Question: 56. Given below are two statements :
Statement I : Aniline does not undergo FriedelCrafts alkylation reaction
Statement II : Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true.
(2) Both Statement I and Statement II are false.
(3) Statement I is correct but Statement II is false.
(4) Statement I is incorrect but Statement II is true.
Answer: Option (1)
Explanation:
Aniline contains an amino group \mathrm{-NH_2},
which is a strong activating group for the benzene ring.
However, during Friedel–Crafts alkylation, a Lewis acid such as \mathrm{AlCl_3} is used.
The amino group of aniline forms a stable complex with \mathrm{AlCl_3},
which reduces the electron density of the benzene ring and deactivates it towards electrophilic substitution.
Therefore, aniline does not undergo Friedel–Crafts alkylation reaction.
Hence, Statement I is true.
Gabriel synthesis is used for the preparation of primary aliphatic amines by the reaction of alkyl halides with potassium phthalimide followed by hydrolysis.
Aryl halides like chlorobenzene do not undergo nucleophilic substitution required in Gabriel synthesis due to the partial double-bond character of the \mathrm{C-Cl} bond and resonance stabilization.
Therefore, aniline cannot be prepared through Gabriel synthesis.
Hence, Statement II is also true.
Since both Statement I and Statement II are correct,
the correct answer is Option (1).