Question

During reduction of aldehydes with hydrazine and ${C_2}{H_5}ONa,$ the product formed is
(A) $R - CH = N - N{H_2}$
(B) $R - C \equiv N$
(C) \[R - \begin{array}{*{10}{c}}
  {} \\
  {C - N{H_2}} \\
  {\begin{array}{*{10}{c}}
  {||} \\
  O
\end{array}}
\end{array}\]
(D) $R - C{H_3}$

Answer 1

Hint- As we know that the reduction of aldehydes generally gives the product known as alcohols. The aldehydes are generally reduced with hydride agents but the reagents given above in the question are hydrazine and sodium ethoxide and the reaction of this with aldehydes, this reaction is known as Wolff Kishner reduction.


Complete step by step solution:
As we know that the Wolff kishner reduction is the reduction of the carbonyl compounds. This is done by heating the carbonyl compounds with hydrazine hydrate and a base.
In the above process the carbonyl group $ - C = o$ gets converted into the $ - C{H_2}$ group through $ - C = NN{H_2}$ and the nitrogen gas evolved out.
This reaction converts the carbonyl groups to methylene groups.
The reaction is given below
The reduction of aldehydes with hydrazine in the presence of sodium ethoxide, the product formed is alkane
$R - CHO + N{H_2} - N{H_2}\xrightarrow{{{C_2}{H_5} - ONa}}R - C{H_3}$

Hence, the correct option is D.

Additional Information- The only structural difference between hydrocarbons and aldehydes is the presence of carbonyl groups in the latter and it is this group that is responsible for the variations in physical and chemical properties. The only difference arises because the carbonyl group is polar, that is the electrons make C=O bond are attracted closer to the oxygen as compared to the carbon. It gives partial negative charge to the oxygen, and partial positive charge to the carbon.
Note- Wolff Kishner reduction is a reduction reaction for aldehydes and ketones; under the conditions of this reaction other reducible groups such as carbon-carbon double bonds and nitro groups usually remain unchanged. Also, to remove the nitrogen gas which is the byproduct of the reaction, you need to use a high boiling point solvent like ethylene glycol.