Question

Treatment of ammonia with excess ethyl chloride will yield?
A. Triethylamine
B. Diethylamine
C. Ethyl amine
D. Tetraethyl ammonium chlorides

Answer 1

Hint: Alkyl halides are derivatives of alkanes in which hydrogen is replaced by halogens. Carbon-halogen bond is a very weak polar bond. This reaction can be done with excess of ammonia or excess of ethyl chloride. Thus the products formed will be different.

Complete step by step answer:
Alkyl halides are classified into primary, secondary and tertiary based on the number of carbons bonded to the carbon with the halogen atom.
When ammonia is treated with ethyl chloride, it undergoes nucleophilic substitution reaction. In this reaction, the nucleophile is ammonia. It is a weak base. The leaving group will be chloride. Negative charge is a stronger nucleophile than an analogous neutral species, i.e. \[{\text{N}}{{\text{H}}_2}^ - > {\text{N}}{{\text{H}}_3}\]. Nucleophilicity increases down the periodic table, i.e. ${{\text{I}}^ - } > {\text{B}}{{\text{r}}^ - } > {\text{C}}{{\text{l}}^ - } > {{\text{F}}^ - }$.
The reaction showing the treatment of ammonia with excess ethyl chloride is given below:
\[{{\text{C}}_2}{{\text{H}}_5}{\text{Cl}}\xrightarrow{{{\text{N}}{{\text{H}}_3}}}{{\text{C}}_2}{{\text{H}}_5}{\text{N}}{{\text{H}}_2}\xrightarrow{{{{\text{C}}_2}{{\text{H}}_5}{\text{Cl}}}}{\left( {{{\text{C}}_2}{{\text{H}}_5}} \right)_2}{\text{NH}}\xrightarrow{{{{\text{C}}_2}{{\text{H}}_5}{\text{Cl}}}}{\left( {{{\text{C}}_2}{{\text{H}}_5}} \right)_3}{\text{N}}\xrightarrow{{{{\text{C}}_2}{{\text{H}}_5}{\text{Cl}}}}{\left( {{{\text{C}}_2}{{\text{H}}_5}} \right)_4}{{\text{N}}^ + }{\text{C}}{{\text{l}}^ - }\]
Ammonia acts as a nucleophile and it attacks the electrophilic carbon of alkyl halide. It displaces the leaving group ${\text{C}}{{\text{l}}^ - }$ and creates a new ${\text{C}} - {\text{N}}$ bond. Excess ethyl chloride deprotonates the nitrogen center forming the secondary amine and thus the reaction goes on.
Generally this reaction produces a mixture of products of primary amines, secondary amines, tertiary amines and quaternary amines. This mixture of products can be avoided by adding excess ammonia to the ethyl chloride.
So we can say that tetraethyl ammonium chloride is formed.

So, the correct answer is Option D .

Note:
When excess ammonia is reacted with ethyl chloride, primary amine is formed. The reaction is given below:
${{\text{C}}_2}{{\text{H}}_5}{\text{Cl}} + {\text{N}}{{\text{H}}_3} \to {{\text{C}}_2}{{\text{H}}_5}{\text{N}}{{\text{H}}_2} + {\text{HCl}}$
                 Excess
Here, the main product is primary amine, ${{\text{C}}_2}{{\text{H}}_5}{\text{N}}{{\text{H}}_2}$.