Why does acylation of $-\mathrm{NH}_2$ group of aniline reduces its activating effect?
Acylation of $-\mathrm{NH}_2$ group of aniline reduces its activity due to resonance of lone pair of nitrogen towards the carbonyl group hence $0-, p^{-}$directive influence of amino group get disturbed.
Explain why $\mathrm{MeNH}_2$ is stronger base than MeOH ?
Basicity of $\mathrm{MeNH}_2$ and MeOH can be explained on the basis of electronegativity of N and O atom. $\mathrm{MeNH}_2$ is stronger base than MeOH because of low electronegativity value of N , it is easy for nitrogen to loose its lone pair readily than compared to MeOH .
What is the role of pyridine in the acylation reaction of amines?
Pyridine being a base, is used to remove the side product i.e., HCl from reaction mixture.
Under what reaction condition (acidic, basic) the coupling reaction of aryl diazonium chloride with aniline is carried out?
In strongly basic conditions, benzenediazonium chloride is converted into diazohydroxide and diazoate as both of which are not electrophilic and do not couple with aniline.
$$\mathrm{{C_6}{H_5}\mathop N\limits^ + \equiv NC\mathop l\limits^ - + \mathop {OH\,S{O_2}C{H_5} - N = N - OH}\limits_{Diazohydroxide}}$$
Similarly, in highly acidic conditions, aniline gets converted into anilinium ion. From this, result aniline is no longer nucleophilic acid and hence will not couple with diazonium chloride. Hence, the reaction is carried out under mild conditions, i.e., $\mathrm{pH}-4-5$
$\underset{\text { Aniline }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2}+\mathrm{H}^{+} \rightarrow \underset{\substack{\text { Anilinium ion } \\ \text { (coupling do not occur) }}}{\mathrm{C}_6 \mathrm{H}_5-\stackrel{+}{\mathrm{N}} \mathrm{H}_3}$
Predict the product of reaction of aniline with bromine in non-polar solvent such as $\mathrm{CS}_2$.
Aniline on reaction with $\mathrm{Br}_2$ in non-polar solvent $\mathrm{CS}_2$ produces 2, 4,6 tribromo aniline.
Aniline has high reactivity towards bromine as it gives the triply substituted product.