehyde. Identify the reaction.
Aldehydes containing at least one \(\alpha\)-hydrogen atom undergo a reaction in presence of dilute aq. NaOH to form \(\beta\)-hydroxy aldehyde. Identify the reaction.
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Dilute alkali with an alpha-hydrogen aldehyde gives a beta-hydroxy aldehyde in the aldol reaction.
Updated On: Nov 12, 2026
- Cannizzaro reaction
- Aldol reaction
- Reformatsky reaction
- Wolff–Kishner reaction
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
An aldehyde with an \(\alpha\)-hydrogen loses that proton in dilute NaOH and forms an enolate.
Step 2: Reaction
The enolate attacks the carbonyl carbon of a second aldehyde molecule. This gives a \(\beta\)-hydroxy aldehyde, such as 3-hydroxybutanal from ethanal.
Step 3: Other reactions
Cannizzaro needs aldehydes with no \(\alpha\)-hydrogen. Reformatsky uses \(\alpha\)-halo esters with zinc. Wolff-Kishner converts C=O to \(\text{CH}_{2}\).
Final Answer:
The reaction is the aldol reaction. \[ \boxed{\text{(B) }\text{Aldol reaction}} \]
An aldehyde with an \(\alpha\)-hydrogen loses that proton in dilute NaOH and forms an enolate.
Step 2: Reaction
The enolate attacks the carbonyl carbon of a second aldehyde molecule. This gives a \(\beta\)-hydroxy aldehyde, such as 3-hydroxybutanal from ethanal.
Step 3: Other reactions
Cannizzaro needs aldehydes with no \(\alpha\)-hydrogen. Reformatsky uses \(\alpha\)-halo esters with zinc. Wolff-Kishner converts C=O to \(\text{CH}_{2}\).
Final Answer:
The reaction is the aldol reaction. \[ \boxed{\text{(B) }\text{Aldol reaction}} \]
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