MCQ
Which will have enantiomer
- ✓$\begin{array}{*{20}{c}}
{C{H_3}C{H_2}CH - C{H_3}} \\
{\,|} \\
{\,\,\,Cl}
\end{array}$ - B$C{H_2}C{H_2}C{H_2}C{H_2}Cl$
- C$C{H_3}C{H_2}C{H_2}CHC{l_2}$
- DNone
$C{H_3}C{H_2} - \mathop {\mathop {\mathop {{C^ \bullet }}\limits_|^| }\limits^H - }\limits_{Cl\;} C{H_3}$
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The possible reagents are :
$(I)\, 2Na/liq.NH_3$ $(II)\, H_2 /Pd/CaCO_3$ (quinoline) $(III)\, 2H_2 / Pd /C$
The correct statement with respect to the above conversion is/are
${I^ - }(aq.) + MnO_4^ - (aq.)\xrightarrow[{weakly\,\,O{H^ - }}]{{Neutral\,\,or}}Y + Mn{O_2}$
$MnO_4^ - (aq.) + M{n^{2 + }}(aq.)\xrightarrow{{ZnS{O_4}}}Z + 4{H^ + }$
Products $X,\,Y$ and $Z$ are respectively :