An organic compound [ A ] ( C ₅ H ₁₂ O ) , on reaction with conc. H ₂ SO ₄ at 443 ~K gives [ B ] as one of the products. Compound [B] undergoes further reaction with O ₃ and Zn / H ₂ O to give Propanone and Ethanal as products. Identify compound [A]

Step-by-step solution

The molecular formula of compound [A] is C₅H₁₂O , which corresponds to a saturated alcohol or ether. Since it reacts with conc. H₂SO₄ at 443 K to form an alkene [B], [A] is an alcohol. The reaction of [B] with O₃ followed by Zn/H₂O (reductive ozonolysis) yields propanone ( CH₃COCH₃ ) and ethanal ( CH₃CHO ). The structure of the alkene [B] can be reconstructed by joining the carbonyl carbons of the products with a double bond. Propanone is CH₃-C(=O)-CH₃ and ethanal is CH₃-CH=O . Removing the oxygen atoms and joining