Question

An alkyl halide reacts with alcoholic ammonia in a sealed tube, the product formed will be
A.A primary amine
B.A secondary amine
C.A tertiary amine
D.All of the above

Answer 1

The reaction of alkyl halide with alcoholic ammonia a nucleophilic substitution reaction occurs. In nucleophilic substitution reaction electron rich compound (nucleophile) replaces a leaving group as the primary substituent in the reaction, as a part of another molecule.

Complete step by step answer:
When an alkyl halide reacts with alcoholic ammonia, it produces primary amine. The reaction doesn't stop here and primary amine further reacts and produces secondary amine. Then after that Secondary amines reacts further and produces tertiary amine. This phenomena is further explained:
${\text{RX + N}}{{\text{H}}_3} \to \mathop {{\text{RN}}{{\text{H}}_2}}\limits_{1^\circ {\text{ amine}}} {\text{ + HX}} \\\ {\text{RN}}{{\text{H}}_2}{\text{ + XR}} \to \mathop {{{\text{R}}_2}{\text{NH}}}\limits_{2^\circ {\text{ amine}}} {\text{ + HX}} \\\ {{\text{R}}_2}{\text{NH + XR}} \to \mathop {{{\text{R}}_3}{\text{N}}}\limits_{3^\circ {\text{ amine}}} {\text{ + HX}} \\\$
In the above reactions Ammonia as a nucleophile reacts with alkyl halides to give primary amines in a nucleophilic substitution reaction. The Yield product of which are often poor (a primary amine),${\text{RN}}{{\text{H}}_2}$ is itself a nucleophile and it further react with more alkyl halide which produces secondary amine which further react and produce tertiary amine.
Hence, the result is mixtures of primary amines, secondary amines and tertiary amines with quaternary ammonium salts.
Therefore, option (D) All the above is the correct answer.

Note:
Secondary haloalkanes can react with ammonia via ${S_N}1$ or ${S_N}2$ reaction mechanism, mostly ${S_N}1$ is favoured, as it is quicker than ${S_N}2$. As the size of the halogen atom increases the rate of reaction increases (as you go down in group 17), because the bond enthalpy of the C-halogen bond decreases which means it is at having higher energy level and is less stable, therefore it requires less energy to break down. The formation of this mixture can be avoided if a large excess of ammonia is used.