How many of the transformations given below would result in aromatic amines?

Solution

Chemistry ma’am · 3 steps

Check each route ends with N on the ring; Gabriel fails because chlorobenzene will not do S_N2.

1. What to look for. An aromatic amine has the -NH₂ group sitting directly on the benzene ring, as in aniline. So ask of each reaction: does it actually run, and does it end with N bonded to the ring?

   Reaction	What happens	Aromatic amine?
   Reaction 1: C₆H₅CONH₂ + Br₂ + NaOH	Hofmann bromamide degradation removes the C=O carbon and gives C₆H₅NH₂	Yes
   Reaction 2: potassium phthalimide + C₆H₅Cl	Gabriel synthesis needs S_N2 attack on the halide. The C–Cl bond of chlorobenzene has partial double-bond character, so there is no reaction	No
   Reaction 3: C₆H₅NO₂ with H₂ and Pd/C	Catalytic hydrogenation reduces -NO₂ to -NH₂, giving aniline	Yes
   Reaction 4: C₆H₅NHCOCH₃ with dil. H₂SO₄ and heat	Acid hydrolysis of the amide removes the acetyl group and frees aniline	Yes

2. Why Gabriel fails. The phthalimide anion must push out Cl⁻ by attacking the carbon from the back. On a benzene ring that backside is blocked by the ring itself, and resonance strengthens the C–Cl bond, so aryl halides never give amines this way.
3. So. Reactions 1, 3 and 4 give aniline, and reaction 2 does not go. The count is 3.

The common slip: 4. Counting the Gabriel reaction with chlorobenzene as making aniline gives 4, but aryl halides do not react with potassium phthalimide.

The idea to keep. Gabriel phthalimide synthesis makes only primary aliphatic amines. Aryl halides do not undergo S_N2, so they never give aromatic amines by this route.

Answer from the NTA final answer key. Solution by PrepWiser.