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.

AcademicChemistryNCERTClass 10

The organic compound, C is ethanoic acid (CH3COOH), as it evolves hydrogen with a pop sound on reacting with sodium.

When (A) ethanoic acid reacts with sodium,  sodium ethanoate (CH3COONa) is formed. Therefore, R is sodium ethanoate.
$2CH_3COOH  +2 Na → 2CH_3COONa + H_2$
(in the presence of Conc H2SO4)

The sweet-smelling compound S (ester) is ethyl ethanoate (CH3COOCH3). Compound A is Ethanol.
$ CH_3COOH  + C_2H_5OH → CH_3COOC_2H_5 + H_2O$
raja
Updated on 10-Oct-2022 13:29:14

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