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\[\underset{\text{Amide}}{\mathop{R\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{-C-}}\,N{{H}_{2}}}}\,\xrightarrow{B{{r}_{2}}/NaOH}\underset{\begin{smallmatrix} \left( {{1}^{\circ }}\,\text{amine} \right) \\ \left( \text{One}\,\,C\,\,\text{less} \right) \end{smallmatrix}}{\mathop{R-N{{H}_{2}}}}\,\]
All other reactions give same number of $$C$$ atoms in the chain of amines as in the reactants.

\[R-X\xrightarrow{KCN}R-CN\xrightarrow{\frac{Na}{{{C}_{2}}{{H}_{5}}OH}}\]       \[R-C{{H}_{2}}N{{H}_{2}}\]

$$N{O_2}$$  group being electron withdrawing that's why it reduces the electron density at $$ortho$$  and $$para$$  - positions. Hence, as compare to $$ortho$$  and $$para$$  the $$meta$$  - position is electron rich on which the electrophile ( nitronium ion ) can easily attacks during nitration.

Schotten-Baumann reaction is a method to synthesise amides from amines and acid chlorides.

In the nitration of benzene in the presence of conc. $${H_2}S{O_4}$$  and $$HN{O_3},$$  nitrobenzene is formed.
$$HN{O_3} + {H_2}S{O_4} \rightleftharpoons $$     $$NO_2^ + + \mathop {HSO_4^ - }\limits_{{\text{Electrophile}}} + \mathop {{H_2}O}\limits_{{\text{Nucleophile}}} $$
If large amount of $$KHS{O_4}$$  is added to this mixture, more $${HSO_4^ - }$$  $$ion$$  furnishes and hence the concentration of $$NO_2^ + ,$$  i.e. electrophile decreases.
As concentration of electrophile decreases, rate of electrophilic aromatic reaction also decreases.

TIPS/Formulae : Toluene has electron - donating methyl group. Hence reacts fastest while others have either electron withdrawing groups $$\left( {{\text{i}}{\text{.e}}\, - COOH\,or\, - N{O_2}\,{\text{etc}}.} \right)$$      or no substituent.

Key Idea $$\frac{{NaN{O_2}}}{{HCl}}$$   causes diazotisation of $$ - N{H_2}$$  group and the diazonium chloride gives a coupling product with active aryl nucleus.