A) Diethyl amine
B) Ethane
C) Tetraethyl ammonium chloride
D) Methyl amine
Correct Answer: C
Solution :
\[{{C}_{2}}{{H}_{5}}Cl\xrightarrow{N{{H}_{3}}}{{C}_{2}}{{H}_{5}}-N{{H}_{2}}\xrightarrow{{{C}_{2}}{{H}_{5}}Cl}{{({{C}_{2}}{{H}_{5}})}_{2}}-NH\]\[\xrightarrow{{{C}_{2}}{{H}_{5}}Cl}{{({{C}_{2}}{{H}_{5}})}_{3}}N\xrightarrow{{{C}_{2}}{{H}_{5}}Cl}\underset{\text{Tetraethyl}\,\text{ammonium}\,\text{chloride}}{\mathop{{{\left[ {{C}_{2}}{{H}_{5}}-\underset{\underset{{{C}_{2}}{{H}_{5}}\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{\overset{\overset{{{C}_{2}}{{H}_{5}}\,\,\,\,\,\,}{\mathop{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,}{\mathop{N\,-{{C}_{2}}{{H}_{5}}}}\,}}\, \right]}^{+}}C{{l}^{-}}}}\,\] If \[N{{H}_{3}}\] is in excess, then \[{{1}^{o}}\] amine will be the main product, if \[{{C}_{2}}{{H}_{5}}Cl\] is in excess then mixture of \[{{1}^{o}},{{2}^{o}},{{3}^{o}}\] and quaternary amine is obtained.
You need to login to perform this action.
You will be redirected in
3 sec
