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

Concept: Aldehydes have the \(-CHO\) group and are easily oxidised, so they give positive Tollens' and Fehling's tests. Ketones lack the \(-CHO\) group and fail both. The iodoform test is positive only for compounds having a \(CH_3CO-\) (methyl carbonyl) group.

Step 1: Acetaldehyde (\(CH_3CHO\)) vs Acetone (\(CH_3COCH_3\)).
Acetaldehyde is an aldehyde; acetone is a ketone. Use an oxidising test:
Tollens' test: Warm each with ammoniacal silver nitrate. Acetaldehyde reduces \(Ag^+\) and gives a shiny silver mirror; acetone gives no reaction.
\[ CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O \]
Fehling's test: Acetaldehyde gives a red precipitate of \(Cu_2O\); acetone does not.
(Note: both give iodoform, so iodoform cannot separate this pair.)

Step 2: Acetaldehyde (\(CH_3CHO\)) vs Formaldehyde (\(HCHO\)).
Both are aldehydes, so both answer Tollens' and Fehling's tests. To tell them apart use the iodoform test (\(I_2\) + NaOH).
Acetaldehyde has a \(CH_3CO-\) group, so it gives a yellow precipitate of iodoform \((CHI_3)\):
\[ CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3\downarrow + HCOONa + 3NaI + 3H_2O \]
Formaldehyde has no \(CH_3CO-\) group, so it gives no iodoform.

Conclusion: Tollens'/Fehling's test separates acetaldehyde from acetone; the iodoform test separates acetaldehyde from formaldehyde.