Question

The product formed is due to-
$C{H}_3-C{H}_2-Br+AgCN\to C{H}_3-C{H}_2-NC+AgBr$
(A) High ionic character of AgCN
(B) Covalent character of AgCN due to larger size of $C{N}^{-}$
(C) Covalent character of AgCN due to Pseudo noble electronic configuration
(D) All of these

Answer 1

Hint: In the above reaction an alkyl halide is reacting with silver cyanide to form an isocyanide. To explain the formation of this product, we should know the characteristic properties of the reagent Silver cyanide that it is an ionic compound or a covalent compound. This is because the ionic or the covalent characteristic of AgCN is responsible for the formation of this product.

Complete answer:
First we will see the difference between ionic compounds and covalent compounds. Ionic compounds on dissolution breaks down into its constituent ions while covalent compounds do not form ions on dissolution.
Here, AgCN is a covalent compound and it does not give its constituent ions on dissolution. AgCN has covalent character due to its pseudo noble electronic configuration.
$A{g}^{+}=n{s}^{2}n{p}^6(n-1)d^{10}$
Due to poor shielding of the ten d shell electrons, there is greater effective nuclear charge which leads to the greater polarizing power of cation and hence greater covalent character in compound.
So when an alkyl halide reacts with AgCN, due to its covalent character, nucleophile cyanide attacks from the nitrogen centre which is active. Carbon centre does not attack on alkyl halide because AgCN does not break down into $A{g}^{+}$ and $C{N}^{-}$ ions.

Option D is the correct answer.

Note:
We should note that in case of covalent compound (AgCN), isocyanide is formed but if we take an ionic compound (KCN) instead of AgCN, then the product formed will be a cyanide. This is due to the fact that ionic compound on dissolution forms its constituent ions and hence, carbon of cyanide ion will attack on the alkyl halide.