Nitrosonium $$ion,\mathop N\limits^ + O$$   from $$HONO$$   is a weak electrophile, hence it can attack only on highly activated benzene nucleus, provided proper position, $$p-$$  or $$o-$$  is free.

Aliphatic amines are more basic than aromatic amines thus methylamine is most basic. Electron donating groups increase the basicity whereas electron withdrawing groups decrease the basicity of the aromatic amines. Thus $$p$$ - methoxyaniline is more basic then aniline which is further more basic then $$p$$ - nitroaniline.

The alkyl groups are electron releasing group $$\left( { + I} \right),$$  thus increases the electron density around the nitrogen thereby increasing the availability of the lone pair of electrons to proton or lewis acid and making the amine more basic. Hence more the no. of alkyl group more basic is the amine. Therefore the correct order is $$N{H_3} < C{H_3}N{H_2} < {\left( {C{H_3}} \right)_2}N$$

Red solution

Wurtz reaction is for the preparation of hydrocarbons from alkyl halide $$RX + 2Na + XR \to R - R + 2NaX$$

Benzaldehyde reacts with primary aromatic amine to form schiff's base
\[\underset{\text{Benzaldehyde}}{\mathop{{{C}_{6}}{{H}_{5}}CH}}\,=O+\underset{\text{Aniline}}{\mathop{{{C}_{6}}{{H}_{5}}N{{H}_{2}}}}\,\to \underset{\text{Benzylidene aniline}}{\mathop{{{C}_{6}}{{H}_{5}}CH=N{{C}_{6}}{{H}_{5}}}}\,\]

\[\begin{align} & C{{H}_{3}}\underset{A}{\mathop{C}}\,\equiv N\xrightarrow[LiAl{{H}_{4}}]{\text{Reduction}} \\ & C{{H}_{3}}-C\underset{B}{\mathop{{{H}_{2}}}}\,-N{{H}_{2}}\xrightarrow{HN{{O}_{2}}}C{{H}_{3}}C{{H}_{2}}OH \\ & {{1}^{\circ }}\,\text{amine(Ethanamine)} \\ & \therefore A\,\,\text{is}\,\,C{{H}_{3}}CN. \\ \end{align}\]

$$Ortho$$  - substituted anilines are weaker bases than anilines regardless of the nature of the substituent whether electron releasing or electron withdrawing. This is called $$ortho$$  effect and is probably due to a combination of steric and electronic factors.

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