{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}} \\
|
\end{array}} \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,OH}
\end{array}$ The reactant $A$ is
- A$C{H_3}COC{H_3}$
- ✓$C{H_3}CHOHC{H_3}$
- C${C_2}{H_5}OH$
- D$C{H_3}COOH$
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$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_2} = CH - C{H_2} - C - H}
\end{array} \to C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$
$(I)$ Only compounds with chiral centers can be optically active
$(II)$ Absence of elements of symmetry is the reason for a molecule's optical activity
$(II)$ All organic compounds contain carbon & hydrogen
$(IV)$ Different cannonical forms of a molecule represent the actual structures of a molecule which has resonace in it

$B _{2}, Li _{2}, C _{2}, C _{2}^{-}, O _{2}^{2-}, O _{2}^{+}$and $He _{2}^{+}$