A compound C (molecular formula, C2H4O2) reacts with Na – 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 NaOH to C, 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.
C is ethanoic acid. It reacts with sodium to form sodium ethanoate. Hence, compound R is sodium ethanoate or sodium acetate. We know that hydrogen gas burns with a pop sound. This reaction can be given by following equation:
2CH3COOH + 2Na → 2CH3COONa + H2
When ethanoic acid reacts with methanol in the presence of an acid, we get (methyl ethanoate) ester which is a sweet smelling substance. Hence, compound S is methyl ethanoate and A is methanol. This reaction can be given as follows:
CH3COOH + CH3OH --Acid → CH3COOH-CH3
When sodium hydroxide is added to ethanoic acid, it gives sodium ethanoate and water.
CH3COOH + NaOH → CH3COONa + H2O
When methyl ethanoate is treated with NaOH solution, it gives back methanol and sodium ethanoate.
CH3COO-CH3 + NaOH → CH3OH + CH3COONa
