Question

Compound A undergoes Rosenmund reduction to give compound B with molecular formula $C_7$H₆O. Compound B does not give Fehling’s test but reacts with conc. NaOH to give C and D. 
Identify A, B, C and D and write all the reactions involved. 
Write one chemical test to distinguish between compound B and propanone.

Hint:

Rosenmund reduction converts acyl chlorides to aldehydes.
Cannizzaro reaction occurs only in aldehydes without $\alpha$-H.
Iodoform test detects methyl ketones and secondary alcohols with –CH\textsubscript{3}CO– group.

Answer 1

Step 1: Compound A undergoes Rosenmund reduction to give compound B.
Rosenmund reduction:
C₆H₅COCl + H₂ (Pd/BaSO₄) → C₆H₅CHO (Benzaldehyde)
Step 2: Compound B (C₆H₅CHO):
- Does not give Fehling’s test (indicates it’s an aromatic aldehyde).
- Reacts with conc. NaOH → undergoes Cannizzaro reaction.
Step 3: Cannizzaro Reaction:
2C₆H₅CHO + NaOH → C₆H₅COONa + C₆H₅CH₂OH
(on acidification): C (Benzoic acid), D (Benzyl alcohol) Final Identifications:
A = Benzoyl chloride (C₆H₅COCl)
B = Benzaldehyde (C₆H₅CHO)
C = Benzoic acid (C₆H₅COOH)
D = Benzyl alcohol (C₆H₅CH₂OH) Distinguishing Test:
Use Iodoform test:
- Propanone gives a positive iodoform test (yellow precipitate of CHI₃)
- Benzaldehyde does not give the iodoform test.