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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)

Chemistry
NCERT
Standard XII

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