Question

Find the reactant R in the following reaction

(1) ${N_2}$
(2) $CHC{l_3}/KOH(alcoholic)$
(3) $N{H_3}$
(4) $KCN$

Answer 1

An extra Carbon atom is being substituted so check of options that have a carbon atom in them. Clearly there are no other reagents so the $- NC$ molecule that is substituted in the place of $- N{H_2}$ must be from the reaction between aniline and the reactant R.

Complete answer:
Given to us, aniline has to be converted to phenyl isocyanide. When we treat aniline with chloroform in presence of a strong base such as alcoholic $KOH$ in this case, phenyl isocyanide is formed. This reaction is known as carbylamine reaction. In this reaction, first we need to attach a carbon to the Nitrogen of amine group in aniline. So when chloroform reacts with aniline, nitrogen atom is attached to the carbon atom eliminating $HCl$ molecule. When this reaction takes place in the presence of alcoholic $KOH$ we get phenyl isocyanide.

**Hence, the reagent R is $CHC{l_3}/KOH(alcoholic)$ i.e. option 2.

Additional information: **
Carbylamine reaction is also known as Hofmann isocyanide synthesis. This process involves the formation of isocyanide by the reaction between an aliphatic or an aromatic primary amide, chloroform and a base. Secondary and tertiary amines do not give this reaction.

Note:
In the reaction given to us, a primary amide is being converted to an isocyanide in the presence of a reagent R. This reaction can be explained by carbylamine reaction or Hoffmann isocyanide synthesis. In this reaction a primary amide is treated with chloroform and a strong base to form an isocyanide.