Chemistry Question 51 – NEET-UG 2023

Let's determine the product for each given reaction:

(1) ${\text{CH}}_3{\text{CONH}}_2\stackrel{{\text{Br}}_2/\text{KOH}}{\to }{\text{CH}}_3{\text{NH}}_2$ (Hofmann Bromamide Degradation). This yields methylamine, which is a primary amine.

(2) $$\begin{gathered} {\text{CH}}_3\text{CN}\stackrel{{\text{(i) LiAlH}}_4 \\ {\text{(ii) H}}_3{\text{O}}^{\oplus }}{\to }{\text{CH}}_3{\text{CH}}_2{\text{NH}}_2 \end{gathered}$$ (Reduction of Nitrile). This yields ethylamine, which is a primary amine.

(3) $$\begin{gathered} {\text{CH}}_3\text{NC}\stackrel{{\text{(i) LiAlH}}_4 \\ {\text{(ii) H}}_3{\text{O}}^{\oplus }}{\to }{\text{CH}}_3{\text{NHCH}}_3 \end{gathered}$$ (Reduction of Isocyanide). This yields dimethylamine, which is a secondary amine.

(4) $$\begin{gathered} {\text{CH}}_3{\text{CONH}}_2\stackrel{{\text{(i) LiAlH}}_4 \\ {\text{(ii) H}}_3{\text{O}}^{\oplus }}{\to }{\text{CH}}_3{\text{CH}}_2{\text{NH}}_2 \end{gathered}$$ (Reduction of Amide). This yields ethylamine, which is a primary amine.

Based on the analysis in Step 2, reactions (1), (2), and (4) all produce primary amines. Reaction (3), the reduction of methyl isocyanide, produces dimethylamine (${\text{CH}}_3{\text{NHCH}}_3$), which is a secondary amine. Therefore, reaction (3) will NOT give a primary amine as the product.