Question

# A compound 'C' (molecular formula, $$C_2H_4O_2$$) 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, $$C_3H_6O_2$$). 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.

Solution

## Compound '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 the following equation:$$2CH_3COOH + 2Na \rightarrow 2CH_3COONa + H_2$$       (C)                                                      (R)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: $$CH_3COOH + CH_3OH \xrightarrow{Acid} CH_3COO - CH_3+H_2O$$          (C)                   (A)                                                (S)When sodium hydroxide is added to ethanoic acid, it gives sodium ethanoate and water. $$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$$          (C)                                                     (R)When methyl ethanoate is treated with NaOH solution, it gives back methanol and sodium ethanoate. $$CH_3COO-CH_3 + NaOH \rightarrow CH_3OH + CH_3COONa$$          (S)                                                      (A)                (R)ChemistryNCERTStandard XII

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More