- ✓Styrene and butadiene
- BIsoprene and butadiene
- CVinyl chloride and Sulphur
- DButadiene.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.


$Mn{O_2}(s)\, + \,4{H^ + }(aq)\, + \,2{e^ - }\, \to \, M{n^{ + 2}}(aq)\, + \,2{H_2}O(l)\,;$ $E_2^o\, = \,1.21\,\,V$
$MnO_4^{ - 1}(aq)\, + \,4{H^ + }(aq)\, + \,3{e^ - }\, \to \, Mn{O_2}(s)\, + \,3{H_2}O(l)\,;$ $E_3^o\,?$
value of $E_3^o$ will be ............ $\mathrm{V}$
The correct order of ${S_{{N^1}}}$ reactivity is
$(I)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow[{5\% \,\,water}]{{95\% \,\,\,acetone}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(II)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow[{10\% \,\,water}]{{90\% \,\,\,acetone}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(III)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow[{20\% \,\,water}]{{80\% \,\,\,acetone}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
$(IV)$ $\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - Cl} \\
{\,\,|} \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow{{100\% \,water}}\begin{array}{*{20}{c}}
{{C_6}{H_5} - CH - OH} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}$
Arrange these reactions in decreasing order of greater proportion of inverted product and select correct answer from the codes given below :