Question

Treatment of ammonia with excess of ethyl iodide will yield

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

Answer 1

Answer: (D)

$\underset{\text{ethyl}\text{iodide}}{C_2{H}_{5}I}+N{H}_3\underset{-HI}{\to }{C}_2{H}_{5}N{H}_2\underset{-HI}{\overset{C_2{H}_{5}I}{\to }}$ $C_2{H}_{5}NH{C}_2{H}_5\underset{-HI}{\overset{C_5{H}_{5}I}{\to }}{(C_2{H}_5)}_{3}N$ $\stackrel{C_2{H}_{5}I}{\to }\underset{\begin{array}{c}\text{tetraethyl}\text{ammonium}\\ \text{iodide}\end{array}}{{(C_2{H}_5)}_4{N}^{+}{I}^{-}}$
If ammonia is in excess, primary amine is the major product.