Question

A compound 'C' , molecular formula C2H4O2 reacts with sodium metal to form a compound 'R' and evolves a gas which burns with a pop sound. Compound 'C' on treatment with an alcohol 'A' in presence of an acid forms a sweet smelling compound 'S' , molecular formula C3H6O2 .On addition of NaOH2C it also gives 'R' and water.'S' on treatment with NaOH solution gives back 'R' and 'A'.Identify C,R,A,S and write down the reactions involved

Answer 1

1)Compound C is Ethanoic acid[Acetic acid]
It reacts with sodium metal to form a compound called R . R is Sodium Ethanoate.
2CH3COOH  +2 Na--------------> 2CH3COONa+H2

                                               conc, H2SO4
2) CH3COOH  +C2H5OH       ------ -------------> CH3COOC2H5 + H20
so compound S is Ester or Ethyl ethanoate and compound A is Ethanol
3)CH3COOC2H5+NaoH--------> CH3COONa    + C2H5OH.
so compound R is again Sodium Ethanoate.
So compound C is Ethanoic acid
A is Ethanol
S is Ester
R is sodium Ethanoate.

Answer 2

Answer:

Hii mates

Explanation:

2Ch3cooh + 2Na ----------> 2CH3COONa + H2

                                            conc.H2so4

2)  CH3COOH + C2H5OH -------------------> CH3COOC2H5 +H2O

3)   CH3COOC2H5 + NaoH ----------->   CH3COONA + C2H5OH

                   

                      !! Hope it will help you !!