Question

Which of the following is obtained by reducing methyl cyanide with $Na$ + $C_2H_5OH$ ?
$CH_3 - CN + 4[H] \to CH_3 - CH_2 - NH_2$
This reaction occurs in the presence of Na/ethanol.
A.Methyl alcohol
B.Acetic Acid
C.Ethyl Amine
D.Methane

Answer 1

Hint: The sodium $(Na^+ )$ in ethyl alcohol $(C_2H_5OH)$ gives out a nascent Hydrogen $(H^+ )$ free and this characteristic of the Sodium $(Na^+ )$ in Ethyl alcohol$(C_2H_5OH)$ makes it a strong reducing agent.

Complete step by step solution:
The presence of Nitrile $( - CN)$ indicates formation of an amine $( - NH_2)$ due to presence of Nitrogen atom. This condition brings about reduction reaction as due to presence of free nascent hydrogen atom the reaction accepts the hydrogen and thus gets reduced. In the above question, the reactants are Methyl cyanide $(CH_3 - CN)$ which shows presence of a Nitrile group. This Nitrile group takes up or accepts the hydrogen atom $(H_2)$ that is set free by the reducing agent sodium $Na^+ $ in ethyl alcohol $(C_2H_5OH)$. Thus, the Nitrile group is hydrogenated or reduced which results in formation of a primary amine $( - NH_2)$ with the same number of carbon atoms in the carbon chain. In the given question, the methyl cyanide is hydrogenated due to presence of the hydrogen atom $(H_2)$ that is set free by the reducing agent sodium $(Na^+ )$ in ethyl alcohol $(C_2H_5OH)$ and results in formation of an amine with 2 carbon atoms in the carbon chain and a amine group $( - NH_2)$ and forms a primary amine namely Ethyl amine.

Hence, the correct answer is option C.

Note: A reaction in which a Nitrile group is reduced to form a primary amine is called Mendius Reduction. In the above reaction, the nitrile group in methyl cyanide gets hydrogenated to form a primary amine.