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 a gas ’C', respectively. ’C' turns lime water milky. The other compound does not react with either Na metal or NaHCO3 but undergoes saponification to give sodium salt of a carboxylic acid and wood spirit, 'D’. Identify A, B, C and D.
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Explanation:
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Answer:
- A is propanoic acid
- B is methyl acetate
- C is carbon dioxide
- D is methanol
Explanation:
The molecular formula is C3H6O2 .
A and B has a molecular formula C3H6O2
Two possible compounds with molecular formula C3H6O2 are propanoic acid (CH3CH2COOH) and methyl acetate (CH3COOCH3).
one of the compound reacts with NaHCO3 and forms co2
CH3CH2COOH + NaHCO3 →→→ CH3CH2COONa + H2O + CO2
Carbon dioxide gas turns lime water milky . The reaction will be :
Ca(OH)2 + CO2 →→CaCO3 + H2O
so compound Is CH3CH2COOH
Methyl acetate on saponification gives sodium acetate and wood spirit i.e., methanol.
CH3COOCH3 + NaOH →→→CH3COONa + CH3OH
so,
- A is propanoic acid
- B is methyl acetate
- C is carbon dioxide
- D is methanol
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