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Carbonyl Compounds: Question 1

Syllabus 17.1

Multiple choice AS 1 mark

A technician has two unlabelled colourless liquids, R and S. One is hexanal, CH3CH2CH2CH2CH2CHO\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}, and the other is hexan-3-one, CH3CH2COCH2CH2CH3\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3, both share the molecular formula C6H12O\text{C}_6\text{H}_{12}\text{O}.

Both liquids form an orange precipitate with 2,4-dinitrophenylhydrazine (2,4-DNPH). When warmed separately with Tollens' reagent, liquid R produces a shiny silver deposit (a silver mirror) on the inside of the test tube, while liquid S shows no visible change.

Which statement correctly identifies liquid R and explains this observation?

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

Step 1: What the 2,4-DNPH result tells us

2,4-DNPH reacts with the carbon-oxygen double bond of any carbonyl compound to form an orange precipitate (a 2,4-dinitrophenylhydrazone). Since both hexanal and hexan-3-one contain a carbonyl group, both give this positive result, so the 2,4-DNPH test alone cannot tell us which liquid is which. It only confirms that R and S are indeed carbonyl compounds.

Step 2: What Tollens’ reagent actually tests for

Tollens’ reagent is a solution of silver nitrate in aqueous ammonia, containing the complex ion [Ag(NH3)2]+[\text{Ag}(\text{NH}_3)_2]^+. On warming, this mild oxidising agent will oxidise an aldehyde to a carboxylic acid, because the aldehyde’s carbonyl carbon still carries a hydrogen atom that can be removed: RCHO+[O]→RCOOH\text{RCHO} + [\text{O}] \rightarrow \text{RCOOH} As the aldehyde is oxidised, the Ag+\text{Ag}^+ ions are reduced to metallic silver, which deposits as a shiny silver mirror on the inside of the test tube.

A ketone, however, has no hydrogen atom on its carbonyl carbon (it is bonded to two carbon-containing groups instead), so there is no easy way to oxidise it further without breaking a carbon-carbon bond. A ketone therefore does not reduce Tollens’ reagent, and no silver mirror forms.

Step 3: Applying this to R and S

Liquid R gives a silver mirror, so R must be the compound that can be oxidised further, the aldehyde, hexanal. Liquid S shows no change, consistent with it being the ketone, hexan-3-one, which cannot reduce Tollens’ reagent.

Why the other options are wrong

  • B reverses the identities: hexan-3-one is a ketone and cannot be oxidised by Tollens’ reagent, so it cannot be R.
  • C is wrong because Tollens’ reagent does distinguish aldehydes from ketones. That is precisely why S (the ketone) shows no change while R (the aldehyde) does.
  • D misstates the chemistry: it is the aldehyde, not the ketone, that reduces Tollens’ reagent, and the relevant feature is the hydrogen atom on the aldehyde’s carbonyl carbon, not the ketone’s carbon substituents.

Final answer

A. R is hexanal; it is oxidised to hexanoic acid, reducing Ag+\text{Ag}^+ to silver and producing the mirror.