The following two reactions give the same foul-smelling product Z. X and Z, respectively, are:
Answer: (A) (A) ; . \textbf{Answer:} (A) converts the alkyl halide to the foul-smelling isocyanide , which is the same product Z formed by the carbylamine reaction.
- A.(A) ; ✓
- B.(B) ;
- C.(C) ;
- D.(D) ;
Correct Answer
(A) (A) ;
Solution & Explanation
\textbf{Answer:} (A) converts the alkyl halide to the foul-smelling isocyanide , 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): . This primary amine then undergoes the carbylamine reaction with chloroform and ethanolic to give the foul-smelling isocyanide: . So (ethyl isocyanide). First path (fixes X): To get the same isocyanide from , the reagent must be silver cyanide: is mainly covalent and the nitrogen lone pair attacks, giving the isocyanide . ( is ionic and would give the nitrile , the wrong product.) Therefore and , i.e. option (A).