A compound A (C2H4O2) reacts with Na metal to form a compound ‘B’ and evolves a gas which burns with a pop sound. Compound ‘A’ on treatment with an alcohol ‘C’ in presence of an acid forms a Sweet smelling compound ‘D’ (C4H8O2). On addition of NaOH to ‘D’ gives back B and C. Identify A, B, C and D write the reactions involved. Explain properly with chemical equation.
Answers
HELLO THERE!
Let us divide this question into some fragments.
In the first fragment, the compound A reacts with sodium (Na) to form B and evolves a gas which burns with a pop sound.
We know, that the gas which burns in air with a pop sound is hydrogen, and when active metals like sodium reacts with acids, hydrogen gas is liberated. Hence, the compound A must be any acid. On analyzing the formula, we see that it contains two carbon atoms, so the compound A is Ethanoic Acid/Acetic acid: CH₃COOH.
Reaction: 2Na + 2CH₃COOH ----> 2CH₃COONa + H₂
In the second fragment, Compound A is treated with an alcohol C in presence of an acid and a sweet smelling compound D is formed, whose molecular formula is given.
Now, we know that A is Acetic acid. And, a carboxylic acid, when reacts with an alcohol, in presence of concentrated sulphuric acid, we get Ester, which is a sweet smelling compound and has general formula R-COO-R, where R represents alkyl group.
On analyzing the formula of the ester, we find it to be CH₃COOC₂H₅, i.e., ethyl acetate or ethyl ethanoate.
COMPOUND D IS ETHYL ACETATE.
And, in place of acid since we have Ethanoic acid, only ethanol can give ethyl acetate. COMPOUND C IS ETHANOL.
Reaction: C₂H₅OH + CH₃COOH -----> CH₃COOC₂H₅ + H₂O.
When NaOH is added to ester, we get back B and C.
Reaction: NaOH + CH₃COOC₂H₅ -----> CH₃COONa + C₂H₅OH
THANKS!
Compound A is CH3COOH - ETHANOIC ACID
It reacts with sodium metal to give Compound B and H2 gas
a ) 2CH3COOH + 2Na -------> 2CH3COONa +H2
COMPOUND A reacts with COMPOUND C [ an alcohol ] and forms a sweet smelling compound D
This reaction is known as Esterification reaction which occurs between alcohol group and carboxylic group to form esters a sweet smelling compound.
b ) C2H5OH + CH3COOH ----> CH3COOC2CH5 +H2O
On reaction of Compound D with NaoH :
CH3COOC2CH5 + NaOH --->CH3COONa + C2H5OH
Hence A --- Ethanoic Acid
B= sodium Acetate
C= Ethanol
D= Esters.