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An organic compound (A) of molecular formula C2H6O liberates H2 gas with metallic sodium and gives (B). (B) on reaction with methyl bromide produces (C) of molecular formula C3H8O. (C) on reaction with excess III produces (D) and (E). Identify A, B, C, D and E and explain the reactions involved

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1. An organic compound (A) reacts with Na metal and liberates H2 gas means it must be an alcohol. From the molecular formula it is identified as ethanol – CH3 – CH2OH (A).

2. Ethanol on reaction with Na metal to produce sodium ethoxide as (B) with liberation of H2 gas.

3. Sodium ethoxide on reaction methyl bromide undergo Williamson’s synthesis to produce methoxy ethane as (C).

4. Methoxy ethane on reaction with excess HI will give Ethyl iodide and Methyl iodide as (D) and (E).

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