Question

An organic compound 'A' having molecular formula C₂H₆O evolves hydrogen gas on treatment with sodium metal and on treatment with red phosphorous and iodine gives compound 'B'. The compound 'B' on treatment with alcoholic KCN and on subsequent reduction gives compound 'C'. The compound 'C' on treatment with nitrous acid evolves nitrogen gas. Write the balanced chemical equations for all the reactions involved and identify the compounds 'A' , 'B' and 'C'.

Answer 1

a is ethanol (C2H5OH), because ethanol reacts with sodium metal to form sodium ethoxide and hydrogen gas.
e.g., C2H5OH + Na-----> C2H5ONa + H2

Now when ethanol is treated with red phosphorus and iodine , firstly phosphorus and iodine combine to form phosphorus iodide (PI3) Ans then, iodide reacts with ethanol to form ethyl iodide.
e.g., 12C2H5OH + P4 + 6I2 ----> 12C2H5I + 4H3PO3

Now, C2H5I is treated with KCN and formed C2H5CN
e.g., C2H5I + KCN ----> KI + C2H5CN

Hence, A => C2H5OH
B => C2H5I
C => C2H5CN

Answer 2

Answer:

2 CH₃COOH + 2 Na ⇒  2 CH₃COONa + H₂O

      ( A )                                   ( B )

CH₃COOH + C₂H₅OH ⇒  CH₃COOC₂H₅ + H₂O

                        ( C )                 ( D )

CH₃COOC₂H₅ + NaOH  ⇒ C₂H₅OH + CH₃COONa

Therefore we identified

A - CH₃COOH

B - CH₃COONa

C - C₂H₅OH

D - CH₃COOC₂H₅