Question

explain the following reactions

(a) carbylamine reaction

(b) sandmeyer reaction

Answer 1

(a) Carbylamine reaction: It is also called the Isocynide test because the product formed will be isocyanide which will have a foul test.

This reaction can be given only by 1-degree amines($\mathrm{R}-{\mathrm{NH}}_2,\mathrm{Ar}-{\mathrm{NH}}_2$)

$\mathrm{R}-{\mathrm{NH}}_2+{\mathrm{CHCl}}_3+\mathrm{KOH}\stackrel{\mathrm{Heat}}{\to }\mathrm{R}-{\mathrm{N}}^{+}{\mathrm{C}}^{-}+3\mathrm{KCl}+3{\mathrm{H}}_2\mathrm{O}$

Mechanism:

• In the first step potassium hydroxide(KOH) breaks down into anion and cation-

$\mathrm{KOH}\to {\mathrm{K}}^{+}+{\mathrm{OH}}^{-}$

• Now the nucleophile $\left({\mathrm{OH}}^{-}\right)$ attacks the chloroform and excretes hydrogen from the chloroform and leaves trichlorocarbene behind-

$$\begin{gathered} {\mathrm{CHCl}}_3+{\mathrm{OH}}^{-}\to :{\mathrm{CCl}}_3+{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Trichloro} \\ \mathrm{carbene} \end{gathered}$$

• Now one chlorine from trichlorocarbene will be released out and it becomes dichlorocarbene.

$$\begin{gathered} {\mathrm{CCl}}_3^{-}\stackrel{-{\mathrm{Cl}}^{-}}{\to }:{\mathrm{CCl}}_2 \\ \mathrm{dichlorocarbene} \end{gathered}$$

• Now dichlorocarbene(electrophile) will attack primary alkylamine and alkyl isocyanide will be formed.

(b) Sandmeyer reaction: In this reaction, the diazonium salt reacts with copper salt in presence of their respective acids to form haloarene.

Intermediate will be the free radical on which the halogen group attacks as an electrophile.