Chloroethane is heated with an ethanolic solution of ammonia in a sealed tube at 373 K. How many of the following compounds can be formed in this reaction?
Ethanamine, N-Ethylethanamine, Ethanenitrile, N,N-Diethylethanamine, Ethanamide, Tetraethylammonium chloride, N-Ethylethanamide
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Chloroethane reacts with ammonia by ammonolysis. Each amine formed reacts again with chloroethane, so the ethyl groups are added one at a time:
\( \displaystyle \ce{C2H5Cl + NH3 -> C2H5NH2 + HCl} \)
\( \displaystyle \ce{C2H5NH2 + C2H5Cl -> (C2H5)2NH + HCl} \)
\( \displaystyle \ce{(C2H5)2NH + C2H5Cl -> (C2H5)3N + HCl} \)
\( \displaystyle \ce{(C2H5)3N + C2H5Cl -> (C2H5)4N^+Cl^-} \)
Formed:
Ethanamine (primary amine)
N-Ethylethanamine (secondary amine)
N,N-Diethylethanamine (tertiary amine)
Tetraethylammonium chloride (quaternary ammonium salt)
Not formed:
Ethanenitrile — needs a cyanide such as KCN.
Ethanamide — amides come from acid derivatives, not from alkyl halides.
N-Ethylethanamide — formed by acylation; no acylating agent is present.
Formed: 4 compounds.
NCERT (2026-27 edition):
9.4 Preparation of Amines – “…an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group. … The reaction is carried out in a sealed tube at 373 K. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt.”
“Ammonolysis has the disadvantage of yielding a mixture of primary, secondary and tertiary amines and also a quaternary ammonium salt.”