Consider the following chemical reactions: (I) Benzamide Br_2, \...

Reaction (I) is Hoffmann bromamide degradation. Benzamide ($C_6H_5CONH_2$) reacts with $Br_2$ and $NaOH$ to give aniline ($C_6H_5NH_2$), which is a primary amine.

Reaction (II) is the reduction of an amide. Benzamide on reduction with $LiAlH_4$ gives benzylamine ($C_6H_5CH_2NH_2$), which is a primary amine.

Reaction (III) is the reduction of an isocyanide. Phenyl isocyanide ($C_6H_5NC$) on reduction with $LiAlH_4$ gives N-methylaniline ($C_6H_5NHCH_3$), which is a secondary amine.

Reaction (IV) is the reduction of a nitrile. Benzonitrile ($C_6H_5CN$) on reduction with $H_2/Ni$ gives benzylamine ($C_6H_5CH_2NH_2$), which is a primary amine.

Thus, reactions (I), (II), and (IV) yield a primary amine as the major product.

Answer: (I), (II) and (IV) only