NOTE: Aromatic amines are less basic than aliphatic amines. Among aliphatic amines the order of basicity is $${{\text{2}}^ \circ } > {1^ \circ } > {3^ \circ }.$$   The electron density is decreased in $${3^ \circ }$$ amine due to crowding of alkyl group over $$N$$ atom which makes the approach and bonding by a proton relatively difficult. Therefore the basicity decreases. Further Phenyl group show - $$I$$ effect, thus decreases the electron density on nitrogen atom and hence the basicity.
∴ dimethylamine ( $${{\text{2}}^ \circ }$$ aliphatic amine ) is strongest base among given choices.
∴ The correct order of basic strength is Dimethylamine > Methyl amine > Trimethyl amine > Aniline.

Under weakly acidic conditions nitrobenzene on electrolytic reduction gives aniline but under strongly acidic conditions gives $$p$$ - aminophenol.

Reduction of nitrobenzene with $$Zn/N{H_4}Cl$$   ( neutral medium ) gives phenyl hydroxyamine.

Only primary aromatic amines undergo diazotisation followed by coupling.

The reaction,
$$RCON{H_2} + B{r_2} + KOH \to RN{H_2}$$
is known as Hofmann bromamide reaction. The mechanism of this reaction is given as :

Key Idea $${1^ \circ }$$ and $${2^ \circ }$$ nitro compounds react with $$HN{O_2}$$  while $${3^ \circ }$$ - nitro compound does not.
The reactions of given compounds with $$HN{O_2}$$  are as follow

Thus, option (C) is incorrect.

$$\mathop {{C_6}{H_5}N{H_2}}\limits_{{\text{Aniline}}} + \mathop {ClCO{C_6}{H_5}}\limits_{{\text{Benzoyl chloride}}} + NaOH \to $$        $$\mathop {{C_6}{H_5}NH - CO{C_6}{H_5}}\limits_{{\text{Benzanilide}}} + NaCl + {H_2}O$$

The reactive species is the nitrosonium ion (electrophile) and the most likely site of attack will be that with the highest electron density.

Such kind of delocalisation is not found in 2, 6-dimethyl derivative due to steric inhibition of resonance.

The basic character of an amine in water is determined by $$(i)$$  electron availability on the $$N$$  atom and $$(ii)$$  the extent of stabilization of the cation ( protonated amine ) due to solvation by hydrogen bonding

By Hoffmann bromamide degradation reaction, the amine formed contains one carbon less than that present in the amide.
\[\underset{2\text{-Phenylpropanamide}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix} \,\,\,{{C}_{6}}{{H}_{5}} \\ |\,\,\,\, \end{smallmatrix}}{\mathop{-CH-}}\,CON{{H}_{2}}}}\,+B{{r}_{2}}+4NaOH\to \]         \[\underset{\text{1-Phenylethanamine}}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix} \,\,\,{{C}_{6}}{{H}_{5}} \\ |\,\,\,\, \end{smallmatrix}}{\mathop{-CH-}}\,N{{H}_{2}}}}\,+N{{a}_{2}}C{{O}_{3}}+\]     \[2NaBr+2{{H}_{2}}O\]