Question
How would you distinguish between propanal () and propanone () using simple chemical tests? Give at least two tests with expected observations.
Solution — Step by Step
Add ammoniacal silver nitrate () to each compound and warm gently.
Propanal (aldehyde): forms a silver mirror on the test tube wall. The aldehyde is oxidised to the carboxylic acid.
Propanone (ketone): no reaction. Ketones do not reduce Tollens’ reagent.
Add Fehling’s reagent (blue copper(II) tartrate complex) and warm.
Propanal: forms a brick-red precipitate of .
Propanone: no change. The blue colour persists.
Add and .
Propanal: no iodoform formed (it has no or group adjacent).
Propanone: yellow precipitate (iodoform). Has group.
So Tollens’ and Fehling’s identify the aldehyde; the iodoform test confirms the methyl ketone.
Final answer: Tollens’ (silver mirror only with propanal), Fehling’s (red ppt only with propanal), iodoform (yellow ppt only with propanone).
Why This Works
Aldehydes have a C-H bond on the carbonyl carbon, making them easily oxidised. Tollens’ and Fehling’s are mild oxidising agents — they oxidise aldehydes but not ketones, which lack that C-H.
The iodoform test is selective for the group (or that oxidises to it). It works for acetaldehyde but not propanal, which has (ethyl group, not methyl).
Alternative Method
Use 2,4-DNP (Brady’s reagent) — both give orange precipitates (any carbonyl), so it doesn’t distinguish them. To then distinguish, follow up with Tollens’ as above. Always pair tests for unambiguous identification.
For NEET, the iodoform test is high-yield. Memorise the substrates that give a positive result: acetaldehyde, all methyl ketones, ethanol, and isopropanol (and any compound with or ).
Common Mistake
Using Tollens’ to test propanone and expecting any reaction. Tollens’ is selective for aldehydes — ketones never give a silver mirror. Using bromine water also fails to distinguish (both are colourless).