- ATrimethyl amine forms a soluble compound with Hinsberg reagent and $KOH$.
- BDimethyl amine reacts with $KOH$ and phenol to form an azo dye
- ✓Methyl amine reacts with nitrous acid and liberates $N_2$ from aq. solution
- DNone of these
$C{{H}_{3}}-N{{H}_{2}}+2HONO\,\to C{{H}_{3}}-ONO+{{N}_{2}}+2{{H}_{2}}O$
while dimethylamine and trimethyl amine form insoluble compound with hinsberg reagent and $KOH$.
${{R}_{2}}NH\,+\,{{C}_{6}}{{H}_{5}}S{{O}_{2}}Cl\,\xrightarrow{O{{H}^{-}}}$ ${{C}_{6}}{{H}_{5}}S{{O}_{2}}N{{R}_{2}}$
${{R}_{3}}N\,+\,{{C}_{6}}{{H}_{5}}S{{O}_{2}}Cl\,\xrightarrow{O{{H}^{-}}}\underset{insoluble}{\mathop{{{R}_{3}}N}}\,$
Azo dyes are not formed by secondary amines.
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$Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e ^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O$
The amount of $Cr ^{3+}$ obtained was $0.104\, g$. The
efficiency of the process(in\%) is
(Take : $F =96000\, C$, At. mass of chromium $=52$ )
