Identify Incorrect Statements Regarding Amine Reactions and Properties

To determine which of the given statements are incorrect, let us analyze each statement individually:

1. Statement A:

   • The first structure is 2-phenylethan-1-amine ($\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$), which is an aliphatic primary amine.
   • The second structure is aniline ($\text{C}_6\text{H}_5\text{NH}_2$), which is an aromatic primary amine.
   • In aniline, the unshared electron pair on the nitrogen atom is in conjugation with the benzene ring and is delocalized through resonance, making it less available for protonation.
   • In 2-phenylethan-1-amine, the nitrogen atom is separated from the aromatic ring by $sp^3$-hybridized carbon atoms, so its lone pair is not delocalized into the ring.
   • Therefore, 2-phenylethan-1-amine is a stronger base than aniline.
   • Thus, Statement A is correct.

2. Statement B:

   • The structure shown is 4-methoxyaniline ($p$-anisidine), which is a primary aromatic amine.
   • Gabriel phthalimide synthesis involves an $S_N2$ nucleophilic substitution reaction between potassium phthalimide and an alkyl halide.
   • Aryl halides do not undergo nucleophilic substitution with potassium phthalimide under ordinary conditions because the $\text{C}-\text{halogen}$ bond has partial double-bond character due to resonance with the benzene ring.
   • Hence, primary aromatic amines cannot be synthesized by Gabriel phthalimide synthesis.
   • Thus, Statement B is incorrect.

3. Statement C:

   • The reactant is 2-phenylacetamide ($\text{C}_6\text{H}_5\text{CH}_2\text{CONH}_2$).
   • Treating an amide with $\text{Br}_2$ and $\text{NaOH}$ undergoes the Hoffmann bromamide degradation reaction, reducing the chain length by one carbon atom: $\text{C}_6\text{H}_5\text{CH}_2\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$
   • The product formed is benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$).
   • In benzylamine, the $-\text{NH}_2$ group is attached to an $sp^3$-hybridized aliphatic carbon atom, making it a primary aliphatic amine, not an aromatic amine.
   • Thus, Statement C is incorrect.

4. Statement D:

   • The reactant is 4-nitroaniline ($\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2$).
   • Step (i): Diazotization with $\text{NaNO}_2/\text{HCl}$ at $0^\circ\text{C}$ yields $4$-nitrobenzenediazonium chloride: $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2 \xrightarrow{\text{NaNO}_2, \text{HCl}, 0^\circ\text{C}} \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+\text{Cl}^-$
   • Step (ii): Heating ($\Delta$) in aqueous medium hydrolyzes the diazonium salt to form 4-nitrophenol: $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_2\text{O}, \Delta} \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{OH} + \text{N}_2 + \text{HCl}$
   • $4$-Nitrophenol is acidic in nature ($\text{p}K_a \approx 7.15$) and readily reacts with aqueous $\text{NaOH}$ to form a water-soluble sodium salt (sodium 4-nitrophenolate): $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{OH} + \text{NaOH} \rightarrow \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{O}^-\text{Na}^+ + \text{H}_2\text{O}$
   • Hence, the product dissolves in $\text{NaOH}$.
   • Thus, Statement D is correct.

Conclusion: The incorrect statements are B and C Only.

Correct Answer: Option C (B and C Only)