2020 H2 Chem P1 Q23 — Carbonyl Compound Solution

The answer is A (T has an aldehyde functional group).

• \(\text{NaBH}_4\) adds 2 H atoms (\(\text{C}_x\text{H}_y\text{O}_z \rightarrow \text{C}_x\text{H}_{y+2}\text{O}_z\)), so T contains exactly one C=O group of an aldehyde or ketone.
• T gains one O atom in a single reaction (\(\text{C}_x\text{H}_y\text{O}_z \rightarrow \text{C}_x\text{H}_y\text{O}_{z+1}\)). This is the oxidation of an aldehyde to a carboxylic acid: \(\text{RCHO} + [\text{O}] \rightarrow \text{RCOOH}\).
• The orange precipitate with 2,4-dinitrophenylhydrazine (a condensation reaction) confirms a C=O group of an aldehyde or ketone.
• No reaction with sodium means T has no -OH group (no alcohol, phenol or carboxylic acid).
So T has an aldehyde functional group.

• B is wrong: a ketone would fit the \(\text{NaBH}_4\) and 2,4-DNPH results, but ketones are not oxidised by adding one O atom in a single reaction, so a ketone cannot form \(\text{C}_x\text{H}_y\text{O}_{z+1}\).
• C is wrong: a primary alcohol would react with sodium to give hydrogen, would not give an orange precipitate with 2,4-DNPH, and is not reduced by \(\text{NaBH}_4\).
• D is wrong: with both an aldehyde and a ketone group, \(\text{NaBH}_4\) would reduce both C=O groups and add 4 H atoms, giving \(\text{C}_x\text{H}_{y+4}\text{O}_z\), not \(\text{C}_x\text{H}_{y+2}\text{O}_z\).