A) diethylamine
B) ethylamine
C) triethylamine
D) tetraethylammonium iodide
Correct Answer: D
Solution :
\[\underset{\text{ethyl}\,\text{iodide}}{\mathop{{{C}_{2}}{{H}_{5}}I}}\,+N{{H}_{3}}\xrightarrow[-HI]{}{{C}_{2}}{{H}_{5}}N{{H}_{2}}\xrightarrow[-HI]{{{C}_{2}}{{H}_{5}}I}\] \[{{C}_{2}}{{H}_{5}}NH{{C}_{2}}{{H}_{5}}\xrightarrow[-HI]{{{C}_{5}}{{H}_{5}}I}{{({{C}_{2}}{{H}_{5}})}_{3}}N\] \[\xrightarrow{{{C}_{2}}{{H}_{5}}I}\underset{\begin{smallmatrix} \text{tetraethyl}\,\text{ammonium} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{iodide} \end{smallmatrix}}{\mathop{{{({{C}_{2}}{{H}_{5}})}_{4}}{{N}^{+}}{{I}^{-}}}}\,\] Note: If ammonia is in excess, primary amine is the major product.
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