Question:
Formaldehyde reacts with KOH to give methanol and potassium formate. What is the name of the reaction?
Formaldehyde reacts with KOH to give methanol and potassium formate. What is the name of the reaction?
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An aldehyde with no \( \alpha \)-hydrogen disproportionates with strong base; name that reaction.
Updated On: Jul 10, 2026
- Perkin reaction
- Wittig reaction
- Cannizzaro reaction
- Reimer-Tiemann reaction
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The Correct Option is C
Solution and Explanation
Step 1: Formaldehyde \( (HCHO) \) has no \( \alpha \)-hydrogen atom.
Step 2: Aldehydes lacking an \( \alpha \)-hydrogen, when warmed with concentrated alkali, undergo self oxidation-reduction (disproportionation).
Step 3: One molecule is reduced to an alcohol (methanol) while another is oxidised to the salt of the acid (potassium formate): \[ 2HCHO + KOH \rightarrow CH_3OH + HCOOK \]
Step 4: This disproportionation of an aldehyde without \( \alpha \)-H is called the Cannizzaro reaction. (Perkin uses anhydrides, Wittig uses ylides, Reimer-Tiemann is phenol formylation, so they do not apply.)
Answer: Cannizzaro reaction.
Step 2: Aldehydes lacking an \( \alpha \)-hydrogen, when warmed with concentrated alkali, undergo self oxidation-reduction (disproportionation).
Step 3: One molecule is reduced to an alcohol (methanol) while another is oxidised to the salt of the acid (potassium formate): \[ 2HCHO + KOH \rightarrow CH_3OH + HCOOK \]
Step 4: This disproportionation of an aldehyde without \( \alpha \)-H is called the Cannizzaro reaction. (Perkin uses anhydrides, Wittig uses ylides, Reimer-Tiemann is phenol formylation, so they do not apply.)
Answer: Cannizzaro reaction.
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