Which of these aldehydes will UNDERGO the Cannizzaro reaction (disproportionation under concentrated NaOH)?

1. The CANNIZZARO reaction is a base-induced DISPROPORTIONATION: one molecule of the aldehyde is OXIDISED to the carboxylic acid (or carboxylate) while another is REDUCED to the alcohol. 2. Crucial requirement: the aldehyde MUST NOT have an $\alpha$-HYDROGEN. (With $\alpha$-H, the molecule undergoes ALDOL condensation instead — a faster, competing reaction.) 3. Formaldehyde ($\mathrm{HCHO}$) has NO $\alpha$-carbon at all, so no $\alpha$-H → Cannizzaro applies. Two HCHO + NaOH → HCOOH(salt) + $\mathrm{CH_3OH}$. 4. Benzaldehyde ($\mathrm{C_6H_5CHO}$): the $\alpha$ position is a benzene-ring carbon with no extractable H → Cannizzaro applies. $2\,\mathrm{C_6H_5CHO} \xrightarrow{\text{NaOH}} \mathrm{C_6H_5COO^-} + \mathrm{C_6H_5CH_2OH}$. 5. Acetaldehyde, propanal, butyraldehyde ALL have $\alpha$-H — they undergo aldol, not Cannizzaro. So options A, B, C are wrong. _Source: NCERT Class 12 Chemistry Part 2, Ch 8, §8.4 (Chemical Reactions — Cannizzaro reaction), p. 13–14._