JEE Mains · Chemistry · STD 12 - 6. Haloalkanes and Haloarenes
In the following reaction sequence The compound \(I\) is \(\mathop I\limits_{({C_3}{H_6}C{l_2})} \xrightarrow{{KOH(aq)}}II\xrightarrow[{(ii){H_2}O/{H^ + }}]{{(i)C{H_3}MgBr}}III\xrightarrow{{Anhy.ZnC{l_2} + Conc.HCl}}\) given turbidity immediately
- A \(\begin{array}{*{20}{c}}
{C{H_2} - CH - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Cl\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\) - B \(\begin{array}{*{20}{c}}
{C{H_2} - C{H_2} - C{H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\) - C \(\begin{array}{*{20}{c}}
{CH - CH - C{H_2} - C{H_3}} \\
{\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\) - D \(\begin{array}{*{20}{c}}
{Cl} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{Cl}
\end{array}\)
Answer & Solution
Correct Answer
(D) \(\begin{array}{*{20}{c}}
{Cl} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{Cl}
\end{array}\)
Step-by-step Solution
Detailed explanation
\(\underset{(I)}{\mathop{\underset{({{C}_{3}}{{H}_{6}}C{{l}_{2}})}{\mathop{\begin{matrix} \begin{matrix} \,\,Cl \\ | \\ \end{matrix} \\ {{H}_{3}}C-C-C{{H}_{3}} \\ | \\ \,\,Cl \\ \end{matrix}}}\,}}\,\) \(\xrightarrow[{(aq.)}]{{KOH}}\)…
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