two compounds a and b have the same molecular formula C3H6O2. One of them reacts with reacts with sodium metal and NaHCO3 to liberate H2 gas and a gas 'C'.'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 carboxylic acid and wood spirit 'D'. Idetify A B C and D
Answers
Answered by
25
A is propanoic acid
C is CO2
B is ethyl formate and can be written as HCOOC2H5
D is methanol
C is CO2
B is ethyl formate and can be written as HCOOC2H5
D is methanol
hardshah1810:
can u state the reaction
for propanoic acid and NaHCO3 it is neutralisation reaction
no not like that
i mean to say equation
???
CH3CH2COOH + NaHCO3 gives CH3CH2COONa (Sodium propanoate) + CO2 + H2O
thank you very much
wlcm
i was blank on this question but your answer helped me
oh that's good
Answered by
9
Welcome dear,
● 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
Hope this helps you...
● 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
Hope this helps you...
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