(i)
The given reaction is an example of a carbylamine test.  

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2+{\mathrm{CHCl}}_3+3\mathrm{alc}. \mathrm{KOH}\stackrel{\mathrm{Carbylamine}}{\to }$
$\mathrm{reaction}3{\mathrm{H}}_2\mathrm{O}+3\mathrm{KCl}+{\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{NC}$
$\mathrm{Aniline} \mathrm{Phenyl} \mathrm{isocyanide}$

(ii)
In this reaction, benzenediazonium salt is converted into benzene as a product.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{Cl}+{\mathrm{H}}_3{\mathrm{PO}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{N}}_2+{\mathrm{H}}_3{\mathrm{PO}}_3+\mathrm{HCl}$
$\mathrm{Benzenediazonium} \mathrm{Benzene}$
$\mathrm{chloride}$

(iii)
In this reaction, aniline gives acid-base reaction with conc. sulphuric acid and  form anilinium ion as a product.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2+\mathrm{conc}.{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{C}}_6{\mathrm{H}}_5\stackrel{+}{\mathrm{N}}{\mathrm{H}}_3\mathrm{HS}{\stackrel{-}{\mathrm{O}}}_4$
$\mathrm{Aniline} \mathrm{Anilinium} \mathrm{hydrogen} \mathrm{sulphate}$

(iv)

In this reaction, benzene diazonium salt reacts with ethanol and forms benzene as the main product.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{Cl} + {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH} \to {\mathrm{C}}_6{\mathrm{H}}_6 + {\mathrm{CH}}_3\mathrm{CHO} + {\mathrm{N}}_2 + \mathrm{HCl}$
$\mathrm{Benzenediazonium} \mathrm{Ethanol} \mathrm{Benzene} \mathrm{Ethanal}$
$\mathrm{chloride}$

(v)
The given reaction is an example of an electrophilic aromatic substitution reaction. The NH₂ group activates the benzene ring and a bromination reaction occurs at the ortho and para positions. 

(vi)
In this reaction, aniline undergoes an acetylation reaction when reacts with acetic anhydride as follows: 

(vii)
In the reaction, benzene diazonium chloride first reacts with HBF₄ , and diazonium fluoroborate is obtained as a product. 
In the next step, diazonium fluoroborate is heated with an aqueous sodium nitrite solution in the presence of copper, the diazonium group is replaced by the –NO₂ group.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2\mathrm{CL}\underset{\left(\mathrm{ii}\right) {\mathrm{NaNO}}_2/\mathrm{Cu},∆}{\overset{\left(\mathrm{i}\right){\mathrm{HBF}}_4}{\to }}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2+{\mathrm{N}}_2+{\mathrm{NaBF}}_4$
$\mathrm{Benzenediazonium} \mathrm{Nitrobenzene}$
$\mathrm{chloride}$