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Reaction of Ethanoic Acid with Metals and Bases

An organic co...

Question

An organic compound A (molecular formula $C_{2} H_{4} O_{2})$ 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 the presence of a little of concentrated sulphuric acid forms a sweet-smelling compound D (molecular formula $C_{2} H_{6} O_{2})$ . Compound D on treatment with NaOH solution gives back B and C. Indentity A, B, C and D.

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Solution

The organic compound, A, with the molecular formula $C_{2} H_{4} O_{2}$ is ethanoic acid $(C H_{3} - C O O H)$ , as it evolves hydrogen with a pop sound on reacting with sodium.

When ethanoic acid reacts with sodium, a compound B called the sodium ethanoate $(C H_{3} - C O O N a)$ is formed.

The alcohol, C, is methanol $(C H_{3} O H)$ which is treated with ethanoic acid in the presence of concentrated sulphuric acid to form a compound D.

The sweet-smelling compound D is methyl ethanoate $(C H_{3} - C O O C H_{3})$ .

The chemical equation for the above-stated reaction is as follows:

$C H_{3} - O H + C H_{3} - C O O H ⟶ C H_{3} - C O O - C H_{3} + H_{2} O$

When ethanoic acid reacts with alcohol in the presence of concentrated sulphuric acid, a sweet-smelling compound, an ester is formed. Methyl ethanoate is a carboxylate ester, which on reaction with sodium hydroxide gives back methanol and Sodium ethanoate.

The chemical equation is as follows:

$C H_{3} - C O O - C H_{3} + N a O H ⟶ C H_{3} - C O O N a + C H_{3} - O H$

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