Choose the correct answer from the options given below:
$B.$ Neutral $FeCl _3$ solution is used to test phenolic compound $(IV)$
$C.$ Alkaline chloroform solution is used to test primary amines $(II)$
$D.$ $2 KI + NaOCl + H _2 O \rightarrow NaCl + I _2+2 KOH$
Potassium iodide and sodium hypochlorite gives $\left( I _2+ KOH \right)$ which is used to test those compounds which have $image$ group (iodoform test). Hence the compound is $image$
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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_3}(excess) + B{r_2}\xrightarrow{{hv}}}
\end{array}$ $\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,Br}
\end{array}}\limits_{(A)} $ + $\mathop {\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_2} - Br}
\end{array}}\limits_{(B)} $
the percentage yields of the products $(A)$ and $(B)$ are expected to be
$2X_2(s) + 3Y^{2+} (aq) \to 2X_2^{3+}(aq) + 3Y(s)$