The reaction of formaldehyde with concentrated NaOH is called

1. syllabus explicitly cites formaldehyde + NaOH as a textbook Cannizzaro example. 2. Mechanism check: formaldehyde ($\mathrm{HCHO}$) has NO $\alpha$-carbon AT ALL — and so no $\alpha$-H. Aldol condensation requires an $\alpha$-H, so it cannot occur. 3. Instead the Cannizzaro disproportionation operates: $2\,\mathrm{HCHO} + \mathrm{NaOH} \to \mathrm{HCOONa} + \mathrm{CH_3OH}$. 4. One molecule of HCHO is REDUCED (to $\mathrm{CH_3OH}$, methanol). Another is OXIDISED (to $\mathrm{HCOONa}$, sodium formate). Two electrons shuttle between the two HCHO molecules. 5. Option A (aldol) is blocked by no $\alpha$-H. Option C (haloform) requires $\mathrm{I_2}$ or $\mathrm{Br_2}$, not present. Option D (Etard) is a different reaction (oxidation of toluene to benzaldehyde with $\mathrm{CrO_2Cl_2}$). _Source: syllabus Chemistry Part 2, Ch 8, §8.4 (Cannizzaro reaction with HCHO), p. 13–14._