$\mathop {R - N{H_2}}\limits_{(A)} \xrightarrow{{Ph - S{O_2}Cl}}$ $\mathop {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
  \,\,\,O \\ 
  {\,\,\,||} 
\end{array}} \\ 
  {R - NH - S - Ph} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,O} 
\end{array}}\limits_{(B)} $ $\xrightarrow{{KOH}}\mathop {Soluble}\limits_{(C)} \xrightarrow{{{C_2}{H_5}I}}$$\mathop {{{\left[ {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
  R \\ 
  | 
\end{array}} \\ 
  {Ph - S{O_2} - N - {C_2}{H_5}} \\ 
  {\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2}{H_5}} 
\end{array}} \right]}^ + }{I^ - }}\limits_{(D)} $

Therefore, $A$ is $R - NH _{2}$

$D$ is $Ph - SO _{2}- NR -\left( C _{2} H _{5}\right)_{2}^{+} I ^{-}$