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  4. Treatment of ammonia with excess of ethyl iodide will yield

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Q. Treatment of ammonia with excess of ethyl iodide will yield

Jharkhand CECEJharkhand CECE 2011

Solution:

$ \underset{\text{ethyl}\,\text{iodide}}{\mathop{{{C}_{2}}{{H}_{5}}I}}\,+N{{H}_{3}}\xrightarrow[-HI]{}{{C}_{2}}{{H}_{5}}N{{H}_{2}}\xrightarrow[-HI]{{{C}_{2}}{{H}_{5}}I} $ $ {{C}_{2}}{{H}_{5}}NH{{C}_{2}}{{H}_{5}}\xrightarrow[-HI]{{{C}_{5}}{{H}_{5}}I}{{({{C}_{2}}{{H}_{5}})}_{3}}N $ $ \xrightarrow{{{C}_{2}}{{H}_{5}}I}\underset{\begin{smallmatrix} \text{tetraethyl}\,\text{ammonium} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{iodide} \end{smallmatrix}}{\mathop{{{({{C}_{2}}{{H}_{5}})}_{4}}{{N}^{+}}{{I}^{-}}}}\, $
If ammonia is in excess, primary amine is the major product.