Why Ammonolysis Gives a Mixture of 1°, 2°, 3° Amines

Because the primary amine (R-NH2) still has a lone pair on nitrogen, so it is also a nucleophile like ammonia. It attacks a fresh molecule of alkyl halide and becomes a secondary amine. The secondary amine does the same and becomes tertiary, and finally a quaternary ammonium salt forms. The reaction cannot naturally stop at one stage because every product is reactive again.

Alkyl groups push electron density onto nitrogen through the +I (inductive) effect. This makes the amine's lone pair more available, so R-NH2 can be a stronger nucleophile than NH3. That is why once a little amine forms, it competes for the alkyl halide and pushes the reaction forward to higher amines.

Take a large excess of ammonia. With so many NH3 molecules around, most collisions are between the alkyl halide and ammonia, not between the alkyl halide and an already-formed amine. This makes the 1° amine the main product. NCERT states this directly: 'primary amine is obtained as a major product by taking large excess of ammonia.'

RI > RBr > RCl. The carbon–halogen bond is weakest in R-I, so iodides react fastest. This is the reactivity order NCERT gives for the reaction of halides with amines. It does not change why a mixture forms; it only affects how fast the substitution happens.

No. Aryl halides (like chlorobenzene) do not undergo easy nucleophilic substitution because the C–X bond has partial double-bond character (resonance) and the ring is electron-rich. So ammonolysis is mainly for alkyl and benzyl halides, not for making aniline. For pure primary amines, Gabriel synthesis or Hoffmann bromamide is preferred.

Because each amine formed still has a lone pair on nitrogen and stays a nucleophile. So R-NH2 reacts with more alkyl halide to give 2°, then 3° amine, and finally a quaternary ammonium salt.

With a large excess of ammonia, the primary (1°) amine is the major product. With excess alkyl halide, the quaternary ammonium salt dominates.

Use Gabriel phthalimide synthesis (only 1° amines, no mixture) or Hoffmann bromamide degradation. These avoid the over-alkylation problem of ammonolysis.

RI > RBr > RCl, because the C–I bond is the weakest and breaks most easily during nucleophilic substitution.

No. Aryl halides resist nucleophilic substitution due to resonance and an electron-rich ring, so ammonolysis is used for alkyl and benzyl halides, not for aromatic amines.