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Ethanal and propanone undergo condensati...

Ethanal and propanone undergo condensation reaction in presence of dil. Alkali to form

A

`CH_3CH(OH) CH_2COCH_3`

B

`CH_3-CO-C(OH)(CH_3)_2`

C

`HOH_2"CCH"_2CH_2COCH_3`

D

`CH_3COCH(CH_3)CH_2OH`

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The correct Answer is:
To solve the problem of what products are formed when ethanal (acetaldehyde) and propanone (acetone) undergo a condensation reaction in the presence of dilute alkali, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: - Ethanal (acetaldehyde) has the formula CH₃CHO. - Propanone (acetone) has the formula CH₃COCH₃. 2. **Understand the Reaction Medium**: - The reaction is carried out in a dilute alkaline medium, which typically involves a base like sodium hydroxide (NaOH). This creates a basic environment conducive for condensation reactions. 3. **Formation of Enolate Ion**: - In the presence of a base, the alpha hydrogen (the hydrogen atom attached to the carbon adjacent to the carbonyl group) of propanone can be deprotonated to form an enolate ion. - The enolate ion from propanone can be represented as: \[ CH_3C(O^-)CH_3 \] 4. **Nucleophilic Attack**: - The enolate ion acts as a nucleophile and can attack the carbonyl carbon of ethanal. This results in the formation of a tetrahedral intermediate. 5. **Formation of the Product**: - The tetrahedral intermediate will rearrange, leading to the formation of a β-hydroxy ketone (an aldol product). The reaction can be summarized as follows: \[ CH_3COCH_3 + CH_3CHO \rightarrow CH_3C(OH)(CH_3)CH_2CHO \] - This product is 4-hydroxy-4-methylpentan-2-one. 6. **Dehydration**: - Under the reaction conditions, the β-hydroxy ketone can undergo dehydration (loss of water) to form an α,β-unsaturated carbonyl compound: \[ CH_3C(OH)(CH_3)CH_2CHO \rightarrow CH_3C(CH_3)=CHCHO + H_2O \] 7. **Final Product**: - The final product of the condensation reaction is 4-methylpent-3-enal (or crotonaldehyde). ### Final Answer: The condensation reaction of ethanal and propanone in the presence of dilute alkali forms 4-methylpent-3-enal (or crotonaldehyde).
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