Question

Mendius reaction involves the:
A. Reduction of aldehydes to give alcohols
B. Reduction of nitriles with sodium and ethanol
C. Oxidation of nitriles
D. Hydrolysis of cyanides

Answer 1

Hint: A reaction in which an organic nitrile is reduced by nascent hydrogen to a primary amine. And here means of reduction is breaking bonds and adding hydrogen molecules.

Complete step-by-step answer:
First of all we know that it is a reduction reaction (in this reduction means breaking of bonds). And in this reaction reduction of nitriles takes place with the help of nascent hydrogen in the presence of sodium and alcohol. In this reaction mainly the reduction of cyano group (\[-C\equiv N\]) takes place by nascent hydrogen. Here nascent hydrogen breaks \[\pi \] bond of the cyano group (\[-C\equiv N\] ). In cyano group there are three bonds between C and N one sigma bond and two \[\pi \] bonds and breaking of \[\pi \] bonds take place.
In Mendius reduction reaction the following reaction takes place:
\[R-C\equiv N+2{{H}_{2}}\xrightarrow[sodium,ethanol]{nascent({{H}_{2}})}RC{{H}_{2}}N{{H}_{2}}\]
“A” is not an answer because it is not reduction of aldehydes.
“B” is the correct answer because in Mendius reaction reduction of nitriles takes place in the presence of sodium and ethanol.
“C” is not an answer because it is a reduction reaction.
“D” is not correct because hydrolysis does not take place.

Note: You have to remember the definition of “Mendius reaction”. You should know the formulas and bonding in the cyano group, in amines and reaction balancing with integer stoichiometric coefficients.