Question: Alkyl cyanides \[(C{H_3}CN)\] when treated with hydrogen in presence of Pt or with \[LiAl{H_4}\] pro...

Question

Alkyl cyanides $(C{H_3}CN)$ when treated with hydrogen in presence of Pt or with $LiAl{H_4}$ produces the same carbon number compound. What is the formula of that compound?

Answer 1

Solution

Lithium aluminium hydride, platinum or palladium with hydrogen are reducing agents. They reduce alkyl cyanide to primary amine. The nucleophilic hydride ion attacks on electrophilic carbon of cyanide and forms an amine group.

Complete step by step answer:
$LiAl{H_4}$ is a strong reducing agent for polar double bonds and a source of ${H^ - }$ . It can reduce aldehydes, ketones, esters, carboxylic acid chlorides, carboxylic acids and carboxylate salts to alcohols. It can also reduce amides and nitriles to amines. In these cases, the partially negative hydrogen of the reagent reacts with the partially positive carbon of the substrate.
Aluminium is a metal having low electronegativity, so the Al-H bond is strongly polarized with Al positive and H negative. In its anionic form, one coordinate covalent bond is present using the lone pair on a hydride ion and forms an empty orbital on aluminium. Four hydrogens are present around the aluminium in anionic form. Oxidation state of aluminium is +3 in $LiAl{H_4}$ .
The alkyl cyanide reacts with $LiAl{H_4}$ in presence of ether followed by the treatment of product with dilute acid. Overall, we can say that the carbon-nitrogen triple bond gets reduced to give a primary amine, that is $- N{H_2}$ group.
Similarly, the carbon-nitrogen triple bond in a cyanide can also be reduced by reaction with hydrogen gas in the presence of a metal catalysts, platinum at high temperature and pressure.
${H_3}C - C \equiv N + 2{H_2}\mathop \to \limits_{Pt}^{LiAl{H_4}} {H_3}C - C{H_2} - N{H_2}$

So, the formula of the compound is ${H_3}C - C{H_2} - N{H_2}$ .

Note:
The compound obtained as product i.e. primary amine cannot be obtained by reduction of alkyl cyanide with other reducing agents such as DIBAL-H ${(i - B{u_2}AlH)_2}$ . An aldehyde is formed in such a case. The chemical equation for this reaction is:
$R - C = N + DIBAL + {H^ + } \to RCHO$
Thus, Lithium aluminium hydride is an unselective reducing agent.