1. An organic compound (A) reacts with Na metal and liberates H2 gas means it must be alcohol. From the molecular formula it is identified as Ethanol (CH3 – CH2OH).
2. Ethanol on oxidation with Cu at 573 K undergoes catalytic dehydrogenation and produces Acetaldehyde as product (B).
3. Acetaldehyde on reaction with CH3 MgBr followed by hydrolysis will give Isopropyl alcohol as (C).
4. Propan – 2 – ol is secondary alcohol and so it gives blue colour in Victor Meyer’s test. (C) on reaction with Cu at 573 K will give Propanone as (D).