Question

Which of the following primary amine is not prepared by Gabriel phthalimide synthesis:
A.$C{H_3} - C{H_2} - N{H_2}$
B.$C{H_3} - N{H_2}$
C.$Ph - N{H_2}$
D.$Ph - C{H_2} - N{H_2}$

Answer 1

The Gabriel synthesis is a chemical reaction that transforms primary alkyl halides into primary amines. Traditionally, the reaction uses potassium phthalimide and, therefore, is also called Gabriel phthalimide synthesis. It uses a nucleophilic substitution reaction to form primary amines from alkyl halide.

Complete answer:
In Gabriel phthalimide synthesis, a base abstract proton from phthalimide gives a nucleophile phthalimide ion which attacks on the unhindered primary alkyl halide. The base hydrolysis of alkylated phthalimide gives the primary unhindered amine and phthalimide ion. Phthalimide ion itself is a bulky ion, so it prefers attack on hindered alkyl halide.
Gabriel phthalimide synthesis can be understood through the following mechanism:


In the given options, $C{H_3} - C{H_2} - N{H_2}$ (ethyl amine), $C{H_3} - N{H_2}$ (methyl amine) and $Ph - C{H_2} - N{H_2}$ ($1 -$phenyl methanamine) are the compounds which can be prepared by Gabriel phthalimide synthesis. Ethylamine and methylamine are simple aliphatic compounds. While in $1 -$phenyl methanamine, the amine part is at the methyl part and phenyl is acting as a functional group and hence should be considered an aliphatic amine.
Hence, $Ph - N{H_2}$ (aniline) is the only compound that cannot be prepared by Gabriel phthalimide synthesis.
Therefore, the correct option is C.

Note:
Gabriel phthalimide synthesis is only used for primary amines as steric hindrances do not allow the reaction for secondary and tertiary aliphatic amines. It is also not applicable for $1^\circ$ aromatic compounds, aryl halide does not undergo nucleophilic substitution reaction. Aromatic compounds usually show electrophilic substitution reactions. Nucleophilic substitution destabilizes the resonance of aromatic compounds.