Correct answer: Phenylamine (option C)Gabriel phthalimide synthesis prepares only primary amines that can be formed through an SN​2 reaction on an alkyl/benzylic halide.Step 1: Phthalimide is treated with alcoholic KOH to form potassium phthalimide:C6​H4​(CO)2​NH+KOH→C6​H4​(CO)2​N−K++H2​OStep 2: The phthalimide anion attacks an alkyl halide R−X by an SN​2 pathway to give N-alkylphthalimide.Step 3: Alkaline hydrolysis (or hydrazinolysis) of the N-alkylphthalimide liberates the primary amine RNH2​.Ethylamine (C2​H5​NH2​) comes from ethyl halide, propylamine (C3​H7​NH2​) from propyl halide, and benzylamine (C6​H5​CH2​NH2​) from benzyl halide.All three of these halides are sp3 (or benzylic) halides that readily undergo SN​2 substitution, so their amines can be prepared this way.Phenylamine (aniline) would require chlorobenzene to react with the phthalimide anion.In chlorobenzene the C−Cl bond has partial double-bond character due to resonance, and the planar aryl carbon is shielded from backside attack, so nucleophilic substitution by SN​2 does not occur.Therefore the phthalimide anion cannot be attached to a benzene ring, and phenylamine cannot be prepared by Gabriel phthalimide synthesis.Phenylamine is instead prepared by reduction of nitrobenzene, e.g. C6​H5​NO2​Sn/HCl​C6​H5​NH2​.Hence the amine that cannot be prepared by the Gabriel phthalimide synthesis is phenylamine, option C.