Question

When excess of ethyl iodide is treated with ammonia, the product is:

• A. ethylamine
• B. diethylamine
• C. triethylamine
• D. tetraethylammonium iodide

Answer 1

The correct option is D tetraethylammonium iodide

The order of reactivity of halides with amines is $RI>RBr>RCl$.

When an excess of ethyl iodide is treated with ammonia, the product formed is tetraethylammonium iodide.

$N{H}_3+C_2{H}_{5}I\left(excess\right)\to C_8{H}_{20}{N}^{+}$

Tetra ammonium iodide is a quaternary salt produced from the ammonolysis of an alkyl halide with ammonia as the nucleophile.

An alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino $(–N{H}_2)$ group. The primary amine thus obtained behaves as a nucleophile and can further react with an alkyl halide to form secondary and tertiary amines and finally quaternary ammonium salt.