See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Cannizzaro Reaction (Cross-Cannizzaro) When an aldehyde lacking alpha-hydrogens is treated with concentrated KOH, it undergoes the Cannizzaro reaction (disproportionation). In a cross-Cannizzaro reaction, an equimolar mixture of two different aldehydes both lacking alpha-hydrogens reacts with concentrated KOH. Step 1: Identify the aldehydes. - 4-methylbenzaldehyde (ArCHO, no alpha-H) - Formaldehyde (HCHO, no alpha-H) Both lack alpha-hydrogens, so the Cannizzaro reaction applies. Step 2: Understand the Cross-Cannizzaro selectivity. In a cross-Cannizzaro reaction between formaldehyde and an aromatic aldehyde, formaldehyde preferentially acts as the hydride donor (reducing agent) because it is more reactive toward nucleophilic addition and is a better hydride source. Formaldehyde is oxidized to formate (HCOO-), which upon acidification gives formic acid (HCOOH). The aromatic aldehyde (4-methylbenzaldehyde) is reduced to the corresponding primary alcohol (4-methylbenzyl alcohol, 4-CH3-C6H4-CH2OH). Step 3: Write the reaction. 4-CH3-C6H4-CHO + HCHO + KOH → 4-CH3-C6H4-CH2OH + HCOOK After acidification: HCOOK → HCOOH So products are: 4-methylbenzyl alcohol + formic acid. Step 4: Evaluate why other options fail. (a) Aldol addition product - not possible since neither aldehyde has alpha-hydrogens for enolization, so aldol condensation cannot occur. (c) Tischenko or oxidation product - 4-methylbenzoic acid + methanol would require a different mechanism; in cross-Cannizzaro, formaldehyde donates hydride (gets oxidized), not the aromatic aldehyde. (d) Shows unreacted 4-methylbenzaldehyde ring with a CH2OH group - this would imply intramolecular or partial reaction, not consistent with the cross-Cannizzaro outcome. Option (b) correctly shows 4-methylbenzyl alcohol + HCOOH, the products of the cross-Cannizzaro reaction between 4-methylbenzaldehyde and formaldehyde. Therefore, the correct answer is B.