JEE Mains · Chemistry · STD 11 - 9. Hydricarbon
The major product (\(Y\)) in the following reactions is \(\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C = CH}
\end{array}\xrightarrow[{{H_2}O}]{{HgS{O_4},{H_2}S{O_4}}}X\) \(\xrightarrow[{(ii)\,conc.{H_2}S{O_4}/\Delta }]{{(i)\,{C_2}{H_5}MgBr,{H_2}O}}Y\)
- A \(\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH - C{H_3}} \\
{\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,{C_2}{H_5}}
\end{array}\) - B \(\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C = CH - C{H_3}} \\
| \\
{\,\,\,\,\,\,C{H_3}}
\end{array}\) - C \(\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}
\end{array}\) - D \(\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C = C{H_2}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}
\end{array}\)
Answer & Solution
Correct Answer
(C) \(\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C = C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}
\end{array}\)
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