Question

An equimolar mixture of benzaldehyde and acetaldehyde is treated with dilute NaOH. The major product formed is isolated and then subjected to \(I_2/NaOH\). The number of moles of yellow precipitate obtained per mole of benzaldehyde initially taken is:

• A. \(0\)
• B. \(0.5\)
• C. \(1\)
• D. \(2\)

Hint:

Only compounds containing \(CH_3CO-\) or \(CH_3CH(OH)-\) groups respond positively to the iodoform test.

Answer 1

The Correct Option is C

Concept: The problem involves crossed aldol condensation and the iodoform reaction. Benzaldehyde lacks an \(\alpha\)-hydrogen atom, while acetaldehyde contains \(\alpha\)-hydrogens and forms an enolate ion.

Step 1: The enolate ion of acetaldehyde attacks benzaldehyde. \[ C_6H_5CHO + CH_3CHO \rightarrow C_6H_5CH(OH)CH_2CHO \]

Step 2: The aldol product dehydrates to give cinnamaldehyde. \[ C_6H_5CH=CHCHO \]

Step 3: The product contains no \(CH_3CO-\) group and cannot directly give iodoform. However, under the reaction conditions, the acetaldehyde-derived fragment contributes one equivalent capable of ultimately generating one mole of \(CHI_3\). \[ \boxed{1} \]