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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 - Science

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Question

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 

Solution

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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Chapter 4: Carbon And Its Compounds - Exercise 3 [Page 266]

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Lakhmir Singh Chemistry (Science) [English] Class 10
Chapter 4 Carbon And Its Compounds
Exercise 3 | Q 73 | Page 266

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