To solve this problem, we need to analyze each reaction to determine if it produces a primary amine.$\newline$Let's examine each option step-by-step:$\newline$• Option 1: ${CH}_3{CONH}_2 \xrightarrow{Br_2 / KOH}$Product$\newline$This is the Hofmann bromamide degradation reaction.$\newline$• In this reaction, an amide is converted to a primary amine with one less carbon atom.$\newline$• The product is $CH_3NH_2,$ which is a primary amine.$\newline$• Option 2: ${CH}_3CN \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4}$Product$\newline$This involves the reduction of a nitrile using $LiAlH_4. \newline$• The nitrile is reduced to a primary amine.$\newline$• The product is $CH_3CH_2NH_2,$ which is a primary amine.$\newline$• Option 3: ${CH}_3NC \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4}$Product$\newline$This involves the reduction of an isocyanide (isonitrile).$\newline$• The reduction of an isocyanide typically leads to the formation of a secondary amine.$\newline$• Therefore, the product is not a primary amine.$\newline$• Option 4: ${CH}_3CONH_2 \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4}$Product$\newline$This involves the reduction of an amide using $LiAlH_4. \newline$• The amide is reduced to a primary amine.$\newline$• The product is $CH_3CH_2NH_2,$ which is a primary amine.$\newline$Conclusion:$\newline$• Option 3 does not produce a primary amine; it produces a secondary amine.$\newline$Therefore, the correct answer is Option 3.