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प्रश्न
An organic compound A (molecular formula C2H4O2) 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 C3H6O2). Compound D on treatment with NaOH solution gives back B and C. Identify A, B, C and
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उत्तर
The organic compound, A, with the molecular formula C2H4O2 is ethanoic acid (CH3COOH), 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 (CH3COONa) is formed.
The alcohol, C, is methanol (CH3OH) 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 (CH3COOCH3).
The chemical equation for the above stated reaction is as follows:
CH3OH + CH3COOH → CH3COOCH3 + H2O
When ethanoic acid reacts with alcohol in the presence of concentrated sulphuric acid, a sweet-smelling cmpound, 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:
CH3COOCH3 + NaOH → CH3COONa + CH3OH
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