A student is given two unlabelled colourless liquids, each with the molecular formula C7H14O: heptanal, CH3CH2CH2CH2CH2CH2CHO, and heptan-4-one, CH3CH2CH2COCH2CH2CH3. (a) A few drops of each liquid are treated separately with 2,4-dinitrophenylhydrazine (2,4-DNPH). State the obse…

Answer: ((a)) Both liquids give an orange precipitate with 2,4-DNPH; this only confirms the presence of a carbonyl group, which both aldehydes and ketones contain, so it cannot distinguish between them. ((b)) Heptanal turns Fehling's reagent from blue to a brick-red precipitate (of Cu2O) on warming, as it is oxidised to heptanoic acid, reducing Cu^2+ to Cu^+; heptan-4-one shows no change (stays blue), since a ketone cannot be oxidised in this way. ((c)) Heptanal produces a silver mirror with Tollens' reagent on warming, being oxidised to heptanoic acid as Ag^+ is reduced to silver; heptan-4-one shows no change. The liquid giving a brick-red precipitate with Fehling's and a silver mirror with Tollens' is heptanal.. Part (a): The 2,4-DNPH test Both heptanal and heptan-4-one contain a carbonyl group, C=O, so both react with 2,4-DNPH to form an orange precipitate (a 2,4-dinitrophenylhydrazone derivative). This confirms that each liquid is an aldehyde or…

AS Chemistry 9701: Using 2,4-DNPH, Fehling's and Tollens' Tests to Identify an Aldehyde

A student is given two unlabelled colourless liquids, each with the molecular formula $\text{C}_7\text{H}_{14}\text{O}$ : heptanal, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$ , and heptan-4-one, $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3$ .

(a) A few drops of each liquid are treated separately with 2,4-dinitrophenylhydrazine (2,4-DNPH). State the observation made with each liquid, and explain why this test alone cannot identify which liquid is which. [2]

(b) Each liquid is then warmed separately with Fehling's reagent. State what is observed with each liquid, and explain each result in terms of oxidation. [3]

(c) Each liquid is instead warmed separately with Tollens' reagent. State what is observed with each liquid, and use your answers to (b) and (c) to identify which liquid is heptanal. [3]

Part (a): The 2,4-DNPH test

Both heptanal and heptan-4-one contain a carbonyl group, $\text{C=O}$ , so both react with 2,4-DNPH to form an orange precipitate (a 2,4-dinitrophenylhydrazone derivative). This confirms that each liquid is an aldehyde or a ketone, but since both classes of compound give this same positive result, the test cannot distinguish which liquid is the aldehyde and which is the ketone.

Part (b): The Fehling’s test

Fehling’s reagent contains $\text{Cu}^{2+}$ ions complexed with tartrate ions in alkaline solution (blue in colour). It is a mild oxidising agent that can oxidise an aldehyde, but not a ketone.

Heptanal (an aldehyde) is oxidised to heptanoic acid on warming: $\text{CH}_3(\text{CH}_2)_5\text{CHO} + [\text{O}] \rightarrow \text{CH}_3(\text{CH}_2)_5\text{COOH}$ As this happens, the blue $\text{Cu}^{2+}$ ions are reduced to $\text{Cu}^+$ , which precipitates as brick-red copper(I) oxide, $\text{Cu}_2\text{O}$ .
Heptan-4-one (a ketone) has no hydrogen atom on its carbonyl carbon and cannot be oxidised without breaking a carbon-carbon bond, so it cannot reduce the $\text{Cu}^{2+}$ ions: the mixture simply remains blue, with no precipitate.

Part (c): The Tollens’ test, and identifying heptanal

Tollens’ reagent is ammoniacal silver nitrate, containing the complex ion $[\text{Ag}(\text{NH}_3)_2]^+$ . Like Fehling’s reagent, it oxidises an aldehyde but not a ketone.

Heptanal is oxidised to heptanoic acid, and the $\text{Ag}^+$ ions are reduced to metallic silver, depositing as a shiny silver mirror on the inside of the test tube.
Heptan-4-one again shows no visible change, since it cannot be oxidised.

Combining parts (b) and (c): the liquid that gives a brick-red precipitate with Fehling’s reagent and a silver mirror with Tollens’ reagent is the one that can be oxidised further. This must be heptanal, the aldehyde. The liquid that shows no change with either reagent is heptan-4-one, the ketone.

Final answers

(a) Both give an orange precipitate; this only shows a carbonyl group is present in both, so it cannot distinguish an aldehyde from a ketone.
(b) Heptanal gives a brick-red $\text{Cu}_2\text{O}$ precipitate (oxidised to heptanoic acid); heptan-4-one shows no change (cannot be oxidised).
(c) Heptanal gives a silver mirror (oxidised to heptanoic acid); heptan-4-one shows no change. Heptanal is therefore identified as the aldehyde.