MCQ
Conc. $H_2SO_4$ cannot be used to prepare $HBr$ from $NaBr$ because it :
- ✓reacts slowly with $NaBr$
- Boxidises $HBr$ into $Br_2$
- Creduces $HBr$
- Ddisproportionates $HBr$
Acidic nature $ \propto EN \propto + O.S.$
For compounds of an element
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$\begin{array}{*{20}{c}}
{C{H_3}C{H_2} - N - C{H_2}C{H_3}} \\
{|\,} \\
{C{H_3}}
\end{array}$
is :
$(I)\,\,CH_3CH_2OH + HCl \xrightarrow{{Anh.\,\,ZnC{l_2}}}$
$(II)\,\,CH_3CH_2OH + HCl \rightarrow$
$(III)\,\,(CH_3)_3COH + HCl \rightarrow$
$(IV)\,\,(CH_3)_2CHOH + HCl \xrightarrow{{Anh.\,\,ZnC{l_2}}}$
