Chemistry - Halogen Derivative Question with Solution | TestHub

In the second sequence of reactions, ${\text{C}}_2{\text{H}}_5{\text{CONH}}_2$ undergoes Hoffmann bromamide degradation with ${\text{Br}}_2$ and $\text{NaOH}$ to form a primary amine with one less carbon atom. Thus, $Y$ is ethylamine (${\text{C}}_2{\text{H}}_5{\text{NH}}_2$).

Ethylamine ($Y$) then reacts with ${\text{CHCl}}_3$ and ethanolic $\text{KOH}$ in a carbylamine reaction to form ethyl isocyanide (${\text{C}}_2{\text{H}}_5\text{NC}$), which is known for its foul smell. Therefore, $Z$ is ${\text{C}}_2{\text{H}}_5\text{NC}$.

In the first reaction, ethyl chloride (${\text{C}}_2{\text{H}}_5\text{Cl}$) reacts with a reagent $X$ to give the same product $Z$ (${\text{C}}_2{\text{H}}_5\text{NC}$). To obtain an isocyanide as the major product from an alkyl halide, the reagent used is silver cyanide ($\text{AgCN}$), because $\text{AgCN}$ is predominantly covalent and the nitrogen atom acts as the nucleophilic center.

Hence, $X=\text{AgCN}$ and $Z={\text{C}}_2{\text{H}}_5\text{NC}$.

Answer: $X=\text{AgCN}$; $Z={\text{C}}_2{\text{H}}_5\text{NC}$