Question

In the following reaction ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\,{\text{ + }}\,{\text{N}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}\, \to \,{\text{A}}\,\mathop \to \limits^{\text{B}} \,{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{3}}}\,{\text{ + }}\,{{\text{N}}_{\text{2}}}$ identify A and B.
A. ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH = NN}}{{\text{H}}_{\text{2}}}\,\,{\text{and}}\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$
B. ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}{\text{ - N}}{{\text{H}}_{\text{2}}}\,\,{\text{and}}\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$
C. ${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - NH - NH - C}}{{\text{H}}_3}\,\,{\text{and}}\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$
D. ${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_2}{\text{ - N}}{{\text{H}}_2}\,\,{\text{and}}\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$

Answer 1

The reaction of aldehyde or ketone in presence of hydrazine gives hydrazone. The nitrogen can be eliminated from the hydrazone to give the product by reacting, the hydrazone with a base.

Complete answer:
The reaction of aldehyde and ketone with hydrazine is known as Wolff kishner reduction. The reaction is used to reduce the aldehyde or ketone.
In Wolff kishner reduction reaction, the aldehyde or ketone reacts with hydrazine to produce a hydrazone as follows:
The reaction of aldehyde or ketone with hydrazine is a condensation reaction because in this reaction a water molecule removes and carbonyl group leaves its double-bonded oxygen and hydrazine leaves its double-bonded nitrogen and forms a condensed product.
${\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}\,{\text{ + }}\,{\text{N}}{{\text{H}}_{\text{2}}}{\text{N}}{{\text{H}}_{\text{2}}}\, \to \,{\text{C}}{{\text{H}}_{\text{3}}}{\text{CH = NN}}{{\text{H}}_{\text{2}}}\,$
So, A in the reaction is ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH = NN}}{{\text{H}}_{\text{2}}}$.
The formed hydrazine reacts with a base to give ethane. Sodium ethoxide is a good base, so the B is ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$. During reaction hydrazone oxidised to form nitrogen gas. The treatment with base causes the carbon-carbon coupling.
The reaction of hydrazine with base sodium ethoxide is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH = NN}}{{\text{H}}_{\text{2}}} + \,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}\, \to \,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{3}}}\,{\text{ + }}\,{{\text{N}}_{\text{2}}}$

Therefore, option (A) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH = NN}}{{\text{H}}_{\text{2}}}\,\,{\text{and}}\,\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$ is correct.

Note: The Wolff kishner reduction is used for carbon-carbon coupling. It is a condensation reaction. Clemmensen reduction is the same as Wolff kishner reduction only the difference is that in Wolff kishner reduction a base is used whereas in Clemmensen reduction the conversion takes place in acidic conditions.