Question

How will you differentiate the following?
(i) Acetaldehyde and acetone
(ii) Acetaldehyde and formaldehyde

Hint:

Aldehydes answer Tollens'/Fehling's, ketones do not; only \(CH_3CO-\) compounds give iodoform, so it separates acetaldehyde from formaldehyde.

Answer 1

Idea: Pick one reagent per pair that reacts with only one member.

Step 1: Distinguishing acetaldehyde and acetone.
Only the aldehyde is a reducing agent. Add Fehling's solution (deep blue) and warm. With \(CH_3CHO\) the blue colour disappears and a brick-red solid \(Cu_2O\) settles out, confirming the aldehyde. With acetone the solution stays blue, showing no \(-CHO\) group. Equivalently, Schiff's reagent turns pink with acetaldehyde but not with acetone.

Step 2: Distinguishing acetaldehyde and formaldehyde.
Both reduce Tollens'/Fehling's, so an oxidising test is useless here. Instead use the iodoform reaction. Only a \(CH_3-C(=O)-\) unit gives iodoform. Acetaldehyde \(CH_3CHO\) carries this unit and forms a pale-yellow crystalline \(CHI_3\) with \(I_2/NaOH\), whereas formaldehyde \(HCHO\) has only one carbon and cannot form iodoform, so nothing precipitates.

Result: Fehling's/Schiff's identifies acetaldehyde against acetone; the iodoform test identifies acetaldehyde against formaldehyde.