An aromatic compound (A) on treatment with aqueous ammonia and heating forms compound (B) which on heating with Br2 and KOH forms a compound (C) of molecular formula C6H7N.
Consider the following statements about the preparation of amines.
(A) On reduction with lithium aluminium hydride followed by treatment with water, an amide R–CO–NH2 gives the amine R–CH2–NH2.
(B) In Gabriel phthalimide synthesis, phthalimide is converted into its potassium salt by treatment with ethanolic potassium hydroxide.
(C) The ammonolysis of an alkyl halide with an ethanolic solution of ammonia is carried out in a sealed tube at 273 K.
(D) The free amine can be obtained from its ammonium salt by treatment with a strong acid.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Primary amines can be made from amides, from phthalimide and from alkyl halides. Each method has its own reagent and conditions.
(A) On reduction with lithium aluminium hydride followed by treatment with water, an amide R–CO–NH2 gives the amine R–CH2–NH2. — True. The C=O group of the amide is reduced to –CH2–, so no carbon atom is lost:
\( \displaystyle \ce{R-CO-NH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2-NH2} \)
(B) In Gabriel phthalimide synthesis, phthalimide is converted into its potassium salt by treatment with ethanolic potassium hydroxide. — True. The N–H hydrogen of phthalimide is removed by KOH, giving potassium phthalimide.
This salt is then heated with an alkyl halide, and alkaline hydrolysis of the N-alkylphthalimide gives the primary amine.
\( \displaystyle \ce{C6H4(CO)2NH + KOH -> C6H4(CO)2N^-K^+ + H2O} \)
(C) The ammonolysis of an alkyl halide with an ethanolic solution of ammonia is carried out in a sealed tube at 273 K. — False. The reaction is carried out in a sealed tube at 373 K.
Ammonia is a gas, so the tube is sealed to keep it in the hot mixture.
\( \displaystyle \ce{R-X + NH3} \xrightarrow{\text{ethanol, 373 K}} \ce{R-NH3^+X^-} \)
(D) The free amine can be obtained from its ammonium salt by treatment with a strong acid. — False. A strong base such as NaOH sets the amine free.
An acid would only keep the amine in its salt form.
\( \displaystyle \ce{R-NH3^+X^- + NaOH -> R-NH2 + H2O + NaX} \)
NCERT (2026-27 edition):
9.4 Preparation of Amines – “4. Reduction of amides: The amides on reduction with lithium aluminium hydride yield amines.”
“5. Gabriel phthalimide synthesis: … Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.”
“2. Ammonolysis of alkyl halides: … This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis. The reaction is carried out in a sealed tube at 373 K.”
“The free amine can be obtained from the ammonium salt by treatment with a strong base.”