Aldehydes, Ketones and Carboxylic Acids: Common Mistakes and Fixes (1)

easy 2 min read

Question

Identify the compound: an organic substance reacts with Tollens’ reagent to give a silver mirror, and also reacts with HCN to give a cyanohydrin, but does NOT react with sodium bicarbonate. The compound has molecular formula C2H4O\text{C}_2\text{H}_4\text{O}. What is it, and explain each test.

Solution — Step by Step

C2H4O\text{C}_2\text{H}_4\text{O}. Possible structures: ethanol (CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}), ethylene oxide, or acetaldehyde (CH3CHO\text{CH}_3\text{CHO}).

Tollens’ reagent ([Ag(NH3)2]+[\text{Ag(NH}_3)_2]^+) oxidises aldehydes to carboxylic acids and itself reduces to silver mirror. Ethanol and ethylene oxide do not have CHO-\text{CHO}, so they fail. Acetaldehyde passes.

So the compound contains an aldehyde group.

Aldehydes and ketones add HCN to form cyanohydrins via nucleophilic addition. Acetaldehyde passes.

NaHCO3\text{NaHCO}_3 reacts with carboxylic acids to release CO2\text{CO}_2. Acetaldehyde does not react — confirming it is NOT a carboxylic acid.

The compound is acetaldehyde (CH3CHO\text{CH}_3\text{CHO}).

Why This Works

Tollens’ is selective for aldehydes (not ketones, because ketones lack the α\alpha-hydrogen on carbonyl required for oxidation in mild conditions). HCN addition works for both aldehydes and ketones. Bicarbonate test detects the acidic COOH-\text{COOH} group.

The combination of these three tests narrows the structure quickly: aldehyde present, carboxylic acid absent — the only fit for C2H4O\text{C}_2\text{H}_4\text{O} is acetaldehyde.

Alternative Method

Degree of unsaturation: C2H4O\text{C}_2\text{H}_4\text{O} has DoU = (2×2+24)/2=1(2 \times 2 + 2 - 4)/2 = 1. One double bond or ring. Among the candidates with CHO-\text{CHO}, only acetaldehyde fits — its C=O\text{C=O} accounts for the one DoU.

Common Mistake

Confusing Tollens’ test with Fehling’s test. Both detect aldehydes, but aromatic aldehydes (like benzaldehyde) pass Tollens’ but FAIL Fehling’s. NEET 2023 had a question that explicitly tested this distinction.

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