An organic compound (A) of molecular formula C_(2)H_(6)O reacts with sodium metal and liberates H_(2) gas. (A) on reaction with alkaline KMnO_(4) gives (B) of formula C_(2)H_(6)O_(2). Sodium salt of (B) on reaction with soda lime gives (C ) a first member of alkane homologous series. Identify A, B, and C.

An organic compound (A) of molecular formula C_(2)H_(6)O reacts with s

1.	An organic compound (A) of molecular formula C_(2)H_(6)O reacts with sodium metal and liberates H_(2) gas. (A) on reaction with alkaline KMnO_(4) gives (B) of formula C_(2)H_(6)O_(2). Sodium salt of (B) on reaction with soda lime gives (C ) a first member of alkane homologous series. Identify A, B, and C.
Answer» Solution :(i) (A) is Ethanol : `CH_(3)-CH_(2)OH`. Ethanol reacts with sodium metal and liberates `H_(2)` gas. `underset("Ethanol (A)")(CH_(3)-CH_(2)OH)+Nararrunderset("Sodium ethoxide")(CH_(3)-CH_(2)ONa)+H_(2)uarr` (ii) Ethanol on reaction with alkaline `KMnO_(4)`, oxidation takes place and the product formed is Ethanoic ACID, `CH_(3)COOH`. It is (B). `CH_(3)-CH_(2)OHunderset(KMnO_(4)//OH^(-))overset(2[O])rarrunderset("Ethanoic acid (B)")(CH_(3)COOH)+H_(2)O` (iii) Sodium salt of ethanoic acid on reaction with sodalime (NaOH + CaO) undergo decarboxylation reaction to give methane `CH_(4)` (C ) the first member of alkane homologous series. `CH_(3)COOHunderset(Delta)overset(NaOH+CaO)rarrunderset("Methane (C)")(CH_(4)uarr+)Na_(2)CO_(3)` `{:("A",CH_(3)-CH_(2)OH,"Ethanol"),("B",CH_(3)-COOH,"Etbanoic acid"),("C",CH_(4),"Methane"):}`