Question

Consider the following reaction sequence: \[ 2CH_3CHO \xrightarrow{\text{Dil. NaOH}} X \] \[ X \xrightarrow{\Delta} Y \] The major product \(Y\) formed in the above reaction is:

• A. Ethanol
• B. Butanal
• C. Crotonaldehyde
• D. Acetic acid

Hint:

Remember: \[ \text{Aldol Condensation} \] always produces a \(\beta\)-hydroxy carbonyl compound initially, which upon heating forms an \(\alpha,\beta\)-unsaturated carbonyl compound.

Answer 1

The Correct Option is C

Concept: Aldehydes containing at least one \(\alpha\)-hydrogen atom undergo aldol condensation in the presence of dilute alkali. The reaction initially forms a \(\beta\)-hydroxy aldehyde (aldol), which on heating loses water to produce an \(\alpha,\beta\)-unsaturated aldehyde. This is one of the most important name reactions from NCERT and has recently become a favorite topic for CUET due to its conceptual nature.

Step 1: Identify whether ethanal can undergo aldol condensation.
The given compound is ethanal: \[ CH_3CHO \] The carbon adjacent to the carbonyl carbon contains hydrogen atoms known as \(\alpha\)-hydrogens. Since ethanal contains \(\alpha\)-hydrogen atoms, it undergoes aldol condensation.

Step 2: Formation of aldol.
Two molecules of ethanal combine to form: \[ CH_3CH(OH)CH_2CHO \] This compound is called: \[ \text{3-Hydroxybutanal} \] or simply aldol.

Step 3: Dehydration of aldol.
Upon heating: \[ CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO +H_2O \] The product formed is: \[ \text{Crotonaldehyde} \]

Step 4: Analysis of options.

• Ethanol is not formed.
• Butanal is not formed.
• Crotonaldehyde is the dehydration product.
• Acetic acid is not produced. Therefore: \[ \boxed{\text{Crotonaldehyde}} \]