When chlorine reacts with $\mathrm{NaOH}$, different products are formed depending upon the temperature and concentration of $\mathrm{NaOH}$.
$
\begin{aligned}
& \mathrm{Cl}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{NaOCl}+\mathrm{H}_2 \mathrm{O} \\
& \text { (Cold and dil.) } \\
& 3 \mathrm{Cl}_2+6 \mathrm{NaOH} \longrightarrow 5 \mathrm{NaCl}+\mathrm{NaClO}_3+3 \mathrm{H}_2 \mathrm{O} \\
&
\end{aligned}
$
(Hot and conc.)
So, $A=\mathrm{NaCl}, B=\mathrm{NaOCl}, C=\mathrm{NaCl}, D=\mathrm{NaClO}_3$