See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The Cannizaro reaction involves a disproportionation of a non-enolizable aldehyde under basic conditions. One molecule is oxidized to a carboxylate and another is reduced to an alcohol. The hydride (H^-) transfer is the key step - the hydride from one aldehyde molecule transfers to the carbonyl carbon of another. The starting material is deuterated formaldehyde: H-C(=O)-D, which can be written as HDCO (formaldehyde with one H and one D on the carbonyl carbon). Step 1: The hydroxide ion (HO^-) attacks one molecule of HDCO to form a tetrahedral intermediate: HD(OH)C-O^- (a gem-diol anion). Step 2: From this intermediate, a hydride (H^-) is transferred to the carbonyl carbon of another HDCO molecule. The key question is which hydrogen transfers - the H or the D. In the Cannizaro mechanism, the hydroxide attacks the carbonyl carbon, and then the hydrogen on that carbon (the aldehydic H) migrates as hydride. In HDCO, the carbon bears both H and D. The hydride transfer is the C-H (or C-D) bond breaking. Step 3: When HO^- attacks HDCO, the intermediate is D(H)(OH)CO^-. The hydride that transfers is the one on the carbon - specifically the H (since primary kinetic isotope effect favors H transfer over D transfer, but more importantly, we analyze the products). If H^- transfers from the oxidized molecule's carbon to the carbonyl carbon of the second HDCO: - The oxidized molecule loses H and becomes D-CO2^- (formate with D: DCO2^-) - The reduced molecule (second HDCO) gains H^- at its carbonyl carbon, giving: CH2D-O^- which upon protonation gives CH2DOH... but wait, the nucleophile is HO^-, so the oxygen source matters. Let me reconsider: The second HDCO receives H^- at its carbonyl carbon. The carbonyl carbon of HDCO already has H and D. Adding H^- gives: C with H, D, H, and O^- → CHD(H)O^- = CH2DO^- → protonation gives CH2DOH. The oxidized molecule: HDCO after losing H^- from carbon, with OH already attached, gives D-C(=O)-O^- = D-CO2^- (deuteroformate). So the products are: D-CO2^- + CH2DOH. Why not option (a) D-CO2^- + CH2DOD: DOD would require D from solvent, but the alcohol oxygen comes from the original carbonyl, not from OD^-. Why not option (b): This would require two different oxidation products. Why not option (d) CHD2OH: This would require D^- transfer instead of H^- transfer, which is disfavored by kinetic isotope effect. The hydroxide oxygen becomes part of the carboxylate (oxidized product), and the carbonyl oxygen of the reduced substrate becomes the alcohol oxygen. The alcohol formed is CH2DOH because H^- (not D^-) transfers preferentially due to kinetic isotope effect, adding H to HDCO's carbonyl carbon to give CH2D-OH. Therefore, the correct answer is C.