The following two reactions give the same foul-smelling product Z. $\ce{C2H5Cl ->[X] Z}$ $\ce{C2H5CONH2 ->[Br2 + NaOH] C2H5NH2 ->[CHCl3 / ethanolic KOH][\Delta] Z}$ X and Z, respectively, are:

Solution & Explanation

\textbf{Answer:} (A) $\ce{AgCN}$ converts the alkyl halide to the foul-smelling isocyanide $\ce{C2H5NC}$, which is the same product Z formed by the carbylamine reaction. \textbf{Solution:} Second path (fixes Z): Propanamide undergoes Hoffmann bromamide degradation to ethylamine (one less carbon): $\ce{C2H5CONH2 ->[Br2/NaOH] C2H5NH2}$. This primary amine then undergoes the carbylamine reaction with chloroform and ethanolic $\ce{KOH}$ to give the foul-smelling isocyanide: $\ce{C2H5NH2 + CHCl3 + 3KOH ->[\Delta] C2H5NC + 3KCl + 3H2O}$. So $\ce{Z = C2H5NC}$ (ethyl isocyanide). First path (fixes X): To get the same isocyanide $\ce{C2H5NC}$ from $\ce{C2H5Cl}$, the reagent must be silver cyanide: $\ce{AgCN}$ is mainly covalent and the nitrogen lone pair attacks, giving the isocyanide $\ce{C2H5Cl ->[AgCN] C2H5NC}$. ($\ce{KCN}$ is ionic and would give the nitrile $\ce{C2H5CN}$, the wrong product.) Therefore $\ce{X = AgCN}$ and $\ce{Z = C2H5NC}$, i.e. option (A).