Question

Give the structures of A, B, and C in the following reaction sequences: (i)\; \(\mathrm{CH_3CH_2I} \xrightarrow[\ ]{\mathrm{NaCN}} A \xrightarrow[\ ]{\;\;HO^-\;/\;H_2O\;\;} B \xrightarrow[\;Br_2/NaOH\;]{} C\)
(ii)\; \(\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow[\ ]{\mathrm{CuCN}} A \xrightarrow[\ ]{\;H_3O^+\;} B \xrightarrow[\ \Delta\ ]{\;NH_3\;} C\)
(iii)\; \(\mathrm{CH_3CH_2Br} \xrightarrow[\ ]{\mathrm{KCN}} A \xrightarrow[\ ]{\mathrm{LiAlH_4}} B \xrightarrow[\;273\,K\;]{\;HNO_2\;} C\)

Hint:

Alkyl halide \(+\) \(\mathrm{CN^-}\) \(\Rightarrow\) nitrile (adds one carbon).
Nitrile \(\xrightarrow[\text{controlled}]{\text{hydrolysis}}\) amide; amide \(\xrightarrow{Br_2/NaOH}\) amine with one carbon less (Hofmann).
\(\mathrm{ArN_2^+}\) \(\xrightarrow{CuCN}\) Ar–CN (Sandmeyer); \(\mathrm{Ar–CN}\) \(\xrightarrow{H_3O^+}\) Ar–COOH; \(\mathrm{Ar–COOH}\) \(\xrightarrow{NH_3,\ \Delta}\) amide.
Primary aliphatic amine \(+\) \(HNO_2\) at \(0{-}5^\circ C\) \(\Rightarrow\) alcohol (deamination).

Answer 1

(i) From ethyl iodide via nitrile and amide (Hofmann bromamide):
\[ \mathrm{CH_3CH_2I \xrightarrow{NaCN} CH_3CH_2CN\;(A)\; \xrightarrow[\,\text{controlled}\,]{HO^-/H_2O} CH_3CH_2CONH_2\;(B) \xrightarrow{Br_2/NaOH} CH_3CH_2NH_2\;(C)} \] \[ \boxed{A=\text{propanenitrile (propionitrile)}, B=\text{propanamide}, C=\text{ethylamine}} \] (ii) Sandmeyer to nitrile, hydrolysis to acid, then ammonolysis:
\[ \mathrm{C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN\;(A) \xrightarrow{H_3O^+} C_6H_5COOH\;(B) \xrightarrow[\ \Delta\ ]{NH_3} C_6H_5CONH_2\;(C)} \] \[ \boxed{A=\text{benzonitrile}, B=\text{benzoic acid}, C=\text{benzamide}} \] (iii) From bromoethane to nitrile, reduction to amine, then nitrous acid:
\[ \mathrm{CH_3CH_2Br \xrightarrow{KCN} CH_3CH_2CN\;(A) \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2\;(B) \xrightarrow[\,273\,K\,]{HNO_2} CH_3CH_2CH_2OH\;(C)} \] \[ \boxed{A=\text{propanenitrile}, B=\text{propylamine}, C=\text{propan-1-ol}} \]