Question

Treatment of ammonia with excess of ethyl chloride will yield.

• A. Diethyl amine
• B. Ethane
• C. Tetraethyl ammonium chloride
• D. Methyl amine

Answer 1

Answer: (C)

Tetraethyl ammonium chloride

Answer 2

$C_{2}H_{5}Cl\overset{\quad NH_{3}\quad}{\rightarrow}C_{2}H_{5} - NH_{2}\overset{\quad C_{2}H_{5}Cl\quad}{\rightarrow}(C_{2}H_{5})_{2} - NH$

$\overset{\quad C_{2}H_{5}Cl\quad}{\rightarrow}(C_{2}H_{5})_{3}N\overset{\quad C_{2}H_{5}Cl\quad}{\rightarrow}\underset{\text{Tetraethylammoniumchloride}}{\left\lbrack C_{2}H_{5} - \underset{\underset{C_{2}H_{5}}{|}}{\overset{\overset{C_{2}H_{5}}{|}}{N - C_{2}H_{5}}} ⥂ \right\rbrack^{+}Cl^{-}}$

If $NH_{3}$ 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.