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Carbonyl Compounds: Question 10

Syllabus 17.1

Multiple choice AS 1 mark

An unknown colourless liquid, T, has the molecular formula C4H8O\text{C}_4\text{H}_8\text{O}. When tested, T gives an orange precipitate with 2,4-dinitrophenylhydrazine (2,4-DNPH), shows no observable change when warmed with Tollens' reagent, and gives a pale yellow precipitate when warmed with alkaline aqueous iodine.

Which compound is T?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Using the molecular formula and the 2,4-DNPH result

T has molecular formula C4H8O\text{C}_4\text{H}_8\text{O}. Option C, butan-1-ol, has formula C4H10O\text{C}_4\text{H}_{10}\text{O} (an alcohol, CnH2n+2O\text{C}_n\text{H}_{2n+2}\text{O}), which does not match, and an alcohol would not react with 2,4-DNPH in any case. C is eliminated.

The remaining three options (butanal, butan-2-one and 2-methylpropanal) are all C4H8O\text{C}_4\text{H}_8\text{O} carbonyl compounds (an aldehyde or a ketone), so each would give an orange precipitate with 2,4-DNPH. This test alone does not distinguish between them.

Step 2: Using the Tollens’ result

T shows no change with Tollens’ reagent, so T cannot be oxidised further. It must be a ketone, not an aldehyde.

  • A, butanal, and D, 2-methylpropanal, are both aldehydes and would each produce a silver mirror with Tollens’ reagent. Both are eliminated.
  • B, butan-2-one, is a ketone: its carbonyl carbon has no hydrogen atom to remove, so it cannot reduce Tollens’ reagent, consistent with the “no change” observation.

Step 3: Confirming with the tri-iodomethane test

Butan-2-one, CH3COCH2CH3\text{CH}_3\text{COCH}_2\text{CH}_3, contains a CH3CO−\text{CH}_3\text{CO}- group (a methyl group bonded directly to the carbonyl carbon), so it gives a positive tri-iodomethane test, a pale yellow precipitate of CHI3\text{CHI}_3, matching the third observation exactly.

Why the other options are wrong

  • A and D are aldehydes and would react with Tollens’ reagent, contradicting the “no observable change” result.
  • C has the wrong molecular formula and would not react with 2,4-DNPH at all.

Final answer

B. T is butan-2-one: a ketone (explaining the DNPH-positive, Tollens-negative results) that contains a CH3CO−\text{CH}_3\text{CO}- group (explaining the positive tri-iodomethane result).