An organic compound (A) of molecular formula C_2H_6O liberates H_2 gas with metallic sodium and gives (B). (B) on reactionwith methyl bromide promide produces (C ) of molecular formula C_3H_8O. (C ) on reaction with excess HI produces (D) and (E ). Identify A,B,C,D and E and explain the reactions involved.s

An organic compound (A) of molecular formula C_2H_6O liberates H_2 gas

1.	An organic compound (A) of molecular formula C_2H_6O liberates H_2 gas with metallic sodium and gives (B). (B) on reactionwith methyl bromide promide produces (C ) of molecular formula C_3H_8O. (C ) on reaction with excess HI produces (D) and (E ). Identify A,B,C,D and E and explain the reactions involved.s
Answer» SOLUTION :(i) An ORGANIC compound (A) reacts with Na metal and liberates `H_2` gas means it must be an alcohol. From the molecular formula it is identified as ethanol `-CH_3-CH_2OH` (A). (II) Ethanol on reaction with Na metal to produce sodium ethoxide as (B) with liberation of `H_2` gas. `2CH_3 - CH_2OH + 2Na to underset("Sodium ethoxide(B)")(2CH_3 - CH_2ONa) + H_2uarr` (iii) Sodium ethoxide on reaction methyl bromide undergo Williamson.s synthesis to produce methoxy ethane as (C). `CH_3 - CH_2ONa + CH_3Ioverset("Ether")to underset("Methoxy ethane(C )")(CH_3 - CH_2 - O - CH_3) + NAI` (iv) Methoxy ethane on reaction with excess HI will give Ethyl iodide and Methyl iodide as (D) and (E). `CH_3 - CH_2 - O - CH_3 + 2HI to underset("Ethyl iodide(D)")(CH_3-CH_2I) + underset("Methyl iodide(E)")(CH_3I) + H_2O`