Question

Two compounds 'A' and 'B' have the same molecular formula C3H6O2. One of them reacts with sodium metal and NaHCO3 to liberate H2 gas and gas 'C' respectively. 'C' turns lime water milky. The other two compound does not react with either Na metal or NaHCO3 but under goes saponification to give salt of carboxylix acid and wood spirit , 'D' . Identify A,B,C and D.

Answer 1

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Answer 2

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● Answer -
A => propanoic acid
B => methyl acetate
C => carbon dioxide
D => methanol

● Explaination -
Given compounds have molecular formula C3H6O2 which can be either propanoic acid or methyl acetate.

Propanoic acid (Compound A) reacts with with NaHCO3 to give sodium propanoate and release CO2 gas (Compound C).
CH3CH2COOH + NaHCO3 ---> CH3COONa + H2O + CO2

Methyl acetate (Compound B) on saponification gives sodium acetate and methanol (Compound D).
CH3COOCH3 + NaOH ---> CH3COONa + CH3OH

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