MCQ
$1^o$ amine and $2^o$ Amine can be differentiated by
- A$CHCl_3/Alc.KOH$
- B$NaNO_2/HCl$
- C$PhSO_2Cl$ then $NaOH$
- ✓All of these
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$X + Y\mathop {\xrightarrow{{NaOH}}}\limits_{5\,^oC} \begin{array}{*{20}{c}}
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - CH - CHO} \\
{\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$
$(X)$ and $(Y)$ will respectively be :
${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{ - 4}}\xrightarrow{{MnO_4^ - /{H^ + }}}F{e^{ + 3}} + C{O_2} + NO_3^ - $
Roasted gold ore $+CN^-+ H_2O \xrightarrow{{{O_2}}} [x] + OH^-;$ $[x] + Zn \rightarrow [y] + Au$
$[x]$ and $[y]$ are :