MCQ
Acetylene can be prepared from
- APotassium fumarate
- BCalcium carbide
- CEthylene bromide
- ✓All of these
$\begin{array}{*{20}{l}}
{CH} \\
{|\,|\,|} \\
{\underbrace {CH}_{Anode}}
\end{array} + 2\,C{O_2} + 2\,KOH + \underbrace {{H_2}}_{{\text{Cathode}}}$
$Ca{C_2} + 2\,{H_2}O \to Ca{\left( {OH} \right)_2} + CH \equiv CH$
$\,\mathop {\mathop {C{H_2}}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{Br\,\,\,\,\,} - \mathop {C{H_2}}\limits_{\mathop {|\,\,\,\,\,\,\,\,\,}\limits_{Br\,\,\,\,} } + 2\,KOH \to CH \equiv CH + 2\,KBr + 2\,{H_2}O$
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$ \text { a } \mathrm{Cl}_2(\mathrm{~g})+\mathrm{b} \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{c} \mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{d} \mathrm{Cl}^{-}(\mathrm{aq}) +\mathrm{e} \mathrm{H}_2 \mathrm{O}(l)$
The values of $a, \ b,\ c$ and $d$ in a balanced redox reaction are respectively :
while $E_{Z{n^{2 + }}/Zn}^o = - 0.76\,V$
Reason : Addition of $HBr$ on $2-$ butene follows Markovnikov rule.