Carbonyl Compounds: Question 7

Worked solution

Step 1: What 2,4-DNPH tests for

2,4-Dinitrophenylhydrazine reacts specifically with the carbonyl group, $\text{C=O}$, of an aldehyde or a ketone, forming an orange precipitate of a 2,4-dinitrophenylhydrazone. This works because the carbonyl carbon in an aldehyde or ketone is bonded only to hydrogen and/or carbon atoms, leaving it strongly electron-poor (δ+) and reactive towards the nucleophilic $\text{-NH}_2$ group of 2,4-DNPH.

Step 2: Why a carboxylic acid behaves differently

A carboxylic acid also contains a $\text{C=O}$ group, but its carbonyl carbon is additionally bonded to an $\text{-OH}$ oxygen atom. A lone pair on this oxygen atom is delocalised (donated) into the carbonyl group, which reduces the positive charge on the carbonyl carbon compared with an aldehyde or ketone. As a result, the carbonyl carbon of a carboxylic acid (or of related groups such as esters and amides) is not electrophilic enough to react with 2,4-DNPH under the conditions of this test, so no orange precipitate forms.

Step 3: Checking each option

• A, butanal: an aldehyde, reacts with 2,4-DNPH to give an orange precipitate.
• B, butan-2-one: a ketone, reacts with 2,4-DNPH to give an orange precipitate.
• C, butanoic acid: a carboxylic acid. Its carbonyl carbon is deactivated by the attached $\text{-OH}$ oxygen, so it does not react with 2,4-DNPH.
• D, 2-methylpropanal: still an aldehyde, despite the branched chain, reacts with 2,4-DNPH to give an orange precipitate.

Why the other options are wrong

• A, B and D are all aldehydes or ketones, and every aldehyde or ketone gives a positive 2,4-DNPH result, so none of them can be the compound that fails the test.

Final answer

C. Butanoic acid does not give an orange precipitate with 2,4-DNPH, because it is a carboxylic acid, not an aldehyde or a ketone.