See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type by examining the substrate and reagent. The substrate is HDCO (formaldehyde with one H replaced by deuterium, i.e., deuterated formaldehyde: H-C(=O)-D). The reagent is H^18O^- (isotopically labelled hydroxide). The products are D-C(=O)-^18O^- (a formate/carboxylate anion bearing the labelled oxygen) and CH2D-OH (a deuterated methanol, acting as the alcohol product). Step 2 - Recognize the Cannizaro reaction. The Cannizaro reaction is the base-catalyzed disproportionation of an aldehyde lacking alpha-hydrogens (no alpha-H). Here, HDCO is a formaldehyde derivative with no alpha-carbon hydrogens (the carbonyl carbon itself bears H and D, but there is no separate alpha-carbon), so it undergoes the Cannizaro reaction. Under base (OH^-), one molecule is oxidized to the carboxylate and another is reduced to the alcohol. Step 3 - Confirm it is the standard (not cross) Cannizaro. Both reactant molecules are the same aldehyde (HDCO), so this is the standard (self) Cannizaro reaction, not a cross Cannizaro (which involves two different aldehydes). This eliminates option (c). Step 4 - Confirm the second descriptor: Disproportionation. In a disproportionation reaction, the same species is simultaneously oxidized and reduced. Here, one HDCO molecule is oxidized to the carboxylate (D-COO^-) and another is reduced to the alcohol (CH2D-OH). This is exactly disproportionation. The Cannizaro reaction is inherently a disproportionation (and also a redox reaction), but the more precise and specific classical term applied to it is 'disproportionation reaction'. Options (b) and (d) incorrectly name it the Tischenko reaction (which produces an ester, not an acid/alcohol pair). Option (c) incorrectly calls it a Cross Cannizaro. Option (a) correctly identifies both names: Cannizaro reaction and Disproportionation reaction. Step 5 - Isotope labelling confirmation. The ^18O from H^18O^- appears in the carboxylate product (D-C(=O)-^18O^-), consistent with the Cannizaro mechanism where hydroxide attacks the carbonyl carbon, and the labelled oxygen is incorporated into the oxidized product. The alcohol CH2D-OH retains the original unlabelled oxygen (from the original C=O of the reduced molecule). This mechanistic detail confirms the Cannizaro (disproportionation) pathway. Therefore, the correct answer is A.