- ✓${C_2}{H_2}$
- B${H_2}$
- C${C_2}{H_4}$
- DNone of these
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(image) $\frac{{conc.\,HN{O_3}}}{{conc.\,{H_2}S{O_4}}}X$
the structure of the major product $'X'$ 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 :