Question

When excess ethyl iodide is treated with ammonia, the product is:
A. Ethylamine
B. Diethylamine
C. Triethylamine
D. Tetraethylammonium iodide

Answer 1

We must know that an ethanolic solution of ammonia when treated with alkyl (or) benzyl halides undergoes nucleophilic substitution reaction. Therefore, it gives quaternary ammonium salt as a product.

Complete answer: We can understand that an ethanolic solution of ammonia when treated with alkyl (or) benzyl halides undergoes nucleophilic substitution reaction. In this nucleophilic substitution reaction, an amino group replaces the halogen atom. The primary amine formed acts a nucleophile and again reacts with alkyl halide to give secondary and tertiary amine. On reacting with excess alkyl halide, a quaternary ammonium salt is obtained.
We know that the reactivity of halides with amines is $RI > RBr > RCl.$
In the ammonolysis reaction of an alkyl halide with ammonia, ammonia acts as nucleophile. The product formed would be quaternary salt of ammonia.
We can now write the reaction as,
$N{H_3} + {C_2}{H_5}I\left( {excess} \right) \to {\left( {{C_2}{H_5}} \right)_4}{N^ + }$
From the above reaction, we can see that when ethyl iodide in excess is made to react with ammonia, the product formed would be tetraethylammonium iodide.
Therefore, the option (D) is correct.

Note:
We must remember that in reaction of alkyl halide with ammonia, the haloalkane is treated with a concentration solution of ammonia in the presence of ethanol. We can also obtain quaternary ammonium salt, by treating alkyl halide with ammonia to form primary amine. The primary amine would react with haloalkane to form a secondary amine. The secondary amine would react with haloalkane to form tertiary amine. The tertiary amine would react with haloalkane to form quaternary salt of ammonium.