Question

two compounds x and y have same molecular formula C2H4O2. one of them reacts with Na metal to liberate H2 and Co2 with NaHCO3. Second one does not react with Na metal and NaHCO3 but undergoes hydrolyis with NaOH to form salt of carboxylic acid and compound Z which is called wood spirit. Identify X,Y and Z .Write chemical equation

Answer 1

Let X be acetic acid.[CH3COOH]
Acetic acid reacts with sodium to liberate H2
the reaction is as follows:
CH3COOH +Na -----> CH3COONa +H2
It reacts with NaHCO3 to liberate CO2
The reaction is as follows:
CH3COOH +NaHCO3 ----> CH3COONa+ CO2
Element Y undergoes Hydrolysis with NaoH to form Carboxylic acid salt . It means Y is an Ester.[Methyl Methanoate.]
The reaction is as follows:
HCOOCH3+ NaOH---->HCOONa +CH3OH
Therefore, Compound Z is Methanol

Answer 2

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