- A${H_2}$
- B${N_2}$
- C${C_2}{H_4}$
- ✓$(b)$ and $(c)$
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$I.$ $CH_3 -CH = CH_2$ $\xrightarrow[{{\text{(CC}}{{\text{l}}_4}{\text{ )}}}]{{{\text{C}}{{\text{l}}_2}}}$ $\begin{array}{*{20}{c}}
{Cl\,\,}\\
{\,\,|\,\,\,\,\,}\\
{C{H_3} - CH - C{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl}
\end{array}$
$II.$ $\begin{array}{*{20}{c}}
O\\
{||}\\
{C{H_3} - C - C{H_3}}
\end{array}$ $\xrightarrow[{{}^\Theta OH}]{{{\text{HCN}}}}$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,OH}\\
|\\
{C{H_3} - C - C{H_3}}\\
|\\
{\,\,\,\,\,CN}
\end{array}$
$III.$ $CH_3-CH_2-CH_3$ $\xrightarrow[{hv}]{{C{l_2}}}$ $\begin{array}{*{20}{c}}
{Cl\,\,}\\
{|\,\,\,\,}\\
{C{H_3} - CH - C{H_3}}
\end{array}$
(At nos. : $L a=57, C e=58, E u=63$ and $Y b=70$ )
$\begin{array}{*{20}{c}}
{{C_2}{H_5}MgBr + {H_2}C - C{H_2}\xrightarrow{{{H_2}O}}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\backslash \,\,\,\,/} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O}
\end{array}A$
Assertion $A$: Permanganate titrations are not performed in presence of hydrochloric acid.
Reason $R$ : Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below.
Given :( $\frac{{2.303RT}}{F} = 0.06)$
$s{n^{ + 2}}\left( {1M} \right) + 2C{l^ - }\left( {2M} \right) \rightleftharpoons s{n_{\left( s \right)}} + C{l_2}\left( {1\,atm} \right)$
Given : ${E^o}_{s{n^{ + 2}}/sn} = - 0.14$ ${E^o}_{C{l_2}/C{l^- }} = 1.4\,V$